Periodic Problem with a Potential Landesman Lazer Condition

Let us consider the nonlinear problem

(1.1)

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

(1.2)

the nonlinearity is bounded by a function and satisfies the following potential Landesman-Lazer condition (see also [7, 8])

(1.3)

where , , and

To obtain our result we use variational approach even if the linearization of the periodic problem (1.1) is a non-self-adjoint operator.

Notation 2.

We will use the classical space of functions whose th derivative is continuous and the space of measurable real-valued 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

(2.1)

where

(2.2)

We say that is a critical point of , if

(2.3)

We see that every critical point of the functional satisfies

(2.4)

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

(2.5)

We denote

(2.6)

then previous equality (2.5) implies and by parts in (2.4) we have

(2.7)

for all Hence there exists a constant such that

(2.8)

on . The condition (1.2) implies and from (2.8) we get Using and differentiating equality (2.8) with respect to we obtain

(2.9)

Thus is a solution to (1.1).

We say that satisfies the Palais-Smale 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 Palais-Smale condition (PS).

Then, the functional has a critical point in .

We define

(3.1)

Assume that the following potential Landesman-Lazer type condition holds:

(3.2)

We also suppose that there exists a function such that

(3.3)

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

(3.4)

by contradiction. Then, assume on the contrary there is a sequence of numbers such that and a constant satisfying

(3.5)

From the definition of and from (3.5) it follows

(3.6)

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

(3.7)

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

(3.8)

and assumption (3.3) implies

(3.9)

Hence and due to compact imbedding we obtain

(3.10)

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 Palais-Smale condition. First, we suppose that the sequence is unbounded and there exists a constant such that

(3.11)

(3.12)

Let be an arbitrary sequence bounded in . It follows from (3.12) and the Schwarz inequality that

(3.13)

From (3.3) we obtain

(3.14)

Put and then (3.13), (3.14) imply

(3.15)

Due to compact imbedding and (3.15) we have in , . Suppose that and set in (3.13), we get

(3.16)

Because the nonlinearity is bounded (assumption (3.3)) and the second integral in previous equality (3.16) converges to zero. Therefore

(3.17)

Now we divide (3.11) by . We get

(3.18)

Equalities (3.17), (3.18) imply

(3.19)

Because , . Using Fatou's lemma and (3.19) we conclude

(3.20)

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

(3.21)

The strong convergence in and the assumption (3.3) imply

(3.22)

If we set , in (3.21) and subtract these equalities, then using (3.22) we have

(3.23)

Hence we obtain the strong convergence in . This shows that satisfies the Palais-Smale condition and the proof of Theorem 3.1 is complete.