BOUNDED SOLUTIONS OF UNILATERAL PROBLEMS FOR STRONGLY NONLINEAR EQUATIONS IN ORLICZ SPACES

In this paper, we prove the existence of bounded solutions of uni- lateral problems for strongly nonlinear equations whose principal part hav- ing a growth not necessarily of polynomial type and a degenerate coercivity, the lower order terms do not satisfy the sign condition and appropriate inte- grable source terms. We do not impose the �2-condition on the considered N-functions defining the Orlicz-Sobolev functional framework. Let be a bounded open subset of R N ; N � 2; and let M be an N-function. In this paper, we establish the existence of bounded solutions for the unilateral problem related to strongly nonlinear equations of the form


Introduction
Let Ω be a bounded open subset of R N , N ≥ 2, and let M be an N-function.In this paper, we establish the existence of bounded solutions for the unilateral problem related to strongly nonlinear equations of the form in the subset Ω.The principal part A is a non everywhere defined elliptic differential operator in divergence form Au = −div a(x, u, ∇u) (1.2) defined from its domain D(A) := u ∈ W 1 0 L M (Ω) : a(x, u, ∇u) ∈ (L M (Ω)) N into W −1 L M (Ω) satisfying, among others, the following condition a(x, s, ξ) where h : R → R + is a continuous decreasing function with unbounded primitive (for instance h(t) = 1 (e+|t| log(e+|t|)) and a(x, s, ξ) = M −1 (M (h(|s|))) M −1 (M(|ξ|)) |ξ| ξ).The Hamiltonian g(x, u, ∇u) does not satisfy the sign condition (i.e.g(x, s, ξ)s ≥ 0) but only grows at most like M (|∇u|), precisely |g(x, s, ξ)| ≤ β(s)M (|ξ|), (1.4) where β : R → R + is a continuous function, while the source term have a suitable summability.Let us note that when h is a nonzero constant and g satisfies the sign condition, Dirichlet problems having lower order terms that behave like M (|∇u|) arise naturally in the Calculus of Variations.For example, if we consider the functional the Euler-Lagrange equation is Let ψ : Ω → R be a measurable function such that K ψ = {v ∈ W 1 0 L M (Ω) : v ≥ ψ a.e. in Ω} is a nonempty set.In fact, we are interested in the existence of bounded solution for the following obstacle problem When M (t) = t p , 1 < p < +∞, and h in (1.3) is a nonzero constant, existence of bounded solution for problem (1.5) have been obtained, using direct method, in [6] with f ≡ 0 and in [8] for quasilinear operators without lower order terms (i.e.β = 0) and data satisfying f ∈ L m (Ω), m > N 2 and then under smallness a condition on the data f in [11] with using symmetrization methods.
In the non standard growth setting, existence basic works for variational inequalities (i.e.where f ∈ W −1 E M (Ω)) were initiated by Gossez and Mustonen in [12] solving the obstacle problem (1.5) in the case g(x, u, ∇u) = g(x, u) by assuming some regularity conditions on the obstacle function ψ.Since, several papers were written on existence of solutions for problem like (1.5) either in the variational case see, for instance, [3] or with L 1 -data see, for instance, [2,4,9].In this latter case, solution is understood as meaning a function u such that It is our purpose, in this paper, to prove the existence of bounded solutions, for unilateral problem associated to (1.1) in the setting of the Orlicz-Sobolev spaces without assuming the ∆ 2 -condition on the N-function M. To this end, we use rearrangement techniques and conditions (3.6) and (3.7), (see [18]), on the source term covering (1.6) in the case of polynomial growth.It is worth recalling here some difficulties we have found in dealing with this kind of problems.First of all, the operator (1.2) does not satisfy the 'coercivity' condition in the setting of Orlicz spaces (see [12]), this is due to the hypothesis (1.3) and the EJQTDE, 2013 No. 21, p. 2 fact that no positive lower bound is assumed on the function h when the unknown has large values.The second difficulty in proving the existence of a solution stems from the fact that g(x, u, ∇u) does not define a mapping from The third one concerns the lower order term; it does not satisfy the well known sign condition (i.e.g(x, s, ξ)s ≥ 0) and so appears the problem of getting the a priori estimates.This hindrance is overcome by using test functions of exponential type, the monotone convergence theorem and a comparison result.As examples of equations (1.1) to which our result can be applied, we give .
The paper is organized as follows: in Section 2 we give some preliminaries and auxiliary results.Section 3 contains the basic assumptions and the main result, while Section 4 is devoted to the proof of the main result.

Preliminaries
2.1 Let M : R + → R + be an N-function, ie.M is continuous, convex, with M (t) > 0 for t > 0, M(t) t → 0 as t → 0 and M(t) t → ∞ as t → ∞.The N-function conjugate to M is defined as M (t) = sup{st − M (t), s ≥ 0}.We will extend these N-functions into even functions on all R. We recall that (see [1]) and the Young's inequality: for all s, t ≥ 0, st ≤ M (s) + M (t).If for some k > 0, we said that M satisfies the ∆ 2 -condition, and if (2.2) holds only for t greater than or equal to t 0 ≥ 0, then M is said to satisfy the ∆ 2 -condition near infinity.Let P and Q be two N-functions, the notation P ≪Q means that P grows essentially less rapidly than Q, that is to say for all ǫ > 0, P (t) Q(ǫt) → 0 as t → +∞.That is the case if and only if Q −1 (t) P −1 (t) → 0 as t → ∞.

2.2
Let Ω be an open subset of R N .The Orlicz class K M (Ω) ( resp. the Orlicz space L M (Ω)) is defined as the set of (equivalence class of) real-valued measurable functions u on Ω such that: EJQTDE, 2013 No. 21, p. 3 Endowed with the norm is a Banach space and K M (Ω) is a convex subset of L M (Ω).We define the Orlicz norm u (M) by where the supremum is taken over all v ∈ E M (Ω) such that v M ≤ 1, for which holds for all u ∈ L M (Ω) (see [17]).The closure in L M (Ω) of the set of bounded measurable functions with compact support in Ω is denoted by E M (Ω).

2.3
The Orlicz-Sobolev space ) is the space of functions u such that u and its distributional derivatives up to 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 ΠL M , we will use the weak topologies σ(ΠL M , ΠE M ) and σ(ΠL M , ΠL M ).The space W 1 0 E M (Ω) is defined as the norm closure of the Schwartz space D(Ω) in W 1 E M (Ω) and the space We say that a sequence {u n } converges to u for the modular convergence in this implies the convergence for σ(ΠL M , ΠL M ).
If M satisfies the ∆ 2 -condition on R + (near infinity only if Ω has finite measure), then the modular convergence coincides with norm convergence.Recall that the norm Du M defined on denotes 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.Recall that an open domain Ω ⊂ R N has the segment property (see [15] p.167) if there exist a locally finite open covering {O i } of the boundary ∂Ω of Ω and corresponding vectors {y i } such that if x ∈ Ω ∩ O i for some i, then x + ty i ∈ Ω for 0 < t < 1.If the open Ω has the segment property then the space D(Ω) is dense in W 1 0 L M (Ω) for the topology σ(ΠL M , ΠL M ) (see [15]).Consequently, the action of a distribution in W −1 L M (Ω) on an element of W 1 0 L M (Ω) is well defined.For an exhaustive treatment one can see for example [1,17].

2.4
We will use the following lemma, (see [10]), which concerns operators of Nemytskii Type in Orlicz spaces.It is slightly different from the analogous one given in [17].
EJQTDE, 2013 No. 21, p. 4 Lemma 2.1.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 for all s ∈ R, where k 1 , k 2 are real constants and c(x . We will also use the following technical lemma which can be found in [16].

2.5
We recall the definition of decreasing rearrangement of a real-valued measurable function u in a measurable subset Ω of R N having finite measure.Let |E| stands for the Lebesgue measure of a subset E of Ω.The distribution function of u, denoted by µ u , is a map which informs about the content of level sets of u, that is The decreasing rearrangement of u is defined as the generalized inverse function of µ u , that is the function u In other words, u * is the (unique) non-increasing, right-continuous function in [0, +∞) equi-distributed with u.Furthermore, for every t ≥ 0 u * (µ t (t)) ≤ t. (2.3) We also recall that (see [18]) (2.4)

Basic assumptions and Main result
Through this paper Ω will be a bounded open subset in of R N , N ≥ 2, satisfying the segment property and M is an N -function twice continuously differentiable and strictly increasing, and P is an N -function such that P ≪M .Let us consider the following convex set where ψ : Ω → R is a measurable function.On the convex K ψ we assume that be the mapping ( non-everywhere defined) given by Au = −div a(x, u, ∇u), where a : Ω × R × R N → R N is a Carathéodory function satisfying, for almost every x ∈ Ω and for all s ∈ R, ξ, η ∈ R N (ξ = η), the following conditions EJQTDE, 2013 No. 21, p. 5 where h : R + → R * + is a continuous decreasing function such that: h(0) ≤ 1 and its primitive H(s) = s 0 h(t)dt is unbounded, (A 4 ) there exist a function c(x) ∈ E M (Ω) and some positive constants k 1 , k 2 , k 3 and k 4 such that where β : R → R + is a continuous function.We assume that the function t → β(t) for all s ∈ R, we have that the function γ is bounded.For what concerns the right hand, we assume one of the following two assumptions: Either or and If Ω has the segment property, assumption (A 2 ) is fulfilled if one of the following conditions is verified: Our main result is the following EJQTDE, 2013 No. 21, p. 6 Theorem 3.2.Suppose that the assumptions (A 1 )-(A 6 ) and either (3.6) or (3.7) are fulfilled.Then, the following obstacle problem has at least one solution.
Before giving the proof of the previous result, the following remarks are in order.
may have no meaning.
Remark 3.4.1)-It is known (see [12]) This follows from assumption (A 2 ) and from the fact that for all u ∈ K ψ one has T n (u) → u for the modular convergence in W 1 L M (Ω).

3)-The assumption (A 1 ) is not a restriction on the obstacle function ψ, instead of it we can assume that
is a nonempty set.Remark 3.5.In light of Remark 3.1 and Remark 3.4, if ψ = −∞ then K ψ = W 1 0 L M (Ω) and problem (3.8) will be reduced to an equation.Hence, our result extends to inequalities the one in [5] stated for equations and also these in [6,7,8].
Remark 3.6.Let M (t) be an N-function.Consider the following equation where β : R → R + is a continuous function such that In what follows, we will use the following real functions of a real variable T k (s) = max(−k, min(k, s)), k > 0, G k (s) = s − T k (s) and φ λ (s) = s exp(λs 2 ), where λ is a positive real number.The following classical lemma turns out to be useful later

Proof of Main result
The proof of Theorem 3.2 is divided into eight steps.EJQTDE, 2013 No. 21, p. 7 Step 1: Approximate problems.For n ∈ N * , Let us denote by m * either N or m according as we assume (3.6) or (3.7).Define f n := T n (f ), A n u := −div a(x, T n (u), ∇u) and g n (x, s, ξ) := T n (g(x, s, ξ)).We can easily check that we have |g n (x, s, ξ)| ≤ |g(x, s, ξ)| and |g n (x, s, ξ)| ≤ n.Let us consider the sequence of approximate problems, Let ν > 1 be large enough.By (3.4) one has Then, Young's inequality enables us to get Let us define the positive real number ρ and the function For each n in N, the function γ n belongs to L 1 (Ω).Thus we have Step 2: Preliminary results.
Lemma 4.1.Let u n be a solution of (4.1).For all t, ǫ in R * + with t > ψ + ∞ , one has the following inequality:  k) .Thanks to [13, lemma 2], the function v belongs to K ψ .Thus, using v as test function in (4.1) and then (3.2) we get 3) Now, we will pass to the limit as k tends to +∞ in (4.3).Observe that the second integral in the left-hand side of (4.3) reads as It follows by applying the monotone convergence theorem, that as k → +∞.In the first integral in the left-hand side of (4.3) the integrand function is nonnegative, so that Fatou's lemma allows us to get while for the remaining terms in (4.3), being g n and f n bounded, we apply the Lebesgue's dominated convergence theorem.Consequently, letting k tends to +∞ in (4.3) we obtain Due to the fact that u + n ≥ ψ + , the function w n vanishes if u n ≤ 0. By virtue of (3.5) we get Hence, (4.4) is reduced to Lemma 4.2.Let u n be a solution of (4.1).For all t, ǫ in R * + , one has the following inequality: ) belongs to K ψ .Thus, the choice of v as test function in (4.1), yields The first integral in the left-hand side of (4.6) is written as dx.
By the monotone convergence theorem, we have as k → +∞.For the seconde integral in the left-hand side of (4.6), we write Applying again the monotone convergence theorem, we obtain as k → +∞.Since g n , f n and γ are bounded, we apply the Lebesgue's dominated convergence theorem for the remaining integrals in (4.6).Hence, letting k tend to +∞ in (4.6), we get Lemma 4.3.Let u n be a solution of (4.1).There exists a constant c 0 , not depending on n, such that for almost every Proof.Being γ bounded, summing up both inequalities (4.2) and (4.5), there is a constant c 0 not depending on n, such that for all t > ψ + ∞ and all ǫ > 0 Using (3.2), dividing by ǫ and then letting ǫ tends to 0 + we obtain (4.7).
Inequality (4.7) allows us to obtain the following comparison result, proved in [5], which is the starting point to obtain uniform estimation in L ∞ for solutions of approximate equations (4.1).
We have for almost every t > ψ + ∞ : Proof.The hypotheses made on the N-function M , allow to affirm that the function is decreasing and convex (see [18]).Hence, Jensen's inequality yields )), using the convexity of C and then letting k → 0 + , we obtain for almost every Recall the following inequality, (see for instance [18]): where m * stands for either N or m according as we assume (3.6) or (3.7).It follows that the sequence {u n } is bounded in W 1 0 L M (Ω).Consequently, there exist a subsequence on {u n }, still denote by {u n }, and a function u ∈ W 1 0 L M (Ω) such that u n ⇀ u weakly in W 1 0 L M (Ω) for σ(ΠL M (Ω), ΠE M (Ω)) (4.16) u n → u in E M (Ω) strongly and a.e. in Ω. (4.17) Step 5: Almost every where convergence of the gradients.Let us begin by the following lemma which will be used in the sequel.To end the proof it is sufficient to show that Ω a(x, T n (u n ), ∇u n ) • ∇u n dx can be estimated by a constant which does not depend on n.To do this, let us use u n as EJQTDE, 2013 No. 21, p. 17 the function h is as above.Then, the assumptions (3.2), (3.3), (3.4), (3.5) of Theorem 3.2 are fulfilled.

Lemma 3 . 7 .
If c and d are positive real numbers such that λ = ( c 2d ) 2 then

By [ 12 ,
Proposition 5] the operator A n satisfies the conditions of [12, Proposition 1] with respect to ψ + .So that in view of the Remark 3.4, by [12, Proposition 1] the variational inequality (4.1) has at least a solution u n .