Existence of Entropy Solutions of the Anisotropic Elliptic Nonlinear Problem with Measure Data in Weighted Sobolev Space

i=1 ∂i ai(x, u, ∇u) is a Leray-Lions anisotropic operator acting from W 1, − p 0 (Ω, − ω ) into its dual W −1, − p ′ (Ω, − ω ) and φi ∈ C0(R,R).

In this study we are using the entropy solutions who was introduced for the first time by P. Benilan et al [7], because the function φ i does not belongs to L 1 loc (Ω) in general, then the problem (1.1) does not admit weak solution. In the case of a datum in µ ∈ L 1 (Ω) + W −1, − → p ′ (Ω) the existence of entropy solutions is treated by A. Salmani, Y. Akdim and H. Redwane in [16]. Moreover, L. Boccardo, T. Gallouet and L. Orsina (see [11]) have considered the case φ ≡ 0. The objective of our paper is to study the anisotropic unilateral nonlinear elliptic problem associated with the nonlinear problem (1.1). More precisely, we establish only the existence of entropy solutions for the following unilateral anisotropic problem, (1.2) in the convex class K ψ := u ∈ W 1, − → p 0 (Ω, − → ω (x)), u ≥ ψ a.e in Ω , where ψ is a measurable function on Ω such that This type of problem has been studied by many authors in recent years, in particular by Y. Akdim, C. Allalou and A. Salmani (see. [4]) have demonstrated the existence of entropy solutions problem like (1.1). In the non weighted case ω i ≡ 1 for any i ∈ {1, ..., N }, Boccardo et al. in [10] studied the existence of weak solutions for nonlinear elliptic problem (1.1) φ i (u) = 0 for i = 1, · · · , N and the right-hand side is a bounded Radon measure on Ω. In addition this, we mention some works in this direction such as [5,8,12,16,6,18]. The uniqueness problem being a rather delicate one, due to acounterexample by J. Serrin (see [17], and [15] for further remarks). One of the motivations for studying (1.1) comes from applications the mathematical modeling of physical and mechanical processes in anisotropic continuous medium. Let us briefly summarize the contents of the paper : after a section devoted to developing the necessary functional setting as Lebesgue space with weighted and the weighted anisotropic Sobolev space, we introduce some useful technical lemmas to prove existence results and basic assumptions of our problem in section 3. In the final section we state the main result and proofs.

Mathematical preliminaries
Throughout this paper Ω is a bounded open subset of R N (N ≥ 2) with smooth boundary. Almost everywhere positive and locally integrable function ω : Ω −→ R will be called a weight. We shall denote by L p (Ω, ω) the set of all measurable functions u on Ω such that the norm Let p 1 , . . . , p N be N real numbers, we define the following vectors − → p = {p 1 , . . . , p N } be a vector of exponent and − → ω = {ω 1 , . . . , ω N } be a vector of weight functions, i.e., every component ω i is a measurable function which is positive a.e. in Ω. Moreover, we assume in all our considerations that for any i = 1, . . . , N, we denote is a Banach space with respect to norm (see [13]) We define the functional space . We pass to limit in

Notion of solutions and main results
In this section we formulate and prove the main result of the paper. Now, we give a definition of entropy solutions for our unilateral elliptic problem (1.1).

Definition 4.1. A measurable function u is said to be an entropy solution for the obstacle problem
for all ϕ ∈ K ψ ∩ L ∞ (Ω). Proof. The proof of Theorem 4.2 will be divided into several steps.
Existence of Entropy Solutions the Anisotropic...

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Step 1: Approximate problems. We consider the following approximate problems where f n = T n (f ) and φ n i (s) = φ i (T n (s)) . We define the operators Φ n of K ψ to W −1, − For the proof of Lemma 4.3, (see "Appendix"). Proof. Thanks to Lemma 4.3 and Theorem 8.2 chapiter 2 in [14], there exists at least one solution to the problem (4.2).
(Ω, − → ω ) and for all η small enough, we get v ∈ K ψ . We choose v as test function in problem (4.2), we have By Young's inequalities, we get for a positive constant λ Existence of Entropy Solutions the Anisotropic...
Proposition 4.6. Let u n be a solution of approximate problem (4.2). Then there exists a measurable function u and a subsequence of u n such that Proof. Using to Proposition 4.5, we obtain Firstly, we will prove that (u n ) n is a Cauchy sequence in measure in Ω. For all λ > 0, we obtain By Hölder's inequality, Lemma 2.1 and (4.7), we have Since the sequence (T k (u n )) n is bounded in (Ω, − → ω ) and strongly in L p − (Ω) as n tends to +∞. Then the sequence (T k (u n )) n is a Cauchy sequence in measure in Ω, thus for all λ > 0, there exists n 0 such that Using (4.8), (4.9) and (4.10), then ∀ λ, ǫ > 0 we have Which implies that (u n ) n is a Cauchy sequence in measure in Ω, then there exists a subsequence denoted by (u n ) n such that u n converges to a measurable function u a.e. in Ω and (Ω, − → ω ) and a.e. in Ω ∀k > 0. (4.11) Secondly, we will show that Existence of Entropy Solutions the Anisotropic...

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Then, Then using the Green's formula, we obtain By the Young's inequality, we get Which implies that According to Lebesgue's theorem, we get Then, we obtain Similarly, we take v = u n − ηT 1 (u n − T m (u n )) + as test function in approximate problem (4.2), we have We define the following function of one real variable: Combining (4.14) and (4.15), we have the second integral in (4.16) converges to zero as n and m tend to +∞.
) as n goes to +∞, then the third integral in (4.16) converges to zero as n and m tend to +∞. We set Φ i (u n ) = Then by Lemma 3.3, we obtain the fourth integral in (4.16) converges to zero as n and m tend to +∞. Using Lebesgue's theorem, we have the first integral on the right hand in (4.16) converges to zero as n and m tend to +∞. Moreover, since weakly in L pi (Ω, ω i ) we get the second integral on the right hand in (4.16) converges to zero as n and m tend to +∞. By Young's inequality, we have Similarly, we have Using (3.1), (4.14), (4.15) and Lebesgue's theorem, we obtain and Combining (4.17)-(4.20), we have the third integral on the right hand in (4.16) converges to zero as n and m tend to +∞. We conclude Similarly, we consider ϕ = u n + (T k (u n ) − T k (u)) − h m (u n ) as test function in approximate problem (4.2), we have

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Let ϕ = u n + T k (u n ) − (1 − h m (u n )) as test function in approximate problem (1.1), we have (4.27) According to (4.14) and (4.15), we have Then the second integral in (4.27) converges to zero as n and m goes to +∞. Since Thanks to Lebesgue's theorem, we obtain Hence the third integral in (4.27) converges to zero as n and m tends to +∞. We set Φ n Then the fourth integral in (4.27) converges to zero as n and m tend to +∞. Using to Lebesgue's theorem, we get the integral on the right hand in (4.27) converges to zero as n and m goes to +∞.
as m tends to +∞ and ∂ i T k (u) − ∈ L pi (Ω, ω i ) thus the second integral on the right hand in (4.27) converges to zero as n and m tend to +∞. Using

Young's Inequality and (3.1), we get
In sight to Lebesgue's theorem and (4.14), we obtain the third integral on the right hand in (4.27) converges to zero as n and m tend to +∞, we conclude (4.29) thanks to (4.14) and (4.15), we have the second integral in the left hand in (4.29) converges to zero as n and m tend to +∞. For the third integral in the left hand in (4.29), we obtain thanks to Lebesgue's theorem and (3.4), we get Then the fourth integral in the left hand in (4.29) converges to zero as n and m tend to +∞. In addition, by the Lebesgue's theorem, we obtain the first integral in the right hand in (4.29) converges to zero as n and m tend to +∞. Using the Young's inequality, we have By Lebesgue's theorem, we have Thanks to Lebesgue's theorem again, we have Thus the second integral in the right hand in (4.29) converges to zero as n and m tend to +∞. Furthermore, by Young's inequality, we obtain According to (3.1) and (4.15), Lebesgue's theorem, the third integral in the right hand in (4.29) converges to zero as n and m tend to +∞.
Hence, the first integral in the right hand in (4.29) satisfies that 1 (n, m), Using (3.1), Young's inequality and Lebesgue's theorem, we have m).
Using (4.28) and (4.30), we have Combining (4.25) and (4.31), the first and the second integrals on the right hand side converge to zero as n and m tend to +∞.
and ∂ i T k (u)(1 − h m (u)) converge to zero in L pi (Ω, ω i ) as n and m tend to +∞. Then, the third integral on the right hand side converge to zero as n and m tend to +∞. Also, since a i (x, T k (u n ), ∇T k (u n ))(1−h m (u)) converges to a i (x, T k (u), ∇T k (u))(1−h m (u)) strongly in L p ′ i (Ω, ω * i ) and ∂ i T k (u n ) ⇀ ∂ i T k (u) weakly in L pi (Ω, ω i ) we obtain the fourth integral on the right hand side converge to zero as n and m tend to +∞. Hence, we get (4.12). By (4.11), (4.12) and Lemma 3.2, we have (Ω, − → ω ) and a.e. in Ω ∀k > 0.

Appendix
Proof of lemma 4.3 Firstly, we will prove that the operator B n is pseudo-monotone. Let (u k ) k be a sequence in We will show that χ = B n u and < B n u k , u k >−→< χ, u > as k → +∞. Since W 1, − → p 0 (Ω, − → ω ) ֒→֒→ L p − (Ω) , then u k → u strongly in L p − (Ω) and a.e. in Ω for a subsequence denoted again (u k ) k .

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Using to Jensen's inequality, we obtain are bounded, then we get We conclude that the operator B n = A + Φ n is coercive.