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

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.