Abstract
In this paper, we study the existence of antiperiodic solutions for a secondorder ordinary differential equation. Using the interaction of the nonlinearity with the Fučík spectrum related to the antiperiodic boundary conditions, we apply the LeraySchauder degree theory and the Borsuk theorem to establish new results on the existence of antiperiodic solutions of secondorder ordinary differential equations. Our nonlinearity may cross multiple consecutive branches of the Fučík spectrum curves, and recent results in the literature are complemented and generalized.
Keywords:
antiperiodic solutions; Fučík spectrum; LeraySchauder degree theory; Borsuk theorem1 Introduction and main results
In this paper, we study the existence of antiperiodic solutions for the following secondorder ordinary differential equation:
where
In what follows, we will consider problem (1.2) directly.
The problem of the existence of solutions of (1.1) under various boundary conditions
has been widely investigated in the literature and many results have been obtained
(see [113]). Usually, the asymptotic interaction of the ratio
has nontrivial solutions, where
we can see that the conditions on the ratio
Note that the study of antiperiodic solutions for nonlinear differential equations
is closely related to the study of periodic solutions. In fact, since
Denote by Σ the Fuc̆ík spectrum of the operator
It is easily seen that the set Σ can be seen as a subset of the Fuc̆ík spectrum of
In this paper, together with the LeraySchauder degree theory and the Borsuk theorem,
we obtain new existence results of antiperiodic solutions of (1.1) when the nonlinearity
Our main result is as follows.
Theorem 1.1Assume that
(i) There exist positive constantsρ,
(ii) There exist connect subset
and
hold uniformly for all
then (1.1) admits a
In particular, if
Simple computation implies that the operator
Corollary 1.2Assume that
holds uniformly for all
Remark It is well known that (1.1) has a
for some
Denote
where
In addition,
for all
for all
This paper is organized as follows. In Section 2, some necessary preliminaries are presented. In Section 3, we give the proof of Theorem 1.1.
2 Preliminaries
Assume that
For
Define an operator
Clearly,
which implies that
Furthermore, we obtain
Note that
using the Parseval equality
which implies that the operator J is continuous. In view of the ArzelaAscoli theorem, it is easy to see that J is completely continuous.
Denote by deg the LeraySchauder degree. We need the following results.
Lemma 2.1 ([[36], p.58])
Let Ω be a bounded open region in a real Banach spaceX. Assume that
Lemma 2.2 ([[36], Borsuk theorem, p.58])
Assume thatXis a real Banach space. Let Ω be a symmetric bounded open region with
3 Proof of Theorem 1.1
Proof of Theorem 1.1 Consider the following homotopy problem:
where
We first prove that the set of all possible solutions of problem (3.1)(3.2) is bounded.
Assume by contradiction that there exist a sequence of number
Set
By (1.4), (3.3) and the fact that f is continuous, there exist
In view of
It is easily seen that
Clearly,
which implies that there exists
Owing to that the sequences
Multiplying both sides of (3.4) by
Taking a superior limit as
By the assumption (ii) and the choice of λ, if
Similarly, we obtain
Note that
It is easy to see that
We claim that
Clearly,
By (3.3), (3.11), we obtain
holds uniformly for
which implies that
Then, together with (3.6), (3.8) and (3.12), we obtain
a contradiction.
Now, we show that (3.9)(3.10) has only a trivial antiperiodic solution. In fact,
if not, we assume that (3.9)(3.10) has a nontrivial antiperiodic solution
with
Take
In fact, by (3.15)(3.16) and the fact that
If there is
Similarly, if
and
then we obtain
where
Since
Then there exist
It is easy to get
Since
which implies that there exists a real number pair
On the other hand, in view of the assumption (ii), by the definition of Σ and
which is contrary to (3.22).
If
In a word, we can see that there exists
Set
Clearly, Ω is a bounded open set in
which implies that
Define
Clearly,
In view of (3.23), it follows that
Hence,
Note that the operator
Now, using Lemma 2.1, we can see that (1.2) has a solution and hence (1.1) has a
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors read and approved the final manuscript.
Acknowledgements
The authors sincerely thank Prof. Yong Li for his instructions and many invaluable suggestions. This work was supported financially by NSFC Grant (11101178), NSFJP Grant (201215184), and the 985 Program of Jilin University.
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