International Journal of Analysis

Volume 2014 (2014), Article ID 320527, 15 pages

http://dx.doi.org/10.1155/2014/320527

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

^{1}Laboratory LAMA, Department of Mathematics, Faculty of Sciences Dhar El Mahraz, University of Fez, Atlas, 1796 Fez, Morocco^{2}Faculty 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

*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*

*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. Approximate Problem**5.1. Existence of a Solution of the 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*

*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