EXISTENCE OF SOLUTIONS FOR UNILATERAL PROBLEMS WITH L 1 DATA IN ORLICZ

This article is concerned with the existence result of the unilateral problem associated to equations of the type Au+ g(x, u,∇u) = f, in Orlicz spaces, where f ∈ L(Ω), the term g is a nonlinearity having natural growth and satisfying the sign condition. Some stability and positivity properties of solutions are proved. AMS Classification : 35J60.


Introduction
Let Ω be a bounded domain in IR N and let Au = −diva(x, u, ∇u) be a Leray-Lions operator defined on its domain D(A) ⊂ W 1 0 L M (Ω), where M is an N -function which satisfies the ∆ 2 -condition and f ∈ L 1 (Ω).
The author's in [6] proved the existence of at least one solution for the following nonlinear Dirichlet problem Where g is a nonlinearity having natural growth with respect to |∇u|, and which satisfies the classical sign condition with respect to u.
It is our purpose, in this paper, to prove an existence theorem for the corresponding obstacle problem.More precisely, we prove the existence of at least one solution for the following unilateral problem ( Where K ψ = {u ∈ W 1 0 L M (Ω)/u ≥ ψ a.e. in Ω.)}, with ψ is a measurable function on Ω such that ψ + ∈ W 1 0 L M (Ω) ∩ L ∞ (Ω), and where T k is the truncation operator at height k > 0, defined onR by T k (s) = max(−k, min(k, s)).
Let us point out that another work in the L p case can be found in [17] in the case of equation, and in [9] in the case of obstacle problems.
Note that this type of equations can be applied in sciences physics.Non-standard examples of M (t) which occur in the mechanics of solids and and M (t) = t log(1 + log(1 + t)) (see [11,12,13,10]) for more details).
This paper is organized as follows, sections 2 contain some preliminaries and some technical lemmas.Section 3 is concerned with basic assumptions and the main result which is proved in section 4, finally, we study the stability and the positivity of solution.
The N -function M conjugate to M is defined by M = R t 0 ā(s) ds, where ā :R + → R + is given by ā(t) = sup{s : a(s) ≤ t}.
The N -function M is said to satisfy the ∆ 2 -condition if, for some k it is readily seen that this will be the case if and only if for every r > 1 there exists a positive constant k = k(r) such that for all t > 0 When (2.1) and (2.2) holds only for t ≥ t 0 for some t 0 > 0 then M is said to satisfy the ∆ 2 -condition near infinity.
We will extend these N -functions as even functions on all IR.Moreover, we have the following Young's inequality ∀s, t ≥ 0, st ≤ M (t) + M (s).
Let P and Q be two N -functions.P << Q means that P grows essentially less rapidly than Q, i.e., for each > 0, P (t) Q( t) → 0 as t → ∞.This is the case if and only if lim t→∞ Q −1 (t) P −1 (t) = 0.

2-2
Let Ω be an open subset of IR N .The Orlicz class K M (Ω) ( resp. the Orlicz space L M (Ω)) is defined as the set of (equivalence classes modulo equality a. e. ) real valued measurable functions u on Ω such that The closure in L M (Ω) of the set of bounded measurable functions with compact support in Ω is denoted by E M (Ω).
The dual of E M (Ω) can be identified with L M (Ω) by means of the pairing Z Ω uv dx, and the dual norm of L M (Ω) is equivalent to k.k M,Ω .

2-3
We now turn to the Orlicz-Sobolev space ] which is the space of all functions u such that u and its distributional derivatives of order 1 lie in Thus, W 1 L M (Ω) and W 1 E M (Ω) can be identified with subspaces of the product of N + 1 copies of L M (Ω).Denoting this product by Q L M , we will use the weak topologies σ( ] denote the space of distributions on Ω which can be written as sums of derivatives of order ≤ 1 of functions in L M (Ω) [resp.E M (Ω)].It is a Banach space under the usual quotient norm (for more details see [1]).
We now introduce the functional spaces we will need later.For an N -function M, τ 1,M 0 (Ω) is defined as the set of measurable functions u : Ω −→ R such that for all k > 0 the truncated functions T k (u) ∈ W 1 0 L M (Ω).We gives the following lemma this is a generalization of Lemma 2.1 [2] in Orlicz spaces.
Lemma 2.1.For every u ∈ τ 1,M 0 (Ω), there exists a unique measurable function Lemma 2.2.Let λ ∈ R and let u and v be two measurable functions defined on Ω which are finite almost everywhere, and which are such that T k (u), T k (v) and T k (u + λv) belong to W 1 0 L M (Ω) for every k > 0 then where ∇(u), ∇(v) and ∇(u + λv) are the gradients of u, v and u + λv introduced in Lemma 2.1.
The proof of this lemma is similar to the proof of Lemma 2.12 [8] for the Sobolev spaces.
We recall some lemmas introduced in [4] which will be used later.
Let M be an N -function and let u ∈ Lemma 2.4.Let F : R → R be uniformly Lipschitzian, with F (0) = 0. We suppose that the set of discontinuity points of F 0 is finite.Let M be an N -function, then the mapping is sequentially continuous with respect to the weak* topology σ( We give now the following lemma which concerns operators of Nemytskii type in Orlicz spaces ( see [4]).
Lemma 2.5.Let Ω be an open subset of R N with finite measure.
Let M, P and Q be N -functions such that Q << P , and let F : Ω×R → R be a Carathéodory function such that, for a.e.x ∈ Ω and all s ∈ R : where k 1 , k 2 are real constants and c(x Then the Nemytskii operator N F defined by

Main results
Let Ω be an open bounded subset of R N , N ≥ 2, with the segment property.
Let M be an N -function satisfying the ∆ 2 -condition near infinity, and let P be an N -function such that P << M. We consider the Leray-Lions operator, Au = −div(a(x, u, ∇u)), Furthermore let g : Ω ×R ×R N → R be a Carathéodory function such that for a.e.x ∈ Ω and for all s ∈ R and all where b : R + →R + is a continuous nondecreasing function, c is a given positive function in L 1 (Ω), and λ > 0. Let the subset convex where ψ : Ω →R is a measurable function on Ω such that In the next section, we will prove the following theorem.Theorem 3.1.Assume that the hypotheses (3.1)-(3.7)holds.Then, there exists at least one solution of the following unilateral problem Remark 3.1.We obtain the same results of our theorem if we suppose that the sign condition (3.4) is fulfilled only near infinity.

Proof of main result
To prove the existence theorem we proceed by steps.STEP 1: A priori estimates.
Let us define | and let us consider the sequence of approximate variational inequalities where f n is a regular function such that f n strongly converges to f in L 1 (Ω).Since g n (x, s, ξ) is bounded and g n (x, s, ξ).s ≥ 0, then by using the Proposition 5 and Remark 6 of [15] (with m = 1 and a 0 (x, s, ξ) = g n (x, s, ξ)) it is easy to verifie that (P n ) has at least one solution. Let The choice of w as a test function in (P n ), we obtain Consequently, we deduce that By using the Young's inequality and the ∆ 2 -condition we have Now, we prove that u n converges to some function u in measure (and therefore, we can always assume that the convergence is a.e. after passing to a suitable subsequence).We shall show that u n is a Cauchy sequence in measure.
Thanks to Lemma 5.7 of [14], there exists two positive constants C 0 3 and C 0 4 such that Then, we deduce by using (4.1) and (4.2) that For every δ > 0, we have strongly in E M (Ω) and a.e. in Ω.
Consequently, we can assume that T k (u n ) is a Cauchy sequence in measure in Ω.
We shall prove that the sequence (a(x, Let w ∈ (E M (Ω)) N be arbitrary.By condition (3.2), we have Using the argument above, we may assume that the first term on the right remains bounded.Moreover, by (3.1), we have Therefore, which implies that the second term on the right in ( 4.5) is also bounded.By the theorem of Banach-Steinhaus, the sequence (a(x, u n , ∇u n )χ {|un−ψ + |≤k} ) remains bounded in L M (Ω).Since k arbitrary, we deduce that (a(x, T k (u n ), ∇T k (u n )) also bounded in L M (Ω).Which implies that, for all k > 0 there exists a function Fix r and let s ≥ r, we have, Now, consider the following function It is well known that where K is a constant which will be used later.
Let k ≥ kψ + k ∞ , we define where h > 2k > 0. For η = exp(−4γk 2 ), we define the following function as v n,h = u n − ηϕ(w n ).(4.9) We take v n,h as test function in (P n ) (for more explication concerned this test function see the appendix II), we obtain, Note that, ∇w n = 0 on the set where |u n | > h + 4k, therefore, setting m = 4k + h, and denoting by ε 1 h (n), ε 2 h (n), ... various sequences of real numbers which converge to zero as n tends to infinity for any fixed value of h, we get, by (4.11), and since for x in the set {x ∈ Ω : |u n (x)| > k}, we have that ϕ(w n )g(x, u n , ∇u n ) ≥ 0, we deduce from (4.11) that Splitting the first integral on the left hand side of (4.12) where |u n | ≤ k and |u n | > k, we can write, using (3.3) and the fact that a(x, s, 0) = 0 for all s ∈ IR.: where C k = ϕ 0 (2k).Since, as n tends to infinity, , the last term in the previous inequality tends to zero for every h fixed.

Now, observe that
The second terms of the right hand side of (4.14) tend to 0. Indeed Moreover, and by using the Lebesgue's theorem.Combining the previous statement, we deduce the result.
The third term of the right hand side of (4.14) tends to the quantity So that (4.13) yields Since the N -function M satisfies the ∆ 2 -condition near infinity, then there exist two positive constants K and K 0 such that Finally, taking K = max(1, k( δ λ )), we deduce (4.16).
For the second term of the left hand side of (4.12), we can estimate as follows let us consider the last integral in the previous inequality It is easy to see that the second term of the right hand side tends to the quantity Reasoning as in (4.14), the third term of the right hand side of (4.18) tends to 0. From (4.17) and (4.18), we obtain Combining (4.12), (4.15) and (4.19), we obtain which implies, by using (4.8), Consequently, from (4.7), we have By passing to the lim sup over n, and letting h, s tend to infinity, we obtain by the same method used in [4] that Again by (4.20), we get The first term of the right hand side of the last inequality tend to The second term of the right hand side of (4.22) tends to 0, since Passing again to the limsup but now over h, and by using that the functions a(x, T k (u), ∇T k (u))∇T k (u), c(x) + K 0 and f belong to L 1 (Ω) and that Using again the fact that a(x, T k (u), ∇T k (u))∇T k (u) ∈ L 1 (Ω) and letting s → ∞ we get, since meas(Ω\Ω s ) → 0, lim sup On the other hand, by Fatou's lemma, Thanks to (4.16), we have and by using (4.23), one obtains, by Vitali's theorem, weakly in (L M (Ω)) N for σ(ΠL M , ΠE M ) on easily see that Now, we need to prove that g n (x, u n , ∇u n ) → g(x, u, ∇u) strongly in L 1 (Ω), (4.27) in particular it is enough to prove that the functions g n (x, u n , ∇u n ) are equiintegrable of g n (x, u n , ∇u n ).To this purpose.We take For any measurable subset E ⊂ Ω, we have In view by (4.24) there exists η(ε) > 0 such that for all E such that meas(E) < η(ε).where k ≥ 0, h ∈ E M (Ω) and b : IR + → IR is a nondecreasing continuous function.(Ω), g(x, u n , ∇u n ) ∈ L 1 (Ω) Then, there exists a subsequence of u n still denoted u n such that u n converges to u almost everywhere and T k (u n ) T k (u) weakly in W 1 0 L M (Ω).Further u is a solution of the unilateral problem (P ).
Proof.We give the proof briefly.
Step 1.A priori estimates We proceed as previous, we take v = ψ + as test function in (P 0 n ), we get

( 4
.25) By Fatou's lemma and the fact that

Remark 4 . 2 .
: Note that we obtain the existence result without assuming the coercivity condition.However one can overcome this difficulty by introduced the functionw n = T 2k (u n − T h (u n ) + T k (u n ) − T k (u)) in the test function (4.9).

Corollary 4 . 1 .
Assume that the hypothesis (3.1)-(3.7)holds.Let f n any sequence of function in L 1 (Ω) that converges to f weakly in L 1 (Ω) and let u n the solution of the following unilateral problem       u n ≥ ψ a.e. in Ω. u n ∈ τ 1,M 0 the same method used in the first step in the proof of Theorem 3.1 there exists a function u (with T k (u) ∈ W 1 0 L M (Ω) ∀k > 0) and a subsequence still denoted by u n such thatT k (u n ) T k (u), n → ∞ weakly in W 1 0 L M (Ω), ∀k > 0.Step 2. Strong convergence of truncationThe choice of v = T h (u n − ηφ(w n )), h > kψ + k ∞ as test function in (P 0 n ), we get, for all l > 0 Z Ω a(x, u n , ∇u n )∇T l (u n − T h (u n − ηφ(w n ))) + Z Ω g(x, u n , ∇u n )T l (u n − T h (u n − ηφ(w n ))) dx ≤ Z Ω f n T l (u n − T h (u n − ηφ(w n ))) dx.