20.dvi @ Applied General TopologyUniversidad Polit�ecnica de ValenciaVolume 2, No. 1, 2001pp. 77 - 98 Attractors of reaction-di�usion equations inBanach spacesJos�e ValeroAbstract. In this paper we prove �rst some abstract theoremson existence of global attractors for di�erential inclusions gener-ated by !-dissipative operators. Then these results are appliedto reaction-di�usion equations in which the Banach space Lp isused as phase space. Finally, new results concerning the fractaldimension of the global attractor in the space L2 are obtained.2000 AMS Classi�cation: 35B40, 35B41, 35K55, 35K57.Keywords: Attractor, asymptotic behaviour, reaction-di�usion equations,fractal dimension. 1. IntroductionThe theory of global attractors for partial di�erential equations have beenapplied to a wide range of equations of parabolic and reaction-di�usion types.It is very important in this theory to choose an appropriate phase space forthe system. The existence, uniqueness and properties of solutions of partialdi�erential equations depend strongly on the phase space.The more common phase space used for reaction-di�usion equations is thespace L2 ( ) (see [3], [4], [5], [6], [7], [13], [17], [18], [21], [25], [26], [28], [29],[31], [32], [33]).However, other spaces like L1 ( ) [1], H ( ) ; 0 < � 1 (see [2], [11], [12],[20] and their bibliography) or C ( ) (see [15]-[16], [27]) have been also usedsuccesfully.As far as we know, the space Lp ( ), where 2 < p < 1; has not beenconsidered so far.In this paper we shall consider the existence of global attractors for di�eren-tial inclusions of the type � x0(t) 2 A(x(t));u(0) = u0;where A is a multivalued !-dissipative operator. 78 Jos�e ValeroThis general framework allows us to apply the obtained results in order toprove the existence of a global attractor for the following reaction-di�usionequation 8<: @u@t � �u + f(u) 3 !u + h; on ]0;T [ � ;u = 0; on ]0;T [ � �;u(0;x) = u0(x) 2 Lp ( ) ;where f : R ! 2R is a multivalued maximal monotone map and 2 � p < 1.Therefore, we extend the results of attractors for reaction-di�usion equationsto the case where the phase space Lp ( ) is used. Moreover, the nonlinearfunction f, which is usually a continuous one, can be in our case multivalued.Hence, our results are valid also for di�erential equations with discontinuousnonlinearities. We obtain also a result (see Theorem 4.4), which is new in thecase where f 2 C1 and p = 2.In the case p = 2 we obtain some estimates of the fractal dimension of theglobal attractor. Such estimates are well known under di�erent conditions onthe function f (see [3]-[5], [7], [18], [25]). In all these papers the function f isat least Lipschitz on any bounded set of R. We shall extend these result byconsidering a function f (s) which is Lipschitz on a �xed bounded set [�a;a]butcan be even discontinuous for s =2 [�a;a].We shall recall now some de�nitions of the theory of dynamical systems (see[20], [22]-[23], [29] for more details). Let Y be a complete metric space with themetric denoted by �(�; �), V : R+ � Y ! Y be a semigroup of operators, i.e.,the following properties holdV (t1;V (t2;x)) = V (t1 + t2;x); for all t1; t2 2 R+;x 2 Y;V (0; �) = I:The map x 7�! V (t;x) is supposed to be continuous for each t 2 R+.Let us introduce the next notation:B(Y ) is the set of all bounded subsets of Y ; +� (x) = [t��V (t;x), +� (A) = [x2A +� (x), where t;� 2 R+, x 2 Y , A 2 B(Y );!(B) = \t�0 +t (B) is the !-limit set.We denote by Z the set of stationary points of V , i.e.,Z = fu 2 Y j V (t;u) = u; for all t 2 R+g:The continuous function L : Y ! R is called a Lyapunov function on Y forV if L(V (t;x)) < L(x) for any t > 0, x 2 Y , x =2 Z.Let us recall the concept of distance between sets. Let A;B � Y . Then thedistance from A to B is determined as follows:d(A;B) = supx2A� infy2B �(x;y)� :De�nition 1.1. The compact set < � Y is called a global attractor of thesemigroup V if the following conditions hold: Attractors of reaction-di�usion equations 79(1) It attracts each set B 2 B(Y ); that is,d(V (t;B);<) ! 0; as t ! +1:(2) It is invariant, i.e., V (t;<) = <, for all t � 0:De�nition 1.2. The semigroup V is called "pointwise dissipative" if there ex-ists a bounded set B1 which attracts each x 2 Y .Remark 1.3. We note that if for any B 2 B(Y ) there exists some T = T(B) 2R+ for which +T (B) 2 B(Y ), then pointwise dissipativeness is equivalent to theexistence of a bounded set B0 such that for each x 2 Y there exists T (x) forwhich V (T;x) 2 B0. In such a case we can take B1 = +T(B0) (B0).De�nition 1.4. The semigroup V is said to be time-continuous if the mapt 7! V (t;x) is continuous for each x 2 Y .Theorem 1.5. ([23, p.107-109] and [19, p.4-5])Let V (t; �) be compact for somet > 0 and let V be pointwise dissipative. Suppose that for any B 2 B(Y )there exists T = T(B) 2 R+ such that +T (B) 2 B(Y ). Then V has the globalattractor <. If V is time-continuous and Y is connected, then < is connected.This paper is organized as follows. In the second section we prove the exis-tence of attractors for abstract di�erential inclusions in Banach spaces generatedby !-dissipative operators. In the third section we apply the abstract resultsof the preceding section to di�erential inclusions generated by subdi�erentialmaps in Hilbert spaces. In the fourth section we prove the existence of globalattractors for reaction-di�usion equations in Lp ( ) spaces. Finally, in the �fthsection we obtain estimates of the fractal dimension of the global attractor ofreaction-di�usion equations in the Hilbert space L2 ( ) :2. Existence of attractors of differential inclusions generatedby !-dissipative operatorsLet X be a real Banach space with its dual denoted by X�. We shall denoteby k�k and k�k� the norms in X and X� respectively. h�; �i will denote pairingbetween the spaces X and X�. Consider the next di�erential inclusion(2.1) ( du(t)dt 2 A(u(t)) ; 0 � t < 1;u(0) = u0 2 X;where A : D (A) � X ! 2X is a multivalued nonlinear operator and u : R+ !X.We remind that the dual operator J : X ! 2X� (may be multivalued) isde�ned as follows:J (x) = nj 2 X� j kxk2 = kjk2� = hx; jio :Consider the next conditions: 80 Jos�e Valero(A1) The operator A is !-dissipative, i.e., for all x1;x2 2 D (A) ; y1 2 A(x1),y2 2 A(x2) ; there exists j (xi;yi) 2 J (x1 � x2) such thathy1 � y2; ji � ! kx1 � x2k2 ;where ! � 0:When ! = 0 the operator A is called dissipative. It is easy to checkthat condition (A1) is equivalent to the dissipativeness of the operatorA � !I.(A2) D (A) � \0<�<�0Im(I � �A), where �0 > 0, �0! < 1:Notice that D (A) is a complete metric space endowed with the usual metric�(x;y) = kx � yk ; for all x;y 2 D (A).If conditions A1 and A2 are satis�ed, then there exists a semigroup of oper-ators V : R+ � D (A) ! D (A) corresponding to the operator A (see [8, p.108]or [14, p.63]) such that V is determined by the next formula(2.2) V (t;x) = limn!1 �I � tnA��n x; x 2 D (A); t 2 R+:Moreover, for any �xed t 2 R+ one has(2.3) kV (t;x) � V (t;y)k � exp(!t)kx � yk ; for all x;y 2 D (A):It follows from inequality (2.3) that the map x 7�! V (t;x) is continuous foreach t 2 R+.A map u : [0;T ] ! X is called a strong solution of inclusion (2.1) on [0;T ]if: (1) u is continuous on [0;T ] and u(0) = u0;(2) u is absolutely continuous on any compact subset of (0;T) ;(3) u is almost everywhere (in short, a.e.) di�erentiable on (0;T) and sat-is�es inclusion (2.1) a.e. on (0;T).Remark 2.1. If u(�);u(0) = u0, is a strong solution of (2.1) and (A1), (A2)hold, then u(t) = V (t;u0) [14, Theorem 3.1], where V is the semigroup de�nedby (2.2).Remark 2.2. Suppose that the space X is re exive, (A1), (A2) hold and theoperator A is closed. Then the map u(t) = V (t;u0) is a strong solution of (2.1)for any T > 0,u0 2 D (A) [14, p.77].We also must consider the next condition:(A3) Im(I � �A) = X,for all � > 0.A dissipative operator which satis�es A3 is called m-dissipative. In such acase A2 is always satis�ed. Moreover, every m-dissipative operator is closed [8,p.75].If A = B + !I with B m-dissipative, then A is closed and (A1), (A2) aresatis�ed. Indeed,Im(I � �A) = Im(I � �B � �!I) = Im((1 � �!) I � �B) = Attractors of reaction-di�usion equations 81= Im�I � �1 � �!B� = X;if 1 � �! > 0. Hence, \0<�<�0Im(I � �A) = X; if �0! < 1:In that case we have the inclusion:(2.4) du(t)dt 2 B (u(t)) + ! u(t) ; u(0) = u0:Remark 2.3. It is easy to prove that if u(t) = V (t;x) is a strong solution of(2.1) for each x 2 D (A); then the set of stationary points Z can be characterizedas follows: Z = fu 2 D (A) j 0 2 A(u)g :It may be proved also that if A = B + !I, where B is m-dissipative, thenthis characterization is also true.First we shall recall the next result, which states the existence of a compactattractor for m-dissipative operators.Theorem 2.4. [30, p.609] Let A be an m-dissipative operator. Suppose that thesemigroup V generated by A is compact for some t0 > 0 and that Z is nonemptyand bounded. Assume also that the next condition is satis�ed:(2.5) If x 2 D (A) and kV (t;x) � vk = kx � vk for all v 2 Z, for all t � 0,then x 2 Z.Then Z is the global attractor of V . Moreover, the following conditions hold:(1) Z is connected if D (A) is connected;(2) Every positive trajectory fx(t) = V (t;x0), t 2 R+, x0 2 D(A)g con-verges to some element z 2 Z, i.e., !(x) = z 2 Z, for all x 2 D(A).Remark 2.5. In fact, m-dissipativeness is too strong. The statement of thetheorem remains valid if (A1) � (A2) hold with ! = 0.Remark 2.6. The proof of Theorem 2.4 is based on the results from [23]. Itis remarked in [19] that in the abstract result of that paper on the connectivityof the global attractor is necessary to supppose the semigroup V to be time-continuous. Since the semigroup V is time-continuous, the statement of point(1) remains valid.To obtain suitable conditions on A and also to study the !-dissipative casewe need strong solutions of (2.1).Further, we shall prove two results on existence of the global attractor forthe semigroup V:Lemma 2.7. Suppose that A satis�es (A1) � (A2) and that u(t) = V (t;u0) isa strong solution of (2.1) for any u0 2 D (A). Suppose also that there existsu 2 Z such that the next condition holds:(2.6)for all x 2 D (A) ; x =2 Z; y 2 A(x) ; there exists j 2 J (x � u) for whichhy;ji < 0: 82 Jos�e ValeroThen the function L(x) = kx � uk is a Lyapunov function for V on D (A).Proof. Suppose that x =2 Z. Since the map x(t) = V (t;x) is continuous on t,there exists an interval [0;s) such that x(t) =2 Z for any t 2 [0;s). Multiplying(2.1) by j 2 J (x(t) � u) and using (2.6), we have�dx(t)dt ;j� < 0 a.e. t 2 (0;s) :By Lemma 1.2 from [8, p.100] we have Ddx(t)dt ;jE = � ddt kx(t) � uk�kx(t) � uk.Hence, � ddt kx(t) � uk�kx(t) � uk < 0 a.e. t 2 (0;s) :By integration we obtainkx(t) � uk < kx(0) � uk ; if t 2 (0;s) .It is easy to see that the last inequality holds for any t > 0. Indeed, let x(s) 2 Z.Then x(t) = x(s) ; for all t � s; andkx(s) � uk = lim�!s kx(�) � uk = inf0<� 0, the map u(t) = V (t;u0) be a strong solution for all u0 2 D (A), Z benonempty and bounded and (2.6) hold for some u 2 Z. Then V has the globalattractor <. It is connected if D (A) is connected.Proof. It follows from Lemma 2.7 that L(x) = kx � uk is a Lyapunov functionfor V on D (A) and also that for some u 2 Z one haskV (t;u0) � uk � ku0 � uk ; for all u0 2 B 2 B�D(A)� :Then, it is obvious that +0 (B) 2 B�D(A)�.Z is nonempty and bounded by assumption. It is well known from [23, The-orems 2.1 and 2.4] that if V is compact for some t > 0, there exists a Lyapunovfunction and +0 (x) is a bounded set, then ! (x) is nonempty, compact andattracts x. Moreover, ! (x) � Z: It follows that Z attracts any x 2 D (A), sothat V is pointwise dissipative.We conclude the proof by applying Theorem 1.5. �Corollary 2.9. Under the conditions of Theorem 2.8, < = Z if we supposethat one of the next conditions holds:(1) (2.6) holds for any u 2 Z;(2) A is m-dissipative.Moreover, in the second case ! (x) = z 2 Z, for all x 2 D (A). Attractors of reaction-di�usion equations 83Proof. Let us prove the �rst statement. Let u 2 <. We must prove that in thiscase u 2 Z.Recall that the function u(�) : R ! X is called a complete trajectory of Vif u(t + t1) = V (t;u(t1)), for all t1 2 R, t 2 R+. This function will be called acomplete trajectory of the point x if u(0) = x.From the de�nition, every global attractor is the union of all bounded com-plete trajectories of V (for a proof see [22, p.10]). Hence, the point u belongsto some bounded complete trajectory fu(t) ; t 2 Rg. Without loss of gener-ality we can put u(0) = u. From each sequence fu(tn)g, tn % +1 (thatis, an increasing sequence of times converging to +1) or tn & �1 (that is,a decreasing sequence of times converging to �1), belonging to a completebounded trajectory fu(t); t 2 Rg, we can choose a subsequence converging tosome stationary point (see [23, Theorem 2.4]). It follows that there exist twosequences fu(tn)g ; tn % +1, fu(tm)g ; tm & �1, such thatlimn!1u(tn) = u1 2 Z; limm!1u(tm) = u2 2 Z:Let us prove by contradiction that u 2 Z. Suppose the opposite. Then by (2.6)and using Lemma 2.7 we haveku(tn) � u2k < ku � u2k < ku(tm) � u2k ; for all tn > 0; tm < 0:Since limtm!�1 ku(tm) � u2k = 0, we obtain u = u1 = u2 2 Z. Thus, < � Z,from which < = Z.Let us prove the second statement. Since L(x) = kx � uk is a Lyapunovfunction for some u 2 Z, (2.5) is immediately satis�ed. Thus, the secondstatement is a consequence of Theorem 2.4. �Remark 2.10. If A = B + !I, where B is m-dissipative (or satis�es (A1) and(A2) with ! = 0), (2.6) can be written in the next way: there exists u 2 Z suchthat(2.7)for all x 2 D (A) ; x =2 Z; y 2 B (x) ; there exists j 2 J (x � u) for whichhy;ji < �! hx;ji :In the preceding theorem we have used conditions providing the existence of acompact attractor. In the next one we shall consider another kind of conditionswithout using the set of stationary points Z.Theorem 2.11. Let A satisfy (A1)-(A2), V be compact for some t0 > 0,u(t) = V (t;u0) be a strong solution for any u0 2 D(A) and the next conditionhold: there exist C > 0; � > 0 such that(2.8) for all u 2 D(A); kuk > C; y 2 A(u); there exists j 2 J(u) for which< y;j >� ��:Then V has the global attractor <. It is connected if D (A) is connected.Proof. First we shall prove that for all x 2 D(A) there exists t(x), for whichV (t;x) 2 B0, where B0 = nu 2 D (A) j kuk � C + �o ; 84 Jos�e Valerowith � > 0.Let x =2 B0, so that kxk > C + �. Suppose that u(t) = V (t;x) =2 B0, for allt � 0. Then, using (2.8) and arguing as in the proof of Lemma 2.7 we getku(t)k2 � ku(0)k2 � 2�t; for all t � 0:For t great enough u(t) 2 B0. The resulting contradiction proves that u(t) 2 B0for some t.Further, let us prove the inclusion +0 (B) 2 B�D(A)�, if B 2 B�D(A)�,B � D(A). For this purpose it su�ces to show that if ku0k � M; u0 2D(A); M > C; then ku(t)k = kV (t;u0)k � M; for all t � 0. Indeed, sup-pose thatku(t1)k > M; for some t1 > 0. Since the map u(�) is continuouson [0;1) it is clear that there exists t2 < t1 such that ku(t2)k = M andku(t)k > M for t2 < t � t1. But in this case we obtain arguing as before thatku(t1)k2 � ku(t2)k2 � 2� (t2 � t1), which is a contradiction.It remains to consider the case where x 2 D(A)nD(A). Let us show �rst thatV (t;x) 2 B0 for some t(x) � 0. Let it not be so, i.e., V (t;x) =2 B0 ; for all t �0. Since x 2 D(A), we can �nd a sequence fxkg � D(A) such that limk!1 xk =x. Let t be �xed. Then using (2.3) we obtain that there exists k0 such thatkV (t;xk)k � C + �=2; for all k � k0. Since for any M > C and x 2 D(A)we have that kV (�;x)k � M; for all � � 0; if kxk � M, we deduce thatkV (�;xk)k � C + �=2; for all � 2 [0; t]; k � k0. Then, using (2.8) again weget kV (t;xk)k2 � kV (�;xk)k2 � 2�(t � �); t > �; for all k � k0:When k ! 1 we obtainkV (t;x)k2 � kV (�;x)k2 � 2�(t � �); t > �:It is easy to see that choosing t great enough we have that V (t;x) 2 B0. Hence,we obtain a contradiction.It remains to prove that +0 (B) 2 B�D(A)� ; for all B 2 B�D(A)�. Weshall prove that V (t;BM) � BM; for all t � 0, whereBM = nv 2 D (A) j kvk � Mo ; M > C:For any B � BM; B � D(A); we have already veri�ed the inclusion V (t;B) �BM; for all t � 0. Suppose that there exist x 2 D(A)nD(A); x 2 BM; andt > 0 such that kV (t;x)k > M. Choosing fxkg � D(A); kx � xkk !k!1 0; andusing (2.3), we have kV (t;x) � V (t;xk)k !k!1 0. Then kV (t;xk)k > M; forall k � k0. The resulting contradiction shows that V (t;BM) � BM and then +0 (B) 2 B�D(A)� ; for all B 2 B�D(A)�.Therefore, +0 (B) 2 B�D(A)� ; for all B 2 B�D(A)� ; and, in view ofRemark 1.3, the semigroup V is pointwise dissipative . We conclude the proofby using Theorem 1.5. �Remark 2.12. In view of Remark 2.2 if X is re exive and A is closed, thenu(t) = V (t;u0) is a strong solution of (2.1) for any u0 2 D(A). Attractors of reaction-di�usion equations 85Remark 2.13. It is easy to see that if there exist � > 0; M � 0 such that(2.9) for all u 2 D (A) ; y 2 A(u) ; there exists j 2 J (u) for whichhy;ji � ��kuk2 + M;then (2.8) holds.3. Applications to inclusions generated by subdifferential mapsLet H be a real Hilbert space where is given the scalar product denoted by(�; �). We identify H with its dual H� and then the dual operator J is theidentity map I. We recall that the multivalued operator A : D(A) � H ! 2His called monotone if(y1 � y2;x1 � x2) � 0; for all x1;x2 2 D(A); y1 2 A(x1); y2 2 A(x2):A monotone operator is called maximal monotone if there does not exist anothermonotone operator B such that Graph(A) � Graph(B), whereGraph(A) = f(x;y) 2 H � H : y 2 A(x)g :Remark 3.1. The operator A is maximal monotone if and only if �A is m-dissipative [8, p.71].Consider the problem(3.10) � dudt 2 �@'(u(t)) + !u(t) + h;u(0) = u0;where ! � 0; @' : H ! H is the subdi�erential of a convex proper lowersemicontinuous function ' : H ! ]�1;+1] and h 2 H. It is well knownthat D(') = D(@') and also that @' is a maximal monotone operator [8,p.54]. Then �@' is m-dissipative in view of Remark 3.1 and �@' + h is alsom-dissipative.For any u0 2 D(@') and T > 0 there exists a unique strong solution of(3.10), u(�) 2 C([0;T ];H) (see [9, p.82] or [21, p.1399]). In view of Remark 2.1we have u(t) = V (t;u0).Let us recall the next well-known criterion of compacity of the semigroup V(see [21, p.1398]). We shall give the proof for the sake of completeness.Lemma 3.2. The semigroup V (t; �) is compact (that is, the map V (t; �) iscompact for any t > 0) if the following property is satis�ed:(L) The level sets BC de�ned byBC = fu 2 D(') j kuk � C; '(u) � Cgare compact in H for every C 2 R+.Moreover, for any B 2 B�D(')� and t > 0 there is R > 0 such thatV (t;B) � BR. 86 Jos�e ValeroProof. We must prove that for every B 2 B(D(')) and any t > 0 the setV (t;B) is precompact in H: Let BN = fu 2 H j kuk < Ng, N > 0. We take anarbitrary u0 2 BN, u0 2 D (') ; u(t) = V (t;u0). It follows from [9, p.82] thatif u0 2 D (') ; thent dudt 2 + t ddt'(u) = t�h + !u; dudt � a.e. on (0;T);and dudt 2 L2 (0;T ;H). Therefore, it follows that '(u(t)) is absolutely continu-ous on [0;T ] [8, p.189]. Hence, integrating by parts we haveZ T0 t dudt 2 dt + T'(u(T)) = Z T0 t(h + !u; dudt )dt + Z T0 '(u(t))dt;and then T'(u(T)) � 12 Z T0 t dudt 2 dt + T'(u(T))(3.11) � 12 Z T0 t(khk + ! kuk)2 dt + Z T0 '(u(t))dt:On the other hand, without loss of generality we can assume that minf'(u) :u 2 Hg = '(x0) = 0. Indeed, let x0 2 D(@'), y0 2 @'(x0): If we introduce thenew function e'(u) = '(u) � '(x0) � (y0;u � x0), the inclusiondudt + @'(u) 3 h + !uis equivalent to dudt + @e'(u) 3 h � y0 + !u = eh + !uand minfe'(u) : u 2 Hg = e'(x0) = 0. It is clear that e' satis�es (L).Hence, since h + !u(t) � du(t)dt 2 @'(u(t)) a.e. on (0;T), we have'(u(t)) � (h + !u(t) � du(t)dt ;u(t) � x0):Integrating over (0;T) we get Z T0 '(u(t))dt� 12 ku(0) � x0k2 � 12 ku(T) � x0k2 + Z T0 (khk + ! ku(t)k)ku(t) � x0kdt� 12 ku(0) � x0k2 + Z T0 (khk + ! ku(t)k)ku(t) � x0kdt:Let v0 2 D (') be �xed. Obviously, there exists a constant C such that kv (t)k �C, for all t 2 [0;T ], where v (t) = V (t;v0). It follows from inequality (2.3) thatku(t) � v (t)k � exp(!T)ku0 � v0k , for 0 � t � T: Attractors of reaction-di�usion equations 87Therefore, there are constants D1; D2 (depending on T and N, but not onu0 2 BN) for which(3.12) ku(t)k � D1, for all t 2 [0;T ] ;(3.13) Z T0 '(u(t))dt � D2 < 1:Using (3.12)-(3.13) in relation (3.11) we obtain that for any T > 0 there existsK > 0 such that '(u(T)) � K.Consider now that u0 2 D ('), u0 2 BN. We take a sequence fun0g � D (')such that un0 ! u0, as n ! 1. It is clear by inequality (2.3) that un (T) !u(T). Since ' is lower semicontinuous, we get'(u(T)) � liminfn!1 '(un (T)) � K:Then V (T;BN) � BR;where R = maxfK;D1g. It follows from (L) that V (T;BN) is precompact inH: �Thus, as a consequence of Theorem 2.11 we obtain the following result, whichis a slight improvement of Corollary 2.1 from [21].Corollary 3.3. Let ' satisfy (L) and let the following condition hold: thereexist �;C > 0 such that for all u 2 D(@'); kuk > C; for all y 2 �@'(u) onehas(3.14) (y;u) � �� � ! kuk2 � (h;u) :Then V has the global attractor <; which satis�es the next property of smooth-ness: there exists C 2 R+ such that < � BC.Proof. It is straightforward to prove by using (3.14) that (2.8) holds. Hence,by Lemma 3.2 all conditions of Theorem 2.11 hold. Finally, since the attractor< is invariant, we have V (t;<) = <; for all t � 0. Taking t > 0 we obtain byLemma 3.2 that < � BC for some C > 0, as claimed. �Further, as a consequence of Theorem 2.8 and Lemma 3.2 we obtain:Corollary 3.4. Let (L) be satis�ed and let the following conditions hold:(1) There exists u 2 Z such that for all x 2 D(@'); x =2 Z; for all y 2@'(x), one has(3.15) (y;x � u) > ! (x;x � u) + (h;x � u) :(2) The set of stationary points Z; i.e., the set of solutions of the inclusion(3.16) @'(u) 3 !u + h; u 2 H;is bounded.Then V has the global attractor <. 88 Jos�e ValeroRemark 3.5. It is possible to transform (3.15) as follows. By the de�nition ofthe subdi�erential map we have(y;x � u) � '(x) � '(u); for all y 2 @'(x):Hence, (3.15) will be satis�ed if'(x) � '(u) > !(x;x � u) + (h;x � u); for all x =2 Z:As a consequence of Corollary 2.9 we have:Corollary 3.6. Let ! = 0; h = 0; and let (L) be satis�ed. If ' is coercive, i.e.,'(u) ! +1, as kuk ! +1, then < = Z 6= ? is the global attractor of V .Proof. Since ' is coercive, we have [8, p.52](3.17) Z = fx j 0 2 @'(x)g = �x 2 D(') j '(x) = miny2D(') '(y)� 6= ?:On the other hand, using again that ' is coercive it is easy to check that Z isbounded. Moreover, in view of Remark 3.5 and since ! and h are equal to zero,in order to prove (3.15) it su�ces to satisfy the next condition: '(x) � '(u) >0; for all x =2 Z; and some �xed u 2 Z: This condition is satis�ed for any u 2 Zin view of (3.17). It remains to apply Lemma 3.2 and Corollary 2.9. �Remark 3.7. From the coerciveness of ' it follows that Z is non-empty andbounded. We can also use these two conditions instead of coerciveness. On theother hand, in Corollaries 3.4 and 3.6, < � BC for some C > 0.4. Applications to reaction-diffusion equationsIn the sequel wil be an open bounded subset of Rn with su�ciently smoothboundary �. Consider the boundary value problem:(4.18) 8><>: @u@t � �u + f(u) 3 !u + h; on ]0;T [ � ;u = 0; on ]0;T [ � �;u(0;x) = u0(x) on ;where ! � 0, f : D(f) � R ! 2R is a maximal monotone multivalued mapsuch that D(f) = R; h 2 Lp( ) and 2 � p < 1.We shall use the Banach space X = Lp ( ) as phase space. De�ne theoperator Bp : D (Bp) ! 2X;Bp (u) = f� 2 Lp ( ) : � (x) 2 �u(x) � f (u(x)) ; a.e. on g ;D (Bp) = nu 2 W2;p ( ) \ W1;p0 ( ) :9y 2 Lp ( ) such that y (x) 2 f (u(x)) a.e. on g :The operator Bp is m-dissipative (see [8, p.87]). Hence, the operator Ap (u)= Bp (u) + !u + h, D (Ap) = D (Bp), satis�es A1{A2.Recall that in Lp ( ) the dual operator F : Lp ( ) ! Lq ( ), 1p + 1q = 1; issingle-valued and de�ned by F (u) = jujp�2ukukp�2Lp : Attractors of reaction-di�usion equations 89Let us denote the semigroup generated by (4.18) in the complete metric spaceD (Bp) � Lp ( ) by Vp.Proposition 4.1. Let f satisfy the next condition: there exist M � 0; " > 0such that for all s 2 D(f); y 2 f(s) one has(4.19) y s � ! jsj2 � " jsj2 � M:Then condition (2.8) holds.Proof. For u 2 D(Ap) and y 2 f(u); a.e. on ; y 2 Lp( ), we obtain by using(4.19) and integrating over thatZ (y(x) � !u(x)) ju(x)jp�2 u(x)kukp�2Lp dx � "Z ju(x)jpkukp�2Lp dx � M Z jujp�2kukp�2Lp dx:Then, using integration by parts, for any � 2 Ap (u) we geth�;F (u)i == Z �uju(x)jp�2 u(x)kukp�2Lp dx �Z (y(x) � !u(x) + h(x)) ju(x)jp�2 u(x)kukp�2Lp dx� �(p � 1)Z jru(x)j2 ju(x)jp�2kukp�2Lp dx � "Z ju(x)jpkukp�2Lp dx+M Z jujp�2kukp�2Lp dx + Z jh(x)j ju(x)jp�1kukp�2Lp dx:Further, H�older inequality impliesh�;F (u)i � �"kuk2Lp + M j j2p kukp�2Lpkukp�2Lp + khkLp kukp�1Lpkukp�2Lp= �"kuk2Lp + M j j2p + khkLp kukLp � �"2 kuk2Lp + M j j2p + 12" khk2Lp ;where j j denotes the Lebesgue measure of . Fix � > 0. Then there existsC > 0 such that if kukLp > C we obtainh�;F (u)i � ��:Thus, condition (2.8) holds. �Remark 4.2. In the case p = 2 condition (4.19) can be weakened by putting" � �1 instead of ", where �1 is the �rst eigenvalue of �� in H10 ( ) :Theorem 4.3. If condition (4.19) is satis�ed and(4.20) 2 � p � 2nn � 2; if n � 3;then the semigroup generated by (4.18) has the global attractor <, which iscompact in Lp ( ) and bounded in H10 ( ) : If p = 2, then < is connected. 90 Jos�e ValeroProof. We have to prove �rst that the semigroup is compact. Consider �rst thecase where p = 2. Denote in this case the semigroup by V2.Note that there exists a proper convex lower semicontinuous map j : R !]�1;+1] such that f = @j, where @j is the subdi�erential of j [8, p.60].Let us determine the function ' : H ! ]�1;+1] ; H = L2( ); by'(u) = � 12 R jru(x)j2 dx + R j(u(x))dx; if u 2 D(');+1; otherwise,where D(') = fu 2 H10( ); j(u) 2 L1( )g. It is well known [8, p.88] thaty 2 @'(u) if and only if u 2 D(@'), y(x) 2 ��u(x) + f(u(x)) a.e. on , andy(�) 2 L2( ), where D(@') = fu 2 H2( ) \ H10( ) jthere exists v 2 L2( ) such that v(x) 2 f(x) a.e. on g;D(') = D(@') = L2( ):Then B2 = @' and we have problem (3.10).We must prove that '(u) satis�es (L). Indeed, since the function j(u) isbounded from below by an a�ne function [8, p.51], for any u 2 BC we haveZ (� + vu(x)) dx + 12 Z jru(x)j2 dx � '(u) � C;where �;v 2 R. Since the norms kukH10( ) and krukL2( ) are equivalent, thepreceding inequality implies that BC is bounded in H10( ). Finally, we use thefact that the inclusion H10( ) � L2( ) is compact [10, p.169].Now Lemma 3.2 implies the compacity of V2 (t; �) for any t > 0. Moreover,since for any bounded set B � L2 ( ) and t > 0 there exists C such thatV2 (t;B) � BC; the set V2 (t;B) is bounded in H10 ( ).Further, let p > 2. It is easy to check using the unicity of solutions thatVp (t;u0) = V2 (t;u0), for any u0 2 D (Bp); t � 0. Hence, for any bounded setB � D (Bp) (in the topology of Lp ( )) and t > 0 we have that Vp (t;B) =V2 (t;B) is bounded in H10 ( ). Since in view of condition (4.20) the injectionH10 ( ) � Lp ( ) is compact (see [10, p.169]), Vp (t;B) is a precompact set.We note that for any u0 2 D (Ap) there exists a unique strong solutionu(�) of inclusion (4.18) (see [8, p.146]), so that since in view of Remark 2.1,u(t) = Vp (t;u0), the existence of the global attractor follows from Proposition4.1 and Theorem 2.11.Further, since the global attractor is bounded in Lp ( ), we get that Vp (t;<)is bounded in H10( ): Therefore, we obtain that < is bounded in H10( ) by theequality Vp (t;<) = <:Finally, if p = 2 we have that D (B2) = D(@') = L2 ( ) : Then since L2 ( )is connected, the global attractor is connected. �Theorem 4.4. Let p = 2; h � 0; 0 2 f(0) and let f satisfy the next condition(4.21) ys � (��1 + !) s2; for all s 2 R; y 2 f(s); Attractors of reaction-di�usion equations 91where �1 is the �rst eigenvalue of �� in H10( ). Moreover, there exists C > 0such that if jsj � C, then(4.22) ys > (��1 + !)s2; for all y 2 f(s):Then the semigroup generated by (4.18) has the global connected attractor <,which is compact in L2 ( ) and bounded in H10 ( ).Proof. We have seen in the proof of the previous theorem that B2 = @' and(L) holds. We must verify that (3.15)-(3.16) hold. It is clear that the functionv(x) � 0 is a stationary point of V2: Let us check (3.15) for this point. It followsfrom (4.21) that for all u 2 D(@'); y 2 f(u) a.e. on , y 2 L2( ); one hasy(x)u(x) � (��1 + !)u2(x); a.e. on ;and so (y;u) = Z y(x)u(x)dx � (��1 + !)kuk2L2 :Therefore,(��u + y;u) � ! kuk2L2 ; for all u 2 D(@'); y 2 L2( );y 2 f(u) a.e. on :It remains to show that this inequality is strict when u =2 Z. Suppose that(��u + y;u) = ! kuk2L2 ; y(x) 2 f(u(x)) a.e. on , y 2 L2( ). This impliesthat y(x)u(x) = (��1 + !) u2(x) a.e. on , and u 2 spanfe1;e2; :::;emg, i.e.,it belongs to the space generated by the eigenfunctions corresponding to �1.Indeed, if y(x)u(x) > (��1 + !) u2(x) on a set 1 � such that j 1j 6= 0, then(y;u) > (��1 + !)kuk2L2 :Hence (��u + y;u) > ! kuk2L2, which is a contradiction. On the other hand,by using the equality y(x)u(x) = (��1 + !) u2(x); we have(��u;u) + (y � !;u) = (��u;u) � �1 kuk2L2 = 0:The last equality can hold only if u 2 spanfe1;e2; :::;emg. Let us suppose thatit is not the case. Then(��u;u) = 1Xi=1 �i iei; 1Xi=1 iei! = 1Xi=1 �i 2i > �1 1Xi=1 2i = �1 kuk2L2 ;and thus we obtain a contradiction. We must check that u 2 Z. We take thepartition = 1 [ 2, where u(x) = 0 a.e. on 1, u(x) 6= 0 a.e. on 2, andde�ne the function �(x) = � 0; on 1;y(x); on 2:Since 0 2 f(0), we have �(x) 2 f (u(x)) a.e. on . On the other hand �(x) =(��1 + !) u(x) a.e. on . Since ��ek = �1ek on ; for all k = 1; :::;m [10,p.192], we get ��u(x) + �(x) = !u(x) a.e. on :Thus, u 2 Z and condition (3.15) holds. 92 Jos�e ValeroNext, we must prove that Z is bounded in L2( ). If u 2 Z; then for somey 2 L2( ); y(x) 2 f(u(x)) a.e. on , we have(��u + y;u) = ! kuk2L2 ;and then, as we have already proved,y(x)u(x) = (��1 + !) u2(x) a.e. on :It follows from this equality that ju(x)j < C a.e. on , because by assumptiony(x)u(x) > �(�1 + !)u2(x) if ju(x)j � C. Thus, Z is bounded in L1( ) andconsequently in L2( ).The properties of the attractor may be proved in the same way as in Theorem4.3. �Remark 4.5. We note, as a particular case, that if ys > ��1s2, for all s 2Rnf0g ; y 2 f(s);then u � 0 is the unique stationary point. It follows fromCorollary 2.9 that < = Z = f0g:5. Dimension of the global attractor of reaction-diffusionequations in the case p = 2We are now interested in the estimation of the fractal dimension of the globalattractor of (4.18) in the Hilbert space L2 ( ) : Such estimates are well knownin the case of a di�erentiable function f (see [3]-[5], [25]). For non-di�erentiablemaps a similar result was obtained in [7] supposing that f is Lipschitz and in[18] in the case where f 2 W1;1loc (R). In all these papers the function f is atleast Lipschitz on any bounded set of R.We shall extend these results by considering a function f (s) which is Lips-chitz on a �xed bounded set [�a;a] but can be even discontinuous for s =2 [�a;a].Recall that the fractal dimension of a compact setA is de�ned bydf(A) = inffd > 0 j �f(A;d) = 0g;where �f(A;d) = lim�!0 �dn�;and n� is the minimum number of balls of radius r � � which is necessary tocover A.First we have to obtain an estimate of the elements of the global attractorin the norm of the space L1 ( ) : For this goal we need to impose a dissipativecondition which is stronger than (4.19).Proposition 5.1. Let us assume that there exist " > 0;r > 2;M � 0 such that(5.23) ys � !s2 � " jsjr � M; for any s 2 R;y 2 f (s) :Let k � 0, u0 2 D (Ak+r) and h 2 L1 ( ). Then u(t) = V2 (t;u0) satis�es:(5.24) ku(t)kLk+2� j j 1k+2 �2 1k+r ��4M" �1r + �4khkL1" � 1r�1� + �"2 (r � 2) t�� 1r�2� ;for all t > 0. Attractors of reaction-di�usion equations 93Proof. Let k � 0 be arbitrary. We note that since u0 2 D (Ak+r) ; we havethat V2 (t;u0) = Vk+r (t;u0) ; so that u(�) 2 C ([0;T ] ;Lk+r ( )) for any T > 0.Denote v (t) = ku(t)kk+2Lk+2. Due to the regularity of u0 we have that u(�) isa strong solution of (4.18), u 2 W1;1 (0;T ;Lk+r ( )) ; for any T > 0; andu(t) 2 W1;k+r0 ( ) \ W2;k+r ( ), for any t 2 [0;T ] (see [8, p.146]). It followsfrom [8, Lemma 1.2, p.100] that 1k+2 ddt kukk+2Lk+2 = R dudt jujk udx. Multiplying(4.18) by jujk u and using the Green formula and (5.23) we obtain1k+2 ddt kukk+2Lk+2 + (k + 1)R jruj2 jujk dx + "R jujk+r dx� M R jujk dx + R h jujk udx:Now the H�older inequalitieskukkLk � kukkLk+2 j j 2k+2 ; kukk+rLk+2 � kukk+rLk+r j jr�2k+2imply1k + 2 ddt kukk+2Lk+2 + " j j2�rk+2 kukk+rLk+2 � M kukkLk+2 j j 2k+2 + Z h jujk udx:Further Young inequality with exponent q = k+rk and coe�cientsa = "4 j j� rk+2 ;Ca = �"4 j j� rk+2��krgives(5.25)1k + 2 ddt kukk+2Lk+2 + 3"4 j j2�rk+2 kukk+rLk+2 � j j�"4��kr M k+rr + Z h jujk udx:Using the H�older inequalitiesZ h jujk udx � khkL1 kukk+1Lk+1 ; kukk+1Lk+1 � kukk+1Lk+2 j j 1k+2we have 1k+2 ddt ku(t)kk+2Lk+2 + 3"4 j j2�rk+2 kukk+rLk+2� j j�"4��kr M k+rr + R h jujk udx� j j�"4��kr M k+rr + khkL1 kukk+1Lk+1� j j�"4��kr M k+rr + khkL1 kukk+1Lk+2 j j 1k+2 :Now applying the Young inequality with exponent q = k+rk+1 and coe�cientsa = "4 j j1�rk+2 , Ca = �"4 j j1�rk+2�k+11�r to the last term of the inequality we get1k+2 ddt kukk+2Lk+2 + "2 kukk+rLk+2 j j2�rk+2� j j��"4��kr M k+rr + �"4��k+1r�1 khkk+rr�1L1 � :Hence,(5.26) ddtv (t) + (v (t))q � �; 94 Jos�e Valerowhere � = (k + 2) j j��"4��kr M k+rr + �"4��k+1r�1 khkk+rr�1L1 �, = "2 (k + 2) j j2�rk+2 .q = k+rk+2.Further we shall use the following version of the Gronwall lemma [29, ChapterIII, p.163]:Lemma 5.2. Let y (t) � 0 be absolutely continuous on (0;1). Suppose thatthere exist q > 1; > 0; � � 0 such that(5.27) ddty (t) + yq (t) � �:Then(5.28) y (t) � �� �1q + ( (q � 1) t)� 1q�1 ; for all t > 0:Thus, Lemma 5.2 impliesku(t)kLk+2 � � j jk+rk+22 ��"4��k+rr M k+rr + �"4��k+rr�1 khkk+rr�1L1 �� 1k+r+�"2 j j2�rk+2 (r � 2) t�� 1r�2� j j 1k+2 � 12 1k+r ��4M" �1r + �4khkL1" � 1r�1� + �"2 (r � 2) t�� 1r�2� ;for all t > 0. �Corollary 5.3. Let u0 2 C10 ( ) ; h 2 L1 ( ) and let (5.23) hold. Then forany � > 0 we have(5.29) kukL1(�;1;L1( )) � �4M" �1r + �4khkL1" � 1r�1 + 1�(r � 2) � "2� 1r�2 :Proof. First we shall show that u0 2 D (Ap) for any p � 2. We take an arbitrarysingle-valued function g (s) such that g (s) 2 f (s), for all s 2 R. Since f (s)is maximal monotone, g (s) is non-decreasing. Any non-decreasing function ismeasurable, so that the composition l (x) = g (u0 (x)) is a measurable selectionof f (u0 (x)). Let us check that l (x) 2 L1 ( ). Since u0 2 C10 ( ), there existsb > 0 such that ju0 (x)j � b, for all x 2 . Hence,g (�b) � g (u0 (x)) � g (b) ; for all x 2 ;so that l (x) 2 L1 ( ). It follows that l (x) 2 Lp ( ), for any p � 2; and thatl (x) 2 f (u0 (x)), a.e. on . Therefore, u0 2 D (Ap), for all p � 2:Proposition 5.1 implies that (5.24) is satis�ed for any k � 0. Passing to thelimit as k ! 1 we obtainku(t)kL1 � �4M" �1r + �4khkL1" � 1r�1 + 1�(r � 2) t"2� 1r�2 ; for all t > 0; Attractors of reaction-di�usion equations 95and then for any � > 0 we getkukL1(�;T;L1( )) � �4M" �1r + �4khkL1" � 1r�1 + 1�(r � 2) � "2� 1r�2 : �Corollary 5.4. Let u0 2 L2 ( ) ; h 2 L1 ( ) and let (5.23) hold. Then in-equality (5.29) holds:Proof. Let un0 2 C10 ( ) be a sequence such that un0 ! u0 in L2 ( ). Then in-equality (2.3) implies that un (t) = V2 (t;un0) converges to u(t) = V2 (t;u0)in C ([0;T ] ;L2 ( )). The sequence un is bounded in L1 (�;T ;L1 ( )) by(5.29). Hence, there exists a subsequence converging to u weakly star inL1 (�;T ;L1 ( )). Therefore, (5.29) holds. �Now we can obtain an estimate of the elements of the global attractor <(which exists in view of Theorem 4.3) in the norm of the space L1 ( ) :Theorem 5.5. Let h 2 L1 ( ) and let (5.23) hold. Then for any y 2 < thefollowing estimate holds:(5.30) kykL1( ) � �4M" �1r + �4khkL1" � 1r�1 :Proof. Let � > 0, y 2 < be arbitrary. We choose � such that�(r � 2) �"2�� 1r�2 < �:Since < is invariant, there exists u0 2 < for which y = u(2�) = V2 (2�;u0).Corollary 5.4 implieskukL1(�;3�;L1( )) � �4M" �1r + �4khkL1" � 1r�1 + �:Since u 2 C (�;3�;L2 ( )) and it is bounded in L1 (�;3�;L1 ( )), we haveu 2 Cw (�;3�;Lq ( )) ; for any2 � q < 1, where Cw denotes the weak topology(see [24, p.275]). Hence,kykLq( ) = ku(2�)kLq( ) � �4M" �1r + �4khkL1" � 1r�1 + �! j j1q ;so that kykL1( ) = ku(2�)kL1( ) � �4M" �1r + �4khkL1" � 1r�1 + �:Since � is arbitrary, (5.30) holds. �Theorem 5.6. Let h 2 L1 ( ) and let (5.23) hold. Suppose that there existsa > 0 such that(5.31) �4M" �1r + �4khkL1" � 1r�1 � a 96 Jos�e Valeroand in the interval [�a;a] the function f (s) is Lipschitz (with Lipschitz constant�).Then there exists K depending on and n for which(5.32) df (<) � K (! + �)n2 :Proof. Let f�Ng1N=1 be the eigenvalues of �� in H10 ( ) ; PN be the ortho-projector to the subspace generated by the eigenfunctions corresponding to the�rst N eigenvalues and QN = I � PN. It follows from [7, Theorem 4] that ifwe �nd t > 0, l 2 [1;+1); � 2 (0; 1p2) such that(5.33) kV2(t;u0) � V2(t;v0)k � l ku0 � v0k ;(5.34) QNV2(t;u0) � QNV2(t;v0) � � ku0 � v0k ;for all u0;v0 2 <, then for any � > 0 such that �p26l�N �p2��� = � < 1 thenext estimate holds(5.35) df(A) � N + �:In view of inequality (2.3) condition (5.33) holds with l (t) = exp(!t) :Further we note that from Theorem 5.5 and the Lipschitz condition of f on[�a;a] it follows that(5.36) kf (u) � f (v)kL2 � � ku � vkL2 , for any u;v 2 <:We take two arbitrary initial conditions u0;v0 2 <. Let now w (t) = u(t) �v (t) ; wN (t) = QNw (t) ; where u(t) = V2 (t;u0) ; v (t) = V2 (t;v0). From (4.18)we can easily obtain12 ddt wN (t) 2L2 + rwN 2L2 + �f (u) � f (v) ;wN� = !�w;wN� :Using the inequality rwN 2L2 � �N+1 wN (t) 2L2 ; (2.3) and (5.36) we obtainddt wN (t) 2L2 + 2�N+1 wN (t) 2L2 � 2(! + �)exp(2!t)kw (0)k2L2 :Multiplying both sides by exp(2�N+1t) and integrating over (0; t) we get wN (t) 2L2� kw (0)k2L2 �exp(�2�N+1t) + !+�!+�N+1 (exp (2!t) � exp (�2�N+1t))�� kw (0)k2L2 �exp(�2�N+1t) + !+�!+�N+1 exp(2!t)�= �2 (t;N)kw (0)k2L2 :Repeating exactly the same proof of Theorem 7 from [7] we can obtain thatfor t = log(p212)�N+1�! there exists a constant D (depending on ) such that ifN = h(D (! + �))n2 i, where [x] denotes the integer part of x, then � (t;N) < 1p2and (12� (t;N) l (t))N < 1, so that �p26l�N �p2��� < 1 for � = N. It followsfrom (5.35) that df(A) � 2N � K (! + �)n2 ; Attractors of reaction-di�usion equations 97with K = 2Dn2 : �Remark 5.7. We note that if f is locally Lipschitz, then it is Lipschitz onany interval [�b;b]. 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Wang, Attractors for reaction-di�usion equations in unbounded domains, Physica D128 (1999), 41-52. Received October 2000Revised version April 2001 J.ValeroUniversidad Cardenal Herrera CEUCampus de ElcheC/Comissari, n 303203 Elche(Alicante), SpainE-mail address: valer.el@ceu.es