Abstract
This paper is devoted to the study of a class of Kirchhoff type problems with critical exponent, concave nonlinearity, and signchanging weight functions. By means of variational methods, the multiplicity of the positive solutions to this problem is obtained.
MSC: 35J20, 35J60, 47J30, 58E50.
Keywords:
Kirchhoff type problem; critical exponent; concave nonlinearity; signchanging weight functions; variational methods1 Introduction and main results
This paper is concerned with the existence and multiplicity of positive solutions for the following problem:
where Ω is a smooth bounded domain in with and the parameters . The weight functions , satisfy the following conditions:
() there exist positive constants and such that and in ;
() there exists such that as .
In (1), if we replace by , it reduces to the following Dirichlet problem of Kirchhoff type:
Problem (2) is related to the stationary analogue of the equation
proposed by Kirchhoff in [1] as an extension of the classical d’Alembert wave equation for free vibrations of elastic strings. Kirchhoff’s model takes into account the changes in the length of the string produced by transverse vibrations. It received great attention only after Lions [2] proposed an abstract framework for the problem. The solvability of the Kirchhoff type problem (2) has been paid much attention to by various authors. The positive solutions of such a problem are obtained by using variational methods [35]. Perera and Zhang [6] obtained a nontrivial solution of problem (2) via the Yang index and the critical group. He and Zou [7] obtained infinitely many solutions by using the local minimum methods and the fountain theorems. Recently, when is a continuous superlinear nonlinearity with critical growth, the existence of positive solutions of the Kirchhoff type problem has been studied [813]. Moreover, the paper [14] considered problem (2) with concave and convex nonlinearities by using a Nehari manifold and fibering map methods, and one obtained the existence of multiple positive solutions. In addition, the corresponding results of the Kirchhoff type problem can be found in [1525], and the references therein.
In the present paper, we deal with problem (1) and consider the existence and multiplicity of positive solutions of problem (1). About the critical growth situation, the aforementioned papers only showed the existence of positive solutions of the Kirchhoff type problem. Moreover, involving the concave and convex nonlinearities, [14] only considered the subcritical growth case. Therefore, our purpose is to extend the result of [14] to critical growth. The main results of this paper extend the corresponding results in [11] and [14].
Before stating our results, we give some notations and assumptions. Let , (), . In addition, we denote positive constants by . The main results of this paper are as follows.
Theorem 1Let, and. Suppose that () and () hold, then there existssuch that problem (1) for allhas at least one positive solution.
Theorem 2Let, and. Suppose that (), (), (), () and () hold, then there existssuch that problem (1) for allhas at least two positive solutions.
Remark 1 Our Theorem 2 extends the results for the critical case of Theorem 1.1 in [11]. Our Theorem 2 shows that we have at least two positive solutions of problem (1), but the authors of the reference only obtain at least one positive solution of problem (1). In addition, the results of Theorem 2.1 in [14] are extended to critical growth.
This paper is organized as follows. In Section 2, we give the local PalaisSmale condition. The proof of Theorems 1 and 2 is provided in Section 3.
2 The local PalaisSmale condition
In this section, we show that the corresponding functional of problem (1) satisfies the condition. Let , the corresponding functional of problem (1) is
It is well known that the critical points of the functional I in are weak solutions of problem (1). By the definition of weak solution u of problem (1), it means that satisfies
Define the best Sobolev constant,
From [26], we know that S is attained when by functions
Definition A sequence is called a sequence of I if and as . We say that I satisfies the condition if any sequence of I has a convergent subsequence.
Lemma 1Let, and. Assume that () and () hold. Ifis asequence ofI, thenis bounded in.
Proof By the Hölder inequality and the Young inequality, it follows from (3) and () that
for any , where . Let be a sequence of I. It follows from (4) that
which implies
Set , we see that is bounded in . □
Lemma 2Let, , and. Assume that () and () hold. Ifis asequence ofI, then there exists a positive constantAdepending ona, q, S, andsuch that
Proof Let be a sequence of I. By Lemma 1, we know that is bounded. Therefore, up to a subsequence, there exists such that converges weakly in , strongly in with and a.e. in Ω. By (), (), and the Dominated Convergence Theorem, we have
Thus, by using also the fact that , we get
from which it follows that in . Since I is , we obtain . In particular, we have , which implies that
It follows from (4) that
Lemma 3Let, and. Assume that () and () hold, thenIsatisfies thecondition with = + + + − , whereAis the positive constant given in Lemma 2.
Proof Let be a sequence of I with . By Lemma 1, we know that is bounded. Up to a subsequence, we may assume that
From Lemma 2, we have . By () and the Dominated Convergence Theorem, we obtain
Let ; by the BrezisLieb lemma [27], one has
and
Assume that , it follows from (6) that
From (3), we have
It follows from (5), (6), and Lemma 2 that
which contradicts the fact that . Therefore, we have , which implies that in . Hence I satisfies the condition with . □
3 The proof of the main results
In this section, we show the proofs of our Theorems 1 and 2. Before we come to the proof of Theorem 1, we first recall the following lemma in [28].
Lemma 4Let, and. Then there existssuch that.
Proof of Theorem 1 Using the hypotheses () and (), it follows from (3) and (4) that
Let , we can find and such that for all
From Lemma 4, we obtain the result that there exists such that
Therefore, one has
Fix ; noticing that , it implies from (8) that there exists small enough such that . Thus we deduce that
By applying the Ekeland’s variational principle in [29], we obtain the result that there exists a sequence of I.
By the expression of , we can choose such that for all . It follows from and Lemma 3 that I satisfies the condition. Therefore, one has a subsequence still denoted by and such that in and
which implies that is a solution of problem (1). After a direct calculation, we derive , which implies . Since , we have . Applying the Harnack inequality [30], we see that is a positive solution of problem (1). The proof of Theorem 1 is completed. □
Lemma 5Let, and. Assume that (), (), (), (), and () hold, then there exists, such that for any, we can findsuch that.
Proof For convenience, we consider the functional defined by
for all . According to () and (), we can choose such a cutoff function that for , for , and , where is a positive constant. Define
According to () and (), similar to the calculation of [31], we have the following estimate (as )
for all . From (9), we have . Note that and for , so is attained for some . By
one has
Therefore, we deduce from (9) that
By the expression of , we can choose such that for all . Using the definitions of I and , from () and (), we have
for all and . It follows that there exist and such that
for all and . Moreover, using the definitions of I and , it follows from () and (10) that
By the above two inequalities, for any , we have
Hence, we can choose such that for all
Therefore, for all and , we have
Set . Let , and , we deduce from (11) and (13) that
□
Proof of Theorem 2 Choose , from the proof of Theorem 1, we have already seen that problem (1) for any has a positive solution with . Now we only need to find the second positive solution of problem (1). According to (), we can see that (4) and (7) hold. It follows from () and Lemma 4 that there exists such that
According to (4), we have
which implies that
Hence, there exists a positive number such that and for any . It implies from (7) that the functional I has the mountain pass geometry. Define
From Lemma 5, we have . Applying Lemma 3, we know that I satisfies the condition. By the Mountain Pass Theorem [32], we obtain the result that problem (1) has the second solution with . After a direct calculation, we derive
which implies that . Hence we have . Since , we have . By the Harnack inequality, we obtain the result that is the second positive solution of problem (1). The proof of Theorem 2 is completed. □
Competing interests
The author declares that they have no competing interests.
Acknowledgements
The author would like to thank the referees for valuable comments and suggestions on improving this paper. This paper was supported by Science and Education Youth culture project in Guizhou Province (Contract Number: Guizhou Provincial Institute of Zi (2012) No. 157). This paper also was supported by Science and Technology Foundation of Guizhou Province (No. J[2013]2141; No. LKM[2011]31).
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