Landesman-Lazer condition revisited: the influence of vanishing and oscillating nonlinearities

In this paper we deal with semilinear problems at resonance. We present a sufficient condition for the existence of a weak solution in terms of the asymptotic properties of nonlinearity. Our condition generalizes the classical Landesman-Lazer condition but it also covers the cases of vanishing and oscillating nonlinearities.


Introduction
Let Ω ⊆ R n be a bounded domain, g : R → R be a bounded continuous function and f ∈ L 2 (Ω). We consider the boundary value problem −∆u − λ k u + g(u) = f in Ω, u = 0 on ∂Ω. (1) Here λ k , k ≥ 1, is the k-th eigenvalue of the eigenvalue problem −∆u − λu = 0 in Ω, By a solution of (1) we understand a function u ∈ H := W 1,2 0 (Ω) satisfying (1) in the weak sense, i.e., holds for any test function v ∈ H.
We use the scalar product (u, v) = Then H =Ĥ ⊕H ⊕H with dimĤ = k − 1, dimH = m, dimH = ∞. Of course, if k = 1 then m = 1 andĤ = ∅. We split an element u ∈ H as u =û +ū +ũ, The purpose of this paper is to introduce rather general sufficient condition of Landesman-Lazer type for the existence of a solution of (1): Here, G(s) = s 0 g(τ ) dτ is the antiderivative of g.
Theorem 1. Assume that either (SC) + or else (SC) − holds. Then the problem (1) has at least one solution.
Remark 1. Note that the sufficient condition which is similar to (SC) + but more restrictive than (SC) + was introduced recently in [1] where the resonance problem with respect to the Fučík spectrum of the Laplacian was studied. In this paper, we benefit from the fact that the resonance occurs at the eigenvalue which allows us to split the underlying function space H into the sum of orthogonal subspaces. In contrast with [1], where such splitting is impossible, we can get rid of the f ⊥ -part of the righthand side f in (SC) ± . This makes our conditions more general and geometrically more transparent.
We also note that verification of (SC) ± does not require the existence of limits g(±∞) at all. As an example we consider g(s) = arctan s + c · cos s with an arbitrary constant c ∈ R. An easy calculation yields that (1) has at least one solution for any f ∈ L 2 (Ω) satisfying for any φ ∈H. On the other hand, the conditions (LL) ± and various generalizations (see, e.g. [12,13]) do not apply in this case if |c| ≥ π 2 . Above mentioned case g(s) = arctan s + c · cos s is covered by the so called potential Landesman-Lazer condition: s . Indeed, l'Hospital's rule implies G − = − π 2 , G + = π 2 and the condition (PLL) + reduces to (4). For the use of (PLL) ± see, e.g. the papers [14][15][16][17][18][19].
The conditions (PLL) ± eliminate the influence of the bounded oscillating term c · cos s which disappears "in an average" as |s| → ∞.
However, the conditions (PLL) ± do not cover the case g(s) = s 1+s 2 + c · cos s, where c ∈ R is an arbitrary constant. Indeed, both conditions are empty, due to the fact G ± = 0. On the other hand, it follows from Theorem 1 that (1) with g given above has a solution for any f ∈ L 2 (Ω) ⊥ . This fact illustrates that our conditions (SC) ± refine also the conditions (PLL) ± and, at the same time, they complement the results from [20] and [21].
has a solution for arbitrary c ∈ R and for any f ∈ L 2 (Ω) satisfying

Preliminaries
In this section we stress some helpfull facts used in the proof of Theorem 1.

Lemma 1.
There exist c 1 > 0, c 2 > 0 such that for any u ∈ H we have and Proof. The inequality (6) follows from the variational characterization of λ k , (7) follows from the Hőlder inequality, the boundedness of g and the fact f ∈ L 2 (Ω).
Lemma 2. There exist c 3 > 0, c 4 > 0 such that for any u ∈ H we have and Proof. The inequality (8) is also a consequence of the variational characterization of λ k , and (9) follows similarly as (7).
Lemma 3. There exist c 5 > 0 such that for any u ∈ H we have Proof. The inequality (10) follows from the Hőlder inequality, the boundedness of g and the fact f ∈ L 2 (Ω).

Proof of Theorem 1
We define the energy functional associated with (1), E : H → R, by Obviously, all critical points of E satisfy (3) and vice versa. We will apply Saddle Point Theorem due to P. Rabinowitz [22]: Assume that (c) E satisfies (P S) condition.
Then the functional E has a critical point in H.
At first we verify the Palais-Smale condition.
Lemma 4. Let us assume (SC) ± . Then E satisfies (P S) condition, i.e., if (E(u n )) ⊂ H is a bounded sequence and ∇E(u n ) → v in H, then there exist a subsequence (u n k ) ⊂ (u n ) and an element u ∈ H such that u n k → u in H.
Proof. In the first step we prove that (u n ) is bounded in L 2 (Ω). Assume the contrary, i.e., u n 2 → ∞. Set v n := un un 2 . Then The second term is equal to − λ k 2 since v n 2 = 1, the last two terms go to zero since (for some c > 0) by the embedding L 2 (Ω) ֒→ L 1 (Ω). Then it follows from (11) that (v n ) is a bounded sequence in H. Passing to a subsequence, if necessary, we may assume that there exists v ∈ H such that v n ⇀ v (weakly) in H and v n → v in L 2 (Ω). For arbitrary w ∈ H, We have (Ω), the boundedness of g and by our assumption u n 2 → ∞. Then it follows from (12) that holds for arbitrary w ∈ H, i.e., v = φ 0 ∈H is an eigenfunction associated with λ k . That is, un un 2 → φ 0 in L 2 (Ω). Now, by the assumption ∇E(u n ) → o and the orthogonal decomposition of H, By Lemma 1 it follows from (13) that with c 1 , c 2 > 0 independent of n. Hence û n is a bounded sequence.
Similarly, we also have By Lemma 2 it follows from (14) that with c 3 , c 4 > 0 independent of n. Hence ũ n is a bounded sequence. Let us split now E(u n ) as follows The boundedness of û n and ũ n implies that A, B and D are bounded terms. On the other hand, (SC) + forces C → +∞ and (SC) − forces C → −∞. In particular, we conclude E(u n ) → ±∞ which contradicts the assumption of the boundedness of (E(u n )). We thus proved that (u n ) is a bounded sequence in L 2 (Ω).
In the second step we select a strongly convergent subsequence (in H) from (u n ). Let us examine again the terms in By the assumption E(u n ) is bounded. The boundedness of the sequence (u n ) in L 2 (Ω) implies that Ω (u n ) 2 dx, Ω G(u n ) dx and Ω f u n dx are bounded independently of n, as well. Therefore, u n 2 = Ω |∇u n | 2 dx must be also bounded. Hence, we may assume, without lost of generality, that u n ⇀ u in H for some u ∈ H, and u n → u in L 2 (Ω). Then we conclude that It follows from Lemma 1 and 3 that On the other hand, we prove that there exists β ∈ R such that inf u∈H + E(u) ≥ β.
Assume the contrary, that is, there exists a sequence (u n ) ⊂ H + such that Then u n 2 → ∞, and for v n := un un 2 (v n ∈ H + ) we have Clearly, by Lemma 3, we have It follows from (17) and (18) that v n is bounded. Passing to a subsequence if necessary, we may assume that there exists v ∈ H + such that v n ⇀ v in H and v n → v in L 2 (Ω). Moreover, by the weak lower semicontinuity of the norm in H. We deduce from (17) - (19) that and hence, from Lemma 2, it follows that v = φ 0 ∈H is an eigenfunction associated with λ k . That is, u n u n 2 → φ 0 in L 2 (Ω). This contradicts (16). By (15) there exists R > 0 such that for D := {u ∈ H − : u ≤ R} the following inequality holds sup u∈ ∂D E(u) < α := β − 1.
Hence, we proved (a) and (b) in Theorem 2.
2. Let us assume that (SC) − holds. In this case we set a critical point of E. Since this is also a solution of (1), the proof of Theorem 1 is finished.