Abstract
In this paper, we study the following problem:
where is a smooth bounded domain, , is the pLaplacian, is the critical Sobolev exponent and is a parameter. By establishing a new deformation lemma, we show that if , then for each , this problem has infinitely many signchanging solutions, which extends the results obtained in (Cao et al. in J. Funct. Anal. 262: 28612902, 2012; Schechter and Zou in Arch. Ration. Mech. Anal. 197: 337356, 2010).
1 Introduction
In this paper, we consider the following problem:
where () is a smooth bounded domain, , is the pLaplacian, is the critical Sobolev exponent and is a parameter.
The first existence result of Problem (1.1) with was obtained by Brezis and Nirenberg in the celebrated paper [1]. In that paper, the authors proved that Problem (1.1) had a positive solution for and or and , where is the first eigenvalue of . After that, many existence results have appeared for (1.1); one can see, for example, [27] and the references therein for case of and [811] and the references therein for case of . In particular, in the case of , the authors in [2] proved that the number of solutions of Problem (1.1) is bounded below by the number of eigenvalues of lying in the open interval , where S is the best Sobolev constant and is the Lebesgue measure of Ω. In [5], the existence of infinitely many signchanging solutions of (1.1) with has been obtained when , and Ω is a ball, while it has been shown in [6] that (1.1) with has infinitely many signchanging radial solutions when , and Ω also is a ball. We remark that the methods used in [5,6] are strongly dependent on the symmetry of the ball Ω. For a general bounded smooth domain Ω, by the method of invariant sets of the descending flow, the authors in [7] have shown that (1.1) with has infinitely many signchanging solutions when and , which extends the main result in [4].
The main purpose of this paper is to try to obtain the existence of infinitely many signchanging solutions of Problem (1.1) for general . In a very recent work [9], the authors have proved that (1.1) has infinitely many solutions for and . However, by using the Picone identity (cf.[12,13]), we see that every nonzero solution of Problem (1.1) is signchanging for , where is the first eigenvalue of (see Lemma 2.1 for more details). Hence, to achieve our purpose, we mainly consider the situation of .
Our main result in this paper is the following.
Theorem 1.1Assume thatand. Then Problem (1.1) has infinitely many signchanging solutions.
Since is the critical Sobolev exponent, in order to overcome the lack of compactness of the embedding of in the Lebesgue space , we follow the ideas of [4,7,9] to consider the following auxiliary problems:
where and is increasing to . It has been shown by [[14], Theorem 1.2] that for every n, Problem () has infinitely many signchanging solutions . Hence, to prove Theorem 1.1, we will show that for every , converges to some signchanging solution of (1.1) as , and that are different. The convergence of can be done with the help of [[9], Theorem 1.2], which we show in Lemma 2.3. To distinguish , we shall establish a new deformation lemma on special sets in ; see Lemma 2.5 for details.
Throughout this paper, we will always respectively denote and by the usual norm in spaces and (). Let C be indiscriminately used to denote various positive constants.
2 Proof of Theorem 1.1
We first consider the case of . Recall that , the first eigenvalue of in , given by , is simple and there exists a positive eigenfunction corresponding to such that for every (cf.[15]). Moreover, by [[16], Proposition 2.1], we know that . On the other hand, we have the following proposition which is the socalled Picone identity.
Proposition 2.1 [[13], Lemma A.6]
Moreover,
and the equality holds if and only iffor some constant.
Lemma 2.1Assume thatis a nonzero solution of (1.1) for. Thenuis signchanging.
Proof By a contradiction, we may assume . By using a standard regularity argument and [[17], Lemmas 3.2 and 3.3], we have for some . Thus, it follows from the strong maximum principle (cf.[18]) that . Now, for every , by applying the above Picone identity (i.e., Proposition 2.1) to and , we see
Noting that u is a solution of (1.1), we have
It follows from the Fatou lemma that
which is impossible since , , and . Therefore, we have proved Lemma 2.1. □
Next, we consider the case of .
It is clear that the corresponding functional of () , given by
is Fréchet differentiable. Let , where is a linearly independent sequence of . It is easy to show that there exists such that for , where (cf. [[14], Lemma 3.9]). We denote
Recall that the genus of a symmetric set A of is defined by
Here, we say that A is symmetric if implies .
By [[14], Theorem 1.2], we know that, for every , has infinitely many critical points, denoted by , in for μ small enough. Moreover,
Lemma 2.2For every, there existssuch thatfor all.
Proof Consider the following auxiliary functional:
where . Since , we may assume for all . Then for all . This means
Note that is the corresponding functional of the following equation:
and the nonlinearity satisfies the assumptions of [[14], Theorem 1.2]. Thus, this equation has a sequence of solutions such that
for μ small enough. For every , the definitions of and , together with (2.2), imply for all . On the other hand, since for every n, is a sequence of solutions for () whose energies satisfy (2.1), it follows that . We complete the proof by choosing . □
By Lemma 2.2 and [[9], Theorem 1.2], we know that for each , there exists such that as in . The next lemma will give more information about .
Lemma 2.3is a signchanging solution of Problem (1.1) for every.
Proof We first prove that is a solution of Problem (1.1) for every . Since as in ,
and
as for every . If we can prove
as for every , then is a solution of (1.1) for is a solution of (). Indeed, a.e. in Ω as since in . By the Egoroff theorem, for every , there exists such that uniformly in and , where is the Lebesgue measure of . This, together with the Lebesgue dominated convergence theorem, implies
On the other hand, for every , we have
For every , by the above inequality and the absolute continuity of the integral, we can take δ small enough such that
for n large enough. By (2.4), for this δ, we have
for n large enough. So (2.3) holds. Moreover, by a similar proof, we can show .
Next, we will show is signchanging for all . Since for each , is a signchanging solution of (), multiplying () by , we obtain , where . Note that , by the Sobolev imbedding theorem, we have . It follows that in as for in as . This gives , i.e., for all . □
Thanks to Lemma 2.3, for some . We claim that for some . Indeed, if not, then as without loss of generality. On the one hand, since is a solution of (1.1), , where . On the other hand, by [[17], Lemma 3.7], we have
for . Note that by a similar proof of [[14], Lemma 3.3], we can see that for μ small enough. Thus, for k large enough. This implies
for k large enough, which contradicts . For the sake of convenience, we denote , , by , , . Note that for every , is compact in (cf. [[9], Theorem 1.2]). It follows from [[19], Proposition 7.5] that there exists such that
Let and . Let . For small enough, we define , then we have the following.
Lemma 2.4Assume that there existssuch thatfornlarge. Then there existssuch thatforand largen.
Proof Assume a contradiction. Then, for every , there exists such that . It is clear that satisfies the (PS) condition for every . Hence there exists such that, up to a subsequence, in as with and . This implies
Thus, by [[9], Theorem 1.2], up to a subsequence, we see that there exists such that in as . Moreover, by using the arguments in the proof of Lemma 2.3, we have and . On the other hand, for large n, since . It follows that . This contradicts the fact that . □
Lemma 2.5Assume that there existssuch thatfor everyand largen. Then there existand an odd continuous mapsuch thatandfor largen.
Proof We first assume . It is clear that there exists such that
where
Let be the local Lipschitz continuous operator obtained in [[14], Lemma 2.1] and let be the solution of the following O.D.E.
Denote to be the maximal interval of existence of .
Claim 1: cannot enter before it enters for small δ, large n and .
Indeed, if the claim fails, then for every , will enter before it enters . Since , there exist such that for and
By [[14], Lemma 2.1], . On the other hand, by the choice of and , we know that for . Thanks to [[17], Lemma 3.8], for large n. This, together with (2.6) and [[14], Lemma 2.1], implies
Claim 2: There exists such that for large n and .
If the claim is not true, then for all . We first consider the case of . In fact, by Claim 1, , i.e., for all . Since for and large n, we must have
On the other hand, by [[14], Lemma 2.1] and [[17], Lemma 5.2], we have
This means for all , which contradicts with (2.7). It follows that there must exist such that for , large n and . Next, we consider the case of . Since for all and large n, it follows from [[14], Lemma 2.1] and [[17], Lemma 5.2] that
Thus, there also exists such that for and . Moreover, we must have for since for all .
Let
Then, by the continuity of , is continuous. Note that is odd and is even, we see that is odd and it is the desired map. The situation of can be proved in a similar way. Therefore, we complete the proof of this lemma. □
Proof of Theorem 1.1 We first consider the case . Thanks to Lemma 2.1 and [[9], Theorem 1.1], (1.1) has infinitely many signchanging solutions. Next, we consider the case of . Since for every , for all , for all . It follows that two cases may occur:
In this case, Problem (1.1) has infinitely many signchanging solutions.
Case 2: There exists such that for all .
In this case, if for every small enough, then Problem (1.1) also has infinitely many signchanging solutions. Otherwise, there exists such that for . Thanks to Lemmas 2.4 and 2.5, there exists such that for small δ and large n. Fix and , the definitions of and give that there exists a large n such that and for small . By the definition of , there exists such that , where , and . It follows that . Thus, . By the choice of δ and , we have . If , then we have
A contradiction. By the properties of gen, we have
This implies . Since is arbitrary, we have , which contradicts with (2.5). □
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors typed, read and approved the final manuscript.
Acknowledgements
The work was supported by the Natural Science Foundation of China (11071180, 11171247) and College Postgraduate Research and Innovation Project of Jiangsu Province (CXZZ110082).
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