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 so-called Picone identity.
Proposition 2.1 [[13], Lemma A.6]
Let be such that , and . Then
Moreover,
and the equality holds if and only if for some constant .
Lemma 2.1 Assume that is a nonzero solution of (1.1) for . Then u is sign-changing.
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,
(2.1)
where .
Lemma 2.2 For every , there exists such that for all .
Proof Consider the following auxiliary functional:
where . Since , we may assume for all . Then for all . This means
(2.2)
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 2.3 is a sign-changing 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
(2.3)
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
(2.4)
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 this δ, since in ,
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 sign-changing for all . Since for each , is a sign-changing 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 . □
Let and , we denote
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 and
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
(2.5)
Let and . Let . For small enough, we define , then we have the following.
Lemma 2.4 Assume that there exists such that for n large. Then there exists such that for and large n.
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 2.5 Assume that there exists such that for every and large n. Then there exist and an odd continuous map such that and for large n.
Proof We first assume . It is clear that there exists such that
(2.6)
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
A contradiction with .
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
(2.7)
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 sign-changing solutions. Next, we consider the case of . Since for every , for all , for all . It follows that two cases may occur:
Case 1: There are such that .
In this case, Problem (1.1) has infinitely many sign-changing 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 sign-changing 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). □