global attractors for multivalued random dynamical systems generated by random differential...

Journal of Mathematical Analysis and Applications 260, 602–622 (2001)doi:10.1006/jmaa.2001.7497, available online at http://www.idealibrary.com on

Global Attractors for Multivalued RandomDynamical Systems Generated by Random Differential

Inclusions with Multiplicative Noise1

T. Caraballo and J. A. Langa

Departamento de Ecuaciones Diferenciales y Analisis Numerico,Universidad de Sevilla, Apdo. de Correos 1160, 41080 Sevilla, Spain

E-mail: [email protected], [email protected]

and

J. Valero

Universidad Cardenal Herrera CEU, Comissari 3, 03203 Elche, Alicante, Spain

Submitted by H. Frankowska

Received August 10, 2000

In this paper we consider a stochastic differential inclusion with multiplicativenoise. It is shown that it generates a multivalued random dynamical system forwhich there also exists a global random attractor. © 2001 Academic Press

Key Words: global attractor; multivalued random dynamical systems; stochasticdifferential inclusions.

1. INTRODUCTION

In our previous paper Caraballo et al. [7], we introduced the concept ofa multivalued random dynamical system (MRDS) by generalizing that ofrandom dynamical systems (see Arnold [2]) in a suitable manner. We alsodeveloped a theory for the existence of global attractors for these multival-ued semiflows which, in essence, seems the natural extension of the theory

1 The first two authors have been partly supported by DGICYT Proyecto PB98-1134. Thethird author has been supported by PB-2-FS-97 grant (Fundacion Seneca de la ComunidadAutonoma de Murcia).

602

0022-247X/01 $35.00Copyright © 2001 by Academic PressAll rights of reproduction in any form reserved.

global attractors 603

developed by Crauel and Flandoli [9]. Finally, the theory was applied toa class of stochastic differential inclusions with additive noise. The mainaim of this paper is to prove similar results to the ones in [7] but foranother important family of stochastic differential inclusions, in fact, theperturbed ones by means of a multiplicative noise. It is worth pointing outthat the way in which we construct the MRDS in [7] is by making a suitablechange of variable which transforms the problem into a family of inclusionsdepending on a parameter. In fact, when one deals with a general stochas-tic differential inclusion (see, e.g., Ahmed [1], Da Prato and Frankowska[10]), it is not always possible to make a change that conduces the problemto another equivalent one depending on a parameter. But there exists twoimportant situations which permit us to do that: when the noise is addi-tive or multiplicative. As the techniques to deal with these two problemsare rather different, we aim to show in this paper how we can construct aMRDS and prove the existence of the global random attractor in the mul-tiplicative case. In Section 2, we collect the definitions and properties ofthe MRDS and also include the sufficient condition ensuring the existenceof the global random attractor. In Section 3, we consider the multivaluedsemiflow generated by a differential inclusion perturbed by a multiplicativenoise and prove that there exists the global random attractor. Finally, someapplications are included in Section 4 to illustrate the theory.

2. MULTIVALUED RANDOM DYNAMICALSYSTEMS AND ATTRACTORS

For the complete details and proofs of the results in this section, thereader is referred to Caraballo et al. [7]. Let �X�dX� be a complete andseparable metric space with the Borel σ-algebra ��X�. Let �� be aprobability space and θt � � → � a measure preserving group of transfor-mations in � such that the map �t� ω� �→ θtω is measurable and satisfying

θt+s = θt ◦ θs = θs ◦ θt θ0 = Id�

The parameter t takes values in � endowed with the Borel σ-algebra ��.Definition 1. A set valued map G� �+ × � × X → C�X� �C�X�

denotes the set of non-empty closed subsets of X) is called a multivaluedrandom dynamical system (MRDS) if is measurable (see Aubin andFrankowska [4, Definition 8.1.1]) and satisfies

(i) G�0�ω� = Id on X;(ii) G�t + s�ω�x = G�t� θsω�G�s�ω�x, for all t� s ∈ �+� x ∈ X,

ω ∈ � (perfect cocycle property).

604 caraballo, langa, and valero

Remark 2. When (ii) holds identically (that is, on a set of measure onewhich does not depend either on t or s), we call G a perfect cocycle. Wecall G a crude cocycle if (ii) holds for fixed s and all t ∈ �+� x ∈ X� �-a�s�(where the exceptional set Ns can depend on s). We call G a very crudecocycle if (ii) holds for fixed s� t ∈ �+, for all x ∈ X��-a�s� (where theexceptional set Ns� t can depend on both s and t).

Remark 3. Throughout this paper all assertions about ω are assumed tohold on a θt invariant set of full measure, where this set does not dependon the time variable t. In order to avoid any confusion we shall write “forall ω ∈ �” instead of “for �-a�a�” when the time variable appears.

Recall the definition of Hausdorff semi-distance between bounded setsof X. For any A�B ⊂ X bounded put dist�A�B� = supy∈A infx∈B dX�y� x�.Definition 4. The MRDS G is said to be upper semicontinuous if for

all t ∈ �+ and ω ∈ � it follows that given x ∈ X and a neighbourhoodof G�t� ω�x��G�t� ω�x�, there exists δ > 0 such that if dX�x� y� < δ thenG�t� ω�y ⊂ ��G�t� ω�x�. On the other hand, G is called lower semicontinu-ous if for all t ∈ �+ and ω ∈ �, given xn → x�n→+∞� and y ∈ G�t� ω�x,there exists yn ∈ G�t� ω�xn such that yn → y.

It is said to be continuous if it is upper and lower semicontinuous.

Definition 5. A closed random set D is a map D� �→ C�X�, which ismeasurable. The measurability must be understood in the sense of Castaingand Valadier [8] for measurable multifunctions, that is, �D�ω��ω∈� is mea-surable if given x ∈ X the map ω ∈ � �→ dist�x�D�ω�� is measurable.

A closed random set D�ω� is said to be negatively (resp. strictly) invariantfor the MRDS G if

D�θtω� ⊂ G�t� ω�D�ω� �resp� D�θtω� = G�t� ω�D�ω��∀ t ∈ �+�ω ∈ ��

Remark 6. This concept of measurability and the previous one areequivalent (see Aubin and Frankowska [4, Theorem 8.3.1]).

Let us assume the following conditions for the MRDS G:

(H1) There exists an absorbing random compact set B�ω�, that is,for �-almost all ω ∈ � and every bounded set D ⊂ X, there exists tD�ω�such that for all t ≥ tD�ω�

G�t� θ−tω�D ⊂ B�ω�� (1)

(H2) G�t� ω�� X → C�X� is upper semicontinuous, for all t ∈ �+

and ω ∈ �.


Recall the definition of the limit set ��D�ω� = �D�ω� of a boundedsubset D ⊂ X as

�D�ω� =⋂T≥0

⋃t≥T

G�t� θ−tω�D� (2)

The following proposition gives some properties of the limit sets.

Proposition 7. Assumeconditions (H1)and (H2)hold.Then, for�-almostallω ∈ � and everyD ⊂ X bounded, it follows:

(i) �D�ω� ⊂ B�ω� is nonvoid and compact.(ii) �D�ω� is negatively invariant, that is, G�t� ω��D�ω� ⊇ �D�θtω�

for all t ∈ �+. If G is lower semicontinuous, then �D�ω� is strictly invariant.(iii) �D�ω� attracts D,

limt→+∞ dist�G�t� θ−tω�D��D�ω�� = 0�

Definition 8. The closed random set ω �→ ��ω� is called a global ran-dom attractor of the MRDS G if �-a.s.

(i) G�t� ω��ω� = ��θtω�, for all t ≥ 0 (that is, it is strictlyinvariant);

(ii) for all bounded D ⊂ X, limt→+∞ dist�G�t� θ−tω�D��ω�� = 0;(iii) ��ω� is compact.

We now have the following theorem on the existence of random attrac-tors for MRDS:

Theorem 9. Let (H1)–(H2) hold, the map �t� ω� ∈ �+ ×� �→ G�t� ω�Dbe measurable for all deterministic bounded sets D ⊂ X, and the map�t� ω� x� ∈ X �→ G�t� ω�x have compact values. Then

��ω� = ⋃D⊂Xbounded

�D�ω�

is a global random attractor for G (measurable with respect to � ). It is uniqueand the minimal closed attracting set.

3. MRDS GENERATED BY A DIFFERENTIAL INCLUSIONWITH MULTIPLICATIVE NOISE

Now, let X be a real separable Hilbert space with the scalar product �·� ·�and the norm �·�. Consider the following stochastic differential inclusionin Stratonovich’s sense{

dudt∈ Au�t� + F�u�t�� + σu�t� ◦ dw�t�

dt� t ∈ �0� T ��

u�0� = u0�(3)


where σ ∈ ��A� D�A� ⊂ X → X is a linear operator and w�t� is a two-sided, i.e., t ∈ �, real Wiener processes with w�0� = 0. Although we couldconsider the more general case of a finite sum of the form

∑ni=1 σiu ◦

�dwi�t�/dt�, we prefer to treat this case for the sake of clarity, bearing inmind that no new difficulties appear in dealing with this more general case.Let us introduce the next conditions:

(A) The operator A is m-dissipative, i.e., ∀ y ∈ D�A��Ay� y� ≤ 0,and Im�A− λI� = X�∀λ > 0.

(F1) F � X → Cv�X�, where Cv�X� is the set of all non-empty,bounded, closed, convex subsets of X.

(F2) The map F is Lipschitz on D�A�, i.e., ∃C ≥ 0 such that ∀ y1� y2 ∈D�A�

distH�F�y1�� F�y2�� ≤ C�y1 − y2��where distH�·� ·� denotes the Hausdorff metric of bounded sets, i.e.,

distH�A�B� = max�dist�A�B�� dist�B�A��Let us consider the Wiener probability space �� defined by

� = �ω ∈ C�� ω�0� = 0��equipped with the Borel σ-algebra � , the Wiener measure �, and the usualuniform convergence on bounded sets of �. Recall that w�t��ω� �= ω�t�.We make the change of variable v�t� = e−σω�t�u�t�. If we denote2 α�t� =α�t� ω� = e−σω�t� then the inclusion (3) turns into{

dvdt∈ Av�t� + α�t�F�α−1�t�v�t��

v�0� = v0 = u0�(4)

We shall define the multivalued map F � �0� T � ×�×X → Cv�X�,F�t� ω� x� = α�t�F�α−1�t�x��

It is easy to obtain from �F2� the existence of constants D1�D2 such that

�F�x��+ ≤ D1 +D2�x��where �F�x��+ = supy∈F�x� �y�. Hence,

�F�t� ω� x��+ ≤ α�t��D1 +D2�α−1�t�x�� = α�t�D1 +D2�x� = n�t� ω� x��

2We will omit ω when no confusion is possible.


It follows that F satisfies the next property:

(F3) For any x ∈ X there exists n�·� ∈ L1�0� T � depending on xand ω such that

�F�t� ω� x��+ ≤ n�t�� a.e. in �0� T ��

On the other hand, it is clear that F satisfies conditions (F1)–(F2) forany fixed t ∈ �0� T � and ω ∈ �, where the constant C does not depend on tor ω.Consider also the equation{

dv�t�dt

= Av�t� + f �t��v�0� = v0�

(5)

where f �·� ∈ L1��0� T ��X�.Definition 10. The function u� �0� T � → X is called a strong solution

of problem (5) if:

(1) u�·� is continuous on �0� T � and u�0� = u0;

(2) u�·� is absolutely continuous on any compact subset of �0� T � andalmost everywhere (a.e.) differentiable on �0� T �;

(3) u�·� satisfies (5) a.e. on �0� T � (hence, u�t� ∈ D�A�, for a.a.t ∈ �0� T �).Definition 11. The continuous function v� �0� T � → X is called an inte-

gral solution of problem (5) if:

(1) v�0� = v0;

(2) ∀ ξ ∈ D�A�,

�v�t� − ξ�2 ≤ �v�s� − ξ�2 + 2∫ t

s�f �τ� +Aξ� v�τ� − ξ�dτ� t ≥ s� (6)

It is well known (see Barbu [5, p. 124]) that any strong solution of prob-lem (5) is an integral solution.

Definition 12. The function v� �0� T � ×�→ X is said to be an integralsolution of problem (4) if for any ω ∈ �:

(1) v�·� = v�·�ω�� 0� T � → X is continuous.

(2) v�0� = v0;

(3) For some selection f ∈ L1��0� T ��X�� f �t� ∈ F�t� ω� v�t�� a.e. on�0� T �, the inequality (6) holds.


If condition (A) holds and f ∈ L1��0� T ��X�, then ∀ v0 ∈ D�A�there exists a unique integral solution v�·� of (5) for each T > 0 (seeBarbu [5, p. 124]). We shall denote this solution by v�·� = I�v0�f �·�.Moreover, for any integral solutions vi�·� = I�vi0�fi�·�� i = 1, 2, the nextinequality holds:

�v1�t� − v2�t�� ≤ �v1�s� − v2�s�� +∫ t

s�f1�τ� − f2�τ��dτ� t ≥ s� (7)

If (A), (F1)–(F3) hold, then ∀ v0 ∈ D�A� there exists at least one integralsolution of (4) for each T > 0 (see Tolstonogov [13, Theorem 3.1]). More-over, for any z�·� = I�z0�g�·�� g�·� ∈ L1��0� T ��X�, and any v0 ∈ D�A�there exists an integral solution v�·� = I�v0�f �·� of (4) such that

�v�t� − z�t�� ≤ ξ�t�� ∀ t ∈ �0� T �� (8)

�f �t� − g�t�� ≤ ρ�t� + 2Cξ�t�� a.e. on �0� T �� (9)

where

ρ�t� = 2 dist(g�t�� F�t� ω� z�t��)�

ξ�t� = �v0 − z0� exp�2Ct� +∫ t

0exp�2C�t − s��ρ�s�ds�

Since T > 0 is arbitrary, each solution can be extended on �0�∞�. Letus denote by ��v0�ω� the set of all integral solutions of (4) such thatv�0� = v0. We define the maps G� �+ ×�×D�A� → P�D�A�� θs� �→ �as

G�t� ω�v0 = �α−1�t�v�t� � v�·� ∈ ��v0�ω��θsω = ω�s + ·� −ω�s� ∈ ��

Proposition 13. Let (A), (F1), (F2) hold. Then G satisfies the cocycleproperty

G�t1 + s�ω�x = G�t1� θsω�G�s�ω�x� ∀ t1� s ≥ 0� x ∈ X�ω ∈ ��Proof. First let y ∈ G�t1 + s�ω�x. Then y = y�t1 + s� = α−1�t1 + s�×

v�t1 + s�, where v�·� ∈ ��x�ω�. It is clear that y�s� ∈ G�s�ω�x. We have toprove that y ∈ G�t1� θsω�y�s�. Define z�t� = α−1�s�α�t + s�y�t + s��∀ t ≥0� g�t� = α−1�s�f �t + s�, a.e. t > 0, where α�t�y�t� = v�t� = I�x�f �t�. Forany r ≤ t� ξ ∈ D�A� we obtain

�z�t� − ξ�2 = �α−1�s�v�t + s� − ξ�2 = α−2�s��v�t + s� − α�s�ξ�2

≤ α−2�s��v�r + s� − α�s�ξ�2 + 2α−2�s�

×∫ t

r�f �τ + s� + α�s�Aξ� v�τ + s� − α�s�ξ�dτ

≤ �z�τ� − ξ�2 + 2∫ t

r�g�τ� +Aξ� z�τ� − ξ�dτ�


On the other hand,

g�t� ∈ α−1�s�α�t + s�F�α−1�t + s�v�t + s��= α−1�s�α�t + s�F�y�t + s��= α−1�s��t + s�F�α�s�α−1�t + s�z�t��

and, as3 α−1�s�α�t + s� = eσ�ω�s�−ω�t+s�� = e−σ�θsω��t�, then

g�t� ∈ e−σ�θsω��t�F�eσ�θsω��t�z�t�� = F�t� θsω� z�t��

Therefore, since z�0� = y�s�, it follows that z�·� ∈ ��y�s�� θsω�. Since y =y�t1+ s� = α�s�α−1�t1+ s�z�t1� = eσ�θsω��t1�z�t1�� we get y ∈G�t1� θsω�y�s�.Hence,

G�t1 + s�ω�x ⊂ G�t1� θsω�G�s�ω�x�

Conversely, let y ∈ G�t1� θsω�G�s�ω�x. Then there exist v1�·� ∈ ��x�ω�and v2�·� ∈��y1�s�� θsω�, y1�s�=α−1�s�v1�s�, such that y = eσ�θsω��t1�v2�t1�.Let

z�t� ={v1�t�� if 0 ≤ t ≤ s,α�s�v2�t − s�� if s ≤ t,

f �t� ={f1�t�� if 0 ≤ t ≤ s,α�s�f2�t − s�� if s ≤ t,

where v1�·� = I�x�f1�·�� v2�·� = I�y1�s��f2�·�. We have to check that for a.a.t ∈ �0� T �� f �t� ∈ F�t� ω� z�t�� = α�t�F�α−1�t�z�t��. If t ≤ s it is obviousthat f �t� ∈ α�t�F�α−1�t�z�t��. If t ≥ s we have

f �t� = α�s�f2�t − s�∈ α�s�e−σ�θsω��t−s�F�eσ�θsω��t−s�v2�t − s��= α�s�e−σ�ω�t�−ω�s��F�eσ�ω�t�−ω�s��v2�t − s��= α�t�F�α−1�t�z�t�� = F�t� ω� z�t��

3Notice that α�t + s�ω� = α�t� θsω�α�s�ω��


It remains to prove that z�·� satisfies (6) for any r ≤ t. If t ≤ s theinequality is evident. If r < s < t we get

�z�t� − ξ�2 = �α�s�v2�t − s� − ξ�2 = α2�s��v2�t − s� − α−1�s�ξ�2

≤ α2�s��v2�0� − α−1�s�ξ�2

+ 2α2�s�∫ t

s�f2�τ − s� + α−1�s�Aξ� v2�τ − s� − α−1�s�ξ�dτ

= �v1�s� − ξ�2

+ 2α2�s�∫ t

s�f2�τ − s� + α−1�s�Aξ� v2�τ − s� − α−1�s�ξ�dτ

≤ �v1�r� − ξ�2 + 2∫ s

r�f1�τ� +Aξ� v1�τ� − ξ�dτ

+ 2∫ t

s�α�s�f2�τ − s� +Aξ�α�s�v2�τ − s� − ξ�dτ

= �z�r� − ξ�2 + 2∫ t

r�f �τ� +Aξ� z�τ� − ξ�dτ�

Finally, the case s ≤ r is similar. Therefore, z�·� ∈ ��x�ω� andy = y�t1 + s� = α−1�t1 + s�z�t1 + s� ∈ G�t1 + s�ω�x�

Hence,

G�t1� θsω�G�s�ω� ⊂ G�t1 + s�ω�x�and the proof is complete.

Let C��0� T ��X�� 0 < T ≤ ∞, be the Banach space of continuous func-tions from �0� T � into X. By πT � C��0� T ′��X� → C��0� T ��X�� T ′ > T , wedenote the projection operator

πT �y�·�� = �y�·� ∈ C��0� T ��X� � y�s� = y�s��∀ t ∈ �0� T ��Lemma 14. Let (A), (F1), (F2) hold. Then for any v0 ∈ D�A��ω ∈ � the

set πT ��v0�ω�� is bounded in C��0� T ��X�. Consequently, for each t ≥ 0,v0 ∈ D�A��ω ∈ � the set G�t� ω�v0 is bounded.

Proof. We have seen before (see (F3)) that there exist constantsD1�D2 ≥ 0 such that ∀x ∈ D�A��∀ y ∈ F�t� ω� x�,

�y� ≤ α�t�D1 +D2�x�� (10)

We take a fixed T > 0. We shall denote by z�·� ∈ C��0� T ��X� the uniqueintegral solution of the equation{

dzdt= Az�t�� 0 ≤ t ≤ T�

z�0� = v0�


Let r0 = max��z�t�� 0 ≤ t ≤ T� and let r�·� ∈ C��0� T �� be the solution tothe equation {

r ′�t� = D1 +D2r�t�� 0 ≤ t ≤ T�r�0� = r0�

Consider an arbitrary solution v�·� ∈ ��v0�ω�� v�·� = I�v0�f �·�. Then itfollows from (7) and (10) that

�v�t�� ≤ �z�t�� +∫ t

0�f �τ��dτ

≤ r0 +∫ t

0�α�τ�D1 +D2�v�τ��dτ� ∀ t ≤ T�

Let K�ω� be such that supt∈�0�T ��α�t�� ≤ K�ω�. Then

�v�t�� ≤ r0 +D1K�ω�t +D2

∫ t

0�v�τ��dτ�

Using the Gronwall Lemma we have that for any t ∈ �0� T � the next inequal-ities are satisfied:{

�v�t�� ≤(r0 + D1

D2

)exp�D2t� − D1

D2= r�t� ω�� if D2 �= 0,

�v�t�� ≤ r0 + D1t� if D2 = 0,(11)

where D1 = D1K�ω�. Hence, πT��v0�ω� is bounded in C��0� T ��X�,∀T ≥ 0, ∀ v0 ∈ D�A��∀ω ∈ �. It is obvious from the definition of G thatthe set G�t� ω�v0 is bounded for each t ≥ 0� v0 ∈ D�A��ω ∈ �.

For T > 0 and bounded B ⊂ D�A�, let us denote ��B�ω� =∪x∈B��x�ω� and

M�B�ω�T �=�f �·�∈L1��0�T ��X� �v�·�=I�x�f �·��v�·�∈πT��B�ω��Lemma 15. Let (A), (F1), (F2) hold. Then for any T > 0�ω ∈ � and any

bounded set B ⊂ D�A� the sets M�B�ω� T � and πT��B�ω� are bounded inL∞��0� T ��X� and C��0� T ��X�, respectively.Proof. Let x ∈ B�T > 0 be arbitrary. In view of Lemma 14, there exists

K1�ω� > 0 such that for any v�·� ∈ ��x�ω�,�v�t�� ≤ K1�ω�� ∀ t ∈ �0� T ��

We take an arbitrary u�·� ∈ ��B�ω�� u�0� = y ∈ B. Then in view of (8)there exists v�·� ∈ ��x�ω� such that

�v�t� − u�t�� ≤ exp�2CT ��x− y�� on �0� T ��


Hence

�u�t�� ≤ �v�t�� + exp�2CT ��x− y�≤ K1�ω� + exp�2CT �K2� on �0� T ��

where K1�ω�, K2 depend on B. We have proved that πT��B�ω� is boundedin C��0� T ��X�.Further, we must prove that M�B�ω� T � is bounded in L∞��0� T ��X�.

Let f �·� ∈M�B�ω� T � be arbitrary. Then, there exist x ∈ B� x�·� ∈ ��x�ω�,such that x�·� = I�x�f �·�� f �t� ∈ F�t� ω� x�t��, a.e. on �0� T �. In view ofproperty (F3)

�f �t�� ≤ α�t�D1 +D2�x�t�� a.e. on �0� T ��Since πT��B�ω� is bounded in C��0� T ��X�, we obtain the required

result.

This semi-distance dist�·� ·�, defined on the Hilbert space X, has thefollowing useful properties, which are easy to check:

(1) ∀Ai�Bi ⊂ X� i = 1� 2�

dist�A1 + B1�A2 + B2� ≤ dist�A1�A2� + dist�B1� B2��(2) ∀A�B�C ⊂ X,

dist�A�B� ≤ dist�A�C� + dist�C�B��(3) ∀A ⊂ X�α�β ∈ �,

dist�αA�βA� ≤ �α− β��A�+�(4) ∀A�B ⊂ X�α ∈ �,

dist�αA�αB� = �α� dist�A�B��Proposition 16. Let (A), (F1), (F2) hold. For any T > 0, ωn → ω0,

un0 → u0

dist�G�t� ω0�u0�G�t� ωn�un0� → 0� as n→∞�

uniformly with respect to t ∈ �0� T �.Proof. Let z ∈ G�t� ω0�u0� t ∈ �0� T �, be arbitrary. Then z = α−10 �t�z�t�,

where z�·� = I�u0�f0�·�, α0�t� = e−σω0�t�, f0�τ� ∈ α0�τ�F�α−10 �τ�z�τ�� =F�τ�ω0� z�τ��, a.e. on �0� T �.Consider now the sequences ωn → ω0, u

n0 → u0. Denote

ρn�τ� = 2 dist�f0�τ�� F�τ�ωn� z�τ��= 2 dist�f0�τ�� αn�τ�F�α−1n �τ�z�τ��


In view of (8) there exist solutions vn�·�= I�un0�fn�·�, f �τ� ∈ F�τ�ωn�vn�τ��, such that

�vn�t� − z�t�� ≤ �un0 − u0� exp�2Ct�

+∫ t

0exp�2C�t − s��ρn�s�ds on �0� T ��

Taking into account that the functions α0�·�� α−1�·�� αn�·�� α−1n �·� are uni-formly bounded on C��0� T ��+�, the convergence αn → α0, α−1n → α−10in C��0� T ��+�, (F2)–(F3), Lemma 14 and the properties of the semi-distance dist�·� ·� cited above, we have

ρn�τ� ≤ 2 dist(F�τ�ω0� z�τ�� F�τ�ωn� z�τ��

)= 2 dist�α0�τ�F�α−10 �τ�z�τ�� αn�τ�F�α−1n �τ�z�τ��≤ 2 dist�α0�τ�F�α−10 �τ�z�τ�� αn�τ�F�α−10 �τ�z�τ��+2 dist�αn�τ�F�α−10 �τ�z�τ�� αn�τ�F�α−1n �τ�z�τ��

≤ 2�α0�τ� − αn�τ�� F�α−10 �τ�z�τ��+

+2Cαn�τ��z�τ�� α−10 �τ� − α−1n �τ��≤ 2�D1 +D2α

−10 �τ��z�τ��α0�τ� − αn�τ��

+2Cαn�τ��z�τ�� α−10 �τ� − α−1n �τ��≤ K1�α0�τ� − αn�τ�� +K2�α−10 �τ� − α−1n �τ��

so that ρn�τ� → 0, as n → ∞, uniformly on �0� T �. We note that K1�K2depend only on u0�ω�ωn, and T .Therefore, for any ε > 0 we can choose N > 0 (which does not depend

on either t or z ∈ G�t� ω�u0� such that ∀n ≥ N�∀ τ ∈ �0� T �

ρn�τ� ≤ε

2T exp�2CT � � �un0 − u0� ≤

ε

2 exp�2CT � �

Hence, �vn�t� − z�t�� ≤ ε�∀n ≥ N�∀ t ∈ �0� T �.Further, vn = α−1n �t�vn�t� ∈ G�t� ωn�un0 and

�vn − z� = �α−1n �t�vn�t� − α−10 �t�z�t��≤ �α−10 �t��vn�t� − z�t�� + ��α−1n �t� − α−10 �t��vn�t��≤ R1ε+ R2ε�

Since N does not depend on either t or z ∈ G�t� ω�u0, we obtain thestatement of the proposition.


Lemma 17. Let (A), (F1), (F2) hold and the semigroup S�t� ·� generatedby the operator A be compact. Then for any T > 0, ω ∈ �, and u0 ∈ D�A�the set πT��u0�ω� is compact and the semiflow G has compact values.

Proof. From Tolstonogov [13, Theorem 3.4], it follows that for eachv0 ∈ D�A�, T > 0 the set πT��u0�ω� is compact in C��0� T ��X�. Hence,the set

K = �y ∈ X� y = v�T �� v�·� ∈ πT��u0�ω��is compact, so that G�T�ω�u0 = α−1�T �K, as the continuous image of acompact set, is compact as well.

Proposition 18. Let (A), (F1), (F2) hold and the semigroup S�t� ·� gener-ated by the operator A be compact. Then for any tn → t, ωn → ω0, u

n0 → u0

one has

dist�G�t� ω0�u0�G�tn�ωn�un0� → 0� as n→∞�

Hence, the map �t� ω� u0� �→ G�t� ω�u0 is lower semicontinuous.

Proof. Take T such that tn < T , t < T�∀n. In view of Lemma 17 the setπT��u0�ω0� is equicontinuous. Then for any ε > 0 there exist N > 0 suchthat ∀n > N�∀ v�·� ∈ πT��u0�ω0�,

�v�tn� − v�t�� < ε�

The continuity of α−1�·� and the compactness of πT��u0�ω0� implies theexistence of constants R1� R2 > 0 such that

�α−1�tn�v�tn� − α−1�t�v�t�� ≤ �α−1�t��v�tn� − v�t��≤ ��α−1�tn� − α−1�t��v�tn��≤ R1ε+ R2ε�

Hence,

dist�G�t� ω0�u0�G�tn�ω0�u0� → 0� as n→∞�

Further, the last inequality and Proposition 16 imply that for any ε > 0there exist N > 0 such that ∀n > N

dist�G�t� ω0�u0�G�tn�ωn�un0� ≤ dist�G�t� ω0�u0�G�tn�ω�u0�+ dist�G�tn�ω�u0�G�tn�ωn�un0�

≤ ε

2+ ε

2= ε�

which completes the proof.


Theorem 19. Let (A), (F1), (F2) hold and the semigroup S�t� ·� gener-ated by the operator A be compact. Then G generates a MRDS.

Proof. Proposition 13 and Lemma 17 imply that the cocycle property issatisfied and that G has compact values. It remains to prove that the mul-tivalued map G� �+ × � ×D�A� → C�D�A�� is measurable with respectto the σ-algebra ��+� ⊗ � ⊗��D�A��. Since this σ-algebra contains allopen sets of the complete separable metric space �+ × � ×D�A� and Ghas closed values, the lower semicontinuity of G proved in Proposition 18implies the measurability of the semiflow. Indeed, the map G is measurableif and only if the inverse image of any open set � ⊂ D�A�

G−1�� = {�t� ω� x� ∈ �+ ×�×D�A�� G�t� ω�x ∩ � �= #}is measurable (see Aubin and Frankowska [4, Theorem 8.3.1]). Since themap G is lower semicontinuous, the inverse image of any open set is openand then measurable (see Aubin and Frankowska [4, p. 40]).

3.1. Existence of the Global Random Attractor

In order to obtain the existence of a compact absorbing set we needmore regularity of the integral solutions. Namely, we shall suppose thateach integral solution of (4) is, in fact, a strong solution of (5).

Proposition 20. Let (A), (F1), (F2) hold. Suppose that each integralsolution of �4�� v�·� = I�u0�f �·� is a strong solution of (5). Let there existconstants δ > 0�M ≥ 0 such that ∀u ∈ D�A�� y ∈ F�u�,

�y� u� ≤ �−δ+ ε��u�2 +M� (12)

where ε ≥ 0 is the biggest constant such that

�Au�u� ≤ −ε�u�2� ∀u ∈ D�A�� (13)

Then there exists a random radius r�ω� > 0 such that for �- almost all ω ∈ �and any bounded set B ⊂ D�A� we can find T �B� = T �B�ω� ≥ 1 for which

�G�−1+ t0� θ−t0ω�u0�+ ≤ r�θ−1ω�� ∀ t0 ≥ T �B��∀u0 ∈ B�Remark 21. We observe that, since A is m-dissipative, �Au�u� ≤ 0,

∀u ∈ D�A�.Proof. We note that for any y ∈ G�r� θsω�u0� y = α−1�r� θsω�v�r�, being

v�·� = I�u0�f �·� an integral solution of{dvdt∈ Av�t� + α�t� θsω�F�α−1�t� θsω�v�t��

v�0� = u0�


which is a strong solution of (5) with

f �t� ∈ α�t� θsω�F�α−1�t� θsω�v�t�� a.e. in �0� T ��After the change of variable z�t� = α�s�ω�v�t�, we obtain that y =α−1�r + s�ω�z�t�, being z�·� the integral solution (in fact, a strong one) ofthe problem {

dzdt= Az�t� + g�t��

z�0� = α�s�ω�u0�(14)

where g�t� = α�t + s�ω�h�t�, and h�t� ∈ F�α−1�t + s�ω�z�t��, a.e. in�0� T �.In our case s = −t0� r = −1+ t0. Multiplying (14) by z�t� we have

12d

dt�z�t��2 = �Az�t�� z�t�� + �g�t�� z�t��

= �Az�t�� z�t�� + �α�t − t0�ω�h�t�� z�t��= �Az�t�� z�t�� + α2�t − t0�ω��h�t�� α−1�t − t0�ω�z�t��

Now by (12), (13) we get

d

dt�z�t��2 ≤ −2ε�z�t��2 + 2�ε− δ��z�t��2 + 2Mα2�t − t0�ω�

and thus,d

dt�z�t��2 ≤ −2δ�z�t��2 + 2Mα2�t − t0�ω�� (15)

Multiplying (15) by exp(2δt) and integrating over (0�−1+ t0) we obtain

�z�−1+ t0��2 ≤ exp�−2δ�−1+ t0��z�0��2

+ 2M exp�−2δ�−1+ t0��∫ −1+t0

0exp�2δs�α2�s − t0�ω�ds

and then, by the change s − t0 = τ,

�z�−1+ t0��2 ≤ exp�−2δ�−1+ t0��z�0��2

+ 2M exp�2δ�∫ −1

−∞exp�2δτ�α2�τ�ω�dτ�

Now, by standard arguments (see, e.g., Crauel and Flandoli [9]) it easilyfollows that the mapping τ �→ exp�2δτ�α2�τ�ω� is pathwise integrable over�−∞�−1�, and similarly exp�2δτ�α2�τ�ω� → 0 as τ→−∞��-a.s.

We take

r21 �θ−1ω� = 1+ 2M∫ −1

−∞exp�−δ�−1− τ��α2�τ�ω�dτ�

r�θ−1ω� = α−1�−1�ω�r1�θ−1ω��


The radius r�θ−1ω� is �-a.s. finite, because of the above considerations.For a bounded set B and almost all ω ∈ �, we choose T �B� = T �B�ω� ≥ 1such that

exp�−2δ�−1+ t0��α2�−t0�ω��u0�2 ≤ 1� ∀ t0 ≥ T �B��∀u0 ∈ B�Since y = α−1�−1�ω�z�−1+ t0� we have

�y� ≤ �α−1�−1�ω�z �−1+ t0�� ≤ r �θ−1ω��for �-a.a. ω ∈ � and any y ∈ G�−1+ t0� θ−t0ω�u0� u0 ∈ B.Theorem 22. Let the conditions of Proposition 20 hold, the semigroup

S�t� ·� generated by the operator A be compact, and the multivalued mapG�1�ω� be compact (that is, it maps bounded sets into precompact ones).Then, G has the minimal global random attractor ��ω�. Moreover, it is mea-surable with respect to � .

Proof. First, since x ∈ X �→ G�t� ω�x is lower semicontinuous, the map�t� ω� �→ G�t� ω�D is measurable for all deterministic bounded setsD ⊂ X� t ≥ 0 and �-a.s. (see Remark 7 in [7]). On the other hand, itfollows from Lemma 17 that G has compact values.Let us define the random ball

B�r�θ−1ω�� ={u ∈ D�A� � �u� ≤ r�θ−1ω�

}�

where r�θ−1ω� is taken from Proposition 20. Let

K�ω� = G�1� θ−1ω�B �r�θ−1ω��The set K�ω� is �-a.s. compact, since the operator G�1� θ−1ω� is compact.It follows from Proposition 20 that for any bounded nonrandom set B and�-a.a. ω ∈ � there exists T �B� = T �B�ω� ≥ 1 such that ∀ t0 ≥ T �B�,

G�t0� θ−t0ω�B ⊂ G�1� θ−1ω�G�−1+ t0� θ−t0ω�B ⊂ K�ω��Therefore, G has the compact random absorbing set K�ω� and we canapply Theorem 9 to ensure the existence of the global random attractor forthe multivalued random semiflow.

4. APPLICATIONS

4.1. The Case of a Subdifferential Map

In order to check when the operatorG�1�ω� is compact we shall considerthe case where the operator −A = ∂ϕ is the subdifferential of a proper


lower semicontinuous function ϕ� X → �−∞�+∞� (being ∂ϕ in our casea linear operator). Inclusion (3) turns into{

dudt∈ −∂ϕ�u� + F�u� + σu ◦ dw �t�

dt� t ∈ �0� T ��

u�0� = uo�(16)

where F � X → 2X is a multivalued map, satisfying (F1)–(F2), D�A� =D�∂ϕ�. It is well known (see Barbu [5, pp. 54 and 71]) that −∂ϕ is anm-dissipative operator. Moreover, D�ϕ� = D�∂ϕ�. Then, ∂ϕ generates anonlinear semigroup of operators S�t� ·�� D�ϕ� → D�ϕ� and the differen-tial inclusion (16) gives the multivalued map G�t� ω�� D�ϕ� → 2D�ϕ�. It isknown (see Haraux [11, p. 1398]) that the semigroup S is compact if thefollowing condition is satisfied:

(H) The level sets

MR = �u ∈ D�ϕ� � �u� ≤ R�ϕ�u� ≤ R�

are compact in X for any R > 0.

If (H) holds, it follows from Lemma 17 that G has compact values.Hence, G�t� ω�� D�ϕ� → K

(D�ϕ�), where K

(D�ϕ�) is the set of all

nonempty compact subsets of D�ϕ�. Theorem 19 implies that G generatesa MRDS.Further we shall recall the next regularity result for solutions of the

inclusion {dvdt∈ −∂ϕ�v� + f �t�� t ∈ �0� T �,

v�0� = v0 ∈ D�ϕ��(17)

Proposition 23. (Brezis [6, p. 82] and Barbu [5, p. 189]). For any f �·� ∈L2��0� T ��X�� v0 ∈ D�ϕ�, there exists a unique strong solution of inclusion(17) such that

v�·� ∈ C��0� T ��X�� √tdv

dt∈ L2��0� T ��X��

ϕ�v�·�� ∈ C��0� T �� ϕ�v�·�� ∈ L1��0� T ��

and ϕ�v�t�� is absolutely continuous on �δ� T ��∀ δ > 0. Moreover,∣∣∣∣dvdt∣∣∣∣2 + d

dtϕ�v�t�� =

⟨f�dv

dt

⟩� a.e. on �0� T �� (18)

If u0 ∈ D�ϕ� then dudt∈ L2��0� T ��X� and ϕ�u� is absolutely continuous

on �0� T �.


We note the next important consequence of the preceding proposition.Let us take an arbitrary integral solution of inclusion (4), v�·� ∈ ��v0� ω��v�·� = I�v0�f �·�. It follows from Lemma 15 that f �·� ∈ L2��0� T ��X�. Then,since the solution of (17) is unique for any v0 ∈ D�ϕ�, it follows that v�·�is a strong solution of (17).Further, Proposition 23 and Lemma 15 allow us to prove an important

property of the map G.

Theorem 24. Let property (H) hold. Then for any bounded B ⊂ X, anyT > 0 and ω ∈ �, there exists R�ω� > 0 such that G�T�ω�B ⊂MR�ω�.

Proof. In view of Lemma 15, the set M�B�ω� T � is bounded inL∞��0� T ��X� and then it is bounded in L2��0� T ��X� for any boundedB ⊂ X�T > 0�ω ∈ �. We take an arbitrary v�·� ∈ ��B�ω�� v�·� =I�v0�f �·�� v0 ∈ B. Consider first that v0 ∈ D�ϕ�. It follows from equality(18) that

t∣∣∣dvdt

∣∣∣2 + td

dtϕ�v� = t

⟨f�dv

dt

⟩� a.e. on �0� T ��

Hence, ∫ T

0t∣∣∣dvdt

∣∣∣2dt + Tϕ�v�T �� =∫ T

0t

⟨f�dv

dt

⟩dt +

∫ T

0ϕ�v�t��dt�

and then,

Tϕ�v�T �� ≤ 12

∫ T

0t∣∣∣dvdt

∣∣∣2dt + Tϕ�v�T ��

≤ 12

∫ T

0t�f �t��2dt +

∫ T

0ϕ�v�t��dt� (19)

On the other hand, there is no loss of generality in assuming thatmin�ϕ�v�� v ∈ X� = ϕ�x0� = 0. Indeed, let x0 ∈ D�∂ϕ�� v0 ∈ ∂ϕ�x0�. If weintroduce the new function ϕ�v� = ϕ�v� − ϕ�x0� − �y0� v − x0�, then theequation

dv

dt+ ∂ϕ�v� % f �t�

is equivalent to

dv

dt+ ∂ϕ�v� % f �t� − y0 = f �t�

and min�ϕ�v�� v ∈ X� = ϕ�x0� = 0. It is clear that ϕ satisfies alsoproperty (H).Hence, since f �t� − dv�t�

dt∈ ∂ϕ�v�t�� a.e. on �0� T �, we have

ϕ�v�t�� ≤⟨f �t� − dv�t�

dt� v�t� − x0

⟩�


Integrating over �0� T � we get∫ T

0ϕ�v�t��dt≤ 1

2�v�0�−x0�2−

12�v�T �−x0�2+

∫ T

0�f �t��v�t�−x0�dt

≤ 12�v�0�−x0�2+

∫ T

0�f �t��v�t�−x0�dt�

Since 0 ∈ −∂ϕ�x0�, it follows from inequality (7) that

�v�t� − x0� ≤ �v�0� − x0� +∫ T

0�f �τ��dτ� 0 ≤ t ≤ T�

It follows from the last two inequalities and Lemma 15 that∫ T

0ϕ�v�t��dt ≤

(�v�0� − x0� +

∫ T

0�f �t��dt

)2

≤ D�ω� <∞� (20)

Since the set M�B�ω� T � is bounded in L∞��0� T ��X� for any boundedset B�D�ω� does not depend on v�·� ∈ ��B�ω�. Using (20) in rela-tion (19) we obtain that, for any T > 0, there exists K�ω� > 0 such thatϕ�v�T �� ≤ K�ω�.Now let v0 ∈ B be arbitrary. We can assume without loss of gener-

ality that B is open and then there exists a sequence vn0 → v0, wherevn0 ∈ D�ϕ�� vn0 ∈ B. In view of (8), for each vn0 there exists an integral solu-tion of (4), vn�·� such that

�vn�t� − v�t�� ≤ �vn0 − v0� exp�2Ct�� ∀ t ∈ �0� T ��Since ϕ�vn�T �� ≤ K�ω��∀n, and ϕ is lower semicontinuous, we get

ϕ�v�T �� ≤ lim infn→∞ ϕ�vn�T �� ≤ K�ω��

On the other hand, by Lemma 15, ��B�ω� is bounded in C��0� T ��X�.Hence,

�v�T �� ≤ L�ω� <∞� ∀ v�·� ∈ ��B�ω��Therefore,

G�T�ω�B ⊂MR�ω��

where R�ω� = max�K�ω�α−1�T�ω�� L�ω�α−1�T�ω��. It follows from (H)that G�T�ω�B is precompact in X.

Corollary 25. Let property (H) hold. Then for any T > 0 and ω ∈ �,G�T�ω� is compact.

Proof. We must prove that for every bounded set B, any T > 0, andω ∈ �, the set G�T�ω�B is precompact in X. But this fact follows immedi-ately from (H) and the previous theorem.


Theorem 26. Let (F1)–(F2), (H), and (12) be satisfied. Then G has theminimal global invariant random attractor ��ω�, which is measurable withrespect to � .

Proof. It is a consequence of Theorem 22 and Corollary 25.

4.2. Reaction-Diffusion Inclusions

Let f � � → Cv�� be a multivalued map. Assume that f is Lipschitz,i.e., ∃C ≥ 0 such that ∀x� z ∈ �

distH�f �x�� f �z�� ≤ C�x− z�� (21)

Let � ⊂ �n be an open bounded subset with smooth boundary ∂�. Con-sider the stochastic inclusion

∂u∂t∈ ;u+ f �u� + h+ σu ◦ dw�t�

dt� on � × �0� T �,

u = 0� on ∂� × �0� T �,u�x� 0� = u0�x� on �,

(22)

where h�·� ∈ L2��. Define the operators A� D�A� → X�F � X → 2X�X =L2��,

Au = ;u�

F�u� = �y ∈ X� y�x� ∈ f �u�x�� + h�x��with D�A� = H2�� ∩ H1

0��. The map −A is the subdifferential of aproper, convex, lower semicontinuous function ϕ and the map F satisfies(F1)–(F2). Moreover, condition (H) is satisfied and D�ϕ� = X (see Melnikand Valero [12, Sect. 3.2.2]). Hence, (22) is a particular case of (16). Weshall assume that there exist M1 ≥ 0� δ > 0 such that ∀ s ∈ ��∀ z ∈ f �s�,

zs ≤ �λ1 − 2δ��s�2 +M1� (23)

where λ1 is the first eigenvalue of −; in H10��.

We obtain the following theorem:

Theorem 27. Let (21), (23) hold. Then, the MRDS generated by (22) hasthe minimal global invariant random attractor ��ω�, which is measurable withrespect to � .

Proof. We have to check that (12) holds. In our case ε = λ1. In viewof (23) for any y ∈ D�A� = H2�� ∩H1

0�� ξ ∈ F�y�,�ξ� y� ≤ �λ1 − 2δ��y�2 +M1µ�� + �h� y� ≤ �λ1 − δ��y�2 +M�

for some M ≥ 0, so that (12) holds. The statement follows fromTheorem 26.


ACKNOWLEDGMENTS

The authors express their thanks to the referee for the helpful and interesting suggestionson the paper. This paper was started while the two first authors were visiting the CEU-SanPablo (Elche). They thank the Institution for financial support and for the kind hospitalitythey received from all people there. Special thanks are given to Jose Valero.

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