We prove the existence of a solution to the periodic nonlinear secondorder ordinary differential equation with damping , , . We suppose that , the nonlinearity satisfies the potential Landesman Lazer condition and prove that a critical point of a corresponding energy functional is a solution to this problem.
1. Introduction
Let us consider the nonlinear problem
where , the nonlinearity is a Caratheodory function and .
To state an existence result to (1.1) Amster [1] assumes that is a nondecreasing function (see also [2]). He supposes that the nonlinearity satisfies the growth condition , for ,, where is the first eigenvalue of the problem, and there exist such that An interval is centered in with the radius where , and is a solution to the problem with .
In [3, 4] authors studied (1.1) with a constant friction term and results with repulsive singularities were obtained in [5, 6].
In this paper we present new assumptions, we suppose that the friction term has zero mean value
the nonlinearity is bounded by a function and satisfies the following potential LandesmanLazer condition (see also [7, 8])
where, , and
To obtain our result we use variational approach even if the linearization of the periodic problem (1.1) is a nonselfadjoint operator.
2. Preliminaries
Notation 2.
We will use the classical space of functions whose th derivative is continuous and the space of measurable realvalued functions whose th power of the absolute value is Lebesgue integrable. We denote the Sobolev space of absolutely continuous functions such that and with the norm . By a solution to (1.1) we mean a function such that is absolutely continuous, satisfies the boundary conditions and (1.1) is satisfied a.e. in .
We denote and we study (1.1) by using variational methods. We investigate the functional , which is defined by
where
We say that is a critical point of , if
We see that every critical point of the functional satisfies
for all .
Now we prove that any critical point of the functional is a solution to (1.1) mentioned above.
Lemma 2.1.
Let the condition (1.2) be satisfied. Then any critical point of the functional is a solution to (1.1).
Proof.
Setting in (2.4) we obtain
We denote
then previous equality (2.5) implies and by parts in (2.4) we have
for all Hence there exists a constant such that
on . The condition (1.2) implies and from (2.8) we get Using and differentiating equality (2.8) with respect to we obtain
Thus is a solution to (1.1).
We say that satisfies the PalaisSmale condition (PS) if every sequence for which is bounded in and (as possesses a convergent subsequence.
To prove the existence of a critical point of the functional we use the Saddle Point Theorem which is proved in Rabinowitz [9] (see also [10]).
Theorem 2.2 (Saddle Point Theorem).
Let , and. Let be a functional such that and
(a)there exists a bounded neighborhood of in and a constant such that ,
(b)there is a constant such that ,
(c) satisfies the PalaisSmale condition (PS).
Then, the functional has a critical point in .
3. Main Result
We define
Assume that the following potential LandesmanLazer type condition holds:
We also suppose that there exists a function such that
Theorem 3.1.
Under the assumptions (1.2), (3.2), (3.3), problem (1.1) has at least one solution.
Proof.
We verify that the functional satisfies assumptions of the Saddle Point Theorem 2.2 on , then has a critical point and due to Lemma 2.1 is the solution to (1.1).
It is easy to see that . Let then and .
In order to check assumption (a), we prove
by contradiction. Then, assume on the contrary there is a sequence of numbers such that and a constant satisfying
From the definition of and from (3.5) it follows
We note that from (3.2) it follows there exist constants , and functions such that , for a.e. and for all , , respectively. We suppose that for this moment . Using (3.6) and Fatou's lemma we obtain
a contradiction to (3.2). We proceed for the case Then assumption (a) of Theorem 2.2 is verified.
(b) Now we prove that is bounded from below on . For , we have
and assumption (3.3) implies
Hence and due to compact imbedding we obtain
Since the function is strictly positive equality (3.10) implies that the functional is bounded from below.
Using (3.4), (3.10) we see that there exists a bounded neighborhood of in , a constant such that , and there is a constant such that .
In order to check assumption (c), we show that satisfies the PalaisSmale condition. First, we suppose that the sequence is unbounded and there exists a constant such that
Let be an arbitrary sequence bounded in . It follows from (3.12) and the Schwarz inequality that
From (3.3) we obtain
Put and then (3.13), (3.14) imply
Due to compact imbedding and (3.15) we have in , . Suppose that and set in (3.13), we get
Because the nonlinearity is bounded (assumption (3.3)) and the second integral in previous equality (3.16) converges to zero. Therefore
Now we divide (3.11) by . We get
Equalities (3.17), (3.18) imply
Because , . Using Fatou's lemma and (3.19) we conclude
a contradiction to (3.2). We proceed for the case similarly. This implies that the sequence is bounded. Then there exists such that in , in , (taking a subsequence if it is necessary). It follows from equality (3.13) that
The strong convergence in and the assumption (3.3) imply
If we set , in (3.21) and subtract these equalities, then using (3.22) we have
Hence we obtain the strong convergence in . This shows that satisfies the PalaisSmale condition and the proof of Theorem 3.1 is complete.
Acknowledgment
This work was supported by Research Plan MSM 4977751301.
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