Existence of solutions for some nonlinear elliptic unilateral problems with measure data

In this paper, we prove the existence of an entropy solution to unilateral problems associated to the equations of the type: Au +H(x;u;ru) div (u) = 2 L 1 ()


Introduction
This paper is devoted to the study of the following nonlinear problem: In Problem (1.1) the framework is the following: Ω is a bounded open subset of IR N , N ≥ 2, and p : Ω → IR + is a continuous function.The operator Au ≡ −div(a(x, u, ∇u)) is a Leray-Lions operator defined on W 1,p(x) 0 (Ω) (this space will be described in Section 2).The function φ is assumed to be continuous on IR with values in IR N and the nonlinear term H(x, s, ξ) satisfies some growth and the sign conditions.The data f and F respectively belong to L 1 (Ω) and (L p (x) (Ω)) N .
The study of problems with variable exponent is a new and interesting topic which raises many mathematical difficulties.One of our motivations for studying (1.1) comes from applications to electrorheological fluids (we refer to [13] for more details) as an important class of non-Newtonian fluids (sometimes referred to as smart fluids).Other important applications are related to image processing (see [8]) and elasticity (see [16]).
Under our assumptions, problem (1.1) does not admit, in general, a weak solution since the term φ(u) may not belong to (L 1 loc (Ω)) N because the function φ is just assumed to be continuous on IR.
In order to overcome this difficulty we use in this paper the framework of an entropy solution (see Definition 3.1).This notion was introduced by Bénilan et al. [1] for the study of nonlinear elliptic problems in case of a constant exponent p(.) ≡ p.
The first objective of our paper is to study the problem (1.1) in the generalized Lebesgue-Sobolev spaces with some general second member µ which lies in L 1 (Ω) + W −1,p (x) (Ω).
The second objective is to treat the unilateral problems, more precisely, we prove an existence result for solutions of the following obstacle problem: u is a measurable function such that u ≥ ψ a.e. in Ω, T k (u) ∈ W 1,p(x) 0 (Ω) and ∀k > 0 where ψ is a measurable function (see assumptions (3.6) and (3.7)), and for any non-negative real number k we denote by T k (r) = min(k, max(r, −k)) the truncation function at height k.
The plan of the paper is as follows.In Section 2, we give some preliminaries and the definition of generalized Lebesgue-Sobolev spaces.In Section 3, we make precise all the assumptions and give some technical results and we establish the existence of the entropy solution to the problem (1.1).In Section 4 (Appendix), we give the proof of Lemma 3.5.

Mathematical preliminaries
In what follows, we recall some definitions and basic properties of Lebesgue and Sobolev spaces with variable exponents.For each open bounded subset Ω of IR N (N ≥ 2), we denote ≤ 1 is a separable and reflexive Banach space, and its dual space is isomorphic to L p (x) (Ω) where 1 p(x) Proposition 2.1 (see [9]).(i) For any u ∈ L p(x) (Ω) and v ∈ L p (x) (Ω), we have and the imbedding is continuous.
3 Main general results

Basic assumptions and some lemmas
Throughout the paper, we assume that the following assumptions hold true: The function a : Ω × IR × IR N → IR N is a Carathéodory function satisfying the following conditions: for every s ∈ IR, ξ ∈ IR N and for almost every x ∈ Ω, where k(x) is a positive function in L p (x) (Ω) and β is a positive constants.
a(x, s, ξ)ξ ≥ α|ξ| p(x) (3.3) for almost every x ∈ Ω and for every s ∈ IR, ξ ∈ IR N , where α is a positive constant such that α ≥ g ∞ .Let H(x, s, ξ) : Ω × IR × IR N → IR N be a Carathéodory function such that for a.e x ∈ Ω and for all s ∈ IR, ξ ∈ IR N the sign and the growth conditions: are satisfied, where g : IR → IR + is continuous increasing positive function that belongs to L ∞ (IR) while γ(x) belongs to L 1 (Ω).
Let ψ be a measurable function such that for the convex set holds.Finally, we suppose that φ ∈ C 0 (IR, IR N ), (3.7) Let p ∈ C + (Ω) be such that there is a vector l ∈ IR N \{0} such that for any x ∈ Ω, (Ω) such that u n u weakly in W 1,p(x) 0 (Ω) and then u n −→ u strongly in W 1,p(x) 0 (Ω).
Remark 3.1 The inequality (3.12) holds true if we assume: there exists a function ξ ≥ 0 such that ∇p∇ξ ≥ 0, |∇ξ| = 0 in Ω (see [6]).Proof.Taking at first the case of F ∈ C 1 (IR) and F ∈ L ∞ (IR).Let u ∈ W 1,p(x) 0 (Ω), and since (Ω), we have u n → u a.e. in Ω and ∇u n → ∇u a.e. in Ω, then F (u n ) → F (u) a.e. in Ω.On the other hand, we have: , (Ω) and we obtain: in Ω, we obtain: (Ω).Let F : IR → IR a Lipschitz uniform function, then F n = F * ϕ n → F uniformly on each compact set, where ϕ n is a regularizing sequence, we conclude that F n ∈ C 1 (IR) and F n ∈ L ∞ (IR), from the first part, we have and thus each term u i can be approximated by a suitable sequence u i k ∈ D(Ω) such that, u i k converges to u i strongly in W 1,p(x) 0 (Ω).Moreover, due to the fact that u i k ∈ C ∞ 0 (Ω), then the Green formula gives On the other hand, gives in view of (3.13): Ω div(u)dx = 0.

Existence of an entropy solution
In this section, we study the existence of an entropy solution of problem (1.1).We now give the definition of an entropy solution Definition 3.1 A measurable function u is an entropy solution to problem (1.1) if for every k ≥ 0: Our main result is Proof of Theorem 3.1.The proof is divided into 4 steps.
Step 1: The approximate problem In this step, we introduce a family of approximate problems and prove the existence of solutions to such problems.
, and f n 1 ≤ f 1 , and we consider the approximate problem: where Assume that (3.1)-(3.10)hold true, then there exists at least one weak solution u n for the approximate problem (P n ).
Proof.Indeed, we define the operator EJQTDE, 2013 No. 43, p. 5 Using the Hölder inequality, we deduce where and C 0 is a constant which depends only on φ, n and p.
We define the operator R n : W (Ω), using the Hölder inequality, we have for all u, v ∈ W (3.15) (Ω) into W −1,p (x) (Ω).Moreover, B n is coercive in the following sense: there exists v 0 ∈ K ψ such that: Proof.See the appendix.
In view of Lemma 3.5, there exists at least one solution u n ∈ W 1,p(x) 0 (Ω) of the problem (P n ), (see [12]).
Step 2: A priori estimate In this step, we establish a uniform estimate on u n with respect to n. Proposition 3.1 Assume that (3.1)-(3.10)hold true.Let u n be a solution of the approximate problem (P n ), then for all k ≥ 0, there exists a constant c(k) (which does not depend on n) such that Taking v as a test function in (P n ) and letting h → +∞, we obtain, for n large enough (n ≥ k+||v 0 || ∞ ): (Ω)) N and using Young's inequality, we obtain From (3.1) and (3.3), we deduce Step 3: Strong convergence of truncations In this step, we prove the strong convergence of truncations.
Proposition 3.2 Let u n be a solution of the problem (P n ), then there exists a measurable function u such that (Ω).
In order to prove Proposition 3.2, we will use the following lemma: Then . By using v as test function in (P n ) and letting l → ∞, we obtain (3.18) Let us define We consider θ(t) = φ(t)χ hk (t) and θ(t) = t 0 θ(s)ds.Then by Lemma 3.4, we obtain Then, the second term of the left side of the inequality (3.18) vanishes for n large enough, which implies that By using Young's inequality, we can deduce that Finally, from (3.3), we deduce (3.17) of Lemma 3.6.
Proof of Proposition 3.2.We will show firstly that (u n ) n is a Cauchy sequence in measure.
Using (3.17) and applying Hölder's inequality and Poincaré's inequality, we obtain that where Finally, for k > 2h > 2||v 0 || ∞ , we have (3.21) Passing to the limit as k goes to infinity, we deduce then, for every ε > 0, there exists k 0 such that For every δ > 0, we have (Ω), then there exists a subsequence (T k (u n )) n such that T k (u n ) converges to η k weakly in W 1,p(x) 0 (Ω) as n → ∞, and by the compact imbedding, we have T k (u n ) converges to η k strongly in L p(x) (Ω) a.e. in Ω.Thus, we can assume that (T k (u n )) n is a Cauchy sequence in measure in Ω, then there exists an n 0 which depends on δ and ε such that By combining (3.23) and (3.24), we obtain EJQTDE, 2013 No. 43, p. 9 Then (u n ) n is a Cauchy sequence in measure in Ω, thus, there exists a subsequence still denoted by u n which converges almost everywhere to some measurable function u, then u n converges to u a.e. in Ω, by Lemma 2.1, we obtain (Ω) and a.e. in Ω. (3.25) ) as test function in (P n ), where (3.26) For every n > m + 1, and by letting l → ∞, we obtain that EJQTDE, 2013 No. 43, p. 10 In view of (3.3) we have The pointwise convergence of u n to u, the bounded character of h m and T k make it possible to conclude that h where (n) tends to 0 as n tends to +∞.Moreover, by using Lebesgue's theorem, we get φ(u n )h m (u n − v 0 ) converges to φ(u)h m (u − v 0 ) strongly in L p (x) (Ω), and since ∇T k (u n ) converges to ∇T k (u) weakly in L p(x) (Ω), we can deduce that Similarly we have On the other hand, remark that Moreover, by using Lebesgue's theorem, we have and since ∇u n → ∇u weakly in (L p(x) (Ω)) N , we have Similarly, we can write where θs (t) = t 0 θ s (z)dz and θ s (z) = φ(z)χ sl (z) with We use the L 1 (Ω) strong convergence of f n and since F ∈ L p (x) (Ω), we have by using Lebesgue's theorem, as first n and then l tends to infinity which implies by Hölder's inequality that On the other hand where Furthermore, by Lemma 3.6 we have and Splitting the first integral on the left hand side of (3.42) where It is easy to see that, the last term of the last inequality tends to zero as n → +∞, which implies that (3.48) Combining (3.46) and (3.48), we obtain By passing to the lim-sup over n and letting m tend to infinity, we obtain lim sup Step 4: Passing to the limit in (P n ) In order to pass to the limit in approximate equation, we now show that In particular, it is enough to prove the equi-integrability of the sequence (H n (x, u n , ∇u n )) n .To this purpose, we take T l+1 (u n ) − T l (u n ) as test function in P n we obtain Let ε > 0 be fixed.Then there exists l(ε) ≥ 1, such that For any measurable subset E ⊂ Ω, we have In view of (3.49), there exists η(ε) > 0, such that for all E such that meas(E) < η(ε).
Finally, by combining (3.51) and (3.52) we have then, we deduce that (H n (x, u n , ∇u n )) n are uniformly equi-integrable in Ω.Let v ∈ K ψ ∩ L ∞ (Ω), we take T l (u n − T k (u n − v)) as test function in (P n ) and letting l to ∞, we can write, for n large enough We get By Fatou's lemma and by the fact that a(x, T k+||v||∞ (u n ), ∇T k+||v||∞ (u n )) converges weakly in (L p (x) (Ω)) N to a(x, T k+||v||∞ (u), ∇T k+||v||∞ (u)), it is easy to see that On the other hand, since F ∈ (L p (x) (Ω)) N , we deduce that the integral Similarly, we have  are bounded, we have where p − = inf x∈Ω p(x) and p + = sup x∈Ω p(x).For p ∈ C + (Ω) , we define the variable exponent Lebesgue space by: L

Lemma 3 . 6
Assume that (3.1)-(3.10)hold true.Let u n be a solution of the approximate problem (P n ).