Abstract and Applied Analysis

Volume 2013, Article ID 792431, 9 pages

http://dx.doi.org/10.1155/2013/792431

## Relaxation Problems Involving Second-Order Differential Inclusions

^{1}Taibah University, Faculty of Applied Science, Department of Applied Mathematics, Al-Madinah, Saudi Arabia^{2}Mathematics Department, Faculty of Science, Helwan University, Cairo, Egypt

Received 12 November 2012; Accepted 12 March 2013

Academic Editor: Malisa R. Zizovic

Copyright © 2013 Adel Mahmoud Gomaa. 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.

#### Abstract

We present relaxation problems in control theory for the second-order differential inclusions, with four boundary conditions, a.e. on ; and, with boundary conditions, a.e. on , where , and is a multifunction from to the nonempty compact convex subsets of . We have results that improve earlier theorems.

#### 1. Introduction

Second-order differential inclusions of three boundary conditions were studied by many authors [1–6], using Hartman-type functions. Such a function was first introduced by [7] for two boundary conditions. Moreover, in [8] we consider second-order differential inclusions with four boundary conditions, where and is a multifunction from to the nonempty compact subsets of , while in [9] we study four-point boundary value problems for differential inclusions and differential equations with and without multivalued moving constraints.

In the present paper, we study relaxation results for the second-order differential inclusions, with four boundary conditions, and, with boundary conditions, where , , and is a multifunction from to the non-empty compact subsets of .

In conjunction with Problem and Problem we also consider the following problems: By , we denote the set of extreme points of .

#### 2. Notations and Preliminaries

Throughout this paper we let and . We will use the following definitions, notations, and summarize some results.

(i) A multifunction from a metric space to the set of all closed subsets of another metric space is lower semicontinuous *(l. s. c.)* at if for every open subset in with there exists an open subset in such that and for all . is *l. s. c.* if it is *l. s. c.* at each .

(ii) is upper semicontinuous *(u. s. c.)* at if for every open subset in and containing there exists an open subset in such that and , for all . is *u. s. c.* if it is *u. s. c.* at each .

(iii) A multifunction from into the set of all closed subsets of is measurable if for all the function is measurable [10–13].

(iv) Let be a measurable space and a separable Banach space. We say that is graph measurable if where is the Borel -field of . For further details we refer to [14–16].

(v) is continuous if it is lower and upper semicontinuous.

(vi) For each , the Hausdorff metric is defined by

It is known that the space is a generalized metric space, if the sets are not bounded (see, for instance, [14, 15]).

(vii) A multifunction is Hausdorff continuous (-continuous) if it is continuous from into the metric space .

(viii) If has compact values in , then is -continuous if and only if it is continuous [14, 17].

(ix) We denote by the nonempty compact convex subsets of .

(x) The Banach spaces , , and endowed with the norms respectively.

(xi) denotes the space equipped with weak norm which is defined by

(xii) is the Sobolev space of functions , and are both absolutely continuous functions so and it is equipped with the norm .

(xiii) Let be a multifunction and .

(xiv) By a solution of (resp., of ) we mean a function such that a.e. on with (resp., ) and .

(xv) By a solution of (resp., of ) we mean a function such that a.e. on with (resp., ) and .

(xvi) In the sequel by (resp., ) we denote the solution set of Problem (resp., of Problem ). Moreover, by (resp., ) we denote the solution set of Problem (resp., of Problem ).

*Definition 1. *Let be a Banach space and let be a metric space. A multifunction has the Scorza-Dragoni property (the SD-property) if for every there exists a closed set such that the Lebesgue measure is less than and is continuous. The multifunction is called integrably bounded on compacta in if, for any compact subset , we can find an integrable function such that , for almost every .

Theorem 2 (see [18]). *Let be a complete metric space, a separable Banach space, the Banach space endowed with the weak topology, , and a compact subset of . Furthermore, let be a multifunction defined by
**If has the SD-property and is integrably bounded on compacta in , then the set
**
is nonempty complete subset of the space . Moreover, where is the space of equivalence classes of Bochner-integrable functions with the norm and
*

Lemma 3 (see [19]). *For such that let , and with respect to the weak norm . Then weakly in . *

Next we state a preliminary lemma, for , which is useful in the study of four boundary problems for the differential equations and the differential inclusions, and moreover we summarize some properties of a Hartman-type function.

Lemma 4 (see [8]). *Let be the function defined as follows: ** as ,
**
when ,
**
lastly if ,
**
Then the following hold. *(i)* If with , then
*(ii)* if , then for all ,
*(iii)*, .*

Let , and let be a linear operator from to defined by such that, for all , If , then clearly . We note that if , then . Moreover, the spectral radius is an eigenvalue of with an eigenvector in [20].

#### 3. Relaxation Theorems

In this section, both Theorems 5 and 7 improve [19, Theorem 4.1] with [21, Theorem 6]. Indeed in [19] Papageorgiou considered and with the two boundary conditions and in [21] Ibrahim and Gomaa study the same problems with three boundary conditions .

Theorem 5. *Let be a multifunction such that*(i)* for each , the multifunction is measurable, *(ii)* a.e. with and , *(iii)* for each ,
with ,*(iv)* the spectral radius, , is less than . ** Then for each solution , there is a sequence converging to in . *

*Proof. *From [9, Theorem 2.1], we obtain . Moreover, we can say that a.e. on for some . Let . Then
where a.e. on . Assume that is a function such that, for each , is the unique solution of the second-order differential equation
Let . It is easy to see that is convex. Let be a sequence in . Hence, with (17) and
Then,
which means that is a compact subset of . Set
where and . Hence, for each , . Assume that and are the Borel -fields of and , respectively. From condition, the function is measurable. Hence, and is measurable in and continuous in that is jointly measurable. Thus, by Aumann's selection theorem, there exists a measurable selection of such that for each . Now we define a multifunction by the following:
with for each . From [22, Proposition 4], is . and clearly has decomposable values. Applying [22, Theorem 3], we have a continuous selection of . Therefore,
From Theorem 2, we find a continuous function such that and for each . Define a multifunction by
Assume that and set a multifunction such that . From Theorem 3.1 in [23], has SD-property. has nonempty convex values. Let be a sequence in for some . So, for each ,
because has closed values in . Therefore, which implies that has compact values in . We can apply Theorem 2 to find a continuous function such that , for all . We see that [24], hence a.e. on . Assume that is the function which for each , . For each , we have and so . Then, is a function from into and also we see that is continuous [19]. Now let , and . Then, for each , the function is a continuous function from the compact set into itself. From Schauder's fixed point theorem, has a fixed point , but [24] so . By passing to a subsequence if necessary, we may assume that in . Then, we obtain
But with respect to the norm from Lemma 3 we get weakly in . So we have
Moreover,
As , we have

Since by assumption (ii), we get . So in and where the closure is taken in which means that . Therefore, the proof is complete if we show that is closed. Indeed if and in , then for . From assumption (iii) and the Dunford-Pettis theorem, is weakly sequentially compact in . So we can say that in . By [25, Theorem 3.1], we get

Moreover, in for and a.e. on . Hence, ; that is is closed in .

Now we consider the following assumptions:, and ; and ; and ;; is defined by

Lemma 6 (see [26]). *If the assumptions hold, then *(i)* for all ,*(ii)*,*(iii)* for each there exists a unique function such that
*(iv)* and .*

*Proof. *(ii) Since
then . Furthermore,
and thus .

Theorem 7. *Assume that the assumptions and hold. Let be a multifunction from to satisfying the following conditions: *(a)* for each , the multifunction is measurable; *(b)* for each , the function is continuous with respect to the Hausdorff metric ; *(c)* for each *(d)* the spectral radius of is less than one. ** Then Problem admits a solution in . *

*Proof. *We can say that a.e. on for some [9]. Let and let be the unique solution of the problem
From Lemma 6, we have . Thus, we define a function such that is the unique solution of . Let
From the Dunford-Pettis theorem, is weakly compact and then is convex and compact subset of . Let . If and , where and , then has SD-property [23]. It is easy to show that is nonempty and convex subset of . If is a sequence in for some , then , where the values of are closed. Therefore, the values of are weakly compact. According to Theorem 5 there exists a continuous function with , for all . Thus, a.e. on [24] which implies a.e. on . If , then and so . Put such that , thus is a continuous function from into [19]. From Schauder's fixed point theorem, there exists such that which means that there is such that .

Theorem 8. *In the setting of Theorem 7, if one replaces condition (b) by the following condition: **(b)′ , a.e. with , and . ** Then is nonempty and where the closure taken in .*

*Proof. *From Theorem 7, we have . Moreover, a.e. on for some . Let . Then
where a.e. on . Assume that is a function such that, for each , is the unique solution of the second-order differential equation
Let . So is convex. Let be a sequence in . Hence, with . Then from Lemma 6,
hence, is a compact subset of . Set
where and . Hence, for each , . Assume that and are the Borel -fields of and , respectively. From condition (i), the function is measurable. Hence, and is measurable in and continuous in that is jointly measurable. Thus, by Aumann's selection theorem, there exists a measurable selection of such that for each . Now we define a multifunction by the following:
with for each . From [22, Proposition 4], is . and clearly has decomposable values. Applying [22, Theorem 3], we have a continuous selection of . Therefore,

From Theorem 2, we find a continuous function such that and for each . Define a multifunction by
As in Theorem 5, let and set a multifunction such that . From [23, Theorem 3.1], has SD-property. has nonempty convex values. Let be a sequence in for some . So, for each ,
because has closed values in . Therefore, which implies has compact values in . We can apply Theorem 2 to find a continuous function such that , for all . We see that [24], hence a.e. on . Assume that is the function which for each , . For each , we have and so . Then, is a function from into and also we see that is continuous [19]. Now let , and . Then, for each , the function is a continuous function from the compact set into itself. From Schauder's fixed point theorem, has a fixed point , but [24] so . Assume that in . From Lemma 6, we obtain
But with respect to the norm and from Lemma 3 we get weakly in . So we have
Moreover, as we have
Since by assumption (ii), , thus from Lemma 6, we get . So in and where the closure is taken in which means that . If and in , then for . From assumption (iii) and the Dunford-Pettis theorem, is weakly sequentially compact in . By [25, Theorem 3.1], we get
Moreover, in for and a.e. on . Hence, ; that is, is closed in .

#### Acknowledgments

The author is deeply indebted and thankful to the deanship of the scientific research and his helpful and distinct team of employees at Taibah University, Al-Madinah Al-Munawarah, Saudia Arabia. This research work was supported by a Grant no. 3029/1434.

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