EXISTENCE RESULT FOR STRONGLY NONLINEAR ELLIPTIC EQUATIONS IN ORLICZ SOBOLEV SPACES

Transcription

1 Proyecciones Vol. 26, N o 2, pp , August Universidad Católica del Norte Antofagasta - Chile EXISTENCE RESULT FOR STRONGLY NONLINEAR ELLIPTIC EQUATIONS IN ORLIC SOBOLEV SPACES A. YOUSSFI UNIVERSITÉ SIDI MOHAMMED BEN ABDALLAH, MAROC Received : July Accepted : May 2007 Abstract In this paper, we prove the existence of solutions for some strongly nonlinear Dirichlet problems whose model is the following div(m 1 M( u ) u u )+um( u ) =f divf in D0 (), where is an open bounded subset of IR N,N 2. We emphasize that no 2 -condition is required for the N-function M. Mathematics Subject Classification : (2000): 46E30, 35J60, 35J65. Key words: Orlicz-Sobolev spaces, strongly nonlinear problems.

2 158 A. Youssfi 1. Introduction Let be a bounded open set of IR N, N 2, and let M be an N-function. Consider the following Dirichlet problem (1.1) A(u)+H(x, u, u) =f, where A(u) := div a(x, u, u) is a Leray-Lions type operator definedonitsdomaind(a) W 1 0 L M() and H is a nonlinearity assumed to satisfy the natural growth condition (1.2) H(x, s, ξ) b( s )(h(x)+m( ξ )) Recently, a large number of papers was devoted to the existence of solutions of (1.1). In the variational framework, that is f W 1 E M (), an existence result was proved in [8] when H depends only on x and u and satisfy the following sign condition H(x, s)s 0, andin[2]whenm satisfies the 2 -condition and H depends also on u and satisfies (1.3) H(x, s, ξ)s 0. The result in [2] was generalized in [7] to N-functions without 2 - condition. In the case where f L 1 (), problem (1.1) was solved in [3] under the so-called coercivity condition (1.4) H(x, s, ξ) βm( ξ ) for s some τ and in [5] assuming the sign condition (1.3) but the result was restricted to N-functions satisfying the 2 -condition (see bellow). The result contained in [5] was then extended in [6] to N-functions without assuming the 2 - condition. The solution u given in this case is such that its truncated

3 Existence result for strongly nonlinear elliptic equations in function T k (u) belongs to the energy space W 1 0 L M() for all k>0, but not the function u it self. Our main goal in this paper, is to prove the existence of a solution in W 1 0 L M() for problems of the kind of (1.1) when the source term has the form f divf with f L 1 () and F E M (), without any restriction on the N-function M. The paper is organized as follows, after giving a background in section 2, in section 3 we list the basic assumptions and our main result which will be proved in six steeps in section Prerequisites 2.1 Let M : IR + IR + be an N-function, ie. M is continuous, convex, with M(t) > 0fort>0, M(t) t 0ast 0and M(t) t as t.the N-function conjugate to M is defined as M(t) = sup{st M(t),s 0}. We recall the Young s inequality: for all s, t 0, If for some k>0, st M(s)+M(t). (2.1) M(2t) km(t) for all t 0, we said that M satisfies the 2 -condition, and if (2.1) holds only for t some t 0, then M is said to satisfy the 2 -condition near infinity. We will extend these N-functions into even functions on all IR. Let P and Q be two N-functions. the notation P Q means that P grows essentially less rapidly than Q, i.e. for all >0, that is the case if and only if P (t) Q( t) 0 as t, Q 1 (t) P 1 (t) 0 as t. 2.2 Let be an open subset of IR N. The Orlicz class K M () (resp. the Orlicz space L M ()) is defined as the set of (equivalence class of) realvalued measurable functions u on such that:

4 160 A. Youssfi µ u(x) M(u(x))dx < (resp. M dx < for some λ>0). λ Endowed with the Luxemburg norm kuk M =inf{λ >0: µ u(x) M dx < }, λ L M () is a Banach space and K M () is a convex subset of L M (). The Orlicz norm is defined on L M () by kuk (M) =sup u(x)v(x)dx, where the supremum is taken over all functions v L M () such that kvk M 1. The two norms k.k M and k.k (M) are equivalent (see [13]). The closure in L M () of the set of bounded measurable functions with compact support in is denoted by E M (). 2.3 The Orlicz-Sobolev space W 1 L M () (resp.w 1 E M ()) is the space of functions u such that u and its distributional derivatives up to order 1 lie in L M () (resp.e M ()). It is a Banach space under the norm kuk 1,M = X kd α uk M. α 1 Thus, W 1 L M () andw 1 E M () canbeidentified with subspaces of the product of (N +1) copiesof L M (). Denoting this product by ΠL M,we will use the weak topologies σ(πl M, ΠE M )andσ(πl M, ΠL M ). The space W 1 0 E M() isdefined as the norm closure of the Schwartz space D() inw 1 E M () and the space W 1 0 L M() astheσ(πl M, ΠE M )closure of D() inw 1 L M (). We say that a sequence {u n } converges to u for the modular convergence in W 1 L M () if,forsomeλ>0, µ D α u n D α u M dx 0 for all α 1, λ this implies convergence for σ(πl M, ΠL M )(see[9,lemma6]). If M satisfies the 2 -condition on IR + (near infinity only if has finite measure), then the modular convergence coincides with norm convergence (see [13, Theorem 9.4]).

5 Existence result for strongly nonlinear elliptic equations in Recall that the norm kduk M defined on W 1 0 L M() isequivalenttokuk 1,M (see [10]). Let W 1 L M () (resp. W 1 E M ()) denotes the space of distributions on which can be written as sums of derivatives of order 1 of functions in L M ()(resp. E M ()). It is a Banach space under the usual quotient norm. If the open has the segment property then the space D() isdense in W 1 0 L M() for the topology σ(πl M, ΠL M ) (see [10]). Consequently, the action of a distribution in W 1 L M () onanelementofw 1 0 L M() iswell defined. For an exhaustive treatments one can see for example [1, 13]. 2.4 We will use the following lemma, (see [6]), which concerns operators of Nemytskii Type in Orlicz spaces. It is slightly different from the analogous one given in [13]. Lemma 2.1. Let be an open subset of IR N with finite measure. let M, P and Q be N-functions such that Q P,andletf : IR IR be a Carathéodory function such that, for a.e.x and for all s IR, f(x, s) c(x)+k 1 P 1 M(k 2 s ), where k 1,k 2 are real constants and c(x) E Q (). Then the Nemytskii operator N f,defined by N f (u)(x) =f(x, u(x)), is strongly continuous from P(E M, k 1 2 )={u L M () :d(u, E M ()) < k 1 2 } into E Q (). We will use the following lemma which can be found in [12], Lemma 2.2. If {f n } L 1 () with f n f L 1 () a.e. in, f n,f 0 a.e. in and f n (x)dx f(x)dx, then f n f strongly in L 1 (). We also use the technical lemma: Lemma 2.3. Let x and y be two nonnegative real numbers and let with θ = y2 4x 2.Then φ(s) =se θs2, xφ 0 (s) y φ(s) x 2, s IR.

6 162 A. Youssfi 3. Assumptions and main result Let be an open bounded subset of IR N, N 2, with the segment property and let M and P be two N-functions such that P M. Let A : D(A) W0 1L M() W 1 L M () be a mapping (non everywhere defined) given by A(u) := div a(x, u, u) where a : IR IR N IR N is a Carathéodory function (i.e., a(x,, ) is continuous on IR IR N for almost every x in and a(,s,ξ)ismeasurable on for every (s, ξ) inir IR N ) satisfying for a.e. x, andforalls IR and all ξ, η IR N, ξ 6= η, (3.1) a(x, s, ξ) a 0 (x)+k 1 P 1 M(k 2 s )+k 1 M 1 M(k 2 ξ ) where a 0 (x) belongs to E M () andk 1,k 2 to IR +, (3.2) (a(x, s, ξ) a(x, s, η)) (ξ η) > 0 (3.3) a(x, s, ξ) ξ M( ξ ) Furthermore, let H : IR IR N IR be a Carathéodory function such that (3.4) H(x, s, ξ) b( s )(M( ξ )+h(x)) for almost x and for all s IR, ξ IR N,withb a real valued positive increasing continuous function and h a nonnegative function in L 1 (), and (3.5) H(x, s, ξ)sgn(s) M( ξ ) for a.e. x, foreveryξ IR N and for every s IR such that s σ, where σ is a positive real number. Consider the following Dirichlet problem: (3.6) A(u)+H(x, u, u) =f div(f ) in, u =0 on, We shall prove the following existence result:

7 Existence result for strongly nonlinear elliptic equations in Theorem 3.1. Assume that f L 1 (), F E M () and (3.1)-(3.5) hold true, then there exists at least a function u solution of (3.6) in the sense that u W0 1L M(), H(x, u, u) L 1 () and a(x, u, u) T k (u v)dx + H(x, u, u)t k (u v)dx = ft k (u v)dx + F T k (u v)dx for every v W 1 0 L M() L () and every k σ. Remark We can replace assumptions (3.3), (3.4) and (3.5) by the following ones: (3.3) 0 a(x, s, ξ) ξ αm with α, λ > 0 and µ (3.4) 0 H(x, s, ξ) b( s ) M with 0 <λ μ and (3.5) 0 H(x, s, ξ)sgn(s) βm with 0 <τ λ and β>0. µ ξ λ µ ξ + h(x) μ µ ξ τ 2. A consequence of (3.3) and the continuity of a with respect to ξ, isthat, for almost every x in and s in IR, a(x, s, 0) = Note that assumption (3.5) gives a sign condition on H only near infinity. 4. In (3.4) we can assume only that b is positive and continuous. Remark 3.2. The solution of (3.6) given by theorem 3.1 belongs to W 1 0 L M() even if F =0, this regularity is due to assumption (3.5).

8 164 A. Youssfi 4. Proofoftheorem3.1 Let {f n } be a sequence of L () functions that converges strongly to f in L 1 (). Let n in IN and let H n (x, s, ξ) = H(x, s, ξ) 1+ 1 n H(x, s, ξ). It s easy to see that H n (x, s, ξ) n, H n (x, s, ξ) H(x, s, ξ) and H n (x, s, ξ)sgn(s) 0for s σ. Since H n is bounded for fixed n, there exists, (see [11, Propositions 1 and 5]), a function u n in W0 1L M() solution of A(u n )+H n (x, u n, u n )=f n divf in, u n =0 on, in the sense (3.7) a(x, u n, u n ) vdx + H n (x, u n, u n )vdx = f n vdx + F vdx for every v W 1 0 L M(). Step1: Estimation in W0 1L M(). For k>0, we denote by T k the usual truncation at level k defined by T k (s) =max( k, min(k, s)) for all s IR. Let us choose v = φ(t σ (u n )) as test function in (3.7), where σ is given by (3.5), φ is the function in lemma 2.3 and b is the function in (3.4). Using (3.3) and the Young s inequality, we obtain

9 Existence result for strongly nonlinear elliptic equations in M( T σ (u n ) )φ 0 (T σ (u n ))dx + H n (x, u n, u n )φ(t σ (u n ))dx φ(σ)kf n k L 1 () + φ 0 (σ) M(2 F )dx + 1 M( T σ (u n ) )φ 0 (T σ (u n ))dx. 2 Since {f n } is bounded in L 1 (), there exists a constant c not depending on n such that 1 M( T σ (u n ) )φ 0 (T σ (u n ))dx + H n (x, u n, u n )φ(t σ (u n ))dx 2 c(φ(σ)+φ 0 (σ)), which we can write, since H n enjoys the same properties of H, 1 M( T σ (u n ) )φ 0 (T σ (u n ))dx + H n (x, u n, u n )φ(t σ (u n ))dx 2 { u n <σ} + H n (x, u n, u n )φ(t σ (u n ))dx { u n σ} c(φ(σ)+φ 0 (σ)). By (3.4) we have H n (x, u n, u n )φ(t σ (u n ))dx { u n <σ} µ b(σ) M( T σ (u n ) )φ(t σ (u n ))dx + φ(σ)khk L 1 (), while using(3.5), we get { u n σ} H n (x, u n, u n )φ(t σ (u n ))dx φ(σ) M( u n )dx. { u n σ} Hence, we obtain R ³ M( T σ(u n ) ) 1 2 φ0 (T σ (u n )) b(σ) φ(t σ (u n )) dx +,φ(σ) M( u n )dx c(φ(σ)+φ 0 (σ)) + b(σ)φ(σ)khk L 1 (). { u n σ}

10 166 A. Youssfi Then, lemma 2.3 with the choice x = 1 2 and y = b(σ), yields 1 M( T σ (u n ) )dx + φ(σ) M( u n )dx 4 { u n σ} c(φ(σ)+φ 0 (σ)) + b(σ)φ(σ)khk L 1 (), which implies that (3.8) M( u n )dx c 0, where c 0 is a constant not depending on n. Thus {u n } is bounded in W0 1L M(), and consequently there exist a function u in W0 1L M() anda subsequence still denoted by {u n } such that (3.9) u n u in W 1 0 L M () for σ(πl M, ΠE M ) and (3.10) u n u in E M () strongly and a.e. in. Step2: {a(x, T k (u n ), T k (u n ))} is bounded in (L M ()) N for all k σ. We will use the Orlicz norm. For that, let ψ (L M ()) N with kψk M 1. For all k σ, we write using (3.2) so that µ a(x, T k (u n ), T k (u n )) a(x, T k (u n ), ψ ) µ T k (u n ) ψ dx 0, k 2 k2 (3.11) 1 a(x, T k (u n ), T k (u n )) ψdx k 2 a(x, T k (u n ), T k (u n )) T k (u n )dx + 1 k 2 a(x, T k (u n ), ψ k 2 ) T k (u n )dx a(x, T k (u n ), ψ k 2 ) ψdx.

11 Existence result for strongly nonlinear elliptic equations in To estimate the first term in the right, we take v = T k (u n )astest function in (3.7) and then use the Young s inequality, the fact that H n (x, u n, u n )T k (u n ) 0ontheset{ u n k} and (3.3), to obtain 1 a(x, T k (u n ), T k (u n )) T k (u n )dx + H n (x, u n, u n )T k (u n )dx 2 { u n k} kkf n k L 1 () + M(2 F )dx. Assumption (3.4) yields H n (x, u n, u n )T k (u n )dx { u n k} µ kb(k) M( u n )dx + khk L 1 () kb(k)(c 0 + khk L 1 ()), where c 0 is the constant in (3.8). Hence, since {f n } is bounded in L 1 (), we deduce that a(x, T k (u n ), T k (u n )) T k (u n )dx λ k, with λ k a constant depending on k. By the Young s inequality, (3.11) becomes 1 k 2 a(x, T k (u n ), T k (u n )) ψdx λ k +(1+2k 1 ) M( T k (u n ) )dx +(1+ 1 )(1 + 2k 1 ) M k k 1 k 2 M( ψ )dx. a(x, T k (u n ), ψ ) k 2 1+2k dx 1

12 168 A. Youssfi By virtue of (3.1) and the convexity of M, weget a(x, T k (u n ), ψ ) M k 2 1+2k dx 1 µ 1 ( a 0 (x) )dx + k 1 M 1 M(kk 2 ) 1+2k 1 We conclude that + k 1 M( ψ )dx. 1+2k 1 a(x, T k (u n ), T k (u n )) ψdx c k for all ψ (L M ()) N with kψk M 1, this means that (3.12) ka(x, T k (u n ), T k (u n ))k (M) c k, for every k σ. Step3: Almost everywhere convergence of the gradients. Since the function u belongs to W0 1L M(), there exists a sequence {v j } D(), (see [9]), which converges to u for the modular convergence in W0 1L M() anda.e.in. For m k σ, wedefine the function ρ m by 1 if s m ρ m (s) = m +1 s if m s m +1 0 if s m +1. Let θ j n = T k (u n ) T k (vj), θ j = T k (u) T k (v j )andz j n,m = φ(θ j n)ρ m (u n ) where φ is the function in lemma 2.3. In what follows, we denote by i (n, j), (i IN), various sequences of real numbers which tend to 0 when n and j respectively, i.e. lim j lim n i(n, j) =0.

13 Existence result for strongly nonlinear elliptic equations in We use z j n,m W 1 0 L M() as test function in (3.7) to get (3.13) <A(u n ),zn,m j > + H n (x, u n, u n )zn,mdx j = f n zn,mdx+ j F zn,mdx. j In view of (3.10), we have zn,m j φ(θ j )ρ m (u) weaklyinl () for σ (L,L 1 )asn,then lim f n z j n n,mdx = fφ(θ j )ρ m (u)dx, and since φ(θ j ) 0weaklyinL () forσ(l,l 1 )asj, we have lim fφ(θ j )ρ m (u)dx =0, j hence, we obtain f n zn,mdx j = 0 (n, j). Thanks to (3.8) and (3.10), we have as n z j n,m φ(θ j )ρ m (u) in W 1 0 L M () for σ(πl M, ΠE M ), which implies that lim F z j n n,mdx = F θ j φ 0 (θ j )ρ m (u)dx + F uφ(θ j )ρ 0 m(u)dx. On the one hand, by Lebesgue s theorem we get lim F uφ(θ j )ρ 0 m(u)dx =0, j on the other hand, we write F θ j φ 0 (θ j )ρ m (u)dx = F T k (u)φ 0 (θ j )ρ m (u)dx F T k (v j )φ 0 (θ j )ρ m (u)dx,

14 170 A. Youssfi so that, by Lebesgue s theorem one has lim F T k (u)φ 0 (θ j )ρ m (u)dx = F T k (u)ρ m (u)dx, j M ³ vj u Let λ>0 such that M λ 0 strongly in L 1 () asj and L 1 (), the convexity of the N-function M allows us to have ³ u λ M Ã Tk (v j )φ 0 (θ j! )ρ m (u) T k (u)ρ m (u) 4λφ 0 (2k) 1 µ 4 M vj u + 1 µ 1+ 1 λ 4 φ 0 M (2k) µ u λ. Then, by using the modular convergence of { v j } in (L M ()) N and Vitali s theorem, we obtain T k (v j )φ 0 (θ j )ρ m (u) T k (u)ρ m (u) in (L M ()) N for the modular convergence, and then lim F T k (v j )φ 0 (θ j )ρ m (u)dx = F T k (u)ρ m (u)dx. j We have proved that F z j n,mdx = 1 (n, j). Since H n (x, u n, u n )z j n,m 0ontheset{ u n >k} and ρ m (u n )=1on the set { u n k}, we have (3.14) <A(u n ),zn,m j > + H n (x, u n, u n )φ(θn)dx j 2 (n, j) { u n k} Now, we will evaluate the first term of the left-hand side of (3.14) by writing

15 Existence result for strongly nonlinear elliptic equations in <A(u n ),zn,m j > = a(x, u n, u n ) ( T k (u n ) T k (v j ))φ 0 (θn)ρ j m (u n )dx + = + and then a(x, u n, u n ) u n φ(θ j n)ρ 0 m(u n )dx a(x, T k (u n ), T k (u n )) ( T k (u n ) T k (v j ))φ 0 (θ j n)dx { u n >k} a(x, u n, u n ) T k (v j )φ 0 (θ j n)ρ m (u n )dx a(x, u n, u n ) u n φ(θ j n)ρ 0 m(u n )dx, (3.15) <A(u n ),zn,m j > = (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (v j )χ s j)) ( T k (u n ) T k (v j )χ s j )φ0 (θn)dx j + a(x, T k (u n ), T k (v j )χ s j)) ( T k (u n ) T k (v j )χ s j)φ 0 (θn)dx j a(x, T k (u n ), T k (u n )) T k (v j )φ 0 (θn)dx j \ s j a(x, u n, u n ) T k (v j )φ 0 (θn)ρ j m (u n )dx { u n >k} + a(x, u n, u n ) u n φ(θn)ρ j 0 m(u n )dx, where by χ s j, s>0, we denote the characteristic function of the subset s j = {x : T k (v j ) s}. For fixed m and s, we will pass to the limit in n andtheninj in the second, third, fourth and five terms in the right side of (3.15). Starting with the second term, we have

16 172 A. Youssfi a(x, T k (u n ), T k (v j )χ s j)) ( T k (u n ) T k (v j )χ s j)φ 0 (θ j n)dx a(x, T k (u), T k (v j )χ s j)) ( T k (u) T k (v j )χ s j)φ 0 (θ j )dx as n, since by lemma 2.1 one has a(x, T k (u n ), T k (v j )χ s j))φ 0 (θ j n) a(x, T k (u), T k (v j )χ s j))φ 0 (θ j ) strongly in (E M ()) N as n,while T k (u n ) T k (u) weakly in (L M ()) N by (3.8). Let χ s denote the characteristic function of the subset s = {x : T k (u) s}. As T k (v j )χ s j T k(u)χ s strongly in (E M ()) N as j, one has as j. Then a(x, T k (u), T k (v j )χ s j)) ( T k (u) T k (v j )χ s j)φ 0 (θ j )dx 0 (3.16) a(x, T k (u n ), T k (v j )χ s j)) ( T k (u n ) T k (v j )χ s j)φ 0 (θn)dx j = 3 (n, j). For the third term of (3.15), by virtue of (3.12) there exist a subsequence still indexed again by n and a function l k in (L M ()) N with k σ such that a(x, T k (u n ), T k (u n )) l k weakly in (L M ()) N for σ(πl M, ΠE M ). Then, since T k (v j )χ \ s j (E M ()) N,weobtain a(x, T k (u n ), T k (u n )) T k (v j )φ 0 (θn)dx j l k T k (v j )φ 0 (θ j )dx \ s j \ s j

17 Existence result for strongly nonlinear elliptic equations in as n. The modular convergence of {v j } allows us to have l k T k (v j )φ 0 (θ j )dx l k T k (u)dx \ s \ s j as j. This, proves that a(x, T k (u n ), T k (u n )) T k (v j )φ 0 (θn)dx j = l k T k (u)dx+ 4 (n, j). \ s j \ s (3.17) As regards the fourth term, observe that ρ m (u n ) = 0 on the subset { u n m +1}, sowehave a(x, u n, u n ) T k (v j )φ 0 (θn)ρ j m (u n )dx = { u n >k} { u n >k} a(x, T m+1 (u n ), T m+1 (u n )) T k (v j )φ 0 (θ j n)ρ m (u n )dx. As above, we obtain a(x, T m+1 (u n ), T m+1 (u n )) T k (v j )φ 0 (θn)ρ j m (u n )dx = { u n >k} { u >k} l m+1 T k (u)ρ m (u)dx + 5 (n, j). Observing that T k (u) = 0 on the subset { u >k}, one has (3.18) a(x, u n, u n ) T k (v j )φ 0 (θn)ρ j m (u n )dx = 5 (n, j) { u n >k} For the last term of (3.15), we have a(x, u n, u n ) u n φ(θn)ρ j 0 m(u n )dx = {m u n m+1} φ(2k) {m u n m+1} a(x, u n, u n ) u n φ(θn)ρ j 0 m(u n )dx a(x, u n, u n ) u n dx.

18 174 A. Youssfi To estimate the last term of the previous inequality, we test by T 1 (u n T m (u n )) W0 1L M() in (3.7), to get a(x, u n, u n ) u n dx {m u n m+1} + H n (x, u n, u n )T 1 (u n T m (u n ))dx { u n m} = f n T 1 (u n T m (u n ))dx + F u n dx. {m u n m+1} Using the fact that H n (x, u n, u n )T 1 (u n T m (u n )) 0 on the subset { u n m} and the Young s inequality, we get 1 a(x, u n, u n ) u n dx 2 {m u n m+1} f n dx + M( F )dx. { u n m} {m u n m+1} It follows that a(x, u n, u n ) u n φ(θn)ρ j 0 m(u n )dx (3.19) Ã! 2φ(2k) f n dx + M( F )dx. { u n m} {m u n m+1} From (3.16), (3.17), (3.18) and (3.19) we obtain (3.20) <A(u n ),zn,m j > (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (v j )χ s j)) ( T k (u n ) T k (v j )χ s j )φ0 (θ j n)dx Ã! 2φ(2k) f n dx + M( F )dx { u n m} {m u n m+1} \ s l k T k (u)dx + 6 (n, j).

19 Existence result for strongly nonlinear elliptic equations in Now, we turn to second term of the left hand side of (3.14). We have H n (x, u n, u n )φ(θ j n)dx { u n k} = H n (x, T k (u n ), T k (u n ))φ(θ j n)dx { u n k} b(k) M( T k (u n ) ) φ(θn) dx j + b(k) h(x) φ(θn) dx j b(k) a(x, T k (u n ), T k (u n )) T k (u n ) φ(θn) dx j + 7 (n, j). Then, (3.21) { u n k} b(k) H n (x, u n, u n )φ(θn)dx j ³ a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (v j )χ s j) ³ T k (u n ) T k (v j )χ s j φ(θn) dx j + b(k) a(x, T k (u n ), T k (v j )χ s j) ( T k (u n ) T k (v j )χ s j) φ(θn) dx j + b(k) a(x, T k (u n ), T k (u n )) T k (v j )χ s j φ(θn) dx j + 7 (n, j). We proceed as above to get b(k) a(x, T k (u n ), T k (v j )χ s j)) ( T k (u n ) T k (v j )χ s j) φ(θn) dx j = 8 (n, j) and b(k) a(x, T k (u n ), T k (u n )) T k (v j )χ s j φ(θn) dx j = 9 (n, j).

20 176 A. Youssfi Hence, we have (3.22) H n (x, u n, u n )φ(θ j n)dx { u n k} ³ b(k) a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (v j )χ s j) ³ T k (u n ) T k (v j )χ s j φ(θn) dx j + 10 (n, j). Combining (3.14), (3.20) and (3.22), we get ³ a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (v j )χ s j) ³ φ T k (u n ) T k (v j )χj s 0 (θn) j b(k) φ(θn) j dx l k T k (u)dx \ s Ã! +2φ(2k) f n dx + M( F )dx + 11 (n, j). { u n m} {m u n m+1} (3.23) Then, lemma 2.3 with x =1andy = b(k), yields ³ a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (v j )χ s j) ³ T k (u n ) T k (v j )χ s j dx 2 l k T k (u)dx \ s Ã! +4φ(2k) f n dx + M( F )dx + 11 (n, j). { u n m} {m u n m+1}

21 Existence result for strongly nonlinear elliptic equations in On the other hand (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (u)χ s )) ( T k (u n ) T k (u)χ s ) dx = ³ a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (v j )χ s j) ³ T k (u n ) T k (v j )χ s j dx + a(x, T k (u n ), T k (u n )) ( T k (v j )χ s j T k (u)χ s )dx a(x, T k (u n ), T k (u)χ s ) ( T k (u n ) T k (u)χ s )dx + a(x, T k (u n ), T k (v j )χ s j) ( T k (u n ) T k (v j )χ s j)dx. We shall pass to the limit in n andtheninj in the last three terms of the right hand side of the above equality. By similar arguments as in (3.15) and (3.21), we obtain a(x, T k (u n ), T k (u n )) ( T k (v j )χ s j T k (u)χ s )dx = 12 (n, j) and and a(x, T k (u n ), T k (u)χ s ) ( T k (u n ) T k (u)χ s )dx = 13 (n, j) (3.24) so that a(x, T k (u n ), T k (v j )χ s j) ( T k (u n ) T k (v j )χ s j)dx = 14 (n, j),

22 178 A. Youssfi (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (u)χ s )) ( T k (u n ) T k (u)χ s )dx (3.25) = (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (v j )χ s j)) ( T k (u n ) T k (v j )χ s j )dx + 15 (n, j). Let r s, we use (3.2), (3.25) and (3.23) to get 0 (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (u))) r ( T k (u n ) T k (u))dx (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (u))) s ( T k (u n ) T k (u))dx = (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (u)χ s )) s ( T k (u n ) T k (u)χ s )dx (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (u)χ s )) ( T k (u n ) T k (u)χ s )dx

23 Existence result for strongly nonlinear elliptic equations in = (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (v j )χ s j)) ( T k (u n ) T k (v j )χ s j)dx + 15 (n, j) 2 l k T k (u)dx \ s Ã! +4φ(2k) f n dx + M( F )dx { u n m} {m u n m+1} + 16 (n, j), Which gives by passing to the limit sup over n and then over j 0 lim sup (a(x, T k (u n ), T k (u n )) a(x, T k (u n ), T k (u))) n r ( T k (u n ) T k (u))dx Ã! 2 l k T k (u)dx +4φ(2k) f dx + M( F )dx. \ s { u m} {m u m+1} Letting s and then m, taking into account that l k T k (u) L 1 (), f L 1 (), M( F ) L 1 (), \ s 0, { u m} 0, and {m u m +1} 0, one has (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 r as n. As in [4], we deduce that there exists a subsequence of {u n } still indexed again by n such that (3.26) u n u a.e. in. Thus, by (3.12) and (3.26) we have (3.27) a(x, T k (u n ), T k (u n )) a(x, T k (u), T k (u)) weakly in (L M ()) N for σ(πl M, ΠE M )andforallk σ.

24 180 A. Youssfi Step4: Modular convergence of the truncations. Going back to (3.23), we write a(x, T k (u n ), T k (u n )) T k (u n )dx a(x, T k (u n ), T k (u n )) T k (v j )χ s jdx + a(x, T k (u n ), T k (v j )χ s j) ( T k (u n ) T k (v j )χ s j)dx Ã! +4φ(2k) f n dx + M( F )dx { u n m} {m u n m+1} +2 a(x, T k (u), T k (u)) T k (u)dx + 11 (n, j). \ s and by (3.24) we get a(x, T k (u n ), T k (u n )) T k (u n )dx a(x, T k (u n ), T k (u n )) T k (v j )χ s jdx Ã! +4φ(2k) f n dx + M( F )dx { u n m} {m u n m+1} +2 a(x, T k (u), T k (u)) T k (u)dx + 17 (n, j). \ s We pass now to the limit sup over n in both sides of this inequality, to obtain lim sup a(x, T k (u n ), T k (u n )) T k (u n )dx n a(x, T k (u), T k (u)) T k (v j )χ s jdx Ã! +4φ(2k) f dx + M( F )dx { u m} {m u m+1} +2 a(x, T k (u), T k (u)) T k (u)dx + lim 17(n, j), \ s n

25 Existence result for strongly nonlinear elliptic equations in in which, we pass to the limit in j to get lim sup a(x, T k (u n ), T k (u n )) T k (u n )dx n a(x, T k (u), T k (u)) T k (u)χ s dx Ã! +4φ(2k) f dx + M( F )dx { u m} {m u m+1} +2 a(x, T k (u), T k (u)) T k (u)dx, \ s letting s and then m, one has lim sup a(x, T k (u n ), T k (u n )) T k (u n )dx a(x, T k (u), T k (u)) T k (u)dx n On the other hand, by Fatou s lemma, we have a(x, T k (u), T k (u)) T k (u)dx lim inf a(x, T k (u n ), T k (u n )) T k (u n )dx. n It follows that a(x, T k (u n ), T k (u n )) T k (u n )dx = lim n By lemma 2.2 we conclude that a(x, T k (u), T k (u)) T k (u)dx. (3.28) a(x, T k (u n ), T k (u n )) T k (u n ) a(x, T k (u), T k (u)) T k (u) strongly in L 1 (), k σ. The convexity of the N-function M and (3.3) allow us to have µ Tk (u n ) T k (u) M a(x, T k(u n ), T k (u n )) T k (u n )+ 1 2 a(x, T k(u), T k (u)) T k (u). Then, by (3.28) we get µ lim sup Tk (u n ) T k (u) M dx =0. E 0 n E 2

26 182 A. Youssfi So that, by Vitali s theorem one has T k (u n ) T k (u) in W 1 0 L M () for the modular convergence, for all k σ. Step5: Equi-integrability of the nonlinearities. As a consequence of (3.10) and (3.26), one has H n (x, u n, u n ) H(x, u, u) a.e. in. We shall prove that the sequence {H n (x, u n, u n )} is uniformly equiintegrable in. Let E be a measurable subset of, for all m σ, wehave H n (x, u n, u n ) dx = H n (x, u n, u n ) dx E + H n (x, u n, u n ) dx. E { u n >m} E { u n m} On the one hand, the use of T 1 (u n T m 1 (u n )) as test function in (3.7), the Young s inequality and (3.3) led to H n (x, u n, u n )T 1 (u n T m 1 (u n ))dx f n dx + M(2 F )dx. {m 1 u n m} { u n m 1} { u n m} Then, assumption (3.5) gives H n (x, u n, u n ) dx { u n m 1} f n dx+ M(2 F )dx. {m 1 u n m} For all >0, one can find an m = m( ) > 1 such that sup H n (x, u n, u n ) dx n { u n m} 2.

27 Existence result for strongly nonlinear elliptic equations in E { u n m} On the other hand, we use (3.3) and (3.4) to get H n (x, u n, u n ) dx H n (x, T m (u n ), T m (u n )) dx E µ b(m) E M( T m (u n ) )dx + h(x)dx E b(m) a(x, T m (u n ), T m (u n )) T m (u n )dx E +b(m) h(x)dx. We use the fact that from (3.28) the sequence {a(x, T m (u n ), T m (u n )) T m (u n )} is equi-integrable and that h L 1 () toobtain lim sup H n (x, u n, u n ) dx =0, E 0 n E { u n m} where E denotes the Lebesgue measure of the subset E. Consequently lim sup H n (x, u n, u n ) dx =0. E 0 n E This proves that the sequence {H n (x, u n, u n )} is uniformly equi-integrable in. By Vitali s theorem, we conclude that H(x, u, u) L 1 () and (3.29) H n (x, u n, u n ) H(x, u, u) strongly in L 1 (). E Step6: Passage to the limit. Let v W 1 0 L M() L (). By [9, Lemma 4], there exists a sequence {v j } D() such that kv j k (N +1)kvk and v j v in W 1 0 L M () for the modular convergence and a.e. in. Let k σ. We go back to approximate equations (3.7) and use T k (u n v j ) as test function to obtain (3.30) a(x, T t (u n ), T t (u n )) T k (u n v j )dx + H n (x, u n, u n )T k (u n v j )dx = f n T k (u n v j )dx + F T k (u n v j )dx,

28 184 A. Youssfi where t = k +(N +1)kvk. The first term in the left-hand side of (7.3.23) is written as a(x, T t (u n ), T t (u n )) T k (u n v j )dx = { u n v j <k} { u n v j <k} Thus, by (3.27) and (3.28) we obtain a(x, T t (u n ), T t (u n )) T k (u n v j )dx as n. Since a(x, T t (u n ), T t (u n )) T t (u n )dx a(x, T t (u n ), T t (u n )) v j dx a(x, T t (u), T t (u)) T k (u v j )dx T k (u n v j ) T k (u v j ) in L () forσ (L,L 1 ), we use (3.29) and the fact that f n f strongly in L 1 () asn, to obtain H n (x, u n, u n )T k (u n v j )dx H(x, u, u)t k (u v j )dx, f n T k (u n v j )dx ft k (u v j )dx, as n. For the last term in the right-hand side of (3.30) we write F T k (u n v j )dx = F u n dx F v j dx. { u n v j <k} { u n v j <k} Hence, by (3.9) we obtain F T k (u n v j )dx F T k (u v j )dx. Therefore, passing to the limit as n in (3.30), we get a(x, u, u) T k (u v j )dx + H(x, u, u)t k (u v j )dx = ft k (u v j )dx + F T k (u v j )dx,

29 Existence result for strongly nonlinear elliptic equations in in which we pass to the limit as j to obtain a(x, u, u) T k (u v)dx + H(x, u, u)t k (u v)dx = ft k (u v)dx + F T k (u v)dx. Which completes the proof of theorem 3.1. Remark 4.1. If the N-function M satisfies the 2 condition, the sequence {a(x, u n, u n )} will be bounded in (L M ()) N. Then, the function u solution of the problem (3.6) is such that: u W0 1L M(), H(x, u, u) L 1 () and a(x, u, u) vdx + H(x, u, u)vdx = fvdx + F vdx, for every v W 1 0 L M() L (). Remark 4.2. We can interpret theorem 3.1 in the following sense: the problem ( u W 1 0 L M (), H(x, u, u) L 1 () div a(x, u, u)+h(x, u, u) =μ admits a solution if and only if μ belongs to L 1 ()+W 1 L M (). Remark 4.3. If we replace (3.1) by the more general growth condition a(x, s, ξ) b 0 ( s )(a 0 (x)+m 1 M(τ ξ )) where a 0 (x) belongs to E M (), τ > 0 and b 0 is a positive continuous increasing function, we can adapt the same ideas to prove the existence of solutions for the problem u W0 1L M(), H(x, u, u) L 1 () and a(x, u, u) T k (u v)dx + H(x, u, u)t k (u v)dx = ft k (u v)dx + F T k (u v)dx for v W0 1L M() L (), by considering the following approximation problems ( un W0 1L M() div a(x, T n (u n ), u n )+H n (x, u n, u n )=f n divf in.

30 186 A. Youssfi As an application of this result, we give div((1 + u ) q exp( u p ) 1 u 2 u)+u(exp( u p ) 1) = f divf in. with p>1 and q>0. References [1] Adams, R.: Sobolev spaces, Academic Press Inc, New York, (1975) [2] Benkirane, A., Elmahi, A.: An existence theorem for a strongly nonlinear problems in Orlicz spaces, Nonlinear Anal. T.M.A. 36, pp (1999). [3] Benkirane, A., Elmahi, A.: strongly nonlinear elliptic equations having natural growth terms and L 1 data, Nonlinear Anal. T.M.A. 39, pp (2000). [4] Benkirane, A., Elmahi, A.: Almost everywhere convergence of the gradients of solutions to elliptic equations in Orlicz spaces and application, Nonlinear Anal. T.M.A. 28, pp (1997). [5] Benkirane, A., Elmahi, A., Meskine, D.: An existence theorem for a class of elliptic problems in L 1, Applicationes Mathematicae 29, 4, pp (2002). [6] Elmahi, A., Meskine, D.: Nonlinear elliptic problems having natural growth and L 1 data in Orlicz spaces, Ann. Mat. Pura Appl. 184, 2, pp (2004). [7] Elmahi, A., Meskine, D.: Existence of solutions for elliptic equations having natural growth terms in Orlicz spaces, Abst. Appl. Anal. 2004, 12, pp (2004). [8] Gossez, J.-P.: A strongly nonlinear elliptic problem in Orlicz-Sobolev spaces, Proc. Sympos. Pure Math., 45, Part 1, (1986). [9] Gossez, J.-P.: Some approximation properties in Orlicz-Sobolev spaces, Stud. Math. 74, pp (1982).