Abstract
Keywords:
elliptic mixed boundary value problem; positive solutions; mountain pass theorem; Sobolev embedding theorem1 Introduction and preliminaries
This paper is concerned with the existence of positive solutions of the following elliptic mixed boundary value problem:
where Ω is a bounded domain in with Lipschitz boundary ∂Ω, , , Γ is a sufficiently smooth dimensional manifold, and ν is the outward normal vector on ∂Ω. We assume , are continuous and satisfy
(S2) For almost every , is nondecreasing with respect to .
(S3) , uniformly in a.e. , where , is the first eigenvalue of (2), , .
(S4) There exists such that for some as and as .
The eigenvalue problem of (1) is studied by Liu and Su in [1]
There exists a set of eigenvalues and corresponding eigenfunctions which solve problem (2), where , as , .
There have been many papers concerned with similar problems at resonance under the boundary condition; see [210]. Moreover, some multiplicity theorems are obtained by the topological degree technique and variational methods; interested readers can see [1117]. Problem (1) is different from the classical ones, such as those with Dirichlet, Neuman, Robin, Noflux, or Steklov boundary conditions.
In this paper, we assume is a closed subspace of . We define the norm in V as , is the norm, is the norm, is the trace operator with for all , that is continuous and compact (see [18]). Furthermore, we define , for (see [1]). Then, by (S3), we obtain
Let Ω be a bounded domain with a Lipschitz boundary; there is a continuous embedding for when , and when . Then there exists , such that
Moreover, there is a continuous boundary trace embedding for when , and when . Then there exists , such that
It is well known that to seek a nontrivial weak solution of problem (1) is equivalent to finding a nonzero critical value of the functional
where , , . Moreover, by (S1) and the Strong maximum principle, a nonzero critical point of J is in fact a positive solution of (1). In order to find critical points of the functional (6), one often requires the technique condition, that is, for some , , ,
It is easy to see that the condition (AR) implies that , that is, must be superlinear with respect to u at infinity. In the present paper, motivated by [19] and [20], we study the existence and nonexistence of positive solutions for problem (1) with the asymptotic behavior assumptions (S3) of f at zero and infinity. Moreover, we also study superlinear of f at infinity with in (S3), which is weaker than the (AR) condition, that is the (AR) condition does not hold.
In order to get our conclusion, we define the minimization problem
then , which is achieved by some with a.e. in Ω; see Lemma 1.
We denote by c, , universal constants unless specified otherwise. Our main results are as follows.
Theorem 1Let conditions (S1) to (S3) hold, then:
(i) If, then the problem (1) has no any positive solution inV.
(ii) If, then the problem (1) has at least one positive solution inV.
(iii) If, then the problem (1) has one positive solutionif and only if there exists a constantsuch thatand, a.e. , whereis the function which achieves Λ.
Corollary 2Let conditions (S1) to (S3) withhold, then:
(i) If, then the problem (1) has no any positive solution inV.
(ii) If, then the problem (1) has at least one positive solution inV.
(iii) If, then the problem (1) has one positive solutionif and only if there exists a constantsuch thatand, a.e. , whereis the eigenfunction of the.
Theorem 3Let conditions (S1) to (S4) withhold, then the problem (1) has at least one positive solution inV.
2 Some lemmas
We need the following lemmas.
Lemma 1If, , , thenand there existssuch thatand. Moreover, a.e. inV.
Proof By the Sobolev embedding function and Fatou’s lemma, it is easy to know that and there exists , which satisfies Λ, that is, . Furthermore, we assume , then could replace by . By the Strong maximum principle, we know a.e. in V. □
Lemma 2If conditions (S1) to (S3) hold, then there existssuch that, , .
Proof By condition (S3), there exists , such that , as . Which implies that , .
By (4) and (5), we obtain
Hence, ; we take ε which satisfies , that is, . Then we take a positive constant β such that as , and is small enough. □
Lemma 3If conditions (S1) to (S3) hold, , is defined by Lemma 1, thenas.
Proof If , is defined by Lemma 1, by Fatou’s lemma, and (S3), we have
Lemma 4Let conditions (S1) and (S2) hold. If a sequencesatisfiesas, then there exists a subsequence of, still denoted bysuch thatfor all, .
Proof Since as , for a subsequence, we may assume that
Then (S2) implies that
It implies that , . Following the same procedures, we obtain , .
For all and positive integer n, by (8), we have
On the other hand, by (8), one has
One has
Combining (9) and (10), we have . □
Lemma 5 (see [21])
SupposeEis a real Banach space, satisfies the following geometrical conditions:
(ii) There existssuch that. Letbe the set of all continuous paths joining 0 ande:
and
3 Proofs of main results
Proof of Theorem 1 (i) If is one positive solution of problem (1), by (3), one has
That is,
It implies that . This completes the proof of Theorem 1(i).
(ii) By Lemma 2, there exists such that with . By Lemma 3, we obtain as . Define
where is given by Lemma 1. Then and by Lemma 3, there exists such that
(14) implies that
Here, in what follows, we use to denote any quantity which tends to zero as .
If is bounded in V, when Ω is bounded and , are subcritical, we can get has a subsequence strong convergence to a critical value of J, and our proof is complete. So, to prove the theorem, we only need show that is bounded in V. Supposing that is unbounded, that is, as . We order
Then is bounded in V. By extracting a subsequence, we suppose is a strong convergence in , is a convergence a.e. , is a weak convergence in V.
We claim that . In fact, by (S1) and (S3), we know , , and there exists such that , . If , is a strong convergence in , and by (15) and (16) we know
It is contradiction with , so .
As follows, we prove satisfies
We order
By (S1) and (S3), there exists such that , , . We select a suitable subsequence and there exists , such that is a strong convergence in , is a strong convergence in , and , , .
It follows from is a strong convergence in that
Hence, is bounded in , in ; is bounded in , in .
By (16), we have
Since is a weak convergence in V, we obtain
We order ; this yields , so . By the Strong maximum principle, we know a.e. in Ω, so a.e. in Ω. Combining (S3) and (3), we obtain
This is a contradiction with . This completes the proof of Theorem 1(ii).
(iii) If , by Lemma 1, there exists some , such that
If u is a positive solution of (1), for the above , we have
We order in (17), and it follows from (18) that
When a.e. in Ω, combining (S2), (S3), and (3), we obtain
Then we must have , a.e. in Ω, also achieves Λ (=1). When , , we have , which achieves Λ.
On the other hand, if for some , and , a.e. , since also achieves Λ. This means is a solution of problem (1) as . This completes the proof of Theorem 1(iii). □
Proof of Corollary 2 Note that when , then . The conclusion follows from Theorem 1. □
Proof of Theorem 3 When , we can replace by in (11) and define c as in (12), then following the same procedures as in the proof of Theorem 1(ii), we need to show only that is bounded in V. For this purpose, let be defined as in (16). If is bounded in V, we know is a strong convergence in , is convergence a.e. , is a weak convergence in V, and .
If , then and . We set , . Obviously, by (16), a.e. in . When in (S3), there exists and n large enough we have , uniformly in . Hence, by (15) and (16), we obtain
Noticing that in and , can be chosen large enough, so and then in Ω.
Then we know , and consequently,
By , as , then it follows Lemma 4 and (13), we obtain
Obviously, (19) and (20) are contradictory. So is bounded in V. This completes the proof of Theorem 3. □
4 Example
In this section, we give two examples on : One satisfies (S1) to (S3) with , but does not satisfy the (AR) condition; the other illustrates how the assumptions on the boundary are not trivial and compatible with the inner assumptions in Ω.
Example 1 Set:
Then it is easy to verify that satisfies (S1) to (S3) with as and as . In addition,
So, for some , , for all t large.
This means does not satisfy the (AR) condition.
Example 2 Consider the following problem:
where is a constant. It is obvious that as . Problem (21) is a case of (1); we can obtain the nontrivial solution: , .
Competing interests
The author declares that he has no competing interests.
Author’s contributions
Li G carried out all studies in this article.
Acknowledgements
The author would like to thank the referees for carefully reading this article and making valuable comments and suggestions.
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