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.