On -Anisotropic Problems with Neumann Boundary Conditions

Abstract

This work is devoted to the study of a general class of anisotropic problems involving -Laplace operator. Based on the variational method, we establish the existence of a nontrivial solution without Ambrosetti-Rabinowitz type conditions.

1. Introduction

The elliptic problems in anisotropic form concerning the Sobolev space with variable exponents have recently attracted the attention of many mathematicians; see [1–13] and the references therein. Such equations arise in connection with the equations describing electromagnetic fields and the plasma physics; see [14, 15] and various applications like those in thermorheological fluids [16], elastic mechanics [17], and image restoration [18]. They also appear in biology; see, for instance, Bendahmane et al. in [19], as a model for the propagation of epidemic diseases in heterogeneous domains.

In the present paper, we study the anisotropic nonlinear elliptic problem of the formwhere () is a bounded open set with smooth boundary and are the components of the outer normal unit vector and for , for all , , where the exponents are continuous functions such that .

We assume that the functions and are Carathéodory and satisfying the following conditions for all .There exists a positive constant such that fulfills for all and all , where (with ) is a nonnegative function and is the mapping which verifies There exists such that for all and all .The monotonicity condition takes place for all and all with .

Example 1. We take and then the operatorbecomes in particular -Laplace operatorThis is why operators (7) are often known as generalized -Laplace type operators.

On the other hand, the anisotropic equations with the variable exponent growth conditions enable the study of equations with more complicated nonlinearities since the differential operator allows a distinct behavior for partial derivatives in various directions.

This paper is organized as follows. In Section 2, we give the necessary notations; we also include some useful results involving the variable exponent Sobolev spaces in order to facilitate the reading of the paper. Finally, in Section 3, we prove the existence of nontrivial solution.

2. Preliminaries and Main Result

We introduce the setting of our problem with some auxiliary results. For convenience, we only recall some basic facts which will be used later; we refer to [20, 21].

For , we introduce the Lebesgue space with variable exponent defined by where This space, endowed with the Luxemburg norm,is a separable and reflexive Banach space. We also have an embedding result.

Proposition 2. Assume that is bounded and , such that in . Then, the embedding is continuous.

Furthermore, the Hölder-type inequalityholds for all and , where is the conjugate space of , with .

Moreover, we denote and, for , we have the following properties:To recall the definition of the isotropic Sobolev space with variable exponent, , we setendowed with the normThe space is a separable and reflexive Banach space.

Now, we consider to be the vectorial function with for all and we put The anisotropic space with variable exponent isand it is endowed with the norm The space is a reflexive Banach space. Furthermore, an embedding theorem takes place for all the exponents that are strictly less than a variable critical exponent, which is introduced with the help of the notations

Proposition 3. Let be a bounded open set for all and for all . If , for all , then one has the compact and continuous embedding .

Remark 4. We make the following notations: Then, by (14), (15), and (16), Thus,

Definition 5. One defines the weak solution for problem (1) as a function satisfying for all .

We suppose the following hypotheses.There exist and with for all , such that verifies for all and all . uniformly for . uniformly for .There exist two positive constants and such that where with . for all and all .

The function , where is an example of functions verifying the assumptions (F1)–(F4). In fact, we have and then we get which means that (F4) is satisfied since we have which is nondecreasing in and then when , so we take . Taking into account that , it follows that Obviously the other assumptions are held.

We report our main result.

Theorem 6. Under conditions (A1)–(A4) and (F1)–(F4), problem (1) has at least a nontrivial weak solution.

The purpose of this work is to improve the results of the above-mentioned papers and many others, without assuming the Ambrosetti-Rabinowitz type conditions (A-R) used, for instance, in [1–3, 10], where in (A-R) there exist , such that for any and we have

In fact, it is known that (F4) is much weaker than the (A-R) condition in the constant exponent case (see, for instance, [22]). We will use the mountain pass theorem with Cerami condition which is weaker than condition used, for example, in [4, 6].

The energy functional corresponding to (1) is defined as ,By a standard argument, we can see that the functional is well defined and of class , with its Gâteaux derivative being described byfor all .

Putting

Proposition 7. (i) By A3, the functional is of type; that is, if and , then in .
(ii) From F1, the functional is weakly strongly continuous; that is, .

The proof of the first assertion (i) is similar to that in [2]. The second assertion is well known.

3. Proof of the Main Result

We will use the mountain pass theorem (see [23–25]), so we start by the condition of geometry in the form of the following lemma.

Lemma 8. (a) There exists with such that as .
(b) There exist , such that for .

Proof. (a) From (F2), we may choose a constant such thatLet large enough and with , and from (A1) and (39) we getwhere is a constant, taking sufficiently large to ensure that which implies that(b) By (A2), for , we haveOn the other side, from (F1), By the continuous embedding from into and ), there exist , such thatfor all . Hence,for all and all .
Therefore, since . Then for sufficiently small, we take such that

Definition 9. A sequence is called a Cerami sequence if is bounded and .

Lemma 10. If , then any sequence of Cerami of is bounded.

Proof. Let be a sequence of . We claim that is bounded; otherwise, up to a subsequence, we may assume that Putting , up to a subsequence, we have in in and in , . Almost everywhere .
Here, two cases appear, when , since we know that , that means Dividing (50) by , by using (A1), a straightforward computation leads to Meanwhile, in view of condition (F2) and Fatou’s lemma, which is contradictory.
In the case when , we choose a sequence satisfying If , since in and , by the continuity of the Nemitskii operator, we see that in as ; therefore,Given , since, for large enough, we have , using (54) with , from assumption (A2), and, considering Remark 4, we getThereby, . On the other hand, we know that , , so we can deduce that It yields Therefore,so we get Moreover,From (A2) and (F4), there exist such thatHence, , which is impossible.

Proof of Theorem 6. According to Lemma 8 and Lemma 10, we are to apply the mountain pass theorem, so seeing that the sequence (in Lemma 10) is strongly convergent to remains and it will be done.
Now, because the Banach space is reflexive (cf. [2, 3]), and regarding the boundedness of in , there exists such that . Since is the sum of type maps and which is weakly strongly continuous (cf. Proposition 7), is also of type. Thus, in .

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The author would like to thank the referee for the suggestions and helpful comments which improved the presentation of the original paper.