Existence of solutions for a fourth order problem at resonance

In this work, we are interested at the existence of nontrivial solutions of two fourth order problems governed by the weighted p-biharmonic operator. The first is the following �(�|�u| p 2 �u) = �1m(x)|u| p 2 u + f(x,u) h in , u = �u = 0 on @, where �1 is the first eigenvalue for the eigenvalue problem �(�|�u| p 2 �u) = �m(x)|u|p 2u in , u = �u = 0 on @. In the seconde problem, we replace �1 bysuch that �1 < � < ¯, whereis given bellow.

1 Introduction and main results 133 2 Preliminaries and proofs of Theorems 135

Introduction and main results
In the present paper, we are concerned with the existence of weak solutions of the following problem where p > 1, Ω is a bounded domain of R N (N ≥ 1) with smooth boundary ∂Ω, ρ ∈ C(Ω), with inf Ω ρ(x) > 0, f : Ω×R −→ R is a bounded Carathéodory function, h ∈ L p ′ (Ω), p ′ = p p−1 , m ∈ C(Ω) is nonnegative weight function and λ 1 design the first eigenvalue for the eigenvalue problem The investigation of existence of solutions for problems at resonance has drawn the attention of many authors, see for example [1,3,6,7,12].In [7], Liu and Squassina study the following p-biharmic problem ∆(∆u| p−2 ∆u) = g(x, u) in Ω u = ∆u = 0 on ∂Ω.Under some conditions on g(x, u) at resonance, the authors established the existence of at least one nontrivial solution.
According to the work of Talbi and Tsouli [10], the eigenvalue problem (1.2) has a nondecreasing and unbounded sequence of eigenvalues, and the first eigenvalue λ 1 is given by where X := W 2,p (Ω) ∩ W 1,p 0 (Ω) is the reflexive Banach space endowed with the norm Since m ∈ C(Ω) and m ≥ 0, λ 1 is positive, simple and isolated.Therefore Moreover, there exists a unique positive eigenfunction ϕ 1 associated to λ 1 , which can be chosen normalized.Let The fact that λ 1 is isolated implies that λ 1 < λ 2 .It can also be shown (see Lemma 2.1) that there exists λ ∈ (λ 1 , λ 2 ] such that for all u ∈ X with In addition, we study the existence of solutions for the following boundary value problem We assume that the function f satisfy the following hypothese: (H) For almost every x ∈ Ω, there exist Let us recall the minimum principle and the saddle point theorem (see [9]).
Then there exists u 0 ∈ X such that Φ(u 0 ) = c.
Problem at resonance 135 Theorem 1.2.Let X be a Banach space.Let Φ : X → R be a C 1 functional that satisfies the Palais-Smale condition, and suppose that X = V ⊕ W, with V a finite dimensional subspace of X.If there exists R > 0 such that then Φ has a least a critical point on X.
Now, we are ready to state our main results.
Then problem (1.1) has at least a weak solution .
, has at least one solution.

Preliminaries and proofs of Theorems
We consider the following energy functional Φ : X → R defined by where It is well known that Φ ∈ C 1 (X, R), with derivative at point u ∈ X is given by for every v ∈ X.
Let denote V = ϕ 1 the linear spans of ϕ 1 and Then we can decompose X as a direct sum of V and W. In fact, let u ∈ X, writing where w ∈ X, and We begin by establishing the existence of λ for which (1.4) holds.
This value is attained in W. To see why this is so, let is bounded in X and therefore, up to subsequence, we may assume that u n ⇀ u weakly in X and u n → u strongly in L p (Ω).
From the strong convergence of the sequence in L p (Ω) we obtain and so that u ∈ W. By the weakly lower semicontinuity of the norm ||.||, we get and hence λ is attained at u. Proof: Let (u n ) be a sequence in X, and c a real number such that: We claim that (u n ) is bounded in X.Indeed, suppose by contradiction that (2.5) By the hypotheses on f, h and (u n ), we obtain then, from (2.5) we deduce that Then v ≡ 0. According to the definition of λ 1 and the weak lower semi continuity of norm, one has This implies that By the definition of ϕ 1 , we deduce that v = ±ϕ 1 .
On the other hand, from (2.3) we have In view of (2.4), for all ε > 0 and n large enough, we have a.e x ∈ Ω, Moreover, the Lebesgue's theorem imply Combining (2.6) and (2.7), we get Dividing by ||u n || the last inequalities, we obtain and passing to the limits, we deduce from (2.9) that which contradicts both (1.7) and (1.8).Thus (u n ) is bounded in X, for a subsequence denoted also (u n ), there exists u ∈ X such that u n ⇀ u weakly in X, and strongly in L p (Ω).From Problem at resonance 139 Using the hypotheses on m, h and f , we see that Consequently, In the same way, we obtain By the uniform convexity of X, it follows that u n → u strongly in X and Φ satisfies the (P S) condition.✷ Lemma 2.3.Assume that (1.6) and (1.7) are satisfied.Then the functional Φ is coercive on X.
Proof: Suppose by contadiction that Φ is not coercive, then there exists a sequence In the proof of lemma 2.2, we have showed that Assume v n → −ϕ 1 (for example).Dividing (2.10) by ||u n ||, we get Passing to the limits, we have which contradicts (1.7).✷ Proof: [Proof of Theorem 1.3].If (1.7) holds, the coerciveness of the functional Φ and the Palais-Smale condition entrain, from theorem 1.1, that Φ attains its minimum, so problem (1.1) admits at least a weak solution in X.
If (1.8) holds, then Φ has the geometry of the saddle point theorem 1.2.Indeed, splitting X = V ⊕ W. Let u ∈ W, using Höder inequality and the properties of F, where C is the embedding constants of Sobolev, .p ′ and .∞ denote the norms in L p ′ (Ω) and L ∞ (Ω) respectively.Then Φ is bounded from below on W, is a consequence of the assumption that p > 1, so that On the other hand, for every t ∈ R, one has where g has been defined by (2.8).From the Lebesgue theorem, it follows that and the limit is negative by (1.8).Analogously, if t tends to −∞, we have the same result with k(x) exchanged with l(x), so that the limit is positive by (1.8 this fact implies (as in proof of lemma 2.2) that v n → v strongly in X.Using again the saddle point theorem, the desired result follows.✷
) is unbounded, and define v n = u n /||u n ||, so that, up to subsequence, (v n ) converges weakly to a function v in X. Dividing (2.4) by ||u n || p−1 , and then taking Φ