Electronic Journal of Qualitative Theory of Differential Equations

In this paper we study the existence of solutions for the generated boundary value problem, with initial datum being an element of L 1 () + W 1,p 0 (,w � ) diva(x,u,ru) + g(x,u,ru) = f divF


Introduction .
Let Ω be a bounded subset of IR N (N ≥ 2), 1 < p < ∞, and w = {w i (x); i = 0, ..., N } , be a collection of weight functions on Ω i.e., each w i is a measurable and strictly positive function everywhere on Ω and satisfying some integrability conditions (see section 2).Let us consider the non-linear elliptic partial differential operator of order 2 given in divergence form It is well known that equation Au = h is solvable by Drabek, Kufner and Mustonen in [7] in the case where h ∈ W −1,p (Ω, w * ).
In this paper we investigate the problem of existence solutions of the following Dirichlet problem Au + g(x, u, ∇u) = µ in Ω. (1.2) where µ ∈ L 1 (Ω) In this context of nonlinear operators, if µ belongs to W −1,p (Ω, w * ) existence results for problem (1.2) have been proved in [2], where the authors have used the approach based on the strong convergence of the positive part u + ε (resp.ngative part u − ε ).The case where µ ∈ L 1 (Ω) is investigated in [3] under the following coercivity condition, Let us recall that the results given in [2,3] have been proved under some additional conditions on the weight function σ and the parameter q introduced in Hardy inequality.
The main point in our study to prove an existence result for some class of problem of the kind (1.2), without assuming the coercivity condition (1.3).Moreover, we didn't supose any restriction for weight function σ and parameter q.
It would be interesting at this stage to refer the reader to our previous work [1].For different appproach used in the setting of Orlicz Sobolev space the reader can refer to [4], and for same results in the L p case, to [10].
The plan of this is as follows : in the next section we will give 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.
Let Ω be a bounded open subset of IR N (N ≥ 2).Let 1 < p < ∞, and let w = {w i (x); 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 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 weighted Sobolev 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, C ∞ 0 (Ω) is dense in W 1,p 0 (Ω, w) and (X, .1,p,w ) is a reflexive Banach space.We recall that the dual of the weighted Sobolev spaces W 1,p 0 (Ω, w) is equivalent to W −1,p (Ω, w * ), where w * = {w * i = w 1−p i }, i = 1, ..., N and p is the conjugate of p i.e. p = p p−1 .For more details we refer the reader to [8].We introduce the functional spaces, we will need later.For p ∈ (1, ∞), T 1,p 0 (Ω, w) is defined as the set of measurable functions u : Ω −→ IR such that for k > 0 the truncated functions T k (u) ∈ W 1,p 0 (Ω, w).We give the following lemma which is a generalization of Lemma 2.1 [5] in weighted Sobolev spaces.
We will define the gradient of u as the function v, and we will denote it by v = ∇u.Lemma 2.2.Let λ ∈ IR and let u and v be two functions which are finite almost everywhere, and which belongs to T 1,p 0 (Ω, w).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 [6] for the non weighted case.
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 the Hardy inequality EJQTDE, 2006 No. 19, p. 3 holds for every u ∈ X with a constant C > 0 independent of u.Moreover, the imbeding determined by the inequality (2.4) is compact.We state the following technical lemmas which are needed later.

Main results .
Let Ω be a bounded open subset of IR N (N ≥ 2).Consider the second order operator A : W 1,p 0 (Ω, w) −→ W −1,p (Ω, w * ) in divergence form where a : Ω×IR×IR N → IR N is a Carathéodory function Satisfying the following assumptions: (H 2 ) For i = 1, ..., N where b : IR + → IR + is a positive increasing function and c(x) is a positive function which belong to L 1 (Ω).Furthermore we suppose that Consider the nonlinear problem with Dirichlet boundary condition We shall prove the following existence theorem Theorem 3.1.Assume that (H 1 ) − (H 3 ) hold true.Then there exists at least one solution of the problèm (P ).

Proof of main results
.
In order to prove the existence theorem we need the following Lemma 4.1 [2].Assume that (H 1 ) and (H 2 ) are satisfied, and let (u n ) n be a sequence in ) We give now the proof of theorem 3.1.STEP 1.The approximate problem.Let f n be a sequence of smooth functions which strongly converges to f in L 1 (Ω).We Consider the sequence of approximate problems: where g n (x, s, ξ) = g(x,s,ξ) ).Note that g n (x, s, ξ) satisfises the following conditions We define the operator G n : X −→ X * by, Thanks to Hölder's inequality, we have for all u ∈ X and v ∈ X, the last inequality is due to (2.3) and (2.4).
Lemma 4.2.The operator Moreover, B n is coercive, in the following sense: This Lemma will be proved below.
In view of Lemma 4.2, there exists at least one solution u n of (4.1) (cf.Theorem 2.1 and Remark 2.1 in Chapter 2 of [11] ). STEP 2. A priori estimates.
and by using in fact that Thank's to Young's inequality and (3.3), one easily has aloways assume that the convergence is a.e. after passing to a suitable subsequence).To prove this, we show that u n is a Cauchy sequence in measure in any ball B R .Let k > 0 large enough, we have Moreover, we have, for every δ > 0, strongly in L q (Ω, σ) and a.e. in Ω.
Consequently, we can assume that T k (u n ) is a Cauchy sequence in measure in Ω.Let ε > 0, then, by (4.4) and (4.5), there exists some k(ε) > 0 such that meas({|u n − u m | > δ} ∩ B R ) < ε for all n, m ≥ n 0 (k(ε), δ, R).This proves that (u n ) n is a Cauchy sequence in measure in B R , thus converges almost everywhere to some measurable function u.Then strongly in L q (Ω, σ) and a.e. in Ω.

STEP 4. Strong convergence of truncations.
We fix k > 0, and let h > k > 0. We shall use in (4.1) the test function ) It follows that Denote 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 will deal with each term of (4.9).First of all, observe that and Splitting the first integral on the left hand side of (4.9) where Setting m = 4k + h, using a(x, s, ξ)ξ ≥ 0 and the fact that ∇w n = 0 on the set where ) and since a(x, s, 0) = 0 ∀s ∈ IR, we have   The second term of the right hand side of the last inequality tends to 0 as n tends to infinity.Indeed.Since the sequence (a(x, On the other hand, the term of the right hand side of (4.16) reads as ) by using the continuity of the Nymetskii operator, while ∂(T k (un)) (4.18) In the same way, we have The second term of the left hand side of (4.9), can be estimated as On the other side, we have As above, by letting n go to infinity, we can easily see that each one of last two integrals of the right-hand side of the last equality is of the form ε 8 h (n) and then which and (4.7) implies that in which, we can pass to the limit as n → +∞ to obtain It remains to show, for our purposes, that the all terms on the right hand side of (4.25) 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 with this term.Let us observe that, if we take φ(T 2k (u n − T h (u n ))) as test function in (4.1) and use (3.3), we obtain and thanks to the sign condition (3.4), we get Using the Young inequality we have so that, since φ ≥ 1, we have again because the norm is lower semi-continuity, we get Consequently, in view of (4.26) and (4.27), we obtain Finally, the strong convergence in L 1 (Ω) of f n , we have, as first n and then h tend to infinity, lim sup Therefore by (4.25), letting h go to infinity, we conclude, which and using Lemma 4.1 implies that STEP 5. Passing to the limit.By using T k (u n − v) as test function in (4.1), with v ∈ W 1,p 0 (Ω, w) ∩ L ∞ (Ω), we get (4.29)By Fatou's lemma and the fact that weakly in For the second term of the right hand side of (4.29), we have On the other hand, we have To conclude the proof of theorem, it only remains to prove g n (x, u n , ∇u n ) → g(x, u, ∇u) strongly in L 1 (Ω), (4.33) in particular it is enough to prove the equiintegrable of g n (x, u n , ∇u n ).To this purpose, we take T l+1 (u n ) − T l (u n ) as test function in (4.1), we obtain Let ε > 0. Then there exists l(ε) ≥ 1 such that For any measurable subset E ⊂ Ω, we have In view of (4.28) there exists η(ε) > 0 such that for all E such that meas E < η(ε).which shows that g n (x, u n , ∇u n ) are uniformly equintegrable in Ω as required.
Thanks to (4.30)-(4.33)we can pass to the limit in (4.29) and we obtain that u is a solution of the problem (P ).This completes the proof of Theorem 3.1.
Remark 4.1.Note that, we obtain the existence result withowt assuming the coercivity condition.However one can overcome this difficulty by introduced the function We will prove that similarly, it is easy to see that (g n (x, u k , ∇u k )) k is bounded in L q (Ω, σ 1−q ), then there exists a function k n ∈ L q (Ω, σ 1−q ) such that It is clear that, for all w ∈ W On the one hand, we have On the other hand, by (4.40) and the fact that a(x, u k , ∇u) → a(x, u, ∇u) strongly in and so, thanks to Lemma 4.1 ∇u n → ∇u a.e. in Ω.
We deduce then that a(x, u k , ∇u k ) → a(x, u, ∇u) weakly in and g n (x, u k , ∇u k ) → g(x, u, ∇u) weakly in L q (Ω, w 1−q i ).
Thus implies that χ = B n u.
Corollary 4.1.Let 1 < p < ∞.Assume that the hypothesis (H 1 ) − (H 3 ) holds, let f n be any sequence of functions in L 1 (Ω) which converge to f weakly in L 1 (Ω) and let u n the solution of the following problem Then there exists a subsequence of u n still denoted u n such that u n converges to u almost everywhere and T k (u n ) T k (u) strongly in W 1,p 0 (Ω, w), further u is a solution of the problem (P ) (with F = 0).EJQTDE, 2006 No. 19, p. 15 Proof.We give a brief proof.
Step 1.A priori estimates.As before we take v = 0 as test function in (P n ), we get Step 3. Passing to the limit.This step is similar to the step 5 of the proof of Theorem 3.1, by using the Egorov's theorem in the last term of (P n ).
Remark 4.2.In the case where F = 0, if we suppose that the second member is nonnegative, then we obtain a nonnegative solution.
Indeed.If we take v = T h (u + ) in (P ), we have On the other hand, thanks to (3.3), we conclude Letting h tend to infinity, we can easily deduce T k (u − ) = 0, ∀k > 0, which implies that u ≥ 0.

( 4 .
35)Finally, by combining (4.34) and (4.35) one easily hasE |g n (x, u n , ∇u n )| dx < ε for all E such that meas E < η(ε), .41) Hence, by the same method used in the first step in the proof of Theorem 3.1 there exists a function u ∈ T 1,p 0 (Ω, w) and a subsequence still denoted by u n such thatu n → u a.e. in Ω, T k (u n ) T k (u) weakly in W 1,p 0 (Ω, w), ∀k > 0.Step 2. Strong convergence of truncation.The choice of v = T h (u n − φ(w n )) as test function in (P n ), we get, for all l > 0 Ω a(x, u n , ∇u n )∇T l (u n − T h (u n − φ(w n ))) dx + Ω g(x, u n , ∇u n )T l (u n − T h (u n − φ(w n )) dx ≤ Ω f n T l (u n − T h (u n − φ(w n ))) dx.Which implies that {|un−φ(wn)|≤h} a(x, u n , ∇u n )∇T l (φ(w n )) + Ω g(x, u n , ∇u n )T l (u n − T h (u n − φ(w n ))) dx ≤ Ω f n T l (u n − T h (u n − φ(w n ))) dx.Letting h tend to infinity and choosing l large enough, we deduceΩ a(x, u n , ∇u n )∇φ(w n ) dx + 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 4 of the proof of Theorem 3.1.

* and lim sup n→∞ Bu n , u n ≤ χ, u , we have Bu n = Bu and Bu
n , u n → χ, u as n → ∞.
Definition 2.1.Let Y be a reflexive Banach space, a bounded operator B from Y to its dual Y * is called pseudo-monotone if for any sequence u n ∈ Y with u n u weakly in Y .Bu n χ weakly in Y Since φ(w n )g n (x, u n , ∇u n ) > 0 on the subset {x ∈ Ω, |u n (x)| > k}, we deduce from (4.8) that .8) EJQTDE, 2006 No. 19, p. 7 Ω a(x, u n , ∇u n in the test function(4.6).