International Journal of Differential Equations

Volume 2013 (2013), Article ID 476781, 14 pages

http://dx.doi.org/10.1155/2013/476781

Research Article

Entropy Solutions for Nonlinear Elliptic Anisotropic Homogeneous Neumann Problem

B. K. Bonzi, S. Ouaro, and F. D. Y. Zongo

Laboratoire d'Analyse Mathématique des Equations (LAME), UFR, Sciences Exactes et Appliquées, Université de Ouagadougou, 03 BP 7021 Ouaga 03, Ouagadougou, Burkina Faso

Received 22 June 2012; Accepted 1 February 2013

Academic Editor: Elena Braverman

Copyright © 2013 B. K. Bonzi 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 prove the existence and uniqueness of entropy solution for nonlinear anisotropic elliptic equations with Neumann homogeneous boundary value condition for -data. We prove first, by using minimization techniques, the existence and uniqueness of weak solution when the data is bounded, and by approximation methods, we prove a result of existence and uniqueness of entropy solution.

1. Introduction

Let be an open bounded domain of with smooth boundary. Our aim is to prove the existence and uniqueness of entropy solution for the anisotropic nonlinear elliptic problem of the form

where the right-hand side and , are the components of the outer normal unit vector.

For the rest of the functions involved in (1), we are going to enumerate their properties after we make some notations.

For any , we set

and we denote

For the exponents, , with for every and for all , we put and . Now, we can give the properties of the rest of the functions involved in (1).

We assume that for , the function is Carathéodory and satisfies the following conditions: is the continuous derivative with respect to of the mapping , , that is, such that the following equality and inequalities holds

for almost every .

There exists a positive constant such that

for almost every and for every , where is a nonnegative function in , with .

There exists a positive constant such that

for almost every and for every , with and

for almost every and for every .

We also assume that the variable exponents are continuous functions for all such that

where .

We introduce the numbers

A prototype example, that is, covered by our assumptions is the following anisotropic equation:

Set , where . Then, we get the following equation.

Actually, one of the topics from the field of PDEs that continuously gained interest is the one concerning the Sobolev space with variable exponents, or depending on the boundary condition (see [1–23]). In that context, problems involving the -Laplace operator

or the more general operator

were intensively studied (see [13]). At the same time, some authors was interested by PDEs involving anisotropic Sobolev spaces with variable exponent when the boundary condition is the homogeneous Dirichlet boundary condition (see [15, 16, 18, 20, 24–26]). In that context, the authors have considered the anisotropic -Laplace operator

or the more general variable exponent anisotropic operator

When the homogeneous Dirichlet boundary condition is replaced by the Neumann boundary condition, one has to work with the anisotropic variable exponent Sobolev space instead of . The main difficulty which appears is that the famous Poincaré inequality does not apply and then it is very difficult to get a priori estimates which are necessary for the proof of the existence result of entropy solutions. Sometimes one can use the Wirtinger inequality which does not apply, in some problems like (1). The first systematic study of anisotropic Neumann problem was done by Fan (see [11]). In a second time, Boureanu and Rădulescu studied an anisotropic nonhomogeneous Neumann problem with obstacle (see [2]). In the two papers, the authors were interested by the existence and multiplicity results of weak solution even if in [2], Boureanu and Rădulescu have showed some conditions under which we can get uniqueness of weak solution. In this paper, we are interested to the existence and uniqueness of entropy solution. For the proof of the existence of entropy solution of (1), we follow [27] and derive a priori estimates for the approximated solutions and the partial derivatives in the Marcinkiewicz spaces and , respectively (see Section 2 or [27, 28] for definition and properties of Marcinkiewicz spaces).

The study of anisotropic problems are motivated, for example, by their applications to the mathematical analysis of a system of nonlinear partial differential equations arising in a population dynamics model describing the spread of an epidemic disease through a heterogeneous habitat.

The paper is organized as follows. In Section 2, we introduce some notations/functional spaces. In Section 3, we prove for the problem (1), the existence and uniqueness of weak solution when the data is bounded, and the existence and uniqueness of entropy solution when the data is in .

2. Preliminaries

In this section, we define Lebesgue, Sobolev, and anisotropic spaces with variable exponent and give some of their properties (see [29] for more details about Lebesgue and Sobolev spaces with variable exponent). Roughly speaking, anistropic Lebesgue and Sobolev spaces are functional spaces of Lebesgue’s and Sobolev’s type in which different space directions have different roles.

Given a measurable function , we define the Lebesgue space with variable exponent as the set of all measurable functions for which the convex modular

is finite. If the exponent is bounded, that is, if , then the expression

defines a norm in , called the Luxembourg norm. The space is a separable Banach space. Moreover, if , then is uniformly convex, hence reflexive, and its dual space is isomorphic to , where . Finally, we have the Hölder type inequality.

Proposition 1 (generalized Hölder inequality, see [10]). (i) For any and , we have

(ii) If ,   for any , then and the imbedding is continuous.

Moreover, the application called the modular of the space is very useful in handling these Lebesgue spaces with variable exponent. Indeed we have the following properties (see [10]). If and then

If, in addition, , then

converges to in measure and .

Now, let us introduce the definition of the isotropic Sobolev space with variable exponent, .

We set

which is a Banach space equipped with the norm

Now, we present a natural generalization of the variable exponent Sobolev space that will enable us to study the problem (1) with sufficient accuracy.

The anisotropic variable exponent Sobolev space is defined as follows:

Endowed with the norm

the space is a reflexive Banach space (see [11, Theorems 2.1 and 2.2]).

We have the following result.

Theorem 2 (see [11, Corollary 2.1]). Let be a bounded open set and for all ,   a.e. in . Then, for any with a.e. in such that

we have the compact embedding

Next, we define

Finally, in this paper, we will use the Marcinkiewicz spaces with constant exponent. Note that the Marcinkiewicz spaces in the variable exponent setting were introduced for the first time by Sanchon and Urbano (see [23]).

Marcinkiewicz spaces contain the measurable functions for which the distribution function

satisfies an estimate of the form

The space is a Banach space under the norm

where denotes the nonincreasing rearrangement of :

We will use the following pseudonorm

which is equivalent to the norm (see [27]).

We need the following Lemma (see [28, Lemma  A.2]).

Lemma 3. Let . Then, for every measurable function on , we have

(i)  Moreover,(ii), for every measurable subset .

In particular, with continuous injection and implies .

The following result is due to Troisi (see [30]).

Theorem 4. Let and let

Then, there exists a constant depending on if  and also on and if  such that

where and .

We will use through the paper, the truncation function at height , that is

We need the following lemma.

Lemma 5. Let be a nonnegative function in . Assume and there exists a constant such that

Then, there exists a constant , depending on , such that

where .

Proof. Consider the following

Step  1 (). Then, obviously we have , for some positive constant . Indeed, since , according to Proposition 1 there exists a positive constant such that

It follows that there exists a positive constant such that

Step  2 (). We get from (37)

Not also that

Therefore, by using (35), we obtain for ,

It follows that

which is equivalent to

Therefore,

Since, we get

which implies that .

For we have

So,

We need the following well-known results.

Theorem 6 (see [31, Theorem ]). Let be a reflexive Banach space and let be Gateaux differentiable over the closed set . Then, the following are equivalent.

(i) is convex over . (ii)We have

where denotes the dual of the space . (iii)The first Gateaux derivative is monotone, that is,

(iv)The second Gateaux derivative of exists and it is positive, that is,

Theorem 7 (see [32, Theorem 1.2]). Suppose is a reflexive Banach space with norm , and let be a weakly closed subset of . Suppose is coercive and (sequentially) weakly lower semicontinuous on M with respect to , that is, suppose the following conditions are fulfilled.

(i) as .(ii)For any , any subsequence in such that weakly in there holds

Then, is bounded from below and attains its infinimun in .

3. Main Results

In the sequel, we denote and .

3.1. Weak Solutions

Let us define first the notion of weak solution.

Definition 8. Let be a measurable function, we say that is a weak solution of problem (1) if belongs to and satisfies the following equation:

for every .

We associate to problem (1) the energy functional , defined by

To simplify our writing, we denote by the functional

We recall the following result (see [15, Lemma 3.4]).

Lemma 9. The functional is well-defined on . In addition, is of class and

for all .

Due to Lemma 9, a standard calculus leads to the facts that is well-defined on and with the derivative given by

for all . Obviously, the weak solutions of (1) are the critical points of ; so by means of Theorem 7, we intend to prove the existence of critical points in order to deduce the existence of weak solutions.

Theorem 10. Assume (4)–(8) and . Then, there exists a unique weak solution of problem (1).

Let us start the proof by establishing some useful lemmas.

Lemma 11. If hypotheses (4)–(8) are fulfilled, then the functional is coercive.

Proof. Let be such that . Using (7), we deduce that

We make the following notations:

We then have

Using (19), (20), and (21), we have

By the generalized mean inequality or the Jensen’s inequality applied to the convex function , we get

thus,

Case  1 (). We have

Therefore,

Case  2  (). Then , and we get

So, we obtain

Then, letting goes to infinity in (66) and (68), we conclude that reaches infinity. Thus, is coercive.

Lemma 12. The functional is weakly lower semicontinuous.

Proof. By [33, Corollary ], it is enough to show that is lower semicontinuous. To this aim, fix and . Since for every is monotone, Theorem 6 yields

Since the map is convex, again by Theorem 6, we have

then (69) becomes

Consider the second term in the right-hand side of (71). By (5) and Hölder type inequality, we have

For the fourth term in the right-hand side of (71), we have

The third term in the right-hand side of (71) gives by using Hölder type inequality

Gathering these inequalities, it follows that

for every such that . Thus, is lower semicontinuous.

Proof of Theorem 10. Consider the following

Step  1. Existence of weak solutions. The proof follows directly from Lemmas 11 and 12 and Theorem 7.

Step  2. Uniqueness of weak solution. Let be two weak solutions of problem (1). Choosing a test function in (54), for the weak solution and for the weak solution , we get

Summing up (76) and (77), we obtain

Thus, by the monotonicity of the functions and , we deduce that almost everywhere.

3.2. Entropy Solutions

First of all, we define a space in which we will look for entropy solutions. We define the space as the set of every measurable function which satisfies for every .

Lemma 13 (see [34, 35]). Let . Then, there exists a unique measurable function such that

where denotes the characteristic function of a measurable set . The functions are called the weak partial gradients of and are still denoted . Moreover, if belongs to , then and coincides with the standard distributional gradient of , that is, .

Definition 14. We define the space as the set of function such that there exists a sequence satisfying

(a) a.e. in ,(b) in , for all .

Definition 15. A measurable function is an entropy solution of (1) if and for every ,

for all .

Our main result in this section is the following.

Theorem 16. Assume (4)–(8) and . Then, there exists a unique entropy solution to problem (1).

Proof. The proof of this Theorem will be done in three steps.

Step  1 (a priori estimates).

Lemma 17. Assume (4)–(8) and . Let be an entropy solution of (1). If there exists a positive constant such that

then

where , for all .

Proof. Take in (80), we have

Since the second term in the previous inequality is nonnegative, it follows that

According to (7), we deduce that

Therefore, defining , we have for all ,

From the previous inequality, the definition of and (81), we have

Lemma 18. Assume (4)–(8) and . Let be an entropy solution of (1), then

for every , with a positive constant. Moreover, we have

and there exists a constant which depends on and such that

Proof. Taking in the entropy inequality (80) and using (7), we obtain

for all . This yields

As , we get from the previous inequality by using Fatou’s lemma

Now, since we have

We deduce that

Indeed,

From aforementioned, we get

Therefore,

which implies

Lemma 19. If is an entropy solution of (1) then there exists a constant such that

Proof. Taking in the entropy inequality (80) and using (7), we get

Note that

Therefore, we deduce according to (101) that

Lemma 20. If is an entropy solution of (1) then

where , , .

Proof. Taking as a test function in the entropy inequality (80), we get

It follows by using (7) that

Therefore,

Lemma 21. If is an entropy solution of (1) then

where , .

Proof. By Lemma 18, we deduce that

and as , it follows by using the Lebesgue dominated convergence theorem that

The proof of the following lemma can be found in [1].

Lemma 22.  Assume (4)–(8) and . Let be an entropy solution of (1), then

where is a positive constant which depends on and .

Step  2 (uniqueness of entropy solution). The proof of the uniqueness of entropy solutions follows the same techniques by Ouaro [20] (see also [35]). Indeed, let and be two entropy solutions of (1). We write the entropy inequality (54) corresponding to the solution , with as test function, and to the solution , with as test function. Upon addition, we get

Define

We start with the first integral in (112). By (7), we have

Using (5) and Proposition 1, we estimate the last integral in (114) as follows:

where

For all , the quantity is finite according to relations (18), (19) and Lemma 20. The quantity converges to zero as goes to infinity according to Lemma 21. Then, the last expression in (115) converges to zero as tends to infinity. Therefore, from (114), we obtain

where converges to zero as tends to infinity. We may adopt the same procedure to treat the second term in (112) to obtain

where converges to zero as tends to infinity.

For the two other terms in the left-hand side of (112), we denote

We have a.e. as goes to infinity and

Then, by the Lebesgue dominated convergence theorem, we obtain

In the same way, we get

Therefore,

Furthermore, consider the right-hand side of inequality (112). We have

Indeed,

so that we are able to apply the Lebesgue dominated convergence theorem. Then, we deduce from relations (112)–(124) after passing to the limit as in (112) the following:

Using (6) and as is monotone, we deduce from (126) that

Since , the following relation is true for any , (cf. [12])

Therefore, from (127), we get that a.e. in , which means that for all , there exists with such that for all ,

Therefore,

Now, using (128) and (130), we obtain

Step  3 (existence of entropy solutions). Let be a sequence of bounded functions, strongly converging to and such that

We consider the problem

It follows from Theorem 10 that problem (133) admits a unique weak solution which satisfies

for all .

Our interest is to prove that these approximated solutions tend, as goes to infinity, to a measurable function which is an entropy solution of the problem (1). We announce the following important lemma, useful to get some convergence results.

Lemma 23. If is a weak solution of (126) then there exist some constants such that

(i), (ii) for  all  .

Proof. (i) is a consequence of Lemmas 19 and 5 by using for all as a test function in (134).

(ii) We first use for all as a test function in (134) to get

Then, let for all , we have for any ,

Using (135) and (i), we get

from which we deduce (ii).

By lemmas 3 and 23, it follows that is uniformly bounded in for some , and in the same way, is uniformly bounded in for some . From this, we get that the sequence is uniformly bounded in , where . Consequently, we can extract a subsequence, still denoted satisfying

By the same way as in the proof of [16, Lemma 3.5] (see also [27]), we prove that

We deduce from (139) that

In order to pass to the limit in relation (134), we need also the following convergence results which can be proved by the same way as in [1].

Proposition 24. Assume (4)–(8), and (132). Let be the solution of (133). The sequence is Cauchy in measure. In particular, there exists a measurable function and a subsequence still denoted by such that in measure.

Proposition 25. Assume (4)–(8), and (132). Let be the solution of (133). The following assertions hold.

(i)For all , converges in measure to the weak partial gradient of .(ii)For all and all , converges to in strongly and in weakly.

We can now pass to the limit in (134). To this end, let . For any , choose as a test function in (134), we get

For the right-hand side of (141), the convergence is obvious since converges strongly to in , and converges weakly- to in and a.e in .

For the second term of (141), we have

The quantity is nonnegative and since for all is continuous; we get

Then, it follows by Fatou’s Lemma that

Let us show that .

We have

If , then .

If , then

Hence, .

Since converges weakly- to in and , it follows that

For the first term of (141), we write it as follows:

The first term of (148) is nonnegative by (7), then by Fatou’s Lemma and (138), we get

According to Proposition 25, the second term of (148) converges to

Combining the previous convergence results, we obtain

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