Existence of solution for a class of biharmonic equations

In this paper, We prove the solvability of the biharmonic problem $$\begin{cases}\Delta^{2}u=f(x,u)+h ~~~~in~~\Omega, &\hbox{}\\ u=\Delta u=0 ~~~~~~on ~~\partial\Omega,\\\end{cases}$$ for a given function $h\in L^2(\Omega)$, if the limits at infinity of the quotients $f(x,s)/s$ and $2F(x,s)/s$ for a.e.$x\in\Omega$ lie between two consecutive eigenvalues of the biharmonic operator $\Delta^2$, where $F(x,s)$ denotes the primitive $F(x,s)=\int_{0}^{s}{f(x,t)dt}$.


Introduction
In this paper, we study a class of biharmonic problem of the form where Ω ∈ R N (N > 4) is a bounded smooth domain, ∆ 2 denotes the biharmonic operator defined by ∆ 2 u = ∆(∆u).Let further f : Ω × R → R be a carathéodory function such that (R) m r (x) := max and h ∈ L 2 (Ω).We will also assume the conditions : uniformly for a.e.x ∈ Ω, with strict inequalities λ i < L(x), K(x) < λ i+1 holding on subsets of positive measure where F (x, s) = s 0 f (x, t)dt and λ i < λ i+1 are two consecutive eigenvalues of the problem ∆ 2 u = λu in Ω, u = ∆u = 0 on ∂Ω.The conditions imposed on f are usually classified as non-resonant or resonant, according as they yield the solvability of problem (1.1) for every h or not.Many papers have been devoted to the obtention of resonant conditions of the second order problem where h ∈ L p (Ω), for some suitable p ≥ 2 is given.See for instance [6], [1] and the references given there.
To the best of our knowledge, the solvability of boundary value problem (1.1) has not been studied till now.The main purpose of this paper is to extend some of the results known in [1], concerning the Dirichlet problem (1.3) to the biharmonic problem (1.1) with Navier boundary condition.
Our main result is the following : The proof is based on variational method, we will use the well-known Rabinowitz saddle point theorem [3].The plan of this paper is the following : in section 2, we prove some preliminary lemmas.In section 3 we give the proof of our main result.

Preliminary lemmas
From the conditions (R) and (f ), it follows that there exist constants a, A > 0 and functions b ∈ L 2 (Ω), B ∈ L 1 (Ω) such that : hence the functional is well-defined and of class C 1 on the space , where .L 2 denote the usual norm in L 2 (Ω).
The derivative I ′ (u) ∈ H * is given by Existence of solution for a class of biharmonic equations 101 for all u, w ∈ H. Thus the critical points of I are precisely the weak solutions u ∈ H of (1.1).Let (u n ) ⊂ H be an unbounded sequence.Then, defining v n by we have v n = 1 and , passing if necessary to a subsequence (still denoted by (v n )), we may assume and , where z ∈ L 2 (Ω).Now, assuming (f ), we obtain that the sequence Lemma 2.1.The function f above satisfies where v and l, k are given in (2.1) and (f ), respectively.
for a.e.x ∈ Ω, where v and K, L are given in (2.1) and (F), respectively.
The next result is a consequence of the equivalence between the unique continuation and the strict monotonicity of the biharmonic operator.Lemma 2.3.(see [4] ) 3. Proof of the theorem 1.1 First we need to study the functional I : H → R defined in the introduction.Throughout this section we will assume that conditions (f ) and (F) hold.
for all v ∈ H, where C is a constant and ε n → 0 as n → +∞.In order to show that (u n ) has a convergent subsequence, it suffices to show that (u n ) remains bounded in H. Suppose by contradiction that u n → +∞ as n → +∞.Then , as we observed in the previous section, (a subsequence of) v n = un un is such that and |v n (x)| ≤ z(x) a.e , where z ∈ L 2 (Ω).Moreover, we have where f satisfies Let us define Then f = m(x)v(x) and, by (3.4) and (3.5) we have so that λ i ≤ m(x) ≤ λ i+1 in view of (f ).Now, we use (3.2) with v = u n and we divide by u n 2 to obtain at the limit so that v ≡ 0, necessarily.On the other hand, for any w ∈ H, we have that from which using (3.3) and the fact that v n ⇀ v weakly in H, we obtain Using Lemma 2.1, we see that v ∈ H is a weak solution of the problem Now, we will distinguish three cases : (i) m(x) ≡ λ i (ii) m(x) ≡ λ i+1 and (iii) λ i ≤ m(x) ≤ λ i+1 with λ i < m(x) and m(x) < λ i+1 on subsets of positive measure.We will see that each case leads to a contradiction.
Case (iii) : Since v ≡ 0, this case can not occur in view of Lemma 2.3.
Since neither one of cases (i), (ii), (iii) can occur, this shows that any (PS) sequence must be bounded, so that the functional I satisfies the Palais-Smale condition.✷ Now, let us consider the decomposition of the space H as H = V ⊕ W where V is the subspace spanned by the eigenfunctions corresponding to λ 1 , ..., λ i and W = V ⊥ .We define the two functionals A and B as follow : We recall the two useful inequalities [5]: and the characterization of the first eigenvalue λ 1 of ∆ 2 on H defined by They will be used in the proof of the the next proposition following the same ideas as in [1].
Proposition 3.2.There exists δ > 0 such that : Proof: (a) First, since w 2 ≥ λ i+1 w 2 L 2 , for all w ∈ W , we have By contradiction if (a) does not hold, then there exists a sequence w n ∈ W such that w n = 1, B(w n ) → 0, and for further subsequence w n ⇀ w ∈ W weakly and w n → w strongly in L 2 (Ω), so that by the weak lower semicontinuity of the convex functional B on W. Therefore, we get B(w) = 0. We claim that w ≡ 0. Indeed , since B(w) = 0 , by (3.13) we get w = 0 on the set On the other hand shows that w is an eigenfunction associated to λ i+1 .Therefore, since w = 0 on the set Ω K of positive measure, using the unique continuation principle we get w ≡ 0. But, then we have for a.e.x ∈ Ω and all s ∈ R. Therefore, we obtain for all w ∈ W , Similarly to (a), we prove that A(v) = 0 implies v ≡ 0, by showing first that v = 0 on the set Ω L = {x ∈ Ω : λ i < L(x)} of positive measure and that v is an eigenfunction associated to λ i .Then by contradiction, we obtain v n ∈ V such that v n = 1 and A(v n ) → 0, where we may assume that v n → v ∈ V in H since V is of finite dimension.Therefore we obtain A(v) = 0 so that v = 0 and consequently Ω L(x)v 2 n → 0 , but then , it follows that for all a.e.x ∈ Ω and all s ∈ R, which implies for all v ∈ V .Since δ − ε λ1 > 0 it follows that I(v) → −∞ as v → +∞, v ∈ V .✷