Non-linear Elliptic Unilateral Problems in Musielak-Orlicz spaces with L 1 data

where A(u) = −div a(x, u,∇u) is a Leray-Lions operator defined on D(A) ⊂ W 1 0Lφ(Ω) → WLψ(Ω) with φ and ψ are two complementary Musielak-Orlicz functions, and g is a non-linearity with sign condition and satisfying, for all s ∈ R, ξ ∈ R and almost all x ∈ Ω, the following natural growth condition: |g(x, s, ξ)| ≤ b(|s|)(a0(x) + φ(x, |ξ|)), where b : R → R is a continuous and non-decreasing function and a0(.) is a given non-negative function in L(Ω). The right-hand side f is assumed to belongs to L(Ω).


Introduction
Let Ω be an open bounded subset of R N (N ≥ 2).Consider the following non-linear Dirichlet problem where A(u) = −div a(x, u, ∇u) is a Leray-Lions operator defined on D(A) ⊂ W 1 0 L ϕ (Ω) → W −1 L ψ (Ω) with ϕ and ψ are two complementary Musielak-Orlicz functions, and g is a non-linearity with sign condition and satisfying, for all s ∈ R, ξ ∈ R N and almost all x ∈ Ω, the following natural growth condition: where b : R → R is a continuous and non-decreasing function and a 0 (.) is a given non-negative function in L 1 (Ω).The right-hand side f is assumed to belongs to L 1 (Ω).

52
M. Ait Khellou and A. Benkirane On Orlicz spaces and in the variational case, it is well known that Gossez and Mustonen solved in [20] the following obstacle problem where K φ is a convex subset in W 1 0 L M (Ω) given by K φ = {v ∈ W 1 0 L M (Ω) : v ≥ φ a.e in Ω}, with φ is a measurable function satisfying some regularity condition.An existence result has been proved in [2] by Aharouch, Benkirane and Rhoudaf where the nonlinearity g depend on x, u and ∇u and without assuming the ∆ 2condition on the N -function.
In the case where f ∈ L 1 (Ω), the unilateral problem corresponding to (1.1) has been studied in [3] by Aharouch and Rhoudaf and in [16] by Elmahi and Meskine without assuming the ∆ 2 -condition on the N -function.
In the framework of variable exponent Sobolev spaces, Azroul, Redwane and Yazough have shown in [6] the existence of solutions for the unilateral problem associated to (1.1) where the second member f is in L 1 (Ω).
In the setting of Musielak-Orlicz spaces and in variational case, Benkirane and Sidi El vally [12] proved the existence of solutions for the obstacle problem (1.2), they generalized the work of Gossez and Mustonen in [20].
The purpose of this paper is to prove, in the setting of Musielak spaces, an existence result for unilateral problem corresponding to (1.1) in the case where f ∈ L 1 (Ω) under the assumption that the conjugate function of the Musielak-Orlicz function ϕ satisfies the ∆ 2 -condition and by assuming dt = ∞ for a.e.x ∈ Ω. (1.3)This assumption (1.3) allows us to use a Poincaré type inequality in the proof of the main result of this work (Theorem 3.3).Remark that this condition corresponds, in the classical Sobolev spaces W 1,p to the case p < N, which is the interesting case in these spaces.Further works for the unilateral problem corresponding to (1.1) in the L p case can be found in [13,14,15].
A function ϕ which satisfies the conditions i) and ii) is called a Musielak-Orlicz function.
For a Musielak-Orlicz function ϕ we put ϕ x (t) = ϕ(x, t) and we associate its nonnegative reciprocal function ϕ −1  x , with respect to t, that is x (t)) = t.The Musielak-Orlicz function ϕ is said to satisfy the ∆ 2 −condition if for some C > 0, and a non negative function h, integrable in Ω, we have ϕ(x, 2t) ≤ Cϕ(x, t) + h(x) for all x ∈ Ω and all t ≥ 0.
In the space L ϕ (Ω) we define the Luxemburg norm by: and the Orlicz norm by where ψ is the Musielak-Orlicz function complementary to ϕ.These two norms are equivalent [22].
The closure in L ϕ (Ω) of the set of bounded measurable functions with compact support in Ω is denoted by E ϕ (Ω).It is a separable space and (E ψ (Ω)) * = L ϕ (Ω) [22].We have E ϕ (Ω) = K ϕ (Ω) if and only if K ϕ (Ω) = L ϕ (Ω) if and only if ϕ satisfy the ∆ 2 -condition (2.1) for large values of t or for all values of t, according to whether Ω has finite measure or not.
We define These functionals are convex modular and a norm on W 1 L ϕ (Ω) respectively.The pair W 1 L ϕ (Ω), u 1 ϕ,Ω is a Banach space if ϕ satisfies the following condition [22]: there exists a constant c > 0 such that inf The space We denote by D(Ω) the Schwartz space of infinitely smooth functions with compact support in Ω and by D(Ω) the restriction of D(R N ) on Ω.The space and the space W 1 0 E ϕ (Ω) as the (norm) closure of the Schwartz space D(Ω) in W 1 L ϕ (Ω).For two complementary Musielak-Orlicz functions ϕ and ψ, we have [22]: i) The Young inequality: ts ≤ ϕ(x, t) + ψ(x, s) for all t, s ≥ 0, x ∈ Ω.

4)
Non-linear Elliptic Unilateral Problems in Musielak spaces

55
We say that a sequence of functions u n converges to u for modular convergence in The following spaces of distributions will also be used: Lemma 2.2.[11] Let Ω be a bounded Lipschitz domain in R N and let ϕ and ψ be two complementary Musielak-Orlicz functions which satisfy the following conditions:

2)]
(ii) There exists a constant A > 0 such that for all x, y ∈ Ω with |x − y| ≤ 1 2 we have

6) (iv)
There exists a constant C > 0 such that ψ(x, 1) ≤ C a.e in Ω. (2.7) Consequently, the action of a distribution S in W −1 L ψ (Ω) on an element u of W 1 0 L ϕ (Ω) is well defined.It will be denoted by S, u .
Then the Nemytskii operator N f defined by The following theorem has already been treated in [5] but we think it is useful to give it again in order to facilitate the reading of this work, it is a Poincaré type inequality in Musielak spaces, for more details see [5].
Theorem 2.6.[5] Let Ω be a bounded Lipschitz domain of R N , and let ϕ be a Musielak-Orlicz function satisfying (1.3) and the conditions (i), (ii), (iii) and (iv) of Lemma 2.2 then there exists a constant C(Ω, ϕ) > 0 such that Proof: Suppose, by contradiction, that for every n ∈ N * , there exists , there exists a subsequence, denoted by (u n k ), weakly convergent in W 1 0 L ϕ (Ω) for the weak * topology σ(ΠL ϕ , ΠE ψ ).By using the compact imbedding ), there exists a function v ∈ L ϕ (Ω), and a subsequence, still denoted by ( , and so (2.9) By combining (2.8) and (2.9), we obtain ∇v = 0, and this imply that v is a constant Non-linear Elliptic Unilateral Problems in Musielak spaces 57

Main result
Let Ω be a bounded Lipschitz domain in R N (N ≥ 2), and let ϕ and γ be two Musielak-Orlicz functions such that γ ≺≺ ϕ and ϕ satisfies the assumption (1.3) and conditions of Lemma 2.2.Given an obstacle measurable function Λ : Ω → R and consider the set be a mapping (not everywhere defined) given by: A(u) = −diva(x, u, ∇u) where ψ is the Musielak function complementary to ϕ which satisfies the ∆ 2 -condition and a : Ω × R × R N → R N is a Carathéodory function satisfying, for a.e x ∈ Ω and for all s ∈ R and all ξ ,ξ where b : R → R is a continuous and non-decreasing function and a 0 (.) is a given non-negative function in L 1 (Ω).Now, assume that for the modular convergence in where a(x, u) is a Carathéodory function such that 0 ≤ µ ≤ a(x, u) ≤ ν, m is the derivative of the Musielak function ϕ with respect to t and g is a continuous function satisfying g(s)s ≥ 0. Then the assumptions (3.1)-(3.5)hold true.(see Remark 3.2 of [16]) Define T 1,ϕ 0 (Ω) to be the set of measurable functions u : , where T k (.) is the truncation at height k > 0, defined by We shall prove the following existence theorem.
Theorem 3.3.Assume that (3.1)-(3.7)hold true, then there exists at least one solution of the following unilateral problem and for all k ≥ 0.
Let (f n ) be a sequence of smooth functions which converges strongly to f in L 1 (Ω) and set g n (x, s, ξ) = T n (g(x, s, ξ)).
Consider the approximate unilateral problems where ., .means the duality between Since g n is bounded for any fixed n > 0, there exists at least one solution u n ∈ K Λ ∩ D(A) of (P n ).(see Proposition 5 of [20] and Theorem 8 of [12]) and by using (3.5), one easily has 3) and by using the fact that {x ∈ Ω : where C η is a positive constant depending on η, thanks to (3.8), we have then from (3.5) and (3.9), we get where C is independent of k.Hence, by using (3.3) we obtain Finally, since k is arbitrary we obtain On the other hand, since ψ (the conjugate of ϕ) satisfies the ∆ 2 -condition then, from proposition 2.1 of [17], there exists ν > 0 and c > 0 such that ϕ(x, t) ≥ c t 1+ν for all t ≥ some t 0 > 0. (3.12) We have then by the Chebyshev, the Poincaré inequality, (3.12) and (3.11) we obtain where C ν,N is a constant from the Poincaré inequality in W 1,1+ν 0 .For any µ > 0, we have (3.13) From (3.11) and by using Theorem 2.6, we deduce that (T k (u n )) n is bounded in W 1 0 L ϕ (Ω) and then we can assume that (T k (u n )) n is a Cauchy sequence in measure in Ω.Let ε > 0, then by (3.13) and the fact that This proves that (u n ) is a Cauchy sequence in measure in Ω, and then converges almost everywhere to some measurable function u.Finally, by Lemma 4.4 of [19], we obtain for all k > 0 , strongly in E ϕ (Ω) and a.e. in Ω. ( Now, we shall prove that (a(x, Let ϑ ∈ E ϕ (Ω) N arbitrary.By using (3.2), we have for every k > 0, where k 3 is defined in (3.1), which gives by (3.10) Since ϑ is arbitrary in E ϕ (Ω) N , choose ω = ϑ k3 − ∇v 0 in the last inequality with ω Lϕ(Ω) N = 1 and we find On the other hand, for β large enough, we have by using (3.1) where C k,v0 is a constant which depends on k and v 0 but not on n.Hence, using the dual norm, one has (a(x, Then, for k > 0 we have which gives by Hölder inequality which implies that, for all k > 0 there exists a function Step 2 : Almost everywhere convergence of the gradients. By assumption (3.6) there exists a sequence

and define
Taking v h n,j = u n − β 2 φ(ω h n,j ) as test function in (P n ), where β 2 = exp(−4δk 2 ) we obtain Observe that ∇ω h n,j = 0 on the set {x ∈ Ω : |u n | > m}, then we have from (3.17) ) dx, using (3.14), we have φ(ω h n,j ) → φ(ω h j ) weakly in L ∞ (Ω) as n → +∞, and then letting j and h to infinity and using Lebesgue theorem we get On the other hand, we have The first term of the right hand side of the last equality can write as Non-linear Elliptic Unilateral Problems in Musielak spaces

65
The second term of the right hand side of (3.19) can write as, using (3.2) Using (3.15) and modular convergence of (v j ), it is easy to see that and since c ′ (.) ∈ L 1 (Ω) we have where χ j s is the characteristic function of the set Ω j s = {x ∈ Ω : as n tends to infinity.
Using the modular convergence of v j , one has For the second term on the right hand side of (3.24) we can write Splitting the first integral on the right hand side of this equality where Non-linear Elliptic Unilateral Problems in Musielak spaces

67
then then, the first term on the right hand side of (3.26) tends to the quantity Concerning the second term on the right hand side of (3.26), it is easy to see that Finally, by combining (3.24), (3.25) and (3.27) we get We now evaluate the second term on the left hand side of (3.18) by writing Non-linear Elliptic Unilateral Problems in Musielak spaces

69
As regards the last term on the last side of this inequality, we have we argue as above to show that Then Combining (3.18), (3.28) and (3.29), we obtain thanks to (3.8), one has (3.30) Now, observe that Passing to the limit in n and j in the last three terms of the right hand side of the last equality gives and Hence (3.32) By passing to the lim sup over n, and letting j, h, s tend to infinity, we obtain As in [8], there exists a subsequence, still denoted by u n , such that ∇u n → ∇u a.e. in Ω. (3.33) Step 3 : Modular convergence of the truncations.Since (3.15) and (3.33), we have l k = a(x, T k (u), ∇T k (u)), which implies by using (3.32) By using Fatou's Lemma we obtain We proceed as above to get Non-linear Elliptic Unilateral Problems in Musielak spaces

73
It follows that By Lemma 2.5, we conclude that strongly in L 1 (Ω).The convexity of the Musielak fonction ϕ and (2.7) allow us to have Then, by (3.34) we get lim So that, by Vitali's theorem one has
As a consequence of (3.14) and (3.33), one has so it suffices to show that g n (x, u n , ∇u n ) is uniformly equi-integrable in Ω.
Let E be a measurable subset of Ω and let m > 0. We have Which shows that g n (x, u n , ∇u n ) is uniformly equi-integrable in Ω.By Vitali's theorem, we conclude that g(x, u, ∇u) ∈ L 1 (Ω) and g n (x, u n , ∇u n ) → g(x, u, ∇u) strongly in L 1 (Ω).
Step 5 : Passage to the limit.
Finally, letting h to the infinity we deduce Elliptic Unilateral Problems in Musielak spaces Non-linear Elliptic Unilateral Problems in Musielak spaces 61 then and denote by χ r the characteristic function of Ω r .Clearly, Ω r ⊂ Ω r+1 and |Ω\Ω r | → 0 as r → ∞.Non-linear Elliptic Unilateral Problems in Musielak spaces 63 Let s ≥ r, we have 0