Existence of solutions of degenerated unilateral problems with L1 data

In this paper, we shall be concerned with the existence result of the Degenerated unilateral problem associated to the equation of the type Au + g(x, u,∇u) = f − divF, where A is a Leray-Lions operator and g is a Carathéodory function having natural growth with respect to |∇u| and satisfying the sign condition. The second term is such that, f ∈ L1(Ω) and F ∈ Πi=1L p(Ω, w1−p ′ i ).


Introduction
Let Ω be a bounded open set of R N , p be a real number such that 1 < p < ∞ and w = {w i (x), 0 ≤ i ≤ N } be a vector of weight functions on Ω, i.e. each w i (x) is a measurable a.e. strictly positive function on Ω, satisfying some integrability conditions (see section 2). Now we consider the obstacle problem associated to the following differential equations Au + g(x, u, ∇u) = f − divF. (1.1) Where A is a Leray-Lions operator from W 1,p 0 (Ω, w) into its dual W −1,p (Ω, w * ) defined by Au = −diva(x, u, ∇u) and where g is a nonlinear lower order term having natural growth (order p) with respect to |∇u|, with respect to |u|, we do not assume any growth restrictions, but we assume only the "sign-condition" g(x, s, ξ)s ≥ 0.
As regards the second member, we suppose that f ∈ L 1 (Ω) and that In the case where F = 0, an existence theorem has been proved in [2] with f ∈ W −1,p (Ω, w), and in [3] with f ∈ L 1 (Ω) and where the nonlinearity g satisfies further the following coercivity condition w i |ξ i | p for |s| > γ. (1.2) Our purpose, in this paper, is to prove an existence result for degenerated unilateral problems associated to (1.1) in the case where F = 0 and without assuming the coercivity condition (1.2). So that, we generalize The previous results given in [3]. Let us point out that another work in the L p case can be found in [6,12] in the case of equation, and in [11] in the case of obstacle problems. This paper is organized as follows, sections 2 containe some preliminaries and some technical lemmas, section 3 is concerned with main results and basic assumptions, in section 4, we prove main results and we study the stability and the positivity of solution.

Preliminaries
Let Ω be a bounded open subset of R N (N ≥ 1). Let 1 < p < ∞, and let w = {w i (x); i = 1, ..., N } , 0 ≤ i ≤ N be a vector of weight functions i.e. every component w i (x) is a measurable function which is strictly positive a.e. in Ω. Further, we suppose in all our considerations that for 0 ≤ i ≤ N w i ∈ L 1 loc (Ω) and w − 1 We define the weighted space with weight γ in Ω as which is endowed with, we define the norm We denote by W 1,p (Ω, w) the space of all real-valued functions u ∈ L p (Ω, w 0 ) such that the derivatives in the sense of distributions satisfy This set of functions forms a Banach space under the norm To deal with the Dirichlet problem, we use the space defined as the closure of C ∞ 0 (Ω) with respect to the norm (2.2). Note that, and (X, . 1,p,w ) is a reflexive Banach space. We recall that the dual space of the weighted Sobolev spaces .., N and p is the conjugate of p i.e. p = p p−1 . For more details we refer the reader to [10]. We now introduce the functional spaces we will need later. For p ∈ (1, ∞), τ 1,p 0 (Ω, w) is defined as the set of measurable functions u : Ω → R such that for k > 0 the truncated functions T k (u) ∈ W 1,p 0 (Ω, w). We gives the following lemma this is a generalization of Lemma 2.1 [4] in weighted spaces. Lemma 2.1: For every u ∈ τ 1,p 0 (Ω, w), there exists a unique measurable function v : Ω → R such that

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,p 0 (Ω, w) 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 [9] for the non weighted case. Now, we state the following assumptions.
is a norm defined on X and is equivalent to the norm (2.2). (Note that (X, u X ) is a uniformly convex (and reflexive) Banach space.
-There exist a weight function σ on Ω and a parameter q, 1 < q < ∞, such that and with q = q q−1 and such that the Hardy inequality holds for every u ∈ X with a constant C > 0 independent of u. Moreover, the imbeding X → L q (Ω, σ) (2.7) determined by the inequality (2.6) is compact. Now, we state the following technical lemmas which are needed later.

Lemma 2.4:[1]: Assume that
The previous lemma, we deduce the following.

Main results
Let A be a nonlinear operator from where a : Ω × R × R N → R N is a Carathéodory function satisfying the following assumptions: where k(x) is a positive function in L p (Ω) and α is a positive constants.
where b : R + → R + is a nonnegative increasing function and c(x) is a positive function which in L 1 (Ω).
where ψ : Ω → R is a measurable function on Ω such that Finally, we assume that and We defined, for s and k in R, k ≥ 0, T k (s) = max(−k, min(k, s)).
For the nonlinear Dirichlet boundary value problem (1.1), we state our main result as follows.

Remarks 3.2:
1. We obtain the same results of our theorem if we suppose that the signe condition (3.4) is only near infinity.
2. The statement of Theorem 3.1 generalizes in weighted case the analogous one in [12] and [11].

Proof of main results
We recall the following lemma wich play an important rôle in the proof of our main result,

Lemma 4.1:[1]
Assume that (H 1 ) and (H 2 ) are satisfied, and let (u n ) be a sequence in W 1,p 0 (Ω, w) such that u n u weakly in W 1,p 0 (Ω, w) and Proof of Theorem 3.1 For the prove of the existence theorem we proceed by steps.
Step 1. A priori estimates Let us defined the following sequence of the unilaterals problems where f n and F n are a regular functions such that f n strongly converges to f in L 1 (Ω) and F n strongly converges to F in Π N i=1 L p (Ω, w 1−p i ). By Theorem 3.1 of [2], there exists at least one solution of (4.1).
The choice ofw as a test function in (4.1) and letting h → +∞, we obtain by Young's inequality and (3.3), one easily has Now, as in [5], we prove that u n converges to some function u locally in measure (and therefore, we can aloways 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 in any ball B R . Let k > 0 large enough, we have We have, for every δ > 0, σ) and a.e. in Ω.
Consequently, we can assume that T k (u n ) is a Cauchy sequence in measure in Ω.
Which implies, by using (3.1), for all k > 0 there exists a function Step 2. Strong convergence of truncation Let k > 0 large enough such that k > ψ + ∞ , we consider the function φ(t) = te γt 2 , with γ > ( b(k) 2α ) 2 (this function is introduce by [7,8]). Thanks to Lemma 1 of [8], we have the following inequality hold for all s ∈ R.
Here, we define w n = T 2k (u n − T h (u n ) + T k (u n ) − T k (u)) where h > 2k > 0. For η = exp(−4γk 2 ), we define the following function as v n,h = u n − ηφ(w n ). (4.7) The use of v n,h as test function in (P n ), we obtain, for all l > 0 we take also l large enough, we have (4.8) 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 (4.8), (4.10) Splitting the first integral on the left hand side of (4.10) where |u n | ≤ k and |u n | > k, we can write, by using (3.3): where C k = φ (2k). Since, when n tends to infinity, we have for all i = 1, ..., N, ∂(T k (u)) ∂x i χ {|un|>k} tends to 0 strongly in L p (Ω, w i ) while, (a i (x, T M (u n ), ∇T M (u n ))) n is bounded in L p (Ω, w 1−p i ) hence the last term in the previous inequality tends to zero for every h fixed as n tends to infinity.
weakly in L p (Ω, w i ), the second term of the right hand side of (4.12) tends to 0 as n → ∞. So that (4.11) yields (4.13) For the second term of the left hand side of (4.10), we can estimate as follows α Ω a(x, T k (u n ), ∇T k (u n ))∇T k (u n )|φ(w n )| dx, (4.14) remark that, we have (4.15) By the Lebesgue's Theorem, we have Moreover, in view of (4.5) the second term of the right hand side of (4.15) tends to The third term of the right hand side of (4.15) tends to 0 since for all i = 1, ..., N a i (x, T k (u n ), ∇T k (u))|φ(w n )| → a i (x, T k (u), ∇T k (u))|φ(T 2k (u−T h (u)))| strongly in L p (Ω, From (4.14) and (4.15), we obtain (4.16) Now, by the strongly convergence of f n and F n and in fact that w n T 2k (u − T h (u)) weakly in W 1,p 0 (Ω, w) and weakly- * in L ∞ (Ω), (4.17) moreover, combining (4.13) and (4.16), we conclude that which and using (4.6), implies that Hence, passing to the limit over n, we get (4.18) It remains to show, for our purposes, that the all term on the right hand side of (4.18) converge to zero as h goes to infinity. The only difficulty that exists is in the last term. For the other terms it suffices to apply Lebesgue's theorem.
We deal now with this term. Let us observe that, if we take u n − ηφ(T 2k (u n − T h (u n ))) as test function in (P n ), we obtain by using (3.3): which yields, thanks to Young's inequalities consequently, thanks to the strong convergence of |w −1 p F n | p and f n , in L 1 (Ω) and letting firstly n → ∞ after h tend to infinity, we obtain Therefore by (4.18), letting h go to infinity, we conclude, which implies that by using Lemma 4.1 Step 3. Passing to the limit We take v ∈ K ψ ∩ L ∞ (Ω) as test function in (P n ), we can write (4.20) By Fatou's lemma and in fact that (4.21) On the other hand, by using the strong convergence of F n and we deduce that the integral Now, we need to prove that g(x, u n , ∇u n ) → g(x, u, ∇u) strongly in L 1 (Ω), (4.22) in particular it is enough to prove the equiintegrable of g(x, u n , ∇u 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. Then there exists l(ε) ≥ 1 such that For any measurable subset E ⊂ Ω, we have In view of (4.19), there exists η(ε) > 0 such that for all E such that |E| < η(ε). Finally, by combining (4.23) and (4.24) one easily has E |g(x, u n , ∇u n )| dx < ε for all E such that |E| < η(ε), which allows us, by using (4.21) and (4.22), we can pass to the limit in (4.20). This completes the proof of Theorem 3.1.

Remark 4.2:
Note that, we obtain the existence result withowt assuming the coercivity condition. However one can overcome this difficulty by introduced the function w n = T 2k (u n − T h (u n ) + T k (u n ) − T k (u)) in the test function (4.7). a(x, u n , ∇u n )∇T l (ηφ(w n )) + Ω g(x, u n , ∇u n )T l (u n − T h (u n − ηφ(w n ))) dx ≤ Ω f n T l (u n − T h (u n − ηφ(w n ))) dx.
Letting h tends to infinity and choosing l large enough, we deduce Ω a(x, u n , ∇u n )∇φ(w n ) + Ω g(x, u n , ∇u n )φ(w n ) dx ≤ Ω f n φ(w n ) dx, the rest of the proof of this step is the same as in step 2 of the proof of Theorem 3.1.
Step 3. Passing to the limit This step is similarly to the step 3 of the proof of Theorem 3.1, by using the Egorov's theorem in the last term of (P n ).

Remark 4.4:
In the case where F = 0, if we suppose that the second mumber are nonnegative, then we obtain a nonnegative solution.