This section is devoted to the proof of the three theorems enunciated in the introduction of this work.
First consider the case for which we prove that the functional ϕ satisfies the Palais-Smale condition.
Lemma 11 Suppose that and conditions (H1) and (H2) hold. Then the functional satisfies condition (PS).
Proof Let and be such that
Then we have
for all sufficiently large k, . Taking in (15), we have for
Adding the last inequality with (14), by assumption (H2), we obtain
which implies that is a bounded sequence in X.
Then, by the compact inclusion , it follows that, up to a subsequence, weakly in X and strongly in . As a consequence, from the inequality
it follows that
Then by (16) it follows that
i.e., strongly in X, which completes the proof. □
Now, we are in a position to prove the main results of this paper.
Proof of Theorem 2 We find by (H1) and (8) that the following inequalities are valid for every :
It is evident that this last expression is strictly positive when , with ρ small enough. Next, let and , with . Then, by (H2) and (3), we have
Since , we conclude that for sufficiently large λ. According to the mountain-pass Theorem 9, together with Lemmas 11 and 7, we deduce that there exists a nonzero classical solution of the problem (P). □
Now consider the case . In the next result we prove that the Palais-Smale condition is also valid.
Lemma 12 Suppose that and conditions (H1) and (H3) hold. Then the functional is bounded from below and satisfies condition (PS).
Proof By , conditions (H1), (H3), and inequality (8), it follows that the functional ϕ is bounded from below:
Further, if is a (PS) sequence, by (17) it follows that is a bounded sequence in X. Then, as in Lemma 11, we conclude that has a convergent subsequence. □
Now we are in a position to prove the next existence result for the problem (P).
Proof of Theorem 3 By assumption, we know that and are odd functions. So are are even functions and the functional ϕ is even. By Lemma 12 we know that ϕ is bounded from below and satisfies condition (PS). Let , be a natural number and define, for any fixed, the set
is homeomorphic to by the odd mapping defined as
Moreover, for , the following inequalities hold:
Clearly is a subset of the m-dimensional subspace
and there exist positive constants and , such that
where is the induced norm of on .
Arguing as in [, pp.16-18], one can prove that there exists , such that
By (H3) we see that for every , , the following inequalities are fulfilled:
Denote . Then by (18)-(21) we have
By the last inequality, it follows that if . Then, by (18), choosing
we obtain for any .
By Clarke’s Theorem 10, there exist at least m pairs of different critical points of the functional ϕ. Since m is arbitrary, there exist infinitely many solutions of the problem (P), which concludes the proof. □
Concerning the problem (P1), one can introduce similarly the notions of classical and weak solutions. In this case it is not difficult to verify that the weak solutions are critical points of the functional defined as
Proof of Theorem 4 By the Poincaré inequalities (7) we find that is an equivalent norm to in X and the functional is convex.
Since the functional
is sequentially weakly continuous, from the fact that the inclusion is compact, we deduce that the functional is weakly lower semi-continuous.
Next, let us see that is bounded from below:
Then, by Theorem 8, there exists a minimizer of , which is a critical point of .
Let u be a weak solution of (P1), i.e., a critical point of . Then
If then . Suppose that u is a nonzero solution and . By (H2′), (7), and (24) it follows that
which is a contradiction. Then, for , the problem (P1) has only the zero solution.
Suppose now that .
Take , . Then
where . For it follows that . Then, since , by (25) it follows that for sufficiently small . In consequence we show that . So we ensure the existence of a nonzero minimizer of , which completes the proof of Theorem 4. □