International Journal of AnalysisVolume 2014 (2014), Article ID 320527, 15 pageshttp://dx.doi.org/10.1155/2014/320527

Research Article

Existence and Nonexistence of a Solution for a Nonlinear -Elliptic Problem with Right-Hand Side Measure

Elhoussine Azroul,¹ Abdelkrim Barbara,¹ and Hicham Redwane²

¹Laboratory LAMA, Department of Mathematics, Faculty of Sciences Dhar El Mahraz, University of Fez, Atlas, 1796 Fez, Morocco²Faculty of Juridical, Economic and Social Sciences, Hassan 1st University, BP 539, Settat, Morocco

Received 4 November 2013; Accepted 24 February 2014; Published 4 May 2014

Academic Editor: Mats Ehrnstrom

Copyright © 2014 Elhoussine Azroul et al. 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 discuss the existence and nonexistence of solution of a nonlinear problem -elliptic-, where is a Radon measure with bounded total variation, by considering the Sobolev spaces with variable exponents. This study is done in two cases: (i) is absolutely continuous with respect to -capacity. and (ii) is concentrated on a Borel set of null -capacity.

1. Introduction

Let be a bounded domain of , let , and let (see definition below). Consider the following nonlinear -elliptic problem: where is a Leray-Lions operator from into its dual . The nonlinear term satisfies a growth condition of the following form: where is a continuous increasing function and and assuming also that , a.e. , , and for , where and are two positive real constants. The second term is a Radon measure on .

We are interested in the existence and nonexistence of solution of the problem . We prove the existence of solution if does not charge the sets of null -capacity, in other words, if and only if is absolutely continuous with respect to -capacity. We give a necessary and sufficient condition that allows us the existence of solution . Thus, if is concentrated on a Borel set of of null -capacity, we will show that the problem admits no solution. Boccardo et al. [1] treated for constant. We can also see other variations of this problem; for example, if the nonlinear term is independent of and behaves as , with , the results of the existence and nonexistence of solution depend on the measure ; see Baras and Pierre [2], Brezis [3], and Gallouët and Morel [4].

2. Preliminaries

Let be a bounded open subset of . The function satisfies the log-Hölder continuity on if with being a positive constant.

Let be log-Hölder continuous such that , where and .

For , we define the variable exponent Lebesgue space normed by . The space is a separable and reflexive Banach space, and its dual space is isomorphic to , where (see [5, 6]).

Proposition 1 (see [5, 6]). (i) For any and , one has the following: (ii) For all such that a.e. in ; then and the embedding is continuous.

Proposition 2 (see [5, 6]). If one denotes , , then

Define now the Sobolev space with variable exponent by normed by Let be the closure of in and let for .

Proposition 3 (see [5, 7]). (i) Assuming that , the spaces and are separable and reflexive Banach spaces. (ii) If and , for any , then embedding is continuous and compact. (iii) The Poincaré inequality: there exists a constant , such that (iv) The Sobolev inequality: there exists another constant , such that

Remark 4. By (iii) of the Proposition 3, we deduce that and are equivalent norms in .

Proposition 5 (see [8]). One denotes the dual space of the Sobolev space by , and, for each , there exist such that , and, for all , one has the following: Moreover,

For all and , the truncation function can be defined by Let be a compact ; the -capacity of over is defined by where is the characteristic function of .

If is open in , the -capacity of from is defined by If is any set of , we define the -capacity variational over by Denote by the space of signed measures on and

3. Measures and -Variational Capacity

In this section we will give some results concerning the measures and -capacity variational capacity of any set of .

Theorem 6. Let ; then there exist and such that

Proof. Let and let be measurable representation quasicontinuous of . Let and . is convex and lower semicontinuous on indeed; let in ; therefore, a.e. in , and hence a.e. in . Then by Fatou’s lemma, we get According to the separability of , there exist and such that , and we have the following: hence, consider the following: Since , then and we have the following: We conclude that .Let ; we have the following: which implies that Hence, and then .Since , , , then Hence, Let ; we get that is absolutely convergent in . For all , we have the following: This shows that . Let be a Borel , such that ; we have the following: And we deduce that , for all , from which . Using the theorem of Radon-Nikodym, then there exists such that, for all , we have , and therefore .Let and ; we have, for any Borel and for all , Therefore, , for all .For , we obtain the following, by monotone convergence theorem: Remark that hence , and since , then .

Theorem 7. Let and let . Then,

Proof. Let , and show that . Indeed, we have and then there exist and such that Let such that , and we prove that . Indeed, for all , there exists in , and .Let , such that in ; consider the sequence ; we have in . Indeed We obtain the following: The Poincaré inequality leads to Since in , then , and we deduce that .Conversely, let , and we can always assume that . And, by Theorem 6, we deduce that where , and . Let be an increasing sequence of compact such that and let be an increasing positive sequence in , with compact supports in . Let Then, converges strongly in and .Let a.e., and let . Let . We have in as and , for is large enough.Choose such that . Then , where We have that converges in and then .Moreover, and then is absolutely convergent in . Then let ; we conclude that

Definition 8. Let be a Radon measure and let be a Borel set of .(i)The restriction of to is the measure defined by  for any Borel set of .(ii)The measure is concentrated on if .

Proposition 9. Let and let with . Then uniquely decomposes as , where

Proof. We assume that is positive (if not, consider and ). We first show the uniqueness. Assume that , where and with and with . We have . Let be a Borel subset of ; if , then , and therefore and ; if we consider and since is focused on such that , then we have Therefore, , which proves the uniqueness.Now we show the existence of . Let there exists a sequence such that is a Borel subset of , , and . Let be a Borel subset of ; we have and .For and , there exist and .It remains to show that . By contradiction, assume that is a Borel subset of such that and . We have and , which contradicts the definition of .

Lemma 10. Let be a positive measure in concentrated on a set such that . Then , , and , such that

Proof. Let ; then for any , there exists which is a compact set in with .Moreover, ; hence, , and then there exists such that on a neighborhood of , , and converges to 0 as goes to .We have , as is concentrated on from which ; therefore , and since on a neighborhood of , we have then the following:

Remark 11. If is compact, we can choose , in view of Lemma 10, and we can deduce that converges to strongly in as goes to .

4. Essential Assumptions

Let be a bounded open subset of and . We consider a Leray-Lions operator from to dual defined by where is a Carathéodory operator, satisfying the following conditions: for a.e. , all , where is a positive function lying in and .

The nonlinear term satisfies the following: where is a continuous growth function and with . Consider the following: for a.e. , all , and, for , , where . We consider the following problem: where .

5. Approximate Problem

5.1. Existence of a Solution of the Approximate Problem

Consider the following approximate problem: where .

Define the operators and of to by

Lemma 12. The operator is bounded.

Proof. For all in , and by the Hölder inequality, where from which , for all fixed.

Lemma 13. The operator is pseudomonotone and coercive in the following sense: when .

Proof. Let ; we have where Hence, operator is bounded and, by Lemma 12, is bounded; then it follows that is bounded.We show that is pseudomonotone: let be a sequence in such that We prove that and converges to as goes to . Indeed, we get that weakly converges to in as tends to ; then converges to , for all , as tends to , and we obtain According to (46) and (49), we conclude that there exist and such that converges to weakly in and converges to weakly in . Hence . We have the following: Hence, In view of (48), we have , and then We deduce that In view of (61) we obtain the following: Finally, On the other hand, we use (64) and we deduce the following: So, in view of [9], we conclude that converges to a.e. in and strongly in ; hence converges to weakly in and converges to weakly in . Since converges weakly to , then hence , and we conclude that the operator is pseudomonotone.We show that is coercive and we have the following: from which where and , which implies that hence is coercive.According to [10], there exists solution of (45).

5.2. Estimation of the Solution Approached

Consider the following approximate problem: where and with .

Lemma 14. The sequence is bounded in .

Proof. Choose as test function in (72); we obtain the following: For , we have . Therefore, for , we have , and hence

Assertion 1. Let . We have Indeed According to of and to (47), we obtain Assertion 1.

Assertion 2. We have the following: with , , and , , and are positive constants.

Proof. In fact, we have with , and since , then . According to the inequality of Hölder, we obtain and therefore where since (Ω); hence , and therefore where therefore According to Assertion 1, (46), and (47), we obtain Assertion 2.

Assertion 3. We have, for all ,

Proof. Choose , and in we have: . Hence, in view of Assertion 2, we obtain and then , for all , which gives By a suitable choice of , we have . Now, we use the result of Stampacchia [11]; then there exists a constant (independent of ) such that , for all ; then , and we obtain Assertion 3.

We prove the following theorem.

Theorem 15. There exists a subsequence of still denoted by and such that converges to strongly in and is a weak solution of the following problem:

Proof. We need the following Lemmas. Lemma 16. Let and be two positive constants, for , and let the function . Then one has .Proof. We have ; then Lemma 17. There exists a positive constant such that, for all , Proof. In view of Assertion 2, we have , and we denote that .Let , where (with ).Choose as test function in (52), and it follows that Hence, consider this implies that and we have this implies that Hence, In view of Lemma 16 we have the following: Hence, Therefore, there exists a positive constant such that There exists a subsequence still denoted by and such that

Lemma 18. One has the following: and converges to strongly in as tends to .

Proof. We choose as a function test in the approximate problem (99) and we obtain This implies that Since , and converges to a.e. in , weakly in , and in weakly-, then We prove that . Indeed, we have the following: Since is bounded in , and in weakly-, then Therefore, .Finally, we obtain We conclude, by [9], that converges to strongly in as tends to .

6. Results of Existence

We state the following two theorems.

Theorem 19. Let and let with . Then there exists a solution of the following problem: in the sense that

Proof. Without loss of generality, we can assume that . Let , and we have By Lemma 18, we have strongly in , and then a.e. in , We use (46) and (49) and passing to the limit, we obtain the following:

Theorem 20. Let . Under the assumptions (45)–(50), there exists a solution of problem (51) in the sense that , and

Proof. Suppose there exists a solution of problem (51), with and . According to (46), we have ; then , and in ; with and by Theorem 7, we have .Conversely, if , by Theorem 7, , and then , where and . Let be a sequence in such that converges to strongly in and let be a sequence in such that converges to strongly in . By Theorem 19, there exists a solution of such that , and, for all , we have We choose , where is given by Lemma 16 and is given by (50); with and , where is given by (49), we obtain which implies that where , et .According to (47) and Young’s inequality, we obtain Hence, consider We can write the following: Using (49), we get and using (50), we obtain Now we use (119) and (120) in (117) and we write the following: In view of Lemma 16, we have , and then Hence, there is such that

Let us show that we have converging to strongly in . We start by proving that Indeed, for , choose as a test function in (113), and using (47) and Young’s inequality, we get We have . , for all , and , for all ; hence, Therefore, and, by (125), we have Since is bounded in , then . We use (127) and (128) and since strongly converges in and strongly converges in , we conclude that Now we show that and Indeed, let be fixed, and choose as test function in (113) where is the function of Lemma 16 with and , and we obtain To simplify writing, we put and .

We have which converges to a.e. in , which converges to in weakly-, and which converges to , and since strongly converges to in , we obtain Since converges to weakly in and converges to strongly in , we obtain Write the first term on the left-hand side of (131) as where . Since is bounded in and converges to zero in weakly- as goes to , then The first term on the right-hand side of (134) is Since converges to weakly in and , then We can write that In view of (50), we have On the other hand, we have Since , converges to a.e. in and in weakly-, the sequence converges to weakly in , and , then Now since is bounded in and converges to zero strongly in , then we have where Therefore, Now using (131) to (141), we conclude that and in view of [9], we have (130).

Now we prove that Indeed, let be a measurable subset of , and then consider We have on the one hand and since then, for , there exists such that Furthermore, , and since converges to strongly in , then there exists such that, for , In view of (150) and (151) we deduce that is equintegrable. Since converges to a.e., then by the Vitali theorem we obtain (146).

Let us show that Since converges to , a.e. in , it suffices to show that the sequence is equintegrable. Let be a measurable subset of ; we have Since , then there exists such that In view of (49), we have since and converges to strongly in ; therefore, there exists such that and , , and this shows that is equintegrable.

Now, let be a solution of the approximate problem; then we have Letting tend to , we conclude that satisfies the following:

7. A Result of Nonexistence

Theorem 21. Let be positive and concentrated in a subset of such that . Let be a sequence of positive functions in such that Assume (49), (50), and that satisfies , a.e. , , and . Let be solution of in the sense that Then, there exists , such that converges to strongly in , the sequence converges to weakly in , and

Proof. Consider the following. Step 1. In view of the proof of Theorem 19 and the assumptions made about and , we have the sequence which is bounded in and then there exists a subsequence, still denoted by , such that converges to weakly in and a.e. in . By (46), there exists such that converges to weakly in .The function is continuous and let such that ; choose as test function in (160), where is given by Lemma 16, and then we obtain Since converges to strongly in as tends to and converges to zero strongly in as tends to , then In view of (49) we have , for , since , and we have Since converges to weakly in and , then We use (47) and (163)–(165) in (162); we conclude that Step 2. Choose as test function in (160); we have the following: We pass to the limit as tends to and tends to zero in (167), and using the same arguments of Step 1, we conclude that Using Step 1, we deduce that , and we have which converges to strongly in . Since the limit is independent of the choice of the subsequence, the sequence converges strongly in ; hence converges to weakly in and converges to a.e. in . Then, we obtain .Now, let , and we obtain Passing to the limit, we have Therefore, consider The proof of Theorem 21 is complete.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The authors would like to thank the anonymous referees for their interesting remarks.

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