Compactness and quasilinear problems with critical exponents

Transcription

1 Dedicated to Professor Roger Temam for his 65 th anniversary. Comactness and quasilinear roblems with critical exonents A. EL Hamidi(1) (1) Laboratoire de Mathématiques, Université de La Rochelle Av. Michel Créeau, LA ROCHELLE Cedex 09 - France J.M. Rakotoson(2) (2) Laboratoire de Mathématiques - Université de Poitiers - SP2MI - Boulevard Marie et Pierre Curie, Téléort 2 BP FUTUROSCOPE CHASSENEUIL Cedex - France. Keywords : Quasilinear roblems, comactness, critical Sobolev exonents. Abstract A comactness result is revised in order to rove the ointwise convergence of the gradients of a sequence of solutions to a general quasilinear inequality (anisotroic or not, degenerate or not) and for an arbitrary oen set. Combining this result with the well-known Brézis-Lieb lemma, we derive simle roofs of Palais-Smale roerties in many otimization roblems esecially on unbounded domains. 1 Introduction. In his recent works [3, 4], the first author observed that the ointwise convergence obtained by the second author for measure data roblems [12, 13, 14, 15], combined with the well-known Brézis-Lieb lemma [2], can be helful for roving the strong convergence of the gradient in the variational roblems with critical exonents even on arbitrary domains. Notice that the concentration-comactness rincile due to Lions [11] and the concentrationcomactness rincile at infinity by Bianchi et al. [1] are widely used to overcome the difficulty due to the lack of comactness. In the case of bounded domains and comact manifolds, the Struwe decomosition is also very useful to recover comactness in nonlinear ellitic roblems with critical exonent [17, 6]. For the reader s convenience and to have a self-contained aer, we rehrase, in a more general framework, comactness results established in [12, 13, 15] and reroduce the roofs given therein (for instance), since they are stated for bounded sets with usual Leray-Lions oerators, and for T-sets. As examles of alications of our result, we recover in articular recent results in [7] on quasilinear ellitic equations involving critical sobolev exonents. There is a large literature on critical exonents see for instance 1

2 [19, 20, 8, 7] and references therein. Two others examles are given, but our rincial result can be alied to a large class of quasilinear ellitic roblems where there holds a lack of comactness. As a corollary of our result, we can state Lemma 1. Let â be a Caratheodory function from IR into satisfying the usual Leray-Lions growth and { monotonicity conditions. Let (u n ) be a bounded sequence of W 1, loc (IRN ) = v L loc (IRN ), v } L loc (IRN ), with 1 < < +, (f n ) be a bounded sequence of L 1 loc (IRN ) and (g n ) be a sequence of W 1, loc ( ) tending strongly to zero. Assume that (u n ) satisfies: (H1) â ( x, u n (x), u n (x) ) ϕ dx = f n ϕ dx+ < g n, ϕ >, ϕ W 1, com( ) = Then { v W 1, ( ), with comact suort 1. there exists a function u such that u n (x) u(x) a.e. in, 2. u W 1, loc (IRN ), 3. there exists a subsequence, still denoted (u n ), such that u n (x) u(x) a.e. in. }, ϕ bounded. This lemma 1 generalizes in articular some results of Boccardo-Murat [18]. Thus, we give an alternate roof for critical exonents equations instead of the concentration-comactness rincile of P.L. Lions [11] and the concentration-comactness at infinity of Bianchi and al. [1] for such roblems. The above lemma will be used to show that suitable Palais-Smale sequences are in fact relatively comact. Let us recall the following Définition 1 (Palais Smale sequence). Let X be a real Banach sace and I : X IR a functional maing which is Gâteaux differentiable. We say that a sequence (u n ) n 0 is a Palais-Smale sequence if: (a) there is a real number c such that lim n + I(u n) = c. 2

3 (b) lim n + I (u n ) = 0 strongly in the dual X. Remark 1. In our alication X will be a reflexive Banach sace, the conditions (a) and (b) will (often) imly the boundedness of (u n ), from which we will have statement 1. and 2. (with additional comactness). The hyothesis (H1) will often follow from (b). 2 Notations and comactness result. Let be an arbitrary oen set of, we shall denote by ω any relatively comact oen subset ω of (that is ω, where ω is the closure of ω). Let 1 < i < +, i = 1,..., N, we set We remark that IL loc () = N = min{ i, 1 i N}, i=1 W 1, loc () = {v N i=1 L i loc (), = ( 1,..., N ), N i=1 L i loc () = L loc (). L i loc () : v IL loc () }. For given q (1, + ), we denote by q := its conjugate exonent. We q 1 shall use the following globally real Lischitz functions: { σ if σ ε For ε > 0, σ IR, S ε (σ) = εsign(σ) otherwise, and σ k := S k (σ) for k 1. We shall consider a nonlinear ma â : IR satisfying: (L1) â(x,, ) is a continuous ma for almost every x and for all (σ, ξ) IR, â(, σ, ξ) is measurable (such a roerty is called Caratheodory roerty), (L2) â mas bounded sets of W 1, loc q () into bounded sets of N i=1 L i loc (), and for almost all x, for all (σ, ξ) in IR, â(x, σ, ξ) ξ 0, 3

4 for almost every x and for all v W 1, loc (), the maing u N â(x, u, v) is continuous from W 1, (ω)-weak into L i (ω)-strong, for all ω, (L3) for almost every x, for all (σ, ξ i ) IR, i = 1, 2, i=1 [â(x, σ, ξ1 ) â(x, σ, ξ 2 ) ][ ξ 1 ξ 2 ] > 0, for ξ1 ξ 2. (L4) if for some[ x, there is a sequence (σ n, ξ 1n ) IR, ξ 2 such that â ( ) ( ) ] [ ] x, σ n, ξ 1n â x, σn, ξ 2 ξ1n ξ 2 and σn are bounded as n + then ξ 1n remains in a bounded set of IR as n +. We start with this result concerning the convergence almost everywhere of the gradients: Theorem 1. Let (u n ) be a bounded sequence of W 1, loc (). Then (i) There is a subsequence still denoted (u n ) and a function u W 1, loc such that u n (x) u(x) a.e. in as n +. (ii) If furthermore, we assume (L1)-(L4) and that ϕ Cc (), k k 0 > 0 : lim su â ( x, u n (x), u n (x) ) ( ϕs ε (u n u k ) ) o(1) n + as ε 0 then there exists a subsequence still denoted (u n ) such that u n (x) u(x) a.e. in. () Remark The function o(1) in (ii) might deend on k and ϕ As in [15], one can give a more general framework using T -sets instead of W 1, loc () but in view of our alications to variational roblems derived from Euler equations, this framework seems not be aroriate The roof of (ii) is the same as in [13, 14, 15], for convenience we reroduce it here. 4

5 2.4. The assumtion (L2) is satisfied if 1 =... = N = and for all ω, there is a constant c ω > 0 and a function a 0 L (ω) such that for almost every x ω, for all (σ, ξ) IR : [ ] â(x, σ, ξ) c ω σ 1 + ξ 1 + a 0 (x). and (L4) is true if â(x, σ, ξ) ξ c 1 ω ξ c 2 ω, c 1 ω > Bounded sets in W 1, loc () will be bounded in { N N } W 1, (ω) = v L i (ω), v L i (ω), for every ω. i= If 1 =... = N, we adot the usual notation for Sobolev saces: W 1, (ω) and we use 2.4. Proof. j=1 (i) Let (ω j ) j 0 be a sequence of bounded relatively comact subsets of such that ω j ω j+1 and + j=0 ω j =. Since (u n ) n 0 is bounded in W 1, (ω j ), by the usual embeddings, we deduce that there is a subsequence u n(j) and a function u in W 1, (ω j ) such that u n(j) (x) n + u(x). We conclude with the usual diagonal Cantor rocess. (ii) Let ϕ C c (u n, u)(x) = Then one has: (), 0 ϕ 1, ϕ = 1 on ω j and su (ϕ) ω j+1, and set [ â ( x, u n (x), u n (x) ) â ( x, u n (x), u(x) )] (u n u)(x). (ii.1) (u n, u)(x) 0 a.e. on (due to (L3)), (ii.2) su (u n, u)dx is finite (since (u n ) is in a bounded set of n ω j+1 W 1, loc () and the growth condition (L2)). Let us show that lim ϕ (u n, u) 1 dx = 0. On one hand, we have n ϕ (u n, u) 1 dx = (u n, u) 1 ϕdx + { u >k} 5 (u n, u) 1 ϕdx. (1) { u k}

6 By the Hölder inequality and noticing that { } meas x ω j+1 : u > k c(j) k, one deduces that (u n, u) 1 c 1 (j) ϕdx. (2) k 1 { u >k} (c m (j) are different constants deending on j and ϕ but indeendent of n, ε and k). While for the second integral, we have (u n, u) 1 ϕdx = (u n, u) 1 ϕdx + (u n, u) 1 ϕdx. (3) { u k} { u k} { u n u ε} { u k} { u n u >ε} Moreover, the second term in the right hand side of the last inequality satisfies { } (u n, u) ϕdx c2 (j)meas x ω j+1 : u n u (x) > ε { u k} { u n u >ε} and since (u n ) converges { to u in measure, we deduce } that, for n sufficiently large, meas x ω j+1 : u n u (x) > ε ε. It follows that lim su (u n, u) 1 ϕdx c2 (j)ε 1 1. (4) n + { u k} { u n u >ε} Setting A ε n,k = ω j+1 { u k} { u n u ε}, we obtain from the Hölder inequality: ( ) (u n, u) 1 1 ϕdx c3 (j) In,k(ε) 1 In,k(ε) 2, (5) with A ε n,k In,k(ε) 1 = A ε n,k { u k} â(x, u n, u n ) (u n u)ϕdx, In,k(ε) 2 = â(x, u n, u) S ε (u n u)ϕdx. Since â(x, u n, u) â(x, u, u) strongly in last statement of (L2)) and S ε (u n u) 0 in we deduce that N L i (ωj+1 ) (by the i=1 N L i (ω j+1 )-weak, i=1 lim n + I2 n,k(ε) = 0, (6) 6

7 while for the term In,k 1 (ε), we get: In,k(ε) 1 â(x, u n, u n ) ( ϕs ε (u n u k ) ) â(x, u n, u n ) ϕs ε (u n u k )dx. (7) Since â(x, u n, u n ) ϕs ε (u n u k )dx c 4(j)ε, (8) the assumtion (ii) imlies then lim suin,k(ε) 1 c 4 (j)ε + o(1) as ε 0. (9) n + Combining relations (5), (6) and (9), it follows: lim su n + (u n, u) 1 ϕdx o(1) as ε 0. (10) A ε n,k From relation (1), (2), (3), (4), and (10), we deduce: ( ) lim su (u n, u) 1 1 ϕdx o(1) (as ε 0) + O. (11) n + k 1 Letting first ε 0 and then k to infinity, we then obtain: (u n, u) 1 ϕdx = 0. lim n + From which we deduce that for a subsequence (u jn ) n 0, (u jn, u)(x) 0 a.e. on ω j. Arguing as in Leray-Lions [10, 9], we deduce from (L4) that u jn (x) u(x) a.e. in ω j. The roof is achieved by the diagonal rocess of Cantor. Proof of Lemma 1 (Corollary of theorem 1) Here, we have 1 =... = N =. Since (u n ) belongs to a bounded set of 7

8 W 1, loc (IRN ), statement (i) of Theorem 1 imlies that there is a function u and a subsequence still denoted by (u n ) such that and u n (x) n + u(x) a.e. in IRN, u W 1, loc (IRN ). Then for all ϕ Cc ( ), ϕs ε (u n u k ) is an element of Wcom(IR 1, N ) and f n ϕs ε (u n u k )dx ε ϕ f n L 1 (ω) c(ϕ)ε, (12) (for every ϕ such that su (ϕ) ω, ω is a comact of ), and < gn, ϕs ε (u n u k ) > gn W 1, (ω) ϕsε (u n u k ). W 1, ( ) Using the fact that ϕs ε (u n u k ) W is bounded indeendently of ε, n, 1, ( ) k and that g n W 1, (ω) 0, it holds: n lim su â(x, u n, u n ) ( ϕs ε (u n u k ) ) dx O(ε). n Finally, Theorem 1 ends the roof. Remark 3. Many extensions of the Theorem 1 can be made (for instance on manifolds or on measure saces). Here, we choose the above framework for the alications we made here. Nevertheless, one can use Theorem 1 for weighted saces choosing correctly the oen set and the ma â. Some results in that direction have been already made by Marchi [16], and also by Fengquan Li, Zhao Huixiu [5] using the method of [12, 13]. 3 Some examles of alications. 3.1 Examle 1. We start by recovering a recent result of H. Ohya [7] using this alternate roof (without concentration-comactness rinciles) to show that we simlify the author s roof. For this, we recall a art of the author s framework. Let 1 < < N, = N and let be an unbounded oen set with N 8

9 smooth boundary if, θ(x), a(x), K(x) be three non negative functions[ with the ] additional regularity that θ C 2 (), a L r () for some N r, + and K is such that e ( )θ(x) K(x) = V (x), being bounded. The author defined the following sets and quantities { } L (θ, ) = u L (); e θ(x) u dx < +, λ 1 = I θ (u) = 1 W 1, (θ, ) = { inf u W 1, (θ,)\{0} { u W 1, 0 () : e θ(x) u dx e θ( u + u ) } < +, / e θ(x) ( u λa(x) u ) dx 1 } e θ(x) a(x) u dx > 0, e θ(x) K(x) u dx, for u W 1, (θ, ). The author showed, under some hyotheses on θ and, that W 1, (θ, ) is embedded continuously (res. comactly) in L q (θ, ) rovided that q [, ] (res. q [, )). Moreover, in this situation the Poincaré inequality holds true, which imlies that the semi-norm ( ) 1 is in fact a norm on W 1, (θ, ). We shall rove the fol- e θ u dx lowing Palais Smale roerty: Theorem 2. Under the above roerty, for every λ < λ 1, any Palais-Smale sequence (u m ) m of I θ on X = W 1, (θ, ) satisfying : (a) I θ (u m ) b θ (b) I θ (u m) 0 in ( W 1, (θ, )) (dual of X) contains a convergent subsequence in W 1, (θ, ) rovided that 0 < b θ < b θ = 1 N S N N V, where S = inf u W 1, 0 ()\{0} { / ( u dx ) } u dx. Remark 4. Theorem 2 corresonds to Theorem 4.1 in [7]. 9

10 Proof of theorem 2 The sequence (u m ) is bounded in X as it was observed in [7]. On one hand one has I θ (u m ) 1 < I θ(u m ), u m >= 1 N u m X λ a(x) u m e θ dx N and alying the Poincaré-Sobolev inequality, there is a ositive constant c such that ) 1 (1 λλ1 u m X N I θ(u m ) 1 < I θ(u m ), u m > c. Thus, e θ(x) K(x) u m dx and e θ(x) a(x) u m dx are bounded indeendently of m. If we set â(x, ξ) = e θ(x) ξ 2 ξ, x, ξ, then â satisfies conditions (L1)-(L4) (since θ C()). With the definition of X, we may then aly the first statement of Theorem 1 to conclude that there is an element u X and subsequence still denoted (u m ) such that u m (x) u(x) a.e. in. Furthermore since S ε (u m u k ) X and S ε (u m u k ) X u m X + u X, we get for all ϕ Cc () : ( ) â(x, u m ) ϕs ε (u m u k ) dx Thus, ( λε ( +ε e θ(x) u m a(x)dx +c I θ(u m ) X ) 1 ( ) 1 ( e θ(x) u m K(x)dx ( u m X + u X ). ) 1 ϕ e θ(x) a(x)dx lim su â(x, u m ) ( ϕs ε (u m u k ) ) dx O(ε). m + ) 1 ϕ e θ(x) K(x)dx Thus, from Theorem 1, assing if necessary to a subsequence, u m (x) m + u(x) a.e. in. 10

11 From Vitali s theorem, we then deduce that u is a critical oint of I θ, that is, φ X, < I θ(u), φ >= 0. At this stage, we aly the Brézis-Lieb lemma [2], that is from the equations, < I θ(u, u >= 0, < I θ(u m ), u m >= o(1), we then have e θ(x) (u u m ) dx = lim m + lim e θ(x) u m u (x)k(x)dx =: l. m + (Note that from the integrability of a and the boundedness of (u m ) in X, we deduce that a(x) u m u e θ(x) dx = 0 (see Proosition 1 below).) lim m + Since V (x) = e ( )θ(x) K(x) is bounded, we deduce l V e θ(x) u m u (x)dx + o(1). From the definition of S, one obtains l V S l. If l = 0, the roof is done. If l 0 then l S N b θ = 1 N N V. Since, lim I θ (u m ) = b θ, thus m u K(x)e θ(x) dx+ 1 l+o(1) (still using the Brézis-Lieb s lemma). N Then b θ 1 N l 1 N S N = b θ, which gives a contradiction. Thus necessarily l = 0. N V In the next examle, we give a classical existence result with a new roof based on our aroach. 3.2 Examle 2. In this aragrah, we are concerned with the existence of (at least) one ositive solution to the ellitic roblem u = λa(x) u 2 u + u 2 u in, (13) 11

12 where the function a satisfies the following conditions: a 0 on, a 0 and a L N ( ). The arameter λ is assumed to be ositive, 1 < < N and N 3. Consider the Euler-Lagrange functional associated to Problem (13) defined by J λ (u) := 1 u dx λ a(x) u dx 1 u dx IR N IR N which is of class C 1 (D 1, ( )). We recall that D 1, ( ) is the comletion of the sace D( ) with resect to the norm ( ) 1 ϕ := ϕ dx. The sace D 1, ( ) can also be seen as D 1, ( ) = { ϕ L ( ) : ϕ L ( ) }. By solutions of Problem (13) we understand critical oints of the functional J λ. Remark that the functional J λ is bounded neither above nor below on D 1, ( ). Then, to find ossible critical oints of J λ, we limit the study to the corresonding Nehari manifold which contains all critical oints of J λ. We recall that the Nehari manifold associated to J λ, denoted by N Jλ, is defined by N Jλ := { ϕ D 1, ( ) \ {0} : J λ(ϕ)(ϕ) = 0 }. ( ) 1 In the sequel, we will set u a, := a(x) u dx. Lemma 2. For every λ > 0, the functional J λ is bounded below on the Nehari manifold N Jλ. Proof. For every u N Jλ, it holds J λ (u) = ( 1 1 ) u = 1 N u, and this ends the roof. As before, due to the integrability of a, the comact embedding D 1, ( ) L loc (IRN ), the Vitali s theorem imlies the following statement: 12

13 Lemma 3. The functional is weakly continuous. by D 1, ( ) IR u a(x) u dx We recall that the Nehari manifold can be characterized more exlicitly N Jλ := { tϕ ; (t, ϕ) (IR \ {0}) (D 1, ( ) \ {0}) : For this reason, we introduce the modified functional J λ (t, u) := J λ (tu), on IR D 1, ( ). } d dt J λ(tϕ) = 0. Since we are interested in ositive solutions to Problem (13), we restrict ourselves in what follows to t > 0. A direct comutation shows that for every u D 1, ( ) \ {0}, there is a unique value λ(u) of λ defined by λ(u) = u u, a, such that for every λ (0, λ(u)), on has t(u, λ)u N Jλ, where We introduce t(u, λ) = λ 1 := ( u λ ) 1 u a,. (14) u inf λ(u) u D 1, ( )\{0} which is not other than the first eigenvalue to the nonlinear eigenvalue roblem u = λa(x) u 2 u in. From Lemma (3), we get clearly that the characteristic value λ 1 is ositive. Now, for every λ (0, λ 1 ) on has more recisely For every λ (0, λ 1 ), we introduce N Jλ = { t(u, λ)u : u D 1, ( ) \ {0} }. α(λ) := inf J λ (u) = inf J λ (t(u, λ)u). u N Jλ u D 1, ( )\{0} 13

14 It is not difficult to see that the functional is 0 homogeneous. Then we get D 1, ( ) \ {0} IR u J λ (t(u, λ)u) α(λ) = inf u S J λ(t(u, λ)u), (15) where S is the unit shere in D 1, ( ). In this articular case, we have in fact ( 1 α(λ) := inf u S 1 ) ( u λ ) u a, u. In what follows, we will write (P S) c to denote a Palais-Smale sequence of J λ with the level c IR. Lemma 4. Let λ (0, λ 1 ). There exists a minimizing sequence of (15) denoted by (u n ) S such that: (i) 0 < lim inf t(u n, λ) lim su t(u n, λ) < +. n n (ii) (t(u n, λ)u n ) is a bounded Palais-Smale sequence for J λ. Proof. Let us denote U n := t(u n, λ)u n, it holds obviously that t(u n, λ) = U n. On one hand, one has Then, ( 1 1 ) ( ) U n λ U n a, = α(λ) + o n (1). U n = λ U n a, + Nα(λ) + o n(1). Alying the Hölder and Young inequalities to U n a,, we conclude that U n is bounded. On the other hand, for every u S, one gets J λ (tu) t (1 λλ1 ) t S, where S is the best Sobolev constant in the embedding D 1, ( ) L ( ). It follows that δ(λ) > 0 such that α(λ) > δ(λ) > 0, (16) 14

15 and consequently lim inf U n n > 0. This ends the claim (i). For every u D 1, ( )\{0} and λ (0, λ 1 ), we have t Jλ (t(u, λ), u) = 0 and tt Jλ (t(u, λ), u) < 0. The imlicit function theorem imlies that t(u, λ) is C 1 with resect to u since J λ is. Let us introduce the C 1 functional J λ defined on S by Then J λ (u) = J λ (t(u, λ), u) J λ (t(u, λ)u). α(λ) = inf u S J λ(u). Using the Ekeland variational rincile on the comlete manifold (S, ) to the functional J λ, there exists a minimizing sequence of (15) denoted by (u n ) S such that: J λ(u n )(ϕ n ) 1 n ϕ n, for every ϕ n T un S, where T un S is the tangent sace to S at the oint u n. Moreover, for every ϕ n T un S, one has J λ(u n )(ϕ n ) = t Jλ (t(u n, λ), u n )t (u n, λ)(ϕ n ) + u Jλ (t(u n, λ), u n )(ϕ n ), = u Jλ (t(u n, λ), u n )(ϕ n ), since t Jλ (t(u n, λ), u n ) 0, where t (u n, λ) denotes the derivative of t(., λ) with resect to its first variable at the oint (u n, λ). Furthermore, let π : D 1, ( ) \ {0} IR ( S ) u u u, := (π u 1 (u), π 2 (u)). Alying Hölder s inequality, we get for every (u, ϕ) ( D 1, ( ) \ {0} ) D 1, ( ): { π 1 (u)(ϕ) ϕ, π 2(u)(ϕ) 2 ϕ u. From (i), there is a ositive constant C such that t(u n, λ) C, n N. 15

16 Then for every ϕ D 1, ( ), there are ϕ 1 n IR and ϕ 2 n T un S such that ϕ 1 n ϕ, ϕ 2 n 2 ϕ and C J λ(t(u n, λ)u n )(ϕ) = t Jλ (t(u n, λ), u n )(ϕ 1 n) + u Jλ (t(u n, λ), u n )(ϕ 2 n), Therefore, We easily conclude that = u Jλ (t(u n, λ), u n )(ϕ 2 n), = J λ(u n )(ϕ 2 n). J λ(t(u n, λ)u n )(ϕ) 1 n ϕ2 n 2 nc ϕ. lim J λ(u n ) = 0 in D 1, ( ), n which achieves the roof. As a consequence of Theorem 4 (see Examle 3 with ρ = ρ 1 = 1 = K, c = = N ), one has : N Lemma 5. Let λ (0, λ 1 ) and (U n ) D 1, ( ) be a Palais-Smale sequence for J λ such that U n U in D 1, ( ). Then, assing if necessary to a subsequence, we get U n U a.e. in. (17) Lemma J λ. If 6. Let λ (0, λ 1 ) and (U n ) D 1, ( ) be a (PS) c sequence for 0 < c < 1 N S N (18) then Problem (13) has a nontrivial solution. Proof. From theorem 4, J λ satisfies the Palais-Smale conditions if 0 < c < 1 N S N. Thus, any weak limit U of Un in D 1, ( ) satisfies then J λ (U) = 0, J λ(u) = c > 0. Consequently, U is nontrivial critical oint of J λ. Lemma 7. For every λ (0, λ 1 ) we have 0 < α(λ) < 1 N S N. 16

17 Proof. Let Ψ ε (x) := ( Nε N 1 (ε + x 1 ) N 2 ) N, x, ε > 0. It is well known that Ψ ε = Ψ ε = SN/. Moreover, J λ (t(ψ ε, λ)ψ ε ) := ( 1 1 Then, using (16) we obtain ) ( Ψε λ Ψ ε a, Ψ ε ) < 1 N SN/. 0 < α(λ) < 1 N SN/. Theorem 3. For every λ (0, λ 1 ), Problem (13) admits at least one ositive solution. Proof. From the receding lemmas, it is clear that Problem (13) ossesses a solution u which is nontrivial, since α(λ) 0. Since J λ is even in u, that u D 1, ( ) imlies that u D 1, ( ) and t(u, λ) = t( u, λ), we conclude that Problem (13) has a nontrivial nonnegative solution. The maximum rincile achieves the roof. 3.3 Examle 3. Similar examles on weighted saces can be made. Let ρ, ρ 1 be two nonnegative { continuous } functions { on and assume } that the closed set F = x : ρ(x) = 0 x : ρ 1 (x) = 0 is of measure zero. We set = \F. For every 1 < < +, we define } Y := D 1, (, ρ) = {u L loc () : u ρ(x)dx < +. 17

18 Let c > so that { / ( ) } S ρ = Inf u ρ(x)dx u c c ρ 1 (x)dx > 0. Y \{0} We set 1 N c = 1 1 c and let { } a L Nc + (, ρ 1 ) = f 0 : f(x) Nc ρ1 (x)dx < +, and the Euler-Lagrange functional defined on Y by I ρ (u) = u ρ(x)dx λ u a(x)ρ 1 (x)dx 1 K(x) u c ρ 1 (x)dx, IR N IR N c where K is a ositive function in L ( ). As reviously, we have ( / ) λ 1ρ = inf u ρ(x) u a(x)ρ 1 (x)dx > 0, (see Proosition 1). Theorem 4. Let λ < λ 1ρ, and (u m ) be a Palais-Smale sequence of I ρ satisfying (a) I ρ (u m ) b ρ, (b) I (u m ) 0 in Y. Then 1. (u m ) contains a weakly convergent subsequence to a function u Y satisfying I ρ(u) = This subsequence is strongly convergent in Y rovided that 0 < b ρ < b ρ = 1 S Nc ρ N c K Nc. We first need the following Proosition 1. Under the above assumtions, if (u m ) m is weakly convergent to u in Y then lim m a(x) u m u ρ 1 (x)dx = 0. 18

19 Proof Since u m c ρ 1 (x)dx is bounded uniformly with resect to m, then for every subset A of A ( a(x) u m u ρ 1 (x) dx c A ) a(x) Nc Nc ρ1 (x)dx < +. But Y W 1, loc () and then Y is comactly embedded in L loc (). Thus using the Vitali s theorem one has, for every bounded set ω in, lim m + ω a(x) u m u ρ 1 (x)dx = 0. (Recall that = \F and meas (F ) = 0). Combining the two relations and the integrability of a, we conclude lim m a(x) u m u ρ 1 (x)dx = 0. Notice that the revious roosition imlies that λ 1ρ > 0. The roof of Theorem 4 is similar to that of Theorem 2, we sketch it as the following: Sketch of the roof of Theorem 4 Setting u m Y = u m ρ(x)dx, it follows that u m Y ( 1 λ ) 1 I ρ (u m ) 1 < I λ 1ρ N c ρ(u m ), u m >. c Thus u m Y remains in a bounded set of IR which imlies that a(x) u m ρ 1 (x)dx and u m c K(x)ρ 1 (x)dx are in a bounded set of IR. Thus u m c ρ 1 (x)dx is bounded. We deduce in articular that (u m ) remains in a bounded set of W 1, loc () (since for every ω, ρ ω = inf ω ρ(x) > 0 and ρ 1ω = inf ω ρ 1 (x) > 0). We may aeal the first statement (i) of Theorem 1: There exist u D 1, (, ρ) and a subsequence still denoted by (u m ) m such that u m (x) u(x) a.e. in. Moreover since S ε (u m u k ) Y, if we set â(x, ξ) = ρ(x) ξ 2 ξ, for all x and ξ, then the conditions (L1)-(L4) are satisfied. Furthermore, 19

20 ϕ C c (), choosing ω such that su ϕ ω, su ω ρ(x) and su ρ 1 (x) ω are finite, it holds then: â ( x, u m (x) ) ( ϕs ε (u m u k ) ) dx c(ϕ)ε+ ( ) I ρ (u m um Y Y + u Y. Consequently, statement (ii) of Theorem 1 is also satisfied so, we conclude that (for a subsequence) u m (x) u(x) a.e. on. From Vitali s theorem, we deduce that ϕ D 1, ( ) < I ρ(u), ϕ >= 0. To conclude that u m u Y 0, we use the Brézis-Lieb lemma to get u m u ρ(x)dx = u ρ(x)dx + u m ρ(x)dx + o(1), u m u c K(x)ρ 1 (x)dx = u c K(x)ρ 1 (x)dx + u m c K(x)ρ 1 (x)dx + o(1) Since < I ρ(u), u >= 0, < I ρ(u m ), u m >= o(1), the above equalities imly that lim u m u m Y = lim u m u c K(x)ρ 1 (x)dx =: l 0 m (noticing that lim m IRN a(x) u m u ρ 1 (x)dx = 0 (see Proosition 1)). Since K L ( ) Thus u m u c K(x)ρ 1 (x)dx K S c ( c c l 0 K S ρ l 0. If l 0 = 0 the roof is done. If l 0 0 then l 0 S Nc ρ K Nc. Since (u m u) ρ(x)dx ) c. lim m I ρ (u m ) = b ρ : b ρ 1 l 0 1 S Nc ρ N c N c K Nc = b ρ, which leads to a contradiction. 20

21 Theorem 5. There exists λ (0, λ 1ρ ) such that the roblem div ( ρ(x) u 2 u ) = ρ 1 (x) ( λa(x) u 2 u + K(x) u c 2 u ) in R N (19) has a nontrivial ositive solution for every λ (λ, λ 1ρ ). Sketch of the roof. Let λ (0, λ 1ρ ) and Φ be a ositive eigenfunction associated to the ositive eigenvalue λ 1ρ. It follows that I ρ (t(φ, λ)φ) = C ( 1 λ ) Nc/, N c λ 1ρ where C is a ositive constant deending on the data of Problem 19 and t(.,.) is defined in the same manner as (14) with minor changes. It is clear that there exists λ (0, λ 1ρ ) such that λ (λ, λ 1ρ ) : 0 < C ( 1 λ ) Nc/ N c λ 1ρ < 1 S Nc ρ N c K Nc. Following the different stes of the revious examle with slight changes, we get the result. Acknowledgments. This work was done during the first author s Délégation CNRS at the Équie Analyse, Géométrie et Modélisation, Université de Cergy-Pontoise. He would like to thank them for the financial suort and the nice working atmoshere during his stay. References [1] G. Bianchi, J. Chabrowski, A. Szulkin On symmetric solutions of an ellitic equation with a nonlinearity involving critical Sobolev exonent. Nonlinear Anal. 25 (1995), no. 1, [2] H. Brézis, E. Lieb A relation between ointwise convergence of functions and convergence of functional Proceedings of the AMS, vol 88, N 3 (July 1983). [3] A. EL Hamidi Sur les conditions nécessaires et suffisantes d existence de solutions our des roblèmes nonlinéaires, Habilitation à Diriger des Recherches, Université de La Rochelle, (2004). 21

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