Entropy Solution for Doubly Nonlinear Elliptic Anisotropic Problems with Robin Boundary Conditions

Ibrango, I.; Ouaro, S.

doi:https://doi.org/10.1155/2015/919608

International Journal of Differential Equations

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Research Article | Open Access

Volume 2015 | Article ID 919608 | https://doi.org/10.1155/2015/919608

Entropy Solution for Doubly Nonlinear Elliptic Anisotropic Problems with Robin Boundary Conditions

I. Ibrango¹and S. Ouaro²

Academic Editor: M. El-Gebeily

Received04 Jul 2015

Accepted21 Oct 2015

Published24 Nov 2015

Abstract

We study in this paper nonlinear anisotropic problems with Robin boundary conditions. We prove, by using the technic of monotone operators in Banach spaces, the existence of a sequence of weak solutions of approximation problems associated with the anisotropic Robin boundary value problem. For the existence and uniqueness of entropy solutions, we prove that the sequence of weak solutions converges to a measurable function which is the entropy solution of the anisotropic Robin boundary value problem.

1. Introduction

The aim of this paper is to study the following nonlinear anisotropic elliptic Robin boundary value problem: where is an open bounded domain of () with smooth boundary and , , is the unit outward normal on , and .

All papers on problems like (1) considered particular cases of function . Indeed, in [1], Bonzi et al. studied the following problems:where . The authors use minimization technics used in [2] or [3] (see also [4, 5]) to prove the existence and uniqueness of entropy solution.

The Robin type boundary conditions in the variable exponents setting are new and interesting problems and were for the first time studied by Boureanu and Radulescu in [3]. The main difficulty for the study of problem in [3] was the definition of an admissible space of solutions. The authors defined the appropriate space and obtained its properties which permit them to use minimization method to prove the existence of weak solutions to the following problem:Since we consider -data instead of the function considered in [3], the suitable notion of solution is the entropy solution introduced by Bénilan et al. in [6] (see also [7]). With the Robin type boundary conditions, the values of the solutions at the boundary must be precise and the notion of solutions considered must include the boundary condition. In this paper, as the function is more general, it is not possible to use minimization technic to get the existence of solution. Therefore, we used the technic of monotone operators in Banach spaces (see [8]) to get the existence of entropy solutions of (1).

For presenting our main result, we first have to describe the data involved in our problem. Let be a bounded domain in () with smooth boundary domain and such that, for any , is a continuous function withFor any , let be a Carathéodory function satisfying the following:(i)There exists a positive constant such thatfor almost every and for every , where is a nonnegative function lying in , with .(ii)For with and for almost every , there exists a positive constant such that(iii)There exists a positive constant such thatfor and for almost every .The hypotheses on are classical in the study of nonlinear problems (see [1, 3]).

The function is such thatThroughout this paper, for any , we assume thatwhere .

We put for all , and for all , We make the following assumption:Note that the function is continuous, defined on with for all in and .

A prototype example that is covered by our assumptions is the following anisotropic -harmonic system:which, in the particular case when for any , is the -Laplace equation.

The rest of the paper is organized as follows. In Section 2, we present some preliminary results. In Section 3, we study the existence and uniqueness of entropy solution.

2. Preliminaries

We recall in this section some definitions and basic properties of anisotropic Lebesgue and Sobolev spaces with variable exponent. Set For any , the variable exponent Lebesgue space is defined by endowed with the so-called Luxemburg norm The -modular of the space is the mapping defined by For any , the following inequality (see [9, 10]) will be used later:For any and with in , we have the following Hölder type inequality:If is bounded and such that for any , then the embedding is continuous (see [11, Theorem 2.8]).

Herein, we need the following anisotropic Sobolev space with variable exponent: is a separable and reflexive Banach space (see [2]) under the norm We need the following embedding and trace results.

Theorem 1 (see [9, Corollary 2.1]). Let be a bounded open set and for all , , a.e. in . Then, for any with a.e. in such that one has the compact embedding

Theorem 2 (see [3, Theorem 6]). Let be a bounded open set with smooth boundary and let , satisfy the conditionThen, there is a compact boundary trace embedding

We introduce the numbers The following result is due to Troisi (see [12]).

Theorem 3. Let ; and Then, there exists a constant depending on if and also on and if such that

In this paper, we will use the Marcinkiewicz space () as the set of measurable functions for which the distribution functionsatisfies an estimate of the formWe will use the following pseudonorm in :For any , the truncation function is defined byIt is clear that and .

In order to simplify the notation, for any , we use instead of for the trace of on .

Set as the set of the measurable functions such that, for any , We define the space as the set of functions such that there exists a sequence satisfyingWe need the following lemma proved in [13].

Lemma 4. Let be a nonnegative function in . Assume and there exists a constant such thatThen, there exists a constant , depending on , such thatwhere

3. Entropy Solutions

The notion of entropy solutions to problem (1) where the data belongs to is the following.

Definition 5. A measurable function is an entropy solution of problem (1) if , , and, for every ,for every .

The existence result is the following theorem.

Theorem 6. Assume that (4)–(12) hold. Then, problem (1) admits at least one entropy solution.

Proof. The proof is done in three steps.

Step 1 (the approximate problem). We define the reflexive space Let be the subspace of defined by where is the trace of in the usual sense, since . In the sequel, we will identify an element with its representative .
For any and , we consider the sequence of approximate problems:where , and we define an operator by where Note thatAssertion 1 (the operator is of type M). (i) The operator is monotone. Indeed, for , we haveTherefore,since for , for almost every , and are monotone.
(ii) The operator is hemicontinuous. Indeed, for every in , let and let be such that . We have in .
Using the Hölder type inequality, there exists such that Let us denote .
Using assumption (5) and [11, Theorems 4.1 and 4.2] we have in . Then, we deduce that is continuous; namely, the operator is hemicontinuous.
Since the operator is monotone and hemicontinuous, then according to Lemma 2.1 in [8], is of type M.
Assertion 2 (the operator is of type M). Indeed, let be a sequence in such thatSince by Fatou’s lemma, we obtain thatand thanks to the Lebesgue dominated convergence theorem, we havefor all in . Consequently, Therefore, we deduce that As in Assertion 1, we prove that the operator is of type M, so we have Thus, it follows that Hence, is of type M.
Assertion 3 (the operator is coercive). Indeed, since thenAccording to (7), we haveDenote We have Using the convexity of the application , , we obtain Then,(i) Assume . Then, (18) gives
So, combining (56) and (61) we get (ii) Assume . Then, combining (56) and (61) we get Consequently, since , the operator is coercive.
Besides, the operator is bounded and hemicontinuous.
Then, for any , we can deduce the existence of a function such that Namely, is a weak solution of problem (39).

We are now going to prove that these approximated solutions tend, as goes to infinity, to a measurable function which is an entropy solution of problem (1). To start with, we establish some a priori estimates.

Step 2 (a priori estimates). Assume (4)–(12) and let be a solution of problem (39). We have the following results.

Lemma 7. There exists a constant such that

Proof. Let us take as a test function in (39). Since using relation (7), we obtainThen, we have Moreover, we have Therefore, we get where

Lemma 8. For any , there exist two constants and such that (i),(ii),

Proof. (i) This is a consequence of Lemmas 4 and 7.
(ii)Let . For any , we have Then, there exists a positive constant such thatLet us consider the function We have for . Thus, if we take in (72) we get If , we have Then Therefore, we deduce that there exists a positive constant such that

Step 3 (existence of entropy solution). Using Lemma 8, we have the following useful lemma (see [13]).

Lemma 9. For , as , one hasIn order to pass to the limit in relation (39), we also need the following convergence results which can be proved as in [7] (see also [1, 13]).

Proposition 10. Assume (4)–(12). If is a weak solution of then the sequence is Cauchy in measure. In particular, there exist a measurable function u and a subsequence still denoted by such that in measure.

Proposition 11. Assume (4)–(12). If is a weak solution of then (i)there exists such that a.e. in and moreover in ,(ii)for all , converges strongly in . Moreover converges to in strongly and in weakly for all ,(iii) converges to some measurable function a.e. in .

We can now pass to the limit in relation (39).

Let and choosing as a test function in (39), we getFor the right-hand side of (79), we havesince converges strongly to in and converges weakly- to in and a.e. in .

For the first term of (79), we have (see [13])We now focus our attention on the second term of (79). We haveNow we show that . Indeed, let us denote by If is a maximal monotone operator defined on , we denote by the main section of ; that is, Remark that as goes to , goes to .

We take as a test function in (39) for the weak solution and (a constant to be chosen later) to getWe have Note also that since the function is nondecreasing with , we have Then, (85) gives which is equivalent to saying We now let go to in the above inequality to obtainChoosing in the above inequality (since is surjective), we obtainFor any , we have Then, (91) gives Hence, for all , we have a.e. in , which implies thatLet us remark that as is a weak solution of (39), then is a weak solution to the following problem:where , , and .

According to (94) we deduce that ThereforeIt follows from (94) and (97) that, for all , which implies

We can now use the Lebesgue dominated convergence theorem to getBy using again the Lebesgue dominated convergence theorem, we obtainFor the third term of (79), let us prove that We have Since the quantity is nonnegative and since for all in , the application is continuous, we have and by Fatou’s lemma, it follows thatWe have Hence, .

Since converges weakly- to in and , it follows thatBy adding (104) and (106), we getSince thus we getCombining (80), (81), (99), (100), and (109) we obtainThen, is an entropy solution of problem (1). That completes the proof of Theorem 6.

We now state the uniqueness result of entropy solution.

Theorem 12. Assume that (4)–(12) hold true and let be an entropy solution of (1). Then, is unique.

Proof. The proof is done in two steps.

Step 1 (a priori estimates). We consider the following.

Lemma 13. Assume (4)–(12) and . Let be an entropy solution of (1). Thenand there exists a positive constant such that

Proof. Let us take in the entropy inequality (36).(i)By the fact that , using (7), we get (111).(ii)Also, using the fact that , relation (36) givesBy (113), we deduce that which imply that or Therefore, So, we obtain Since the function is nondecreasing, then Consequently, there exists a constant such that

Lemma 14. Assume (4)–(12) and let . If is an entropy solution of (1), then there exists a constant which depends on and such thatand a constant which depends on and such that

Proof. (i) For any , relation (112) gives Therefore, that is, (ii) See [7] for the poof of (122).

Lemma 15. Assume (4)–(12) and let . If is an entropy solution of (1), thenwhere and

Proof. Since the function is surjective, according to Lemma 14-(121), we have and as , it follows by using the Lebesgue dominated convergence theorem that

Lemma 16. Assume (4)–(12) and let . If is an entropy solution of (1), then there exists a positive constant such thatwhere , ,

Proof. Let as test function in the entropy inequality (36). We get Thus, and using (7), we have Consequently,

Step 2 (uniqueness of entropy solution). Let and be two entropy solutions of (1). We write the entropy inequality corresponding to the solution , with as test function, and to the solution , with as test function to getUpon addition, we getDefine the following sets: We start with the first integral in (135). We have Then, we obtainAccording to (5) and the Hölder type inequality, we have where Thanks to relation (18) and Lemma 16, the quantity is finite for all .
According to Lemma 15, the quantity converges to zero as goes to infinity. Consequently, the last integral of (138) converges to zero as goes to infinity. Then,with
We may adopt the same procedure to treat the second term in (135) to obtainwith
For the other terms in the left-hand side of (135), we denote We have Then, by the Lebesgue dominated convergence theorem, we obtain Then,In the same way, we getNow, consider the right-hand side of inequality (135). We have By the Lebesgue dominated convergence theorem, we obtainAfter passing to the limit as goes to in (135), we getSince , , and are monotone, then According to (6), we deduce from (153) that For fixed in , is nondecreasing and vanishes at . Then, Now, by inequality above and (152), we deduce that for all there exists with such that, for all , Therefore,As , the following relation is true for any , (cf. [14]): From inequality above and (157), we get Finally as it follows that That completes the proof of Theorem 12.

Conflict of Interests

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

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Copyright

Copyright © 2015 I. Ibrango and S. Ouaro. 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.

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