Quasilinear degenerated equations with L 1 datum and without coercivity in perturbation terms

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1 Electronic Journal of Qualitative Theory of Differential Equations 2006, No. 19, 1-18; htt:// Quasilinear degenerated equations with L 1 datum and without coercivity in erturbation terms L. AHAROUCH, E. AZROUL and A. BENKIRANE Déartement de Mathématiques et Informatique Faculté des Sciences Dhar-Mahraz B.P 1796 Atlas Fès, Morocco Abstract In this aer we study the existence of solutions for the generated boundary value roblem, with initial datum being an element of L 1 () + W 1, (,w ) diva(x,u, u) + g(x,u, u) = f divf where a(.) is a Carathéodory function satisfying the classical condition of tye Leray-Lions hyothesis, while g(x, s, ξ) is a non-linear term which has a growth condition with resect to ξ and no growth with resect to s, but it satisfies a sign condition on s. 1. Introduction Let be a bounded subset of IR N (N 2), 1 < <, and w = {w i (x); i = 0,...,N}, be a collection of weight functions on i.e., each w i is a measurable and strictly ositive function everywhere on and satisfying some integrability conditions (see section 2). Let us consider the non-linear ellitic artial differential oerator of order 2 given in divergence form Au = div(a(x, u, u)) (1.1) It is well known that equation Au = h is solvable by Drabek, Kufner and Mustonen in [7] in the case where h W 1, (,w ). In this aer we investigate the roblem of existence solutions of the following Dirichlet roblem Au + g(x,u, u) = µ in. (1.2) AMS Subject Classification: 35J15, 35J20, 35J70. Key words and hrases: Weighted Sobolev saces, boundary value roblems, truncations, unilateral roblems EJQTDE, 2006 No. 19,. 1

2 where µ L 1 () + L (,w 1 i ). In this context of nonlinear oerators, if µ belongs to W 1, (,w ) existence results for roblem (1.2) have been roved in [2], where the authors have used the aroach based on the strong convergence of the ositive art u + ε (res. ngative art u ε ). The case where µ L 1 () is investigated in [3] under the following coercivity condition, g(x,s,ξ) β w i ξ i for s γ, (1.3) Let us recall that the results given in [2, 3] have been roved under some additional conditions on the weight function σ and the arameter q introduced in Hardy inequality. The main oint in our study to rove an existence result for some class of roblem of the kind (1.2), without assuming the coercivity condition (1.3). Moreover, we didn t suose any restriction for weight function σ and arameter q. It would be interesting at this stage to refer the reader to our revious work [1]. For different aroach used in the setting of Orlicz Sobolev sace the reader can refer to [4], and for same results in the L case, to [10]. The lan of this is as follows : in the next section we will give some reliminaries and some technical lemmas, section 3 is concerned with main results and basic assumtions, in section 4 we rove main results and we study the stability and the ositivity of solution. 2. Preliminaries Let be a bounded oen subset of IR N (N 2). Let 1 < <, and let w = {w i (x);0 i N}, be a vector of weight functions i.e. every comonent w i (x) is a measurable function which is strictly ositive a.e. in. Further, we suose in all our considerations that for 0 i N We define the weighted sace with weight γ in as which is endowed with, we define the norm w i L 1 loc() and w 1 1 i L 1 loc(). (2.1) L (,γ) = {u(x) : uγ 1 L 1 ()}, ( u,γ = )1 u(x) γ(x)dx. We denote by W 1, (,w) the weighted Sobolev sace of all real-valued functions u L (,w 0 ) such that the derivatives in the sense of distributions satisfy u L (,w i ) for all i = 1,...,N. This set of functions forms a Banach sace under the norm ( u 1,,w = u(x) w 0 dx + u ) 1 w i (x) dx. (2.2) EJQTDE, 2006 No. 19,. 2

3 To deal with the Dirichlet roblem, we use the sace X = W 1, 0 (, w) defined as the closure of C0 () with resect to the norm (2.2). Note that, C 0 () is dense in W 1, 0 (,w) and (X,. 1,,w ) is a reflexive Banach sace. We recall that the dual of the weighted Sobolev saces W 1, 0 (,w) is equivalent to W 1, (,w ), where w = {wi = w1 i }, i = 1,...,N and is the conjugate of i.e. = 1. For more details we refer the reader to [8]. We introduce the functional saces, we will need later. For (1, ), T 1, 0 (,w) is defined as the set of measurable functions u : IR such that for k > 0 the truncated functions T k (u) W 1, 0 (,w). We give the following lemma which is a generalization of Lemma 2.1 [5] in weighted Sobolev saces. Lemma 2.1. For every u T 1, 0 (, w), there exists a unique measurable function v : IR N such that T k (u) = vχ { u <k}, almost everywhere in, for every k > 0. 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, 0 (, w). Then, (u + λv) = u + λ v a.e. in, where u, v and (u + λv) are the gradients of u, v and u + λv introduced in Lemma 2.1. The roof of this lemma is similar to the roof of Lemma 2.12 [6] for the non weighted case. Definition 2.1. Let Y be a reflexive Banach sace, a bounded oerator B from Y to its dual Y is called seudo-monotone if for any sequence u n Y with u n u weakly in Y. Bu n χ weakly in Y and lim su Bu n,u n χ,u, we have n Bu n = Bu and Bu n,u n χ,u as n. Now, we state the following assumtions. (H 1 )-The exression ( N u X = u ) 1 w i (x) dx, (2.3) 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 sace. -There exist a weight function σ on and a arameter q, 1 < q <, such that the Hardy inequality ( ( )1 u q q N σ(x) dx C u ) 1 w i (x) dx, (2.4) EJQTDE, 2006 No. 19,. 3

4 holds for every u X with a constant C > 0 indeendent of u. Moreover, the imbeding determined by the inequality (2.4) is comact. We state the following technical lemmas which are needed later. X L q (,σ) (2.5) Lemma 2.3 [2]. Let g L r (,γ) and let g n L r (,γ), with g n,γ c,1 < r <. If g n (x) g(x) a.e. in, then g n g weakly in L r (,γ). Lemma 2.4 [2]. Assume that (H 1 ) holds. Let F : IR IR be unifomly Lischitzian, with F(0) = 0. Let u W 1, 0 (,w). Then F(u) W 1, 0 (,w). Moreover, if the set D of discontinuity oints of F is finite, then { F(u) F (u) u = a.e. in {x : u(x) / D} 0 a.e. in {x : u(x) D}. From the revious lemma, we deduce the following. Lemma 2.5 [2]. Assume that (H 1 ) holds. Let u W 1, 0 (,w), and let T k (u),k IR +, be the usual truncation, then T k (u) W 1, 0 (, w). Moreover, we have 3. Main results T k (u) u strongly in W 1, 0 (,w). Let be a bounded oen subset of IR N (N 2). Consider the second order oerator A : W 1, 0 (,w) W 1, (,w ) in divergence form Au = div(a(x,u, u)), where a : IR IR N IR N is a Carathéodory function Satisfying the following assumtions: (H 2 ) For i = 1,...,N 1 a i (x,s,ξ) βwi (x)[k(x) + σ 1 s q + 1 w j (x) ξ j 1 ], (3.1) j=1 [a(x,s,ξ) a(x,s,η)](ξ η) > 0 for all ξ η IR N, (3.2) a(x,s,ξ)ξ α w i (x) ξ i. (3.3) where k(x) is a ositive function in L () and α,β are ositive constants. Assume that g : IR IR N IR is a Carathéodory function satisfying : (H 3 ) g(x,s,ξ) is a Carathéodory function satisfying g(x,s,ξ).s 0, (3.4) EJQTDE, 2006 No. 19,. 4

5 g(x,s,ξ) b( s )( w i (x) ξ i + c(x)), (3.5) where b : IR + IR + is a ositive increasing function and c(x) is a ositive function which belong to L 1 (). Furthermore we suose that µ = f divf, f L 1 (), F L (,w 1 i ), (3.6) Consider the nonlinear roblem with Dirichlet boundary condition u T 1, 0 (,w), g(x,u, u) L 1 () a(x,u, u) T k (u v)dx + g(x,u, u)t k (u v)dx (P) ft k (u v)dx + F T k (u v)dx v W 1, 0 (,w) L () k > 0. We shall rove the following existence theorem Theorem 3.1. Assume that (H 1 ) (H 3 ) hold true. Then there exists at least one solution of the roblèm (P). Remark 3.1. If w i = σ = q = 1, the result of the receding theorem coincides with those of Porretta (see [10]). 4. Proof of main results In order to rove 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 W 1, 0 (, w) such that 1) u n u weakly in W 1, 0 (,w), 2) then, [a(x,u n, u n ) a(x,u n, u)] (u n u)dx 0 u n u in W 1, 0 (,w). We give now the roof of theorem 3.1. STEP 1. The aroximate roblem. Let f n be a sequence of smooth functions which strongly converges to f in L 1 (). We Consider the sequence of aroximate roblems: u n W 1, 0 (,w), a(x,u n, u n ) v dx + v W 1, 0 (,w). = g n (x,u n, u n )v dx f n v dx + F v dx (4.1) EJQTDE, 2006 No. 19,. 5

6 where g n (x,s,ξ) = g(x,s,ξ) 1+ 1 g(x,s,ξ) θ n(x) with θ n (x) = nt 1/n (σ 1/q (x)). n Note that g n (x,s,ξ) satisfises the following conditions g n (x,s,ξ)s 0, g n (x,s,ξ) g(x,s,ξ) and g n (x,s,ξ) n. We define the oerator G n : X X by, G n u,v = and Au,v = g n (x,u, u)v dx a(x,u, u) v dx Thanks to Hölder s inequality, we have for all u X and v X, g n (x,u, u)v dx ( n ( g n (x,u, u) q σ q q C n v X, σ q /q σ q /q dx ) 1 ( q dx )1 v q q σ dx ) 1 q v q,σ (4.2) the last inequality is due to (2.3) and (2.4). Lemma 4.2. The oerator B n = A + G n from X into its dual X is seudomonotone. Moreover, B n is coercive, in the following sense: < B n v,v > v X + if v X +,v W 1, 0 (,w). This Lemma will be roved 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 Chater 2 of [11] ). STEP 2. A riori estimates. Taking v = T k (u n ) as test function in (4.1), gives a(x,u n, u n ) T k (u n )dx + g n (x,u n, u n )T k (u n )dx = f n T k (u n )dx + F n T k (u n )dx and by using in fact that g n (x,u n, u n )T k (u n ) 0, we obtain a(x,u n, u n ) u n dx ck + F n T k (u n )dx. { u n k} Thank s to Young s inequality and (3.3), one easily has α T k(u n ) w i (x) dx c 1 k. (4.3) 2 x i STEP 3. Almost everywhere convergence of u n. We rove that u n converges to some function u locally in measure (and therefore, we can EJQTDE, 2006 No. 19,. 6

7 aloways assume that the convergence is a.e. after assing to a suitable subsequence). To rove this, we show that u n is a Cauchy sequence in measure in any ball B R. Let k > 0 large enough, we have k meas({ u n > k} B R ) = T k (u n ) dx T k (u n ) dx { u n >k} B R B R ( ) 1 ( ) 1 T k (u n ) w 0 dx w 1 q 0 dx BR ( c 0 T ) 1 k(u n ) w i (x) dx x i c 1 k. 1 Which imlies that meas({ u n > k} B R ) c 1 Moreover, we have, for every δ > 0, k 1 1 k > 1. (4.4) meas({ u n u m > δ} B R ) meas({ u n > k} B R ) + meas({ u m > k} B R ) +meas{ T k (u n ) T k (u m ) > δ}. (4.5) Since T k (u n ) is bounded in W 1, 0 (,w), there exists some v k W 1, 0 (,w), such that T k (u n ) v k weakly in W 1, 0 (, w) T k (u n ) v k strongly in L q (,σ) and a.e. in. Consequently, we can assume that T k (u n ) is a Cauchy sequence in measure in. 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 roves that (u n ) n is a Cauchy sequence in measure in B R, thus converges almost everywhere to some measurable function u. Then T k (u n ) T k (u) weakly in W 1, 0 (,w), T k (u n ) T k (u) strongly in L q (,σ) and a.e. in. STEP 4. Strong convergence of truncations. We fix k > 0, and let h > k > 0. We shall use in (4.1) the test function { vn = φ(w n ) = T 2k (u n T h (u n ) + T k (u n ) T k (u)), w n (4.6) with φ(s) = se γs2,γ = ( b(k) α )2. It is well known that It follows that φ (s) b(k) α φ(s) 1 2 s IR, (4.7) a(x,u n, u n ) w n φ (w n ) dx + g n (x,u n, u n )φ(w n ) dx = f n φ(w n ) dx + F φ(w n ) dx. (4.8) EJQTDE, 2006 No. 19,. 7

8 Since φ(w n )g n (x,u n, u n ) > 0 on the subset {x, u n (x) > k}, we deduce from (4.8) that a(x,u n, u n ) w n φ (w n ) dx + f n φ(w n ) dx + { u n k} g n (x,u n, u n )φ(w n ) dx F φ(w n ) dx. (4.9) Denote by ε 1 h (n),ε2 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 f n φ(w n ) dx = fφ(t 2k (u T h (u))) dx + ε 1 h (n) (4.10) and F φ(w n ) dx = F T 2k (u T h (u))φ (T 2k (u T h (u))) dx + ε 2 h (n). (4.11) Slitting the first integral on the left hand side of (4.9) where u n k and u n > k, we can write, a(x,u n, u n ) w n φ (w n ) dx = a(x,t k (u n ), T k (u n ))[ T k (u n ) T k (u)]φ (w n ) dx { u n k} + a(x,u n, u n ) w n φ (w n ) dx. { u n >k} (4.12) Setting m = 4k + h, using a(x,s,ξ)ξ 0 and the fact that w n = 0 on the set where u n > m, we have { u n >k} and since a(x,s,0) = 0 { u n k} a(x,u n, u n ) w n φ (w n ) dx φ (2k) s IR, we have { u n >k} a(x,t m (u n ), T m (u n )) T k (u) dx, a(x,t k (u n ), T k (u n ))[ T k (u n ) T k (u)]φ (w n ) dx = a(x,t k (u n ), T k (u n ))[ T k (u n ) T k (u)]φ (w n ) dx. (4.13) (4.14) Combining (4.13) and (4.14), we get a(x,u n, u n ) w n φ (w n ) dx a(x,t k (u n ), T k (u n ))[ T k (u n ) T k (u)]φ (w n ) dx φ (2k) a(x,t m (u n ), T m (u n )) T k (u) dx. { u n >k} (4.15) EJQTDE, 2006 No. 19,. 8

9 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,t m (u n ), T m (u n ))) n is bounded in L (,w 1 i ) while T k (u)χ un >k tends to 0 strongly in L (,w i ), which yields a(x,u n, u n ) w n φ (w n ) dx (4.16) a(x,t k (u n ), T k (u n ))[ T k (u n ) T k (u)]φ (w n ) dx + ε 3 h. On the other hand, the term of the right hand side of (4.16) reads as a(x,t k (u n ), T k (u n ))[ T k (u n ) T k (u)]φ (w n ) dx = [a(x,t k (u n ), T k (u n )) a(x,t k (u n ), T k (u))] + [ T k (u n ) T k (u)]φ (w n ) dx a(x,t k (u n ), T k (u)) T k (u n )φ (T k (u n ) T k (u)) dx a(x,t k (u n ), T k (u)) T k (u)φ (w n ) dx. (4.17) Since a i (x,t k (u n ), T k (u))φ (T k (u n ) T k (u)) a i (x,t k (u), T k (u))φ (0) strongly in L (,w 1 by using the continuity of the Nymetskii oerator, while (T k(u n)) (T k(u)) weakly in L (,w i ), we have a(x,t k (u n ), T k (u)) T k (u n )φ (T k (u n ) T k (u)) dx (4.18) = a(x,t k (u), T k (u)) T k (u)φ (0) dx + ε 4 h (n). In the same way, we have a(x,t k (u n ), T k (u)) T k (u)φ (w n ) dx (4.19) = a(x,t k (u), T k (u)) T k (u)φ (0) dx + ε 5 h (n). Combining (4.16)-(4.19), we get a(x,u n, u n ) w n φ (w n ) dx [a(x,t k (u n ), T k (u n )) a(x,t k (u n ), T k (u))] [ T k (u n ) T k (u)]φ (w n ) dx + ε 6 h (n). The second term of the left hand side of (4.9), can be estimated as ( g n (x,u n, u n )φ(w n ) dx { u n k} { u b(k) c(x) + w i T k(u n ) n k} x i b(k) c(x) φ(w n ) dx a(x,t k (u n ), T k (u n )) T k (u n ) φ(w n ) dx. + b(k) α ) (4.20) φ(w n ) dx (4.21) EJQTDE, 2006 No. 19,. 9 i )

10 Since c(x) belongs to L 1 () it is easy to see that b(k) c(x) φ(w n ) dx = b(k) c(x) φ(t 2k (u T h (u))) dx + ε 7 h(n). (4.22) On the other side, we have a(x,t k (u n ), T k (u n )) T k (u n ) φ(w n ) dx = [a(x,t k (u n ), T k (u n )) a(x,t k (u n ), T k (u))] + + [ T k (u n ) T k (u)] φ(w n ) dx a(x,t k (u n ), T k (u n )) T k (u) φ(w n ) dx a(x,t k (u n ), T k (u))[ T k (u n ) T k (u)] φ(w n ) dx. (4.23) 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 g n (x,u n, u n )φ(w n ) dx { u n k} [a(x,t k (u n ), T k (u n )) a(x,t k (u n ), T k (u))] (4.24) [ T k (u n ) T k (u)] φ(w n ) dx +b(k) c(x) φ(t 2k (u T h (u))) dx + ε 9 h (n). (4.9)-(4.11), (4.20) and (4.24), we get [a(x,t k (u n ), T k (u n )) a(x,t k (u n ), T k (u))] [ T k (u n ) T k (u)](φ (w n ) b(k) α φ(w n) ) dx b(k) c(x) φ(t 2k (u T h (u))) dx + fφ(t 2k (u T h (u))) dx, + F T 2k (u T h (u))φ (T 2k (u T h (u))) dx + ε 10 h (n), which and (4.7) imlies that [a(x,t k (u n ), T k (u n )) a(x,t k (u n ), T k (u))] [ T k (u n ) T k (u)] dx 2b(k) c(x) φ(t 2k (u T h (u))) dx + 2 fφ(t 2k (u T h (u))) dx, +2 F T 2k (u T h (u))φ (T 2k (u T h (u))) dx + ε 11 h (n), in which, we can ass to the limit as n + to obtain lim su [a(x,t k (u n ), T k (u n )) a(x,t k (u n ), T k (u))][ T k (u n ) T k (u)] dx n 2b(k) c(x) φ(t 2k (u T h (u))) dx + 2 fφ(t 2k (u T h (u))) dx, +2 F T 2k (u T h (u))φ (T 2k (u T h (u))) dx. (4.25) EJQTDE, 2006 No. 19,. 10

11 It remains to show, for our uroses, 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 aly 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 α u n w i φ (T 2k (u n T h (u n )))dx {h u n 2k+h} x i + g n (x,u n, u n )φ(t 2k (u n T h (u n ))) dx F u n φ (T 2k (u n T h (u n ))) dx {h u n 2k+h} f n φ(t 2k (u n T h (u n ))) dx, and thanks to the sign condition (3.4), we get α u n w i φ (T 2k (u n T h (u n )))dx {h u n 2k+h} x i F u n φ (T 2k (u n T h (u n ))) dx {h u n 2k+h} + f n φ(t 2k (u n T h (u n ))) dx. Using the Young inequality we have α 2 {h u n 2k+h} so that, since φ 1, we have u n w i φ (T 2k (u n T h (u n )))dx x i f n φ(t 2k (u n T h (u n ))) dx + c k {h u n } T 2k (u T h (u)) x w i dx i T 2k (u T h (u)) again because the norm is lower semi-continuity, we get T 2k (u T h (u)) c k lim inf n w 1.F dx, w i φ (T 2k (u T h (u)))dx, (4.26) w i φ (T 2k (u T h (u)))dx T 2k (u T h (u)) x w i dx i T 2k (u n T h (u n )) x w i dx i T 2k (u n T h (u n )) x w i φ (T 2k (u n T h (u n )))dx i (4.27) c k lim inf n EJQTDE, 2006 No. 19,. 11

12 Consequently, in view of (4.26) and (4.27), we obtain T 2k (u T h (u)) w i φ (T 2k (u T h (u)))dx w 1 F dx lim inf n c k + lim inf n {h u n } f n φ(t 2k (u n T h (u n ))) dx. Finally, the strong convergence in L 1 () of f n, we have, as first n and then h tend to infinity, hence lim su u w i φ (T 2k (u T h (u)))dx = 0, h {h u 2k+h} x i lim F T 2k (u T h (u))φ (T 2k (u T h (u))) dx = 0. h Therefore by (4.25), letting h go to infinity, we conclude, lim [a(x,t k (u n ), T k (u n )) a(x,t k (u n ), T k (u))][ T k (u n ) T k (u)] dx = 0, n which and using Lemma 4.1 imlies that T k (u n ) T k (u) strongly in W 1, 0 (, w) k > 0. (4.28) STEP 5. Passing to the limit. By using T k (u n v) as test function in (4.1), with v W 1, 0 (,w) L (), we get a(x,t k+ v (u n ), T k+ v (u n )) T k (u n v) dx + g n (x,u n, u n )T k (u n v) dx = f n T k (u n v) dx + F T k (u n v) dx. By Fatou s lemma and the fact that a(x,t k+ v (u n ), T k+ v (u n )) a(x,t k+ v (u), T k+ v (u)) (4.29) weakly in L (,w 1 i ) one easily sees that a(x,t k+ v (u), T k+ v (u)) T k (u v) dx lim inf a(x,t n k+ v (u n ), T k+ v (u n )) T k (u n v) dx. (4.30) For the second term of the right hand side of (4.29), we have F T k (u n v) dx F T k (u v) dx as n. (4.31) EJQTDE, 2006 No. 19,. 12

13 since T k (u n v) T k (u v) weakly in L (,w i ), while F L (,w 1 i ). On the other hand, we have f n T k (u n v) dx ft k (u v) dx as n. (4.32) To conclude the roof of theorem, it only remains to rove g n (x,u n, u n ) g(x,u, u) strongly in L 1 (), (4.33) in articular it is enough to rove the equiintegrable of g n (x,u n, u n ). To this urose, we take T l+1 (u n ) T l (u n ) as test function in (4.1), we obtain g n (x,u n, u n ) dx f n dx. { u n >l+1} { u n >l} Let ε > 0. Then there exists l(ε) 1 such that g n (x,u n, u n ) dx < ε/2. (4.34) { u n >l(ε)} For any measurable subset E, we have ( g n (x,u n, u n ) dx b(l(ε)) c(x) + w i (T ) l(ε)(u n )) E E x i + g n (x,u n, u n ) dx. { u n >l(ε)} dx In view of (4.28) there exists η(ε) > 0 such that ( ) b(l(ε)) c(x) + w i (T l(ε)(u n)) x E i dx < ε/2 for all E such that meas E < η(ε). (4.35) Finally, by combining (4.34) and (4.35) one easily has g n (x,u n, u n ) dx < ε for all E such that meas E < η(ε), E which shows that g n (x,u n, u n ) are uniformly equintegrable in as required. Thanks to (4.30)-(4.33) we can ass to the limit in (4.29) and we obtain that u is a solution of the roblem (P). This comletes the roof 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 w n = T 2k (u n T h (u n ) + T k (u n ) T k (u)) in the test function (4.6). Proof of Lemma 4.2. From Hölder s inequality, the growth condition (3.1) we can show that A is bounded, and by using (4.2), we have B n bounded. The coercivity folows from (3.3) and (3.4). it remain to EJQTDE, 2006 No. 19,. 13

14 show that B n is seudo-monotone. Let a sequence (u k ) k W 1, 0 (,w) such that and lim su B n u k,u k χ,u. k We will rove that u k u weakly in W 1, 0 (,w), B n u k χ weakly in W 1, (,w ), χ = B n u and B n u k,u k χ,u as k +. Since (u k ) k is a bounded sequence in W 1, 0 (,w), we deduce that (a(x,u k, u k )) k is bounded in L (,w 1 i ), then there exists a function h L (,w 1 i ) such that a(x,u k, u k ) h weakly in L (,w 1 i ) as k, similarly, it is easy to see that (g n (x,u k, u k )) k is bounded in L q (,σ 1 q ), then there exists a function k n L q (,σ 1 q ) such that g n (x,u k, u k ) k n weakly in L q (,σ 1 q ) as k. It is clear that, for all w W 1, 0 (,w), we have Consequently, we get χ,w = χ, w = lim B nu k,w k + = lim a(x,u k, u k ) w dx k + + lim g n (x,u k, u k ).w dx. k + On the one hand, we have g n (x,u k, u k ).u k dx and, by hyotheses, we have therefore { lim su k h w dx + k n.w dx w W 1, 0 (,w). (4.36) a(x,u k, u k ) u k dx + k n.u dx as k, (4.37) h u dx + } g n (x,u k, u k ).u k dx k n.u dx, lim su a(x,u k, u k ) u k dx h u dx. (4.38) k EJQTDE, 2006 No. 19,. 14

15 By virtue of (3.2), we have (a(x,u k, u k ) a(x,u k, u))( u k u) dx > 0. (4.39) Consequently a(x,u k, u k ) u k dx hence lim inf k This imlies by using (4.38) lim k + a(x,u k, u) u dx + a(x,u k, u k ) u dx a(x,u k, u) u k dx, a(x,u k, u k ) u k dx a(x,u k, u k ) u k dx = By means of (4.36), (4.37) and (4.40), we obtain B n u k,u k χ,u as k +. h u dx. h u dx. (4.40) On the other hand, by (4.40) and the fact that a(x,u k, u) a(x,u, u) strongly in L (,w 1 ) it can be easily seen that i and so, thanks to Lemma 4.1 We deduce then that lim (a(x,u k, u k ) a(x,u k, u))( u k u) dx = 0, k + u n u a.e. in. a(x,u k, u k ) a(x,u, u) weakly in L (,w 1 i ), and g n (x,u k, u k ) g(x,u, u) weakly in L q (,w 1 q i ). Thus imlies that χ = B n u. Corollary 4.1. Let 1 < <. Assume that the hyothesis (H 1 ) (H 3 ) holds, let f n be any sequence of functions in L 1 () which converge to f weakly in L 1 () and let u n the solution of the following roblem u n T 1, 0 (,w), g(x,u n, u n ) L 1 () a(x,u n, u n ) T k (u n v) dx + g(x,u n, u n )T k (u n v) dx (P n) f n T k (u n v) dx, v W 1, 0 (,w) L (), k > 0. 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, 0 (,w), further u is a solution of the roblem (P) (with F = 0). EJQTDE, 2006 No. 19,. 15

16 Proof. We give a brief roof. Ste 1. A riori estimates. As before we take v = 0 as test function in (P n ), we get w i T k(u n ) dx C 1 k. (4.41) x i Hence, by the same method used in the first ste in the roof of Theorem 3.1 there exists a function u T 1, 0 (,w) and a subsequence still denoted by u n such that u n u a.e. in,t k (u n ) T k (u) weakly in W 1, 0 (,w), k > 0. Ste 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 a(x,u n, u n ) T l (u n T h (u n φ(w n ))) dx + g(x,u n, u n )T l (u n T h (u n φ(w n )) dx f n T l (u n T h (u n φ(w n ))) dx. Which imlies that { u n φ(w n) h} a(x,u n, u n ) T l (φ(w n )) + g(x,u n, u n )T l (u n T h (u n φ(w n ))) dx 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 deduce a(x,u n, u n ) φ(w n ) dx + g(x,u n, u n )φ(w n ) dx f n φ(w n ) dx, the rest of the roof of this ste is the same as in ste 4 of the roof of Theorem 3.1. Ste 3. Passing to the limit. This ste is similar to the ste 5 of the roof 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 suose that the second member is nonnegative, then we obtain a nonnegative solution. Indeed. If we take v = T h (u + ) in (P), we have a(x,u, u) T k (u T h (u + )) dx + g(x,u, u)t k (u T h (u + )) dx ft k (u T h (u + )) dx. EJQTDE, 2006 No. 19,. 16

17 Since g(x,u, u)t k (u T h (u + )) 0, we deduce a(x,u, u) T k (u T h (u + )) dx and remark also that by using f 0 we have ft k (u T h (u + )) dx On the other hand, thanks to (3.3), we conclude α Letting h tend to infinity, we can easily deduce {u h} ft k (u T h (u + )) dx, ft k (u T h (u)) dx. w i T k(u ) dx ft k (u T h (u)) dx. x i {u h} T k (u ) = 0, k > 0, which imlies that u 0. References [1] L. Aharouch, Y. Akdim and E. Azroul, Quasilinear degenerate ellitic unilateral roblems, AAA 2005:1 (2005) DOI: /AAA [2] Y. Akdim, E. Azroul and A. Benkirane, Existence of Solution for Quasilinear Degenerated Ellitic Equations, Electronic J. Diff. Equ., Vol. 2001, N 71, (2001) [3] Y. Akdim, E. Azroul and A. Benkirane, Existence Results for Quasilinear Degenerated Equations vias Strong Convergence of Truncations, Revista Matemática Comlutence, Madrid, (2004), 17; N 2, [4] A. Benkirane, A. Elmahi, and D. Meskine, An existence theorem for a class of ellitic roblems in L 1, Alicationes Mathematicae., 29, 4, (2002) [5] P. Bénilan, L. Boccardo, T. Gallouet, R. Gariey, M. Pierre and J. L. Vázquez, An L 1 -theory of existence and uniqueness of nonlinear ellitic equations.,, Ann. Scuola Norm. Su. Pisa 22 (1995), [6] G. Dalmaso, F. Murat, L. Orsina and A. Prignet, Renormalized solutions of ellitic equations with general measure data, Ann. Scuola Norm. Su Pisa Cl. Sci 12 4 (1999) [7] P. Drabek, A. Kufner and V. Mustonen, Pseudo-monotonicity and degenerated or singular ellitic oerators, Bull. Austral. Math. Soc. Vol. 58 (1998), [8] P. Drabek, A. Kufner and F. Nicolosi, Non linear ellitic equations, singular and degenerate cases, University of West Bohemia, (1996). EJQTDE, 2006 No. 19,. 17

18 [9] P. Drabek, A. Kufner and F. Nicolosi, On the solvability of degenerated quasilinear ellitic equations of higher order, Journal of differential equations, 109 (1994), [10] A. Porretta, Existence for ellitic equations in L 1 having lower order terms with natural growth, Portugal. Math. 57 (2000), [11] J. Lions, Quelques méthodes de résolution des roblèmes aux limites non linéaires, Dunod, Paris (1969). L. Ahrouch ( l aharouch@yahoo.fr) E. Aazroul ( azroul elhoussine@yahoo.fr ) Déartement de Mathématiques et Informatique, Faculté des Sciences Dhar-Mahraz, B.P Atlas, Fès, Maroc. (Received October 15, 2005) EJQTDE, 2006 No. 19,. 18