#### Abstract

We consider the following problem: where is a bounded set in () with a smooth boundary, , , , and and belong to for some . In this paper, we assume that a.e. in and without sign condition and then we prove the existence of at least two bounded solutions under the condition that and are suitably small. For this purpose, we use the Mountain Pass theorem, on an equivalent problem to with variational structure. Here, the main difficulty is that the nonlinearity term considered does not satisfy Ambrosetti and Rabinowitz condition. The key idea is to replace the former condition by the* nonquadraticity condition at infinity*.

#### 1. Introduction and Main Result

Let be a bounded set in () with a smooth boundary . In this paper, we are concerned with the following elliptic problem:where is the -Laplacian operator, , , , and and belong to for some .

In the literature, there are many results concerning the existence, the uniqueness, and the multiplicity of solutions for models like under various assumptions on and . At first, it is important to mention that the sign of plays a crucial role in the problem regarding uniqueness, as well as existence, of bounded solutions. In this setting, we refer to ([1]) for more details. In the coercive case, that is, a.e. in for some , Boccardo, Murat, and Puel ([2â€“4]) proved the existence of bounded solutions for more general divergence form problems with quadratic growth in the gradient by using the sub and supersolution method. Moreover, Barles and Murat ([5]) and Barles et at. ([6]) have treated the uniqueness question for similar problems. Notice that if we allow a.e. in , then Ferone and Murat ([7, 8]) observed that finding solutions to becomes rather complex without imposing some strong regularity conditions on the data. For the particular case , there had been many contributions ([9â€“11]). However, for that may vanish only on some parts of , the uniqueness of solutions was left open until the recent paper authored by Arcoya et at. ([12]). This last result was proved for and and under the following condition:where . For a related uniqueness result see also Arcoya et at. ([13]).

The case where a.e. in , the question of nonuniqueness has been being an open problem given by Sirakov ([14]) and it has received considerable attention by many authors. Moreover, it should be pointed out that the sign of and whether is a function or a constant generate additional difficulties for solving . In this setting, Jeanjean and Sirakov ([1]) showed the existence of two bounded solutions assuming that , , and are in for some and satisfyingwhere depends only on , and is the optimal constant in Sobolevâ€™s inequality. Here, is allowed to change sign. Shortly after, this result was extended by Coster and Jeanjean ([15]) for is a bounded function such that by using the degree topological method.

Finally, in the case where is allowed to change sign and with a.e. in , Jenajean and Quoirin ([16, Theorem 1.1]) showed the existence of two bounded positive solutions when ; is a positive constant, and and are suitably small.

We would also like to mention that all the above quoted multiplicity results were restricted to the Laplacian operator with quadratic growth in the gradient, i.e. , and for . Moreover, it is interesting to mention that when is allowed to change sign the solutions are positive.

In this work, we prove the multiplicity of bounded solutions for the problem by assuming the following assumption:

Now, we give a brief exposition of the proof of our multiplicity result. At first, without loss of generality, we solve the problem by restricting it to the case is a positive constant. For is a negative constant, we replace by in , then we conclude. Next, we observe that the problems of type do not have a variational formulation due to the presence of the -gradient term. To overcome this difficulty, we perform the Kazdan-Kramer change of variable, that is, . Thus, we obtain the following equivalent problem :whereand

We mean by bounded weak solutions of the functions satisfyingfor any Obviously, if is a solution of , then is a solution of . Hence, the solutions obtained here are not necessarily positive (compare with ([16])).

One of the most fruitful ways to deal with is the variational method, which takes into account that the weak solutions of are critical points in of the -functionalwith and .

In this work, to obtain the two critical points for , we use the Mountain Pass Theorem to show one critical point and the standard lower semicontinuity argument to show the other. For the first one, according to the famous paper by Ambrosetti and Rabinowitz ([17]), the most important step is to show that satisfies the Palais-Smale condition at the level (see Definition 3). The fulfillment of this condition relies on the well-known Ambrosetti-Rabinowitz condition (() for short), namely,

Unfortunately, this condition is somewhat restrictive and not being satisfied by many nonlinearities . However, many researches have been made to drop the . We refer, for instance, to [18â€“22]. Notice that the nonlinearity considered here does not satisfy (). Moreover, since we do not assume any sign condition on , the fulfillment of the Palais-Smale condition turns out more delicate (see, e.g., [16, 23]). To the best of our knowledge, only Jenajean and Quoirin ([16]), recently, proved the Palais-Smale condition under the assumptions changes sign and is positive, and without assuming (). In their proof, for and , the authors based one of the arguments on the positivity of and the explicit determination of a function ;In our situation, as is allowed to change sign and the analog of their function cannot be computed explicitly, due to our general consideration of and ( and ); hence, their arguments cannot be adapted.

The key point to show the Palais-Smale condition in this paper is to prove that , among other conditions, satisfies the following (see Lemma 7):

The condition is a variant of the well-known* nonquadraticity condition at infinity*, which was introduced by Costa and MalgalhÃ£es ([18]) and is given as follows:

Observe that since , then is weaker than . Moreover, it should be noted that was considered by Furtado and Silva in their recent paper ([21]). Our result follows by using similar arguments.

Concerning the existence of the second critical point handled by the standard lower semicontinuity argument, we look for a local minimum in for the functional . Indeed, we observe that takes positive values in a large sphere, due to its geometrical structure (see Proposition 6), and .

Now we state the main result of this paper.

Theorem 1. *Assume that is satisfied. If and are suitably small, then the functional has at least two critical points. Hence, the problem has at least two bounded weak solutions.*

The paper is organized as follows. In Section 2 we recall some preliminary results and show that the functional has a geometrical structure. In Section 3 we prove our main result, Theorem 1.

*Notation*. Through this paper, we use the following notations.(1)The Lebesgue norm in is denoted by for . The norm in is denoted by . The HÃ¶lder conjugate of is denoted by .(2)The spaces and are equipped with PoincarÃ© norm and the dual norm , respectively.(3)We denote by the ball of radius centered at in and its boundary.(4)We denote by any positive constants that are not essential in the arguments and that may vary from one line to another.

#### 2. Preliminaries and Geometry of the Functional

In this section, we recall the standard definitions of Palais-Smale sequence at the level and Palais-Smale condition at the level for and we prove that the functional defined in (6) has a geometrical structure.

Let us define the level at as follows:where is the set of continuous paths joining and , where is defined in Proposition 6 below.

*Definition 2. *Let be a Banach space with dual space and is a sequence in . We say that is a Palais-Smale sequence at the level for if

*Definition 3. *We say that satisfies the Palais-Smale condition at the level if any Palais-Smale sequence at the level for possesses a convergent subsequence.

In order to prove that has a geometrical structure, we need some properties of , which we gather in the following lemma without proof.

Lemma 4. (1)* as , where if and if .*(2)* as , for all .*(3)* and as , for all *

Lemma 5. (1)*If , then we have *â€‰*for all , and for all .*(2)*If , then we have *â€‰*for all , and for all .*

*Proof. *By using Lemma 4(1), there exists such that for all we haveLet . If , then we haveMoreover, simple calculation yieldsNow, if , then we have , where Hence,By combining (14) and (16), (1) holds. To prove the property (2), we use Lemma 4(2) and the same previous argument.

Proposition 6. *Assume that holds. If and are suitably small, then the functional has a geometrical structure; that is, satisfies the following properties:*(i)*There exists such that for all in , , where .*(ii)*There exists such that and .*

*Proof. *(i) To prove this lemma we distinguish two cases on . Firstly, if , then by using Lemma 5(2) and HÃ¶lderâ€™s inequality, we getWe choose with close to such that , which exists due to the assumption . Obviously, . Thus, by using Sobolevâ€™s embedding we getMoreover, from the definition of the function in (4), we haveUsing Sobolevâ€™s embedding, we getBy the definition of in (6), we deduce thatNow, let in . Then, we haveWe take sufficiently large, such that and (which are sufficiently small by hypothesis), thenSecondly, , we choose again as above such that . Then, by using Lemma 5(1) and Sobolevâ€™s embedding, we getNow, as the first case, we getFinally, we summarize the two cases and get(ii) To prove the second property, we show that as . For this, let be a positive function such that . By the definition of in (6), we have From inequality (19), we getMoerever, by Lemma 4(3), we get Thus, we deduce the desired result.

Finally, we stress that since has a geometrical structure, then the existence of a Palais-Smale sequence at the level for is ensured. This can be observed directly from the proof given in ([17]), or alternatively using Ekelandâ€™s variational principle ([24]).

#### 3. Proof of Theorem 1

Recall from introduction that the proof of our main result is divided into two steps as follows. In the first step, we show the existence of the first critical point for the -functional by using the Mountain Pass Theorem due to Ambrosetti-Rabinowitz ([17]). Precisely, we show that the functional satisfies the Palais-Smale condition at the level . In the second step, we show the existence of the second critical point of on (which is a local minimum) by using the lower semicontinuity argument. Moreover, we are going to see that these critical points are not the same. Finally, we show that any solution of problem is bounded.

##### 3.1. First Critical Point: Palais-Smale Condition

In this subsection, we prove that satisfies the Palais-Smale condition at the level . Precisely, we show that any Palais-Smale sequence at the level for is bounded in , and, then, it has a strongly convergent subsequence.

They key point to prove the boundedness of the Palais-Smale sequence at the level in is to show that verifies the nonquadraticity condition at infinity . Indeed, we have the following lemma.

Lemma 7. *The function defined in (3) verifies the nonquadraticity condition at infinity .*

*Proof. *To prove , we show that is increasing and unbounded for sufficiently large. We recall that . Then, by simple calculations, we getwhere Thus, is increasing for large enough. Moreover, is unbounded. Indeed, by contradiction, if is bounded, then there exists a positive constant such thatIn addition, from the definition of and using integration by parts on , we getBy choosing , we obtainWhen , we obtain and . Hence, we have a contradiction. As a conclusion, the function verifies .

Lemma 8. *Let be a Palais-Smale sequence at the level for in . Then, is bounded in .*

*Proof. *Let be a Palais-Smale sequence at the level for in . We prove by contradiction that is bounded in . We assume that is unbounded in , that is, .

For all integer , we defineWe are going to prove that and also is bounded, which is the desired contradiction.*(a) Showing that *: We set ; then is bounded in . Hence, there exists a subsequence denoted again such that converges weakly and strongly to in and in for some , respectively. Moreover, also converges to almost everywhere in . Recall that is Sobolev conjugate.*Now, we claim by contradiction that ** a.e. in *.

Since is Palais-Smale type sequence, then we haveHence,for all and for some as . We divide both sides of (36) by , to obtainOn the one hand, since converges weakly to in and by the inequality (19), then for large enough the second and the third terms of the right-hand side of (37) are bounded.

On the other hand, if in , then in . Now, we choose such that in and in , with . Since in , then by using Lemma 4(3), we obtainHence, by using Fatouâ€™s lemma in (37) we obtain the unbounded term in the left-hand side of (37). Hence, the claim ( a.e. in .)

Since , then there exists such that , for large enough. Moreover, we haveIn what follows, we treat only the case . The other case follows with similar arguments. From Lemma 5(2), we have , where . Since , for some and converges strongly to in with , then, we obtaindue to a.e. in By HÃ¶lderâ€™s inequality, we getHence, by choosing large enough, we deduce that , as *(b) Showing that ** is bounded*: To prove that is bounded, we distinguish two cases: and .*The Case*â€‰â€‰. Here, we only handle the proof for . The other case follows as in the proof of Proposition 6(i). By the definition of and , we have , which means thatBy the definition of in (6), we havewhere the function is defined in and . Moreover, from Lemma 5(2), we have By choosing and as in the proof of Proposition 6(i), we getBy inequality (19) and Sobolevâ€™s embedding, we getThen, by (43), (45), and (46), we obtainfor all , where is independent of . Thus, is bounded, which contradicts the fact that is unbounded (see* (a)*).*The Case *. Here, we are proceeding the technique inspired by [21]. To this end, we need the following technical lemma.**Lemma ****9****.*** Let ** the nonnegative function defined as**with **. Then, we have*(i)(ii)*for any positive function*â€‰â€‰â€‰* in*â€‰â€‰â€‰* and*â€‰â€‰,*Proof*. Obviously we have (i). To prove (ii), we follow the same approach given in [21] for the case and , which can be immediately generalized for any positive function and . â–¡

Now, we resume the proof of Lemma 8. From Lemma 7, we have , for large enough and some (which will be chosen later). Moreover, if , then from Lemma 4(2), we have for sufficiently small,Then, by the continuity of , we have for all Letting , then we haveLet us handle the two terms and , respectively. By using (51), we getFor the term , we haveBy setting , we obtain