ON A PROBLEM OF LOWER LIMIT IN THE STUDY OF NONRESONANCE

We prove the solvability of the Dirichlet problem { −∆pu = f(u) + h in Ω, u = 0 on ∂Ω for every given h, under a condition involving only the asymptotic behaviour of the potential F of f with respect to the first eigenvalue of the p-Laplacian ∆p. More general operators are also considered.


Introduction
This paper is concerned with the existence of solutions for the problem where Ω is a bounded domain of IR N , N ≥ 1, ∆ p denotes the p-Laplacian ∆ p u = div(|∇u| p−2 ∇u), 1 < p < ∞, f is a continuous function from IR to IR and h is a given function on Ω.
A classical result, essentially due to Hammerstein [H], asserts that if f satisfies a suitable polynomial growth restriction connected with the Sobolev imbeddings and if then problem (P 2 ) is solvable for any h.Here F denotes the primitive F (s) = s 0 f (t) dt and λ 1 is the first eigenvalue of -∆ on H 1 0 (Ω).Several improvements of this result have been considered in the recent years.
In 1989, the case N=1 and p=2 was considered in [Fe,O,Z].It was shown there that (P 2 ) is solvable for any h ∈ L ∞ (Ω) if If N ≥ 1 and p=2, [F,G,Z] showed later that (P 2 ) is solvable for any h ∈ L ∞ (Ω) if ) 2 , where R(Ω) denotes the radius of the smallest open ball B(Ω) containing Ω.This result was extended to the p-laplacian case in [E,G.1],where solvability of (P p ) was derived under the condition Note that (F 4 ) reducer to (F 3 ) when p = 2.
Observe that for N > 1 and p = 2, ( π 2R(Ω) ) 2 < λ 1 , and a similar strict inequality holds when 1 < p < ∞.One of our purposes in this paper is to show that the constants in (F 3 ) and (F 4 ) can be inproved a little bit.
Denote by l(Ω) = l the length of the smallest edge of an arbitrary parallelepiped containing Ω.In the first part of the paper we assume where Observe that for N = 1, C p = λ 1 is the first eigenvalue of -∆ p on Ω =]0, l[.In particular: C 2 = ( π l ) 2 , and we recover the result of [ Fe,O,Z].It is clear that (F 5 ) is a weaker hypothesis than (F 4 ).The difference between (F 5 ) and (F 4 ) is particularly important when Ω is a rectangle or a triangle.However C p (l) < λ 1 when N > 1, and the question raised above remains open.
In the second part of the paper we investigate the question of replacing ∆ p by the second order elliptic operator where |∇u| 2 a = 1≤i,j≤N a ij (x) ∂u ∂x i ∂u ∂x j .Observe that the method used in [F,G,Z], and [E,G.1]exploits the symmetry of the Laplacian or p-laplacian.It is not clear whether it can be adapted to more general second order elliptic operators like A p above.
While this paper was being completed, we learned of a work by P.Omari and Grossinho (Cf. [GR,O.1],[GR,O.2]),where a result of the same type as ours is established in the case of the linear operator A 2 (u).The authors in [GR,O.2]also consider parabolic operators.

The case of the p-laplacian
In this section we will consider the problem (P p ) where Ω is a bounded domain of IR N , N ≥ 1, 1 < p < ∞, f is a continuous function from IR to IR and h ∈ L ∞ (Ω).
Denote by [AB] the smallest edge of an arbitrary parallelepiped containing Ω. Making an orthogonal change of variables, we can always suppose that AB is parallel to one of the axis of IR N .So Ω ⊂ P = N j=1 [a j , b j ] with, for some i, |AB| = b i − a i = min 1≤j≤N {b j − a j }, a quantity which we denote by l(Ω) = l. where x in Ω A lower solution α is defined by reversing the inequalities above.

Lemma 1. Assume that (P p ) admits an upper solution β and a lower solution
Proof.This lemma is well known when p = 2 (see, e.g., [F.G.Z]).We sketch a proof in the general case By a simple fixed point argument and the results of Di Benedetto [B], there is a solution We claim that α(x) ≤ u(x) ≤ β(x) in Ω, which clearly implies the conclusion.
Lemma 2. Let a < b and M > 0, and assume Then there exists Then there exists Accepting for a moment the conclusion of these two lemmas, let us turn to the Proof of Theorem 1.By Lemma 1 it suffices to show the existence of an upper solution and a lower solution for (P p ).Let us describe the construction of the upper solution (that of the lower solution is similar).
By Lemma 2 there exists β 1 : , and we have: The proof of Theorem 1 is thus complete.
Proof of Lemma 2. The proof of Lemma 3 follows simiarly.

Suppose now inf
Thus f (s) + K ≥ 1 for all s ≥ 0. Define g : IR → IR by: For that purpose we will need the following three lemmas.
In particular (F + ) implies lim sup Proof.Let be a positive number such that lim inf s→+∞ pG(s) s p < ρ < C p .Then lim sup s→+∞ (K(s)) = +∞ where K(s) = ρ|s| p − pG(s).Let w n be the smallest number in [0, n] such that max 0≤s≤n K(s) = K(w n ); it is easily seen that w n is increasing with respect to n.Since and therefore lim sup for all ρ such that lim inf To prove that this condition holds, suppose by contradiction that ∀n = 1, 2, ... ∃u n ∈ ∂B(0, n), ∃λ n ∈ [0, 1] such that: u n = λ n T d (u n ).The latter relation means ( 1) Therefore u n ∈ C 1 (I) and we have successively (3) . By (4) we have Let us distinguish two cases.First if ∃d > 0 such that u d (b) ≥ 0, then the conclusion of Lemma 6 clearly follows.So we can assume that ∀d > 0 : and so it is sufficient to show that lim sup d→+∞ δ d > a+b 2 .Processing as in the proof of Lemma 5 we obtain (p − 1) Integrating from a to δ d and changing variable s = u d (t), one gets, (p − 1) Now one easily deduces from Lemma 4 that lim sup We will show that this function β fulfills the conditions of Lemma 2. To see this it is sufficient to show that: , so that the conclusion follows from the sign of u on [a, a+b  2 ].Proof of (b).
Changing variable u = s + b−a 2 , this implies 3. The case of a more general operator.
Let Ω be a bounded domain in IR N and let A p be an elliptic operator of the form where (a ij (x)) 1≤i,j≤N are real-valued L ∞ (Ω) functions verifying a ij (x) = a ji (x) for all i, j and ( * ) 1≤i,j≤N x ∈ Ω and f or all ξ ∈ IR N .
We now consider the problem Note that A p is defined from W 1,p 0 (Ω) to W −1,p (Ω).Note also that ( * ) implies that for each i, a i,i (x) > 0 a.e. in Ω.We suppose that: We observe that (A 0 ) holds in particular when a i,i i = 1, ..., N , are fixed constants.
Denote by b = b i and a = a i where [a i , b i ] is an edge of an arbitrary parallelepiped containing Ω such that [a i , b i ] is parallel to the x i -axis and by Then (P ) has a solution u ∈ W 1,p 0 (Ω) ∩ L ∞ (Ω) for any h ∈ L ∞ (Ω).The proof of Theorem 2 follows as in Theorem 1. Upper and lower solutions are defined for A p in the same way as in definition 1 relative to ∆ p .

Lemma 7. Assume that (P ) admits an upper solution β and a lower solution α, then (P ) admits a solution
The proof of Lemma 7 follows similar lines as Lemma 1.It sufficis to remark that -A p is strictly monotone.
Proof of Theorem 2. Let us describe the construction of the upper solution (that of lower solution is similar).Let g be the continuous function defined by and denote G(s) = s 0 g(t) dt.
It is clear that (f 0 ) is not verified in the example of comment 1 above.
ρ < C p , which clearly implies the lemma.Lemma 5. Let d > 0 and consider the mapping T d defined by T d (u)(t) = d − space C(I).Then T d has a fixed point.Proof.Clearly by Ascoli's theorem T d is compact.The proof of Lemma 5 uses a homotopy argument based on the Leray Schauder topological degree.So T will have a fixed point if the following condition holds: ∃r > 0 such that (I −λT d )(u) = 0 ∀u ∈ ∂B(0, r) ∀λ ∈ [0, 1], where ∂B(0, r) = {u ∈ C(I); u ∞ = r}.
1)g(0) .Letting n → +∞, we get a contradiction.Let us denote by u d ∈ C(I) a fixed point of the mapping T d of Lemma 5. Lemma 6. ∃d > 0 such that u d (t) ≥ 0 ∀t ∈ [a, a+b 2 [.Proof.We know that u d is decreasing and that u d (a) = d for all d > 0.

2 .
The problem (P p ) has at least one solution for any given h ∈ L ∞ (Ω) if we assume that: |s| p < λ 1 .
and denote G(s) = s 0 g(t) dt for all s in IR.It is easy to see that g(s) ≥ 1 ∀s ∈ IR and that Now it is clearly sufficient to prove the existence of a function β 1 s→+∞ pF (s) s p < C p .