Abstract
In this paper, we use variational methods to prove two existence of positive solutions of the following mixed boundary value problem:
One deals with the asymptotic behaviors of
MSC: 35M12, 35D30.
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
(S1)
(S2) For almost every
(S3)
(S4) There exists
The eigenvalue problem of (1) is studied by Liu and Su in [1]
There exists a set of eigenvalues
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
Let Ω be a bounded domain with a Lipschitz boundary; there is a continuous embedding
Moreover, there is a continuous boundary trace embedding
It is well known that to seek a nontrivial weak solution of problem (1) is equivalent
to finding a nonzero critical value of the
where
It is easy to see that the condition (AR) implies that
In order to get our conclusion, we define the minimization problem
then
We denote by c,
Theorem 1Let conditions (S1) to (S3) hold, then:
(i) If
(ii) If
(iii) If
Corollary 2Let conditions (S1) to (S3) with
(i) If
(ii) If
(iii) If
Theorem 3Let conditions (S1) to (S4) with
2 Some lemmas
We need the following lemmas.
Lemma 1If
Proof By the Sobolev embedding function
Lemma 2If conditions (S1) to (S3) hold, then there exists
Proof By condition (S3), there exists
By (4) and (5), we obtain
Hence,
Lemma 3If conditions (S1) to (S3) hold,
Proof If
So,
Lemma 4Let conditions (S1) and (S2) hold. If a sequence
Proof Since
For any fixed
Then (S2) implies that
It implies that
For all
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,
(i)
(ii) There exists
and
Then there exists a sequence
3 Proofs of main results
Proof of Theorem 1 (i) If
That is,
It implies that
(ii) By Lemma 2, there exists
where
(14) implies that
Here, in what follows, we use
If
Then
We claim that
It is contradiction with
As follows, we prove
We order
By (S1) and (S3), there exists
It follows from
Hence,
By (16), we have
Since
We order
This is a contradiction with
(iii) If
If u is a positive solution of (1), for the above
We order
which implies that
When
Then we must have
On the other hand, if for some
Proof of Corollary 2 Note that when
Proof of Theorem 3 When
If
Noticing that
Then we know
By
Obviously, (19) and (20) are contradictory. So
4 Example
In this section, we give two examples on
Example 1 Set:
Then it is easy to verify that
So, for some
This means
Example 2 Consider the following problem:
where
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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