Abstract and Applied Analysis

Volume 2013 (2013), Article ID 427908, 6 pages

http://dx.doi.org/10.1155/2013/427908

Research Article

Existence Results for Constrained Quasivariational Inequalities

V. V. Motreanu

Department of Mathematics, Ben Gurion University of the Negev, Be’er Sheva 84105, Israel

Received 18 June 2013; Accepted 4 September 2013

Academic Editor: Rodrigo Lopez Pouso

Copyright © 2013 V. V. Motreanu. 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 deal with a constrained quasivariational inequality under a general form. We study existence of solutions in two situations depending on whether the set of constraints is bounded or possibly unbounded.

1. Introduction and Statement of Main Results

Let be a real reflexive and separable Banach space assumed to be compactly embedded in a Banach space . We denote by the dual space of , by the dual space of , by the duality brackets between and , by the duality brackets between and , by the norm of , and by the norm of . Given a function , we denote by the effective domain of .

In this paper we deal with the following problem

We describe the data entering problem (1):

(i) is a nonempty, convex, closed subset;(ii) is a (possibly nonlinear) operator; (iii) is such that, for all , the function is convex with ; moreover, we will denote by the convex subdifferential of ; that is,

(iv) is a locally Lipschitz function, and the notation stands for its generalized directional derivative in the sense of Clarke [1]; that is,

 In addition, we will denote by the generalized gradient of ; that is,

(v).

Problem (1) is called a constrained quasivariational problem. Typically, we can choose to be the Sobolev space defined as the closure of in for a bounded domain (), to be the Lebesgue space for (where if and if ), , (the negative Laplacian operator), where is convex in the second variable (then ), and where is locally Lipschitz in the second variable. Constrained quasivariational problems were extensively studied; we refer, for example, to [2–5] and to the references therein. We point out three aspects which make our approach natural and general. First, we deal with the general setting of a pair of Banach spaces instead of focusing on spaces of functions; in particular, our results can be applied to problems with different boundary conditions. Second, the set of constraints may be unbounded. Third, the form of the studied problem allows both variational and hemivariational constraints as it involves both a convex term and a generalized directional derivative ; this type of problems models important processes in mechanics and engineering (see [6, 7]).

In this paper, we consider the following hypotheses on the data described above:

for every sequence with in , for some , one has

whenever , in , in , one has and

given , if satisfy , and

 then .

Remark 1. We emphasize certain situations when hypotheses – are satisfied.

(a) Hypothesis is satisfied, for instance, if is weakly strongly continuous, that is, is continuous from endowed with the weak topology to endowed with the norm topology.

(b) Note that is satisfied, for instance, for , any closed, convex subset , and defined by , where is the Laplacian operator, with () a bounded domain. Indeed, let a sequence with in , for some . Using the weak lower semicontinuity of the norm, we can write

for all . Here, are the duality brackets for the pair and denotes the scalar product on . Whence holds in this case.

(c) Hypothesis is fulfilled in the case where is sequentially weakly lower semicontinuous, is weakly closed, and is weakly strongly continuous on its effective domain for all .

(d) If is strongly monotone, that is, there exists a constant such that

and is bounded on in the sense that

with a positive constant , where is the best constant satisfying , for all (which exists by the continuity of the embedding of in ), then condition is satisfied.

(e) If is strictly monotone and is Gâteaux differentiable and regular (see [1, Definition ]), then condition is satisfied. In particular, if is strictly monotone and is continuously differentiable, then is satisfied.

In this paper, we distinguish two cases depending on whether the set is bounded or not necessarily bounded. The following result concerns the former situation.

Theorem 2. Assume that conditions – are satisfied and that the closed, convex set is bounded in . Then problem (1) has at least one solution.

Remark 3. Note that the existence of a solution of problem (1), which is the conclusion of Theorem 2, forces the intersection to be nonempty, where the notation stands for the diagonal of the set ; that is, . The nonemptiness of this intersection is not directly implied by the hypotheses –, nor by the assumption made that for all . However, Theorem 4 below incorporates hypothesis which assumes in particular that .

Now, we deal with the case where is not assumed to be bounded. In this case, we additionally suppose the following:

there exist an element with for all and a real such that

there exists a constant such that we have

 for all with , where and are as in .

We state now our main result for problem (1) dealing with the case where the set is possibly unbounded.

Theorem 4. Assume that conditions – are satisfied. Then problem (1) has at least a solution.

The rest of the paper is organized as follows. In Section 2, we present the proof of Theorem 2, where we apply a version of the Schauder fixed point theorem. In Section 3, we give the proof of Theorem 4, which is actually based on Theorem 2.

2. Proof of Theorem 2

For each , we consider the auxiliary problem

Our first purpose, accomplished in Lemma 6 below, is to show that problem (13) has a unique solution. To do this, we need Fan’s lemma (see [8, page 208]) which we recall in the following statement.

Theorem 5. Let be a Hausdorff topological vector space, let be a nonempty subset of , and let be such that

(i) is a nonempty, closed subset of , for all ; (ii) for all ; (iii)there is for which is compact.

Then .

Lemma 6. Assume that hypotheses – are fulfilled and that the closed, convex set is bounded in . Then, for every , problem (13) has a unique solution.

Proof. Fix . Consider the set-valued mapping defined by

for all . We show that the assumptions of Theorem 5 are satisfied for endowed with the weak topology, , and .

For every , we clearly have ; hence is nonempty.

We check that is weakly compact for every . To this end, we first prove that is sequentially weakly closed in . Let a sequence with in , for some . Taking into account that is compactly embedded in it follows that in . Using the first part of assumption , we have that . As , we know that

Passing to the as , we find

Here we made use of the weak convergence in , the continuity of on , and the second part of . Combining with , we obtain that , thereby is sequentially weakly closed in .

Using that is reflexive and separable and is bounded, convex, and closed, we deduce that is metrizable and weakly compact (see, e.g., [9, pages 44–50]). Since and using that is sequentially weakly closed, we derive that is weakly compact whenever . Therefore conditions (i) and (iii) in Theorem 5 are fulfilled.

We focus now on the verification of condition (ii) in Theorem 5. Arguing by contradiction, we suppose that there exist and such that . The convexity of the set and of the function ensures that . Then the assertion that reads as

Let

It is clear that for all . The convexity of the functions and implies that is a convex subset in . We infer that , so , which is obviously impossible. This contradiction justifies condition (ii) in Theorem 5. Thus all the assumptions of Theorem 5 are satisfied.

Applying Theorem 5, we obtain

This ensures the existence of an element satisfying

for all . The above inequality being also satisfied if , we conclude that is a solution of problem (13).

It remains to show that the solution of problem (13) is unique. If are solutions of (13), then we have that , , and

Letting in the first inequality and in the second one and then adding the obtained relations, we arrive at

By assumption , we conclude that . The proof is complete.

Denote by the unique solution of problem (13) corresponding to . Lemma 6 guarantees that exists and is unique. We define by

Lemma 7. Assume that hypotheses – are fulfilled and that the closed, convex set is bounded in . Then, the map given in (23) is sequentially weakly continuous.

Proof. Let a sequence such that in for some . We need to show that as . To do this, it suffices to check that, for any relabeled subsequence , there is a subsequence of weakly converging to .

By the compactness of the embedding of in , we have that in . Denote, for simplicity, . The definition of yields and

Since is bounded, and is reflexive, we know that along a subsequence, denoted again by , we have

for some . The first part of yields . Moreover, the compactness of the embedding of in implies that in . Letting in (24), by means of , , the convergences and in , and the upper semicontinuity of on , we get

This means that is a solution of problem (13). Lemma 6 ensures that is the unique solution of (13). Thus, by (23), we have . Taking into account (25), it follows that as up to a subsequence. This completes the proof.

Remark 8. As noted in the proof of Lemma 6, the closed, bounded, convex subset is metrizable for the weak topology. Therefore, Lemma 7 implies that is weakly continuous.

We need the following version of the Schauder fixed point theorem (see [10, page 452]).

Theorem 9. Suppose that

(i) is a reflexive, separable Banach space; (ii)the map is sequentially weakly continuous; (iii)the set is nonempty, closed, bounded, and convex.

Then has a fixed point.

We are now in position to prove Theorem 2.

Proof of Theorem 2. In view of Lemma 7 and the assumptions on and , we may apply Theorem 9 which shows that the map admits a fixed point ; that is, . Using the definition of (see (23)), we deduce that is a solution of problem (1).

3. Proof of Theorem 4

It suffices to prove Theorem 4 when the set is unbounded because for a bounded set the result is true according to Theorem 2. Let . Let be an integer such that , where is the element entering . We claim that Theorem 2 can be applied with replaced by whenever .

Note that , so for all , all (using the first part of ). Thus, for all , all . Since is convex and closed in , it turns out that is convex, closed, and bounded in , for all .

We check that assumptions – of Theorem 2 remain valid when is replaced by with . Towards this, we fix some . If satisfies in and in , then assumption (for ) implies . On the other hand, the weak convergences ensure that

Hence, . The second part of for and conditions and for hold because , , and have been imposed for , which contains . Thus it is permitted to apply Theorem 2 for in place of , with any .

Applying Theorem 2, we find a sequence in such that , , and

for all , all . Letting (see ) in (28), we obtain

for all . By the definition of the convex subdifferential , we have

Then, invoking the growth condition for in , we see that

Recall that

(see [1, Proposition (b)]). This fact combined with the growth condition for the generalized gradient as stated in enables us to write

for all . By the continuity of the embedding , the inequality above leads to

where is a constant. Combining (29), (31), and (34) yields

for all . Relation (35) ensures that the sequence is bounded in ; indeed, if we suppose that we have along a (relabeled) subsequence, then it is seen from (35) that there is a constant such that

which contradicts hypothesis .

By the reflexivity of , there exists a subsequence of , denoted again by , such that

for some . Using hypothesis with , we derive that .

It remains to show that verifies the inequality in problem (1). Let an arbitrary element and let such that . Then for each and so from (28), we have that

The compactness of the embedding and (37) guarantee that in as . Then the upper semicontinuity of on implies

Assumptions and ensure that

Passing to the as in (38) and using (39) and (40), we get that satisfies the inequality in (1). Since was chosen arbitrarily in , we conclude that solves problem (1). The proof of Theorem 4 is complete.

Acknowledgment

This work is funded by a Marie Curie Intra-European Fellowship for Career Development within the European Community’s 7th Framework Program (Grant Agreement no. PIEF-GA-2010-274519).

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