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The Scientific World Journal
Volume 2013 (2013), Article ID 591620, 7 pages
Existence Results for Differential Inclusions with Nonlinear Growth Conditions in Banach Spaces
Department of Mathematics, College of Sciences, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
Received 24 September 2013; Accepted 4 November 2013
Academic Editors: N. Henderson, Z. Mukandavire, and X. Song
Copyright © 2013 Messaoud Bounkhel. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
In the Banach space setting, the existence of viable solutions for differential inclusions with nonlinear growth; that is, a.e. on , , , , (*), where is a closed subset in a Banach space , , , , is an upper semicontinuous set-valued mapping with convex values satisfying , , where , with , and . The existence of solutions for nonconvex sweeping processes with perturbations with nonlinear growth is also proved in separable Hilbert spaces.
The first motivation of the study of the concept of differential inclusions comes from the development of some studies in Control Theory and Optimization; see, for instance, [1–3] and the references therein. Many works investigated the existence of solutions and topological properties of solution sets for first- and second-order differential inclusions [2, 4–8]. For example, in , the authors proved an existence result for the inclusion by assuming that the set-valued mapping is included in the subdifferential of convex lower semicontinuous (l.s.c) function . This result has been extended in many ways by many authors (see, e.g., [9–11] and the references therein. We state one of them from , in which the author proved an existence result of viable solutions in the finite dimensional setting for the differential inclusion: where is included in the subdifferential of a regular (not necessary convex) function , is closed subset in , and is a continuous set-valued mapping. The infinite dimensional case of (2) has been studied in . A very important type of differential inclusions that will be considered in this work is the following: where is the normal cone to at . This differential inclusion is called Sweeping Process Problem (SP) and has been introduced and studied by Moreau in 1960s in the convex case . This differential inclusion (SP) models a phenomena from elastoplasticity; see the excellent books in  and the references therein. Since the works , many works extended in different ways the sweeping process problem. In , the author introduced some new techniques from which many results can be derived, essentially the existence of a solution of (SP) for , where is a fixed nonconvex closed set and is a mapping with finite variation. Another important study of the inclusion (3), with a nonconvex set , has been realized by the author in  who proved, in the finite dimensional setting, the existence of solution for (3) whenever the set-valued mapping has a closed graph. Here is the Clarke normal cone and denotes the closed unit ball of . The main example of such sets provided in  is that of complements of open convex sets. In [17, 18] the authors independently proved the existence of a solution of (3) for general nonconvex sets moving in a Lipschitz way in a finite dimensional space. Note that the results in  are given for -convex sets in Hilbert spaces and under some compactness assumptions on . In , the authors proved an existence result in Hilbert spaces with a regular set-valued mappings of the perturbed sweeping process problem defined as follows: where is an upper semicontinuous set-valued mapping with convex compact values. The class of inclusions (4) appears in particular in mathematical economy. It corresponds for (independent of ) to modeling planning procedures introduced by Henry  for convex and also considered by  for tangentially regular.
To the best of our knowledge no existing works studied the existence of solutions for differential inclusions with nonlinear growth. The main purpose of the paper is to prove the existence of solutions for (*) in Banach spaces and for (4) in separable Hilbert spaces, under the nonlinear growth condition of the set-valued mapping ; that is, when is an upper semi-continuous set-valued mapping with closed convex values satisfying , where , with , and and is a convex compact set. The paper is organized as follows. After recalling the needed concepts in Section 2, we prove in Section 3 the existence of viable solutions for (*) in Banach spaces. In Section 4, we prove the existence of solution for (SP) with perturbations having nonlinear growth conditions with prox-regular values of in separable Hilbert spaces.
This section is devoted to recall some notations and concepts needed in the paper.
Definition 1. Let be a Banach space, let be a nonempty closed subset of , and let . The Bouligand tangent cone is defined by
where is the usual distance function associated with .
Recall from  the original definition of the class of uniformly -prox-regular sets in Hilbert spaces as the class of all closed sets satisfying the following definition. Many equivalent definitions of this class have been used for different applications; see, for example, [5, 19, 22].
Definition 2. Let be a Hilbert space. For a given , a subset is uniformly -prox-regular if and only if for all , the distance function is at .
Example 3. (1) Any convex set is uniformly -prox-regular with .
(2) The union of two disjoint convex sets is not convex but it is uniformly -prox-regular with , where is the distance between the two sets. More examples, details, and characterizations of this class of sets in Hilbert spaces can be found in [5, 19, 22].
A set-valued mapping is said to be upper semicontinuous (u.s.c) at provided for every , there exists such that We say that is u.s.c. on whenever it is u.s.c on all . Obviously, the upper semicontinuity coincides with the continuity for single-valued mappings. The following proposition proves the u.s.c. of set-valued mappings with closed graphs under the compactness assumption on the closure of the range. For its proof, we refer the reader to Proposition 1.2 in Deimling .
Proposition 4. Let be a nonempty closed subset in and let be a set-valued mapping with closed values. If the graph of is closed and is compact, then is upper semicontinuous.
3. Nonlinear Variants of Gronwall Inequalities
Before starting this section, we refer the reader to the nice book in  on Gronwall inequalities and applications. We recall from  the following variant of Gronwall inequality that can be also found in .
Lemma 5. Let be a positive differentiable function satisfying the inequality where the functions and are continuous on and (with ) is a constant. Then for all , where is chosen so that the expression is positive in the subinterval .
In the following lemma we extend Lemma 5 to the case of negative exponent . To the best of our knowledge no such results on Gronwall inequalities with decreasing right hand side of the inequality (7) which is the case when the exponent is assumed to be negative.
Lemma 6. Let be a positive differentiable function satisfying (7) with and assume that is nonnegative. Then for all .
Proof. Multiplying (7) by we obtain Let Then and . Rearranging the above inequality and multiplying it by yield Let . Then and and hence (12) becomes Integrating this inequality over we get Thus, for all , and hence the proof is finished.
4. Solutions of Differential Inclusions with Nonlinear Growth
The two following consequences of Lemmas 5 and 6 are the key tools in this section. In these lemmas we take the case , and we separate the results in two cases depending on the exponent . The first case is and the second case is .
Lemma 7. Let be a positive differentiable function satisfying (7) with and assume that , for all . Let , for all and let satisfying the inequality Then for all .
Lemma 8. Let be a positive differentiable function satisfying (7) with . Let , for all . Then for all .
Let be a set-valued mapping satisfying the following nonlinear growth: where , , is a closed nonempty set in , , and with . Clearly, when , this assumption coincides with the well known linear growth; that is, Our main aim in this paper is to prove the existence of absolutely continuous solutions under the nonlinear growth condition (20) of for any . To ensure the viability of the solution on the set , we need the following classical tangential condition: where is the Bouligand tangent cone to at .
The following proposition is a main tool in our next proofs.
Proposition 9. Assume that is a mapping from to satisfying Then is bounded by on the interval , where when and for , where is given as in Lemma 7.
Proof. Let . Since is absolutely continuous on , then the derivatives and exist a.e. on and satisfy where is the normalized duality mapping (for the definition we refer to ). For such , we have Take the functions and as in Lemma 7 satisfying , for all . Then by Lemmas 7 and 8 we get the conclusion of the proposition.
In all what follows let and be as in Proposition 9. We recall from Deimling [2, Theorem 9, Page 117] the following existence result for u.s.c. set-valued mappings with values contained in a compact set.
Theorem 10. Let be a Banach space, a nonempty closed set, , and satisfying the following:(a) is u.s.c. with closed convex values;(b) on , for some convex compact set in and ;(c) on .Then for every , there exists an absolutely continuous mapping such that We start now by proving the following proposition needed in the proof of the main result.
Proposition 11. Let be a closed subset in and let be an upper semicontinuous set-valued mapping with closed convex values and let be such that , and let be a continuous function such that for and for . Let be a set-valued mapping defined on as follows: If satisfies the nonlinear growth on ; that is, on , for some , with , and is a convex compact set in , then is upper semicontinuous on with closed convex values.
Proof. Clearly, has closed convex values. Let . For any and any with , we have by the convexity of the following: where and for any and any with , we have . Then . Then, by Proposition 4, it is sufficient to prove that the graph of is closed. To do that, we fix with and we have to prove that ; that is . By definition of we have First, we assume the existence of a subsequence of such that . In this case we have for large enough and so is bounded which ensures that and so . On the other hand, by continuity of and the convergence of to , we obtain and so . Thus, we get . Assume now that there exists some and such that for all . Then by continuity of we have . Thus, by upper semicontinuity of , we get ; that is, . This completes the proof of the closedness of the graph of and hence the proof is achieved.
Now, we are ready to prove our main existence result under the nonlinear growth condition in Banach spaces.
Theorem 12. Let be a Banach space, a nonempty closed set, and satisfying the following: (a) is u.s.c. on with being closed convex for all ;(b) on , for some , and with , and for some convex compact set in ;(c) on .Then for every , there exists an absolutely continuous mapping such that where when and for and is as in Lemma 7.
Proof. Let be such that and assume that . Let and let be as in Proposition 9 with instead of the function . Then where is defined as in Proposition 9. Set and let be a continuous function such that for and for . Define now the set-valued mapping on as follows: By Proposition 11, the set-valued mapping inherits the convexity and the upper semicontinuity of the set-valued mapping with , where . We have to check that satisfies the tangential condition on . Let and let . Then and so the tangential condition is satisfied from (c). Assume now that and with . Then by (c) there exists some such that . Let . Clearly, and , since is a cone. So, . Therefore, the tangential condition is satisfied for on all . Consequently, all the assumptions (a), (b), and (c) in Theorem 10 with the compact set instead of are satisfied and hence for every there exists a solution on of associated with the set-valued mapping defined above, that is, an absolutely continuous mapping such that a.e. on , , and on . Let us prove that is the desired solution for (31). Clearly and so we get Assume that . Using Proposition 9, we get a.e. on and so by the definition of and we get that is, is a solution of (31) on .
5. Nonconvex Sweeping Process with Perturbations Having Nonlinear Growth Conditions
Our purpose, in this section, is to use the techniques developed previously to extend some existing results, in separable Hilbert spaces, of nonconvex sweeping processes with perturbations from the case of perturbation with linear growth to the case of perturbation with nonlinear growth. For this end let be a separable Hilbert space, let (), and let be a set-valued mapping satisfying the following Lipschitz condition for any and any : We start with the following existence result which is a consequence of Theorem 4.1 in .
Theorem 13. Let be a separable Hilbert space and let . Assume that is -prox-regular for every and that the assumption (36) holds. Let be a set-valued mapping with convex compact values in such that is u.s.c. on . Assume that for all , for some compact set in . Then, for any , the sweeping process (SPP) with the perturbation has at least one Lipschitz continuous solution; that is, there exists an Lipschitz continuous mapping such that and , a.e. on .
Using the techniques from the previous section and Theorem 13 we prove our main result in this section.
Theorem 14. Let . Assume that is -prox-regular for every and that the assumption (36) holds. Let be a set-valued mapping with convex compact values. Assume also that has nonlinear growth; that is, there exist a positive continuous function , a convex compact set , and such that Assume further that the following conditions on the constants , , , , and are satisfied: Then for any , there exists a Lipschitz continuous mapping satisfying the following sweeping process with a perturbation:
Proof. Assume without loss of generality that . By our assumptions on the constants , , and we have after simple computations So we can find some positive number such that Let . Then Define then the function to be a continuous function such that for and for and define the set-valued mapping on as follows: Clearly inherits the convexity of the values from . Also, for any , we have and for any , we have . Thus, , for any with . Consequently, the upper semicontinuity follows from the u.s.c. of and Proposition 11. Applying now Theorem 13 with and we get a Lipschitz continuous mapping such that with Now, let us check that is a solution of (SPP) with . Clearly, we have We use now the choice of and the assumptions on the constants , , to deduce from (43) that which ensures that and hence which yields that and so . This means that is a solution of (40) and hence the proof is complete.
The author extends his appreciation to the Deanship of Scientific Research at King Saud University for funding the work through the research group project no. RGP-VPP-024.
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