We define

Assume that the following potential Landesman-Lazer 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 Palais-Smale 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 Palais-Smale condition and the proof of Theorem 3.1 is complete.