Classification of low energy sign-changing solutions of an almost critical problem

Transcription

1 Journal of Functional Analysis Classification of low energy sign-changing solutions of an almost critical problem Mohamed Ben Ayed a, Khalil El Mehdi b, Filomena Pacella c, a Département de Mathématiques, Faculté des Sciences de Sfax, Route Soukra, Sfax, Tunisia b Faculté des Sciences et Techniques, Université de Nouakchott, BP 506 Nouakchott, Mauritania c Dipartimento di Matematica, Università di Roma La Sapienza, P. le A. Moro, 0085 Roma, Italy Received 9 October 006; accepted 7 May 007 Available online 7 July 007 Communicated by H. Brezis Abstract In this paper we make the analysis of the blow up of low energy sign-changing solutions of a semilinear elliptic problem involving nearly critical exponent. Our results allow to classify these solutions according to the concentration speeds of the positive and negative part and, in high dimensions, lead to complete classification of them. Additional qualitative results, such as symmetry or location of the concentration points are obtained when the domain is a ball. 007 Elsevier Inc. All rights reserved. Keywords: Blow-up analysis; Sign-changing solutions; Nodal domains; Critical exponent. Introduction In this paper we consider the following semilinear elliptic problem with subcritical nonlinearity: M. Ben Ayed and F. Pacella are supported by MURST, Project Metodi Variazionali ed Equazioni Differenziali non lineari. K. El Mehdi is supported by the University of Roma La Sapienza, IHÉS-Schlumberger Fondation programme and by FNRS, Project 003/006. * Corresponding author. addresses: mohamed.benayed@fss.rnu.tn M. Ben Ayed, khalil@univ-nkc.mr K. El Mehdi, pacella@mat.uniroma.it F. Pacella /$ see front matter 007 Elsevier Inc. All rights reserved. doi:0.06/j.jfa

2 348 M. Ben Ayed et al. / Journal of Functional Analysis { u = u ε u in, u = 0 on, where is a smooth bounded domain in R n, n 3, ε is a positive real parameter and = n/n is the critical Sobolev exponent for the embedding of H 0 into L. Problem is related to the limiting problem when ε = 0 which exhibits a lack of compactness and gives rise to solutions of which blow up as ε 0. In the last decades there have been many works devoted to the study of positive solutions of problem. In sharp contrast to this, very little study has been made concerning the signchanging solutions. For details one can see [5] and the references therein. The existence of sign-changing solutions of for any ε 0, has been proved in [4,6] and []. On the other hand, when ε = 0, problem becomes delicate. Pohozaev showed in [6] that if is starshaped, problem with ε = 0 has no solutions whereas Clapp and Weth proved in [] that problem with ε = 0 has a solution on domains with small holes and on some contractible domains with an involution symmetry. In view of this qualitative change in the situation when ε = 0, it is interesting to study the following question: what happens to the solutions of as ε 0? In this paper, we are mainly interested in the study of the behavior of low energy sign-changing solutions of. The first part of this paper is devoted to analyze the asymptotic behavior, as ε 0, of sign-changing solutions of whose energy converges to the value S n/, S being the Sobolev constant for the embedding of H 0 into L. We prove that these solutions blow up at two points which may coincide and which are the limit of the concentration points a ε, and a ε, of the positive and negative part of the solutions. Moreover, we make a precise study of the location of these blow-up points. More precisely, we have Theorem.. Let n 3 and let u ε be a family of sign-changing solutions of which satisfies u ε := u ε S n/ as ε 0.. Then, the set \{x u ε x = 0} has exactly two connected components. Furthermore, there exist two points a ε,, a ε, in one of them can be chosen to be the global maximum point of u ε and there exist two positive reals μ ε,, μ ε, such that u ε Pδ aε,,μ ε, + Pδ aε,,μ ε, 0, as ε 0,. μ ε,i da ε,i, +, μ ε,i ε, as ε 0, for i {, },.3 where Pδ a,μ denotes the projection of δ a,μ on H0, that is, P δ a,μ = δ a,μ in, Pδ a,μ = 0 on, and δ a,μ x = β n μ n / + μ x a n /. Here β n is a constant chosen so that δ a,μ = δ n+/n a,μ β n = nn n /4.

3 M. Ben Ayed et al. / Journal of Functional Analysis Note that, for the supercritical case that is for ε<0, a recent result [9] shows that there is no sign-changing solution u ε with low energy which satisfies. and.3. Now, our aim is to give a complete classification of the solutions satisfying.. To this aim, we divide this kind of solutions into two categories: the ones for which the positive and negative part blow up with the same rate hypothesis.5 below and the ones for which these rates are note comparable hypothesis.9 below. In the first case we are able to prove that the concentration points of the positive and negative part of a solution of this type are distinct and away from the boundary and we characterize their limits in terms of the Green s function and of its regular part. Moreover, when the domain is a ball we prove that the limit concentration points are antipodal with respect to the center of the ball, the solution is axially symmetric with respect to the line joining these points and the nodal surface intersects the boundary. In the second case, i.e. when.9 holds, we are able to prove that, if n 4, the positive and negative part of the solution concentrate at the same point i.e. we have bubble tower solutions and we get a precise estimate of the blow-up rates, in terms of the distance of the concentration points from the boundary of the domain. Moreover, if n 6 we prove that the unique limit of the concentration points is away from the boundary and it is a critical point of the Robin s function. As far as we know this is the first time that the phenomenon of different concentration points converging to the same point is analyzed for critical exponent problems. Indeed this never happens in the case of positive solutions, see [4]. Note that solutions of both type exist, at least in symmetric domains. Indeed for the first type of solutions it is enough to take a positive solution in a symmetric cap of the domain and reflect it by antisymmetry. In the second case bubble tower solutions have been recently found by Pistoia and Weth [5]. To describe more precisely our results, we introduce some notations. We denote by G the Green s function of the Laplace operator defined by x Gx,. = c n δ x in, Gx,. = 0 on, where δ x is the Dirac mass at x and c n = n ω n, with ω n denoting the area of the unit sphere of R n. We denote by H the regular part of G, that is, Hx,x = x x n Gx,x for x,x. For x = x,x \ Γ, with Γ ={y, y/y }, we denote by Mx the matrix defined by Then we have Mx = m ij i,j, where m ii = Hx i,x i, m = m = Gx,x..4 Theorem.. Let n 3 and let u ε be a family of sign-changing solutions of satisfying. and let a ε,, a ε, be the concentration points obtained in Theorem.. Assume that there exists a positive constant η such that η max u ε / min u ε η..5 Then, a ε, and a ε, are two global extremum points of u ε and we have

4 350 M. Ben Ayed et al. / Journal of Functional Analysis μ ε,i = uε a ε,i /n ε/ /β /n n for i =,..6 In addition, there exists a positive constant γ such that, for ε small, More precisely, we have a ε, a ε, γ, da ε,i, γ for i =,..7 c ε/c / u ε a ε,i β n /Λ i, a ε,i a i with a a, for i =,,.8 where Λ i is a positive constant, c = β n n+/n δ R n 0, and c = n βn/n n Furthermore, a, a, Λ, Λ is a critical point of the function Log + x x dx + x n+. R n Ψ : \ Γ 0, R; a, Λ := a,a,λ,λ t ΛMaΛ logλ Λ, where Ma is the matrix defined by.4. Remark.3. The assumption.5 allows us to prove that the distance between the concentration points is bounded from below by a positive constant. The low energy positive solutions of have to blow up at a critical point of the Robin s function ϕx = Hx,x, see [3] and [7]. 3 A similar matrix to that involved in the above function Ψ plays also a crucial role in the characterization of the concentration points for the positive solutions of, see [3]. The next result describes the asymptotic behavior, as ε 0, of low energy sign-changing solutions of satisfying.5 outside the limit concentration points. Theorem.4. Let n 3 and let u ε be a family of sign-changing solutions of satisfying. and.5. Then there exist positive constants m and m such that ε / u ε u := m Ga,. m Ga,. in C loc \{a, a } as ε 0. To prove the previous theorem we will make use of the analysis of the blow-up points as done in [4]. The second part of this paper is devoted to the study of symmetry properties of low energy sign-changing solutions of of the first type when is a ball. We shall prove the following results. Theorem.5. Assume that n 3. Let be the unit ball and let u ε be a family of sign-changing solutions of satisfying. and.5. Then, up to a rotation of, the limit concentration points a i s and the reals Λ i s, defined in Theorem., satisfy

5 M. Ben Ayed et al. / Journal of Functional Analysis where t is the unique solution of gt = a = a = a := 0,...,0,t, / Λ = Λ =, Ha,a + Ga, a t t n t n + t + t = 0, for t 0,. n The characterization of the points a i s and the reals Λ i s allows us to improve the result of Theorem.4 and therefore we can prove that the nodal surface intersects the boundary. In fact, we have Theorem.6. Under the assumptions of Theorem.5, we have the following. a The constants m and m defined in Theorem.4 are equal. b The nodal surface of u ε intersects the boundary. Observe that the limit function Ga,. Ga,., defined in Theorem.4 with m = m,is symmetric with respect to any hyperplane passing through the points a and a. Furthermore, it is antisymmetric with respect to the hyperplane passing through the origin and which is orthogonal to the line passing through the points a and a. Moreover, it changes sign once. Following the idea of [0], we can prove, for ε small, that the functions u ε satisfy also the symmetry property. More precisely, we have Theorem.7. Let n 3 and let be a ball and u ε be a family of sign-changing solutions of satisfying. and.5. Then, for ε sufficiently small, the concentration points a ε, and a ε, of u ε, given by Theorem., are bounded away from the origin and they lie on the same line passing through the origin and u ε is axially symmetry with respect to this line. Moreover the points a ε, and a ε, lie on different sides with respect to T and all the critical points of u ε belong to the symmetry axis and u ε ν T x > 0 x T, where T is any hyperplane passing through the origin but not containing a ε, and where ν T is the normal to T, oriented towards the half space containing a ε,. Concerning the antisymmetric property, it is easy to construct a family of solutions u ε which are antisymmetric by reflecting the positive minimizing solution on the half ball and hence interesting questions arise: Let u ε be a family of solutions satisfying the assumptions of our theorems, are the solutions u ε antisymmetric? What about the uniqueness? A further investigation in this direction is in progress. The last part of this paper is devoted to the study of the case where the assumption.5 is removed, that means the case when we can have sign-changing bubble tower solutions. We recall that Pistoia and Weth [5] have constructed such solutions in symmetric domains. Without loss of

6 35 M. Ben Ayed et al. / Journal of Functional Analysis generality, we can assume, in the case where the assumption.5 is removed, that the following holds: Our main result reads: max u ε min u ε as ε 0..9 Theorem.8. Let n 4 and let u ε be a family of sign-changing solutions of satisfying. and.9. Let a ε,, a ε, be the concentration points and μ ε,, μ ε, the speeds of the concentration points obtained in Theorem.. Then, a there exists a positive constant c such that, for ε small, c μ ε, μ ε, μ ε, d ε, cμ ε,, μ ε, a ε, a ε, 0, a ε, a ε, d ε, 0, c ε μ ε, d ε, n cε, μ ε,μ ε, a ε, a ε, c, d ε, d ε, as ε 0, where d ε,i = da ε,i, for i =, ; b the nodal surface of u ε does not intersect the boundary. Furthermore, if n 6, we have μ ε, μ ε, a ε, a ε, 0, d ε,i 0 and a ε,i a for i =,,.0 where a is a critical point of the Robin s function ϕx = Hx,x. Let us mention that we are not able to extend the results of Theorem.8 to the dimension 3 because of serious technical difficulties. Also the restriction to n 6 to deduce that the unique limit concentration point is away from the boundary is due to some technical difficulties but we think that the same result should be true also in lower dimensions, at least for n = 4, 5. Moreover in the case of the ball since the only critical point of the Robin s function is the center, from Theorem.8 we deduce, for n 6, that the bubble tower concentration point is the center, with negative and positive part blowing up with a prescribed rate. This makes us to conjecture that this solution should be radial and hence should be the only solution of this type. The outline of the paper is the following. Section is devoted to the proof of Theorems. and.. We prove Theorems.4.7 in Section 3. Finally, Section 4 is devoted to the proof of Theorem.8.. Proof of Theorems. and. First, we deal with the proof of Theorem.. Regarding the connected components of \{x u ε x = 0}, let be one of them. Multiplying by u ε and integrating on, we derive that

7 M. Ben Ayed et al. / Journal of Functional Analysis u ε S n/ + o,. where we have used Hölder s inequality and the Sobolev embedding. Since u ε S n/ as ε goes to 0, we deduce that there are only two connected components. The following lemma shows that the energy of the solution u ε converges to S n/ in each connected component. In fact we have Lemma.. Let u ε be a family of sign-changing solutions of satisfying.. Then i ii u + ε S n/, u + ε n n S n/, iii u ε 0 as ε 0, iv u ε S n/ as ε 0, u ε n n S n/ as ε 0, M ε,+ := max u+ ε +, M ε, := max u ε where u + ε = maxu ε, 0 and u ε = max0, u ε. + as ε 0, Proof. The proof is the same as that of Lemma. of [8], so we omit it. Now, we are going to prove Theorem.. Proof of Theorem.. Arguing as in the proof of Theorem. of [8], we obtain that there exist a ε,, a ε,, μ ε, and μ ε, such that, as ε 0, u + ε Pδ a ε,,μ ε, 0, u ε Pδ a ε,,μ ε, 0, μ ε,i da ε,i, +, for i =,.. Next, we will prove that one of the points is the global maximum point of u ε. Arguing as in the proof of Lemma. of [8], we obtain M /n ε/ ε da ε, 0 asε 0,.3 where a ε satisfies u ε a ε = u ε := M ε. Without loss of generality, we can assume that u ε a ε >0. Now, let us define ũ ε X := M ε u ε aε + X/M /n ε/ ε, for X ε := M /n ε/ ε a ε. By Lemma., the limit domain of ε, denoted by Π, has to be the whole space R n or a half space and by.3, it contains the origin. Since ũ ε is bounded in ε, using the standard elliptic theory, it converges in Cloc Π to a function u satisfying

8 354 M. Ben Ayed et al. / Journal of Functional Analysis u = u u in Π, u0 =, u= 0 on Π, and Π u S n/..4 Observe that, any sign-changing solution w of.4 satisfies w > S n/. Thus we derive that u is positive. It follows that Π has to be R n and u = δ / n 0,β n. We deduce that ũ ε > 0inany compact subset of ε. But we have u ε = 0on +, where + := {x : u ε x > 0}. Hence u + ε Mε εn / Pδ aε,μ ε 0 and M /n ε/ ε da ε, + as ε 0,.5 where μ ε := M /n ε/ ε M εn / ε /βn /n. Now, we observe that S n/4 + o = Mε εn / Pδ aε,μ ε u + ε Mεn / ε Pδ aε,μ ε + u + ε = S n/4 + o..6 Thus, Lemma. and.6 imply that Mε ε goes to as ε 0. The proof of our theorem is thereby completed. The goal of the sequel of this section is to prove Theorem.. We start by the following proposition which gives a parametrization of the function u ε. It follows from the corresponding statements in []. Proposition.. Let n 3 and let u ε be a family of sign-changing solutions of satisfying.. Then the following minimization problem min { u ε α Pδ a,λ + α Pδ a,λ, α i > 0, λ i > 0, a i } has a unique solution α,α,a,a,λ,λ up to permutation. In particular, we can write u ε as follows u ε = α Pδ a,λ α Pδ a,λ + v, where v H0 such that v 0 as ε 0 and V 0 : v,ϕ =0 for ϕ { Pδ ai,λ i, Pδ ai,λ i / λ i, Pδ ai,λ i / a j i,i=,, j n}, where a j i denotes the jth component of a i..7 Remark.3. For each i =,, the point a i is close to a ε,i and each parameter λ i satisfies λ i /μ ε,i is close to, where a ε,i, μ ε,i are defined in Theorem. and a i, λ i are defined in Proposition..

9 M. Ben Ayed et al. / Journal of Functional Analysis As usual in this type of problems, we first deal with the v-part of u ε, in order to show that it is negligible with respect to the concentration phenomena. Using the following estimate see Section 3. of [9] δ ε i x = βn ε λ εn / i + O ε log + λ i x a i = + o.8 since λ ε i asε 0 and arguing as in Lemma 3.3 of [8] we derive the following lemma. Lemma.4. The function v defined in Proposition. satisfies the following estimate where v cε + c { i i + ε λ i d i n log ε n /n if n<6, + ε n+/n λ i d i n+/ εn log ε n+/n if n 6, λ ε = + λ n/ + λ λ a a. λ λ Now, arguing as in the proof of Propositions 3.4, 3.5 and 3.6 of [8] and using.8 we obtain the following results. Proposition.5. Assume that n 3 and let α i, a i and λ i be the variables defined in Proposition.. We then have α4/n ε i = O ε + βn ελεn / i n Ha i,a i α i c λ n i cε + c λ i d i n + ε + v,.9 α j c λ i ε logλ k d k k=, k=, + n Ha,a λ i λ λ n / λ k d k n n + ε n log ε if n 4, + ε λ k d k log ε /3 if n = 3, where i, j {, } with i j, and c, c are defined in Theorem.. Proposition.6. Let α i, a i and λ i be the variables defined in Proposition.. a For n 4, we have Ha i,a i α i λ n + α j ε H a,a a i i λ i a i a i λ λ n / = O λ k d k n + ε n n log ε + εε log ε n n + k=, where i, j {, } with i j. α i n c ε ε λ i d i n.0,.

10 356 M. Ben Ayed et al. / Journal of Functional Analysis b For n = 3, we assume that a a then we have for i {, } λ i Ha i,a i a i + λ i ε a i H a,a a i λ λ / where η, η are two positive parameters chosen such that = o k=, Ba,η Ba,η = and Ba i,η i, for i =,. λ k η k,. Note that the proof of Propositions.5 and.6 are based on some integral estimates proved in [] and [8]. To deal with dimension 3, we prove the following lemma. Lemma.7. Assume that n = 3 and assume further.5 holds. Then there exists a positive constant c 0 such that a a c 0 maxd,d, where the points a i s are defined in Proposition. and d i = da i,. Proof. Arguing by contradiction, we assume that a a =omaxd,d. This implies that d /d asε 0. Now, choosing η = η = a a /4, we see that and Ba,η Ba,η = and Ba i,η i, for i =,, λ i η i c λ λ a a cε, Ha i,a i λ a c i i λ i d i = o λ λ a a = o ε, Ha,a λ i λ λ / a c i λ i d i = o ε, λ i λ λ / = a i a a λ i λ λ / a a cε, where we have used the fact that λ and λ are of the same order. Applying. and the above estimates, we derive a contradiction and therefore our lemma follows. Next we prove the following crucial lemmas. Lemma.8. Under the assumptions of Theorem., there exists a positive constant c 0 > 0 such that the variable a i, defined in Proposition., satisfy where d i = da i,. i c 0 d d c 0 ; ii c 0 a a d i c 0, for i =,,

11 M. Ben Ayed et al. / Journal of Functional Analysis Proof. On the one hand, using.0, we have ε = O λ j d j n + ε..3 On the other hand, using.5 an easy computation shows that ε = λ λ a a n / + O ε n/n,.4 ε λ i := n λ i ε λ j n ε λ = n i λ λ a a n / + oε..5 Thus, using.9,.3 and.5, the estimate.0 becomes λ n i Ha i,a i λ n i + Ga,a λ λ n / c ε = o ε + c λ j d j n..6 Now we claim that Ha i,a i + G a,a a i λ i a i λ λ n / = o ε n /n + λ j d j n..7 For n 4,.7 follows immediately from.9,. and.3. For n = 3, choosing η i = minc 0, d i /4 in. where c 0 is the positive constant defined in Lemma.7, Claim.7 follows from.. Now we are going to prove Claim i. Arguing by contradiction, we assume, for example, that d = od. Using.6 for i = and i =, we get λ d n = oε..8 Using.4,.8 and the fact that Ha,a / a k cdk d d n/, it is easy to obtain G λ λ n / a,a λ k a cεn /n for k =,..9 k Clearly,.7,.8 and.9 give a contradiction. Thus, we derive that d and d are of the same order. Hence Claim i is proved. Regarding Claim ii, arguing by contradiction, we assume that d = o a a. In this case, it is easy to obtain G λ λ n / a,a a c λ λ n / a a n + c d a a n = o d λ d n..0 Thus,.7 and.0 give again a contradiction and we derive that d / a a is bounded below. Hence, by Lemma.7, the proof is completed for n = 3.

12 358 M. Ben Ayed et al. / Journal of Functional Analysis It remains to prove that d / a a is bounded above for n 4. To this aim, we argue by contradiction and we assume that a a =od. Therefore, λ n Ha,a a c λ d n = o + Ha,a λ λ λ n / a λ λ a a n /.. We now observe that ε λ a c λ λ λ n / a a n c λ λ a a.. n / Hence,.7,. and. give a contradiction, and therefore a a /d is bounded below. Finally, using Claim i, the proof of Claim ii is completed. Now, we will prove that the concentration points are in a compact set of and they are far away of each other. Lemma.9. There exists a positive constant d 0 such that a a d 0 ; d i d 0 for i =,. Proof. The proof is the same as that of Lemma 3.8 of [8], so we omit it. Now we are ready to prove Theorem.. Proof of Theorem.. Without loss of generality we can assume that M ε := max u ε M ε, := min u ε. Hence.5 holds. Now, let b ε be such that M ε, := u ε b ε. Using.5 and arguing as in the proof of.5 we can prove that u ε M εn / ε, Pδ bε /n ε/,μ ε, 0 and M ε, db ε, as ε 0,.3 where μ ε, := M /n ε/ ε, /βn /n and := {x : u ε x < 0}. Hence a ε, and a ε, can be chosen as a ε and b ε which are two global extremum points of u ε. Regarding, Claim.7, it follows from Lemma.9. Therefore each a i converges to a i with a a. Now, let us introduce the following change of variable λ n / i c ε / = Λ i. c Note that,.6 and.7 imply, for i, j =, with j i,

13 M. Ben Ayed et al. / Journal of Functional Analysis Ha i,a i Λ i + Ga,a Λ j Λ i = oλ i,.4 Ha i,a i a i Λ i + Ga,a a i Λ Λ = o Λ i..5 Since each a i converges to a i with a a, thus the functions H, G and its derivatives are bounded. Therefore, from.4 and.5, it is easy to see that for each i =,, Λ i is bounded above and below. Hence, each Λ i converges to Λ i > 0 up to a sequence which implies.8 see.6 and Remark.3. Passing to the limit in.4 and.5, we get Ha i, a i Λ i + Ga, a Λ j Λ i = 0,.6 Ha i, a i Λ i a + Ga, a Λ Λ = 0, i a i.7 where i, j =, with j i. Equations.6 and.7 imply that Ψa, a, Λ, Λ = 0. Hence a, a, Λ, Λ is a critical point of Ψ. 3. Proof of Theorems.4.7 Regarding Theorem.5, it follows immediately from Theorem. and the following lemma. Lemma 3.. Let be a ball and assume that n 3. Then, up to a rotation of, the function Ψ, defined in Theorem., has only one critical point X := a,b,x,y. It satisfies where t is the unique solution of gt = a = b = 0,...,0,t with t > 0, / x = y =, Ha,a + Ga, a t t n t n + t + t = 0, for t 0,. n Proof. Let a,a,x,x be a critical point of Ψ. Then, for i, j =, with j i, we derive Multiplying 3. by x i, we get Ha i,a i x i + Ga,a x j = x i, 3. x i Ha,a a Recall that when is the unit ball, we have Ga,a j + x j a=ai a = a=ai Ha,a x = Ha,a x. 3.3

14 360 M. Ben Ayed et al. / Journal of Functional Analysis Thus, Ga, b = Ha,a = Ha,a a a b n a b, a,b n / = a. 3.5 n n a a, 3.6 n G n b a n b b a a, b = a a b n a b a,b n/ First, using 3., 3.6 and 3.7, it is easy to prove that a i 0fori =,. Without loss of generality, we can assume that a = 0,...,0,γ, where γ is a constant. Taking the jth component for j =,...,n of the vector defined by 3., with i =, it follows that a = 0,...,0,γ, where γ is a constant. Hence a and a lie on the same line passing through the origin. It remains to prove that γ = γ. Using 3., 3.3, 3.6 and 3.7, we get γ i Ha γi,a H a,a / γ j γ i + γ γ n γ j = 0, 3.8 γ γ n for i, j =, with j i. Adding 3.8 for i = and i =, we derive γ + γ γ γ γ n/ γ = γ + γ. 3.9 n/ γ γ n Thus, if γ + γ 0, 3.9 implies that γ γ = γ γ which implies that γ = γ and therefore a = a which is a contradiction. Thus γ + γ = 0, that means a = a = 0,...,0,t, with t is the unique solution of gt = t t n t n + t + t = 0, for t 0,, n where we have used 3.8. Now using 3.3, 3.5 and the fact that the reals x i s are positive, it is easy to obtain that x = x. Using again 3. we derive that / x = x = 3.0 Ha,a + Ga,a which completes the proof of our lemma. Next we are going to prove Theorem.4.

15 M. Ben Ayed et al. / Journal of Functional Analysis Proof of Theorem.4. Observe that, by Theorems. and., we know that u ε can be written as Pδ aε,,μ ε, Pδ aε,,μ ε, + v with v 0, u ε a ε, = max u ε, u ε a ε, = min u ε and μ ε,i a ε, a ε,,fori =,. Furthermore, the concentration speeds satisfy.8. Set h ε := max dx,s n / u ε x where S ={a ε,,a ε, }. It is easy to prove that h ε is bounded if not, we can construct another blow-up point and therefore the energy of u ε becomes bigger than 3S n/ which gives a contradiction. Let d ε, = da ε,, + and d ε, = da ε,,, where + ={x : u ε x > 0} and ={x : u ε x < 0}. We need to prove that d ε,i 0asε 0. Arguing by contradiction, assume that d ε, d ε, and d ε, 0. We define the following function w ε X := d α ε ε, u εa ε, + d ε, X for X ε,+ := d ε, + a ε,, where α ε = n /4 εn. An easy computation shows that w ε = w p ε ε, w ε > 0 in ε,+, w ε = 0 on ε,+, where p = = n + /n. Observe that B0, ε,+ and w ε > 0in ε,+. Since h ε is bounded and d n / ε, u ε a ε, see.5 and.3, using [4], we deduce that 0 is an isolated simple blow-up point of w ε, which means that the spherical averages of w ε centered at a ε,, suitably weighted, have exactly one critical point for ε sufficiently small. Hence, we have w ε 0w ε y c y n, for all y /. 3. By standard elliptic theories, we derive that w ε 0w ε converges in Cloc Π to a function w satisfying w = 0 inπ \{0}, w= 0 on Π, where Π is the limit domain of ε,+. Since 0 is an isolated simple blow-up point of w ε we deduce that 0 is a nonremovable singularity and therefore w = cg Π, where G Π is the Green s function and c is a positive constant. Now, using Pohozaev identity in the form of Corollary. of [4] we obtain cε + o B0,σ w p+ ε ε where c is a positive constant and σ p + ε B0,σ w p+ ε ε = B0,σ Bσ,x,w ε, w ε dx, 3. Bσ,x,w ε, w ε = n w w ε ε ν σ w ε wε + σ. ν Observe that, using 3., we obtain

16 36 M. Ben Ayed et al. / Journal of Functional Analysis w ε 0σ B0,σ ε + o w ε 0 w p+ ε ε cw ε 0 +ε p σ εn n 0 asε 0, 3.3 B0,σ w p+ ε ε cd α ε ε, εu εa ε, 0 asε 0, 3.4 where we have used.8 and the fact that α ε n / and d ε, 0asε 0. For the last term in 3., an easy computation shows lim ε 0,σ 0 B0,σ B σ,x,w ε 0w ε,w ε 0 w ε dx = chπ 0, Clearly, 3.,..., 3.5 and the fact that Π R n yield a contradiction. Hence d ε, 0as ε 0 and therefore a is an isolated simple blow-up point of u ε. The same holds for a.now, arguing as in the proof of 4.0 of [7], the result follows. Now, we are going to prove Theorem.6. Proof of Theorem.6. We start by proving Claim a. By Theorem.4, the points a i s are two isolated simple blow-up points of u ε. Thus, as in 3., we derive that cε + o Ba ε,i,σ u p+ ε ε σ p + ε Ba ε,i,σ u p+ ε ε = Ba ε,i,σ for i =,, where c is a positive constant independent of i. As in 3.3, we have σu ε a ε,i and using.8, we have Ba ε,i,σ Bσ,x,u ε, u ε dx, 3.6 u p+ ε ε cu ε a ε,i +ε p σ εn n 0 asε 0, 3.7 εu ε a ε,i Ba ε,i,σ u p+ ε ε S n/ εu ε a ε, S n/ c β n c Λ i as ε It remains to study the right side integral of 3.6. Using again.8 and Theorem.4, we derive that u ε a ε,i u ε m i Ga i,. m j Ga j,. 3.9 where j i and m, m are two positive constants satisfying m /m = m /m. Thus

17 lim ε 0,σ 0 Ba ε,i,σ M. Ben Ayed et al. / Journal of Functional Analysis B σ,x,u ε a ε,i u ε,u ε a ε,i u ε dx = c m i Ha i, a i m j Ga, a, for i, j =, with i j and where c is a positive constant independent of i and j. Using 3.9 and 3.0 and the fact that Λ = Λ see Theorem.5, we deduce that m Ha, a m Ga, a = m Ha, a m Ga, a. Hence since a = a, using 3.5, we derive that m m Ha, a Ga, a = It is easy to verify that Ha, a Ga, a and therefore we obtain that m = m which implies that m = m. The proof of Claim a is thereby completed. It remains to prove Claim b. Arguing by contradiction and assuming that the set {x, u ε x = 0} does not intersect the boundary. Thus u ε / ν does not change sign which implies that u/ ν does not change sign also, where u is defined in Theorem.4. Now, since m = m see Claim a, an easy computation shows that u ν x dx = m Ga,x Ga,x dx = 0, ν which implies a contradiction. Thus the result follows. Now, we are going to prove Theorem.7. Proof of Theorem.7. According to Theorem.5, we know that, for ε close to 0, both points a ε,i are bounded away from the origin and they lie on different sides with respect to T, where T is any hyperplane passing through the origin but not containing a ε,. Arguing now as in the proof of Lemma 4. of [8], we see that the points a ε,i lie on the same line passing through the origin. Lastly, the proof of the other statements of Theorem.7 is exactly the same as that of Theorem.5 of [8], so we omit it. This ends the proof of our result. 4. Proof of Theorem.8 Throughout this section, c stands for a generic constant depending only on n and whose value may change in every step of the computations. To prove Theorem.8, we need a delicate analysis and careful estimates. First, for ε sufficiently small, Proposition. implies that u ε can be uniquely written as u ε = α,ε Pδ a,ε,λ,ε α,ε Pδ a,ε,λ,ε + v ε, 4. where v ε H 0, v ε 0asε 0 and v ε satisfies.7.

18 364 M. Ben Ayed et al. / Journal of Functional Analysis To simplify the notations, we write α i, λ i, Pδ i, a i and v instead of α i,ε, λ i,ε, Pδ ai,ε,λ i,ε, a i,ε and v ε, respectively. We will also use the following notations: λ d i = da i, and ε = + λ n/ + λ λ a a. λ λ Now, we prove the following crucial lemmas: Lemma 4.. Assume that n 4 and ε = λ λ a a n/ + oε as ε Then, for ε small, there exists a positive constant c such that i d c d cd and ii λ c λ cλ. Proof. On the one hand, using 4., we derive that.5 holds. Thus, using.9,.0,.3 and.5, we see that.6 holds. On the other hand,.9,. and.3 imply that.7 holds. Now, arguing by contradiction we assume, for example, that λ d = oλ d. Using.6 for i = and i =, we obtain.8. Using 4. and.8, we get G λ λ n / a,a λ a cεn /n. 4.3 Clearly,.6 for i =, 4.,.8 and 4.3 give a contradiction. Thus, we derive that λ d and λ d are of the same order. Assume, for example, that d = od. In this case, it is easy to obtain that a a d d d /. Thus.0 holds. Obviously,.6 and.0 give a contradiction and therefore Claims i and ii are proved. Lemma 4.. Let n 4 and assume that.9 holds. Then, for ε small, we have a a =od and d = d + od. Proof. Observe that Remark.3 implies that λ /λ + as ε 0. Thus, using Lemma 4., we derive that there exists a positive constant c such that λ λ cλ λ a a. Thus λ a a c. Now,sinceλ d + as ε 0, we derive that a a =od and therefore d = d + od. Remark 4.3. Notice that assumption λ /λ + as ε 0 and Lemma 4. imply that ε c λ λ n / for ε small.

19 M. Ben Ayed et al. / Journal of Functional Analysis Lemma 4.4. Let n 4 and assume that.9 holds. Then, there exists a positive constant c such that, for ε small, i ii iii c ε λ d n cε, c ε ε cε, c λ λ λ d cλ. Proof. Using.0 for i =, we see that c ε c ε = o λ d n + ε + ε. Thus ε = c ε + o c λ d n + ε. 4.4 Using.3,.5 and.0 for i = and for i =, we derive that c Ha,a λ n n c ε ε λ Ha,a + o = o λ λ n + ε. 4.5 But, by Remark 4.3, we know that there exists c>0 such that c λ λ ε /n. 4.6 Clearly, 4.4, 4.5, 4.6 and the fact that c Ha,a d n imply Claims i and ii. Now, it follows from Claim i and Lemma 4. that the following holds: n c λ c ε λ λ d n cε c where c, c and c are positive constants. Therefore Claim iii follows. λ λ n, Lemma 4.5. Let n 4 and assume that.9 holds. Then, there exists a positive constant c such that, for ε small, i λ λ a a c, ii λ a a cε /n 0 as ε 0.

20 366 M. Ben Ayed et al. / Journal of Functional Analysis Proof. Using. for i = and Lemma 4.4, we obtain λ n H n a,a + n λ a a ε n a = o λ d n = o ε n n. 4.7 Now, from 4.6 we deduce that c λ λ a a ε n n λ a a ε n/n λ λ a a ε n n. 4.8 Arguing by contradiction, assume that λ λ a a +. Using 4.7, 4.8 and the fact that H a,a a c n n cε λ d n, λ n we obtain a contradiction and therefore Claim i follows. Finally, Claim ii follows from Claim i and Remark 4.3. Now, to deal with the case of n 6, we need the following crucial proposition which improves the estimate.. Proposition 4.6. Let n 6 and assume that.9 holds. Then we have ε = o ε n+/n λ a. Proof. For sake of simplicity, we will use the following notations: Pδ i = Pδ ai,λ i, δ i = δ ai,λ i, θ i = δ i Pδ i, u ε = α Pδ α Pδ, ϕ = λ Pδ a, ψ = λ δ a. Multiplying by ϕ and integrating on, we obtain α δ p ϕ α δ p ϕ = u ε p ε u ε ϕ, 4.9 where p = n + /n. For the left-hand side of 4.9, it follows from [] that δ p δ p Pδ = λ a Pδ = λ a R n δ c λ n p Ha,a a δ + O λ a + O p δ c λ d n δ + δ p λ, 4.0 θ a

21 M. Ben Ayed et al. / Journal of Functional Analysis = c ε + O λ a a ε n+ n λ a + o λ d n λ d n = c ε + o ε n+/n λ a, 4. where we have used Lemmas 4., 4.4 and 4.5. For the other integral in 4.9, an easy expansion implies that u ε p ε u ε ϕ = u ε p ε u ε ϕ + p u ε p vϕ + O v + ε v + v p ε θ +ε / log ε n n = u ε p ε u ε ϕ + p u ε p ε vϕ + o ε n+/n, 4. where we have used in the last inequality Lemmas.4 and 4.4 and the fact that θ cλ d n/. For the last integral in 4., arguing as in in [8], we obtain u ε p vϕ c v ε n+ n log ε n+ n + logλ d λ d n+/ = o ε n+ n, 4.3 where we have used in the last inequality Lemmas.4 and 4.4. It remains to study the first integral in the right-hand side of 4.. Denoting f = α δ α δ, we observe that u ε p ε u ε ϕ = f p ε fψ + p ε f p ε α θ α θ ϕ R n + O f p ε θ λ a + f p θ + θ ϕ + O Now, using Lemma 4.4, we can write θ p + θ p λ θ p ψ Ba,d λ d n λ d n+ [ θ a λ d n+ + θ n/n δ n/n + θ +θ f + O λ d n+ Ba,d c δ θ p + θ p ] p+ λ d n ψ + θ λ a. 4.4 = o ε n+ n ; 4.5 c logλ d λ d n = o ε n+ n ; 4.6

22 368 M. Ben Ayed et al. / Journal of Functional Analysis θ p ψ Ba,d c d n+ θ p/ δ p/ x a δ n/n + d λ n + c λ d n λ d n+ Ba,d c θ p δ = o ε n+ n. 4.7 Note that, since p 0, we get f p θ + θ θ p ϕ + θ p ψ + θ λ a = o ε n+ n, 4.8 θ +θ f where we have used Now, using Lemmas 4. and 4.4, we derive that f p ε θ λ a + f p ε δ p θ ϕ λ d n + δp c λ d n λ d n = o ε n+ n. 4.9 Using again Lemmas 4. and 4.4, we obtain f p ε θ ϕ δ p θ + δ p θ x a δ n λ d n λ d n n + c + λ d n + δ p θ c λ d n λ θ a λ d n = o ε n+ n. 4.0 Now, we deal with the first integral in the right-hand side of 4.4. To this aim, we set δ := δ a,λ. We note that, using Lemma 4.5, we derive that We now introduce the following sets: where δ = δ + O λ a a δ. 4. A := { { x R n : δ ε /6 δ } = x R n : x a β }, λ λ A := { { x R n : δ ε /6 δ } = x R n : x a A 3 := R n \ A A, β λ λ β := λ ε /3n λ / λ ε /3n λ / ε /3n as ε 0. },

23 M. Ben Ayed et al. / Journal of Functional Analysis We will estimate the first integral in the right-hand side of 4.4 on each set A i for i =,, 3. First, we write f p ε fψ = α δ p ε ψ + O A A A p δ δ n n x a. 4. Recall that ε satisfies 4.6. Hence we get p δ δ n n x a ε n 4 6n δ n n n δ n+ x a A A n 4 6n ε λ n x a =o ε n+ n, 4.3 n+ n δ A p ε δ ψ = p ε δ R n = R n δ ψ p ε δ ψ A A 3 p ψ + O εε log ε n n A + O λ a a A A 3 p δ ψ, 4.4 where we have used the evenness of δ, the oddness of ψ,.8 and 4.. Note that, as in 4., we get A A 3 p δ ψ ε 6n A A 3 n+ n δ x a δ n+ n ε 6n ε n+ n log ε. 4.5 Combining , ii of Lemma 4.5 and using the estimate F of [], we obtain A f p ε p ε c ε fψ = α + o ε n+ n λ a. 4.6 Secondly, we write f p ε fψ = α δ p ε ψ p ε α δ p ε α δ ψ A A A 6 + O δ δ n x a. 4.7 A Note that

24 370 M. Ben Ayed et al. / Journal of Functional Analysis p ε A A 6 δ δ n x a ε p ε δ δ ψ = p p δ R n n 5 6n δ ψ p A n+ n δ δ n+ n A A 3 δ x a =o ε n+ n, 4.8 p δ ψ + O εε log ε n n. 4.9 Using the evenness of δ, the oddness of ψ and 4., we get p p δ ψ = O λ a a δ ψ A A 3 δ A A 3 δ Arguing as in 4.3, we obtain p δ A A 3 δ ψ = o ε n+ n. 4.3 Thus, using the evenness of δ, the oddness of ψ, the estimate F of [] and combining , we derive that f p ε p ε c ε fψ = α α + o ε n+ n λ a. 4.3 A Lastly, denoting by f = α δ α δ, and using the evenness of f, we obtain A 3 f p ε fψ = O A 3 f p δ δ ψ + δ δ f δ δ p ψ Note that the set {x: δ δ f } implies that δ = δ + o. Hence it is contained in F := {x: x a cλ λ / }. Therefore, as in 4.3, we get δ δ f δ δ p ψ c λ a a n+ n F x a δ p = o ε n+ n Now, arguing as in 4.5, we find f p δ δ ψ c λ a a p δ + δ p n δ x a δ n = o ε n+ n A3 A 3 Combining , we obtain f p ε n+ fψ = o ε A 3 n δ n n

25 M. Ben Ayed et al. / Journal of Functional Analysis Clearly, our proposition follows from , 4.6, 4.3 and Now, we are ready to prove Theorem.8. Proof of Theorem.8. According to Remark.3 and Lemmas 4., 4.4 and 4.5, Claim a follows. Now we will prove.0 in the case where n 6. On the one hand, observe that Proposition 4.6 and Remark 4.3 imply ε λ a = λ ε λ λ a = o ε n n On the other hand, Remark 4.3 implies that ε λ a = cλ n a a ε n c λ λ a a ε n n Clearly, from 4.37, 4.38 and Remark.3, we get the first claim of.0. Now, using 4.37, Lemma 4.4 and. for i =, we derive that This implies that λ n H a,a = o a λ d n. d 0 and H a a,a 0 asε 0. This completes the proof of.0. It remains to prove Claim b. Observe that, by Theorem. and assumption.9, we have u ε = Pδ aε,,μ ε, Pδ aε,,μ ε, + v with v 0 and u ε a ε, = max u ε First we claim that m>0 s.t. h ε := max x x a ε, n / uε x m In fact, if h ε +, then, using the method of R. Schoen [4], we can construct a concentration point b ε with a concentration speed c u ε b ε /n ε/ and the function u ε becomes close to Observe that, since h ε +, we derive that Pδ aε,,μ ε, Pδ bε,c u ε b ε /n ε/. c u ε b ε /n ε/ b ε a ε,, which contradicts the conclusions of Theorem.8. Hence our claim is proved.

26 37 M. Ben Ayed et al. / Journal of Functional Analysis Now, let and we introduce the following function w ε X := μ n / ε, The function w ε satisfies := \ B a ε,,μ εn /4 ε,, u ε aε, + μ εn /4 ε, X for X ε := μ εn /4 ε, a ε,. { wε = w ε ε w ε in ε, w ε = 0 on ε. Observe that, using 4.40 we derive that w ε x m for each x := ε \ B0,. Hence w ε converges in C loc Rn \ B0, to a function w satisfying w = w w in R n \ B0,. But from 4.39, we deduce that w has to be δ /n 0,β n where β n is defined in Theorem.. Thus w< c <0in B0, which implies that w ε < 0 in the same set. Hence u ε < 0in B ε := Ba ε,, μ εn /4 ε,. Now since μ ε, da ε,, see Theorem. we derive that B ε is contained in a compact set of. Finally, using the fact that u ε a ε, >0, a ε, B ε since μ ε, a ε, a ε, 0 and \{x : u ε x = 0} has exactly two connected components, we deduce that the nodal surface does not intersect the boundary of. Hence Claim b follows. This completes the proof of our theorem. Acknowledgments This work has begun while the first and the second authors were visiting the Mathematics Department of the University of Roma La Sapienza. They would like to thank the Mathematics Department for its warm hospitality. Part of it was accomplished when K. El Mehdi was visiting the Institut des Hautes Études Scientifiques IHÉS at Bures-sur-Yvette in France. He is grateful to the Institute for the excellent atmosphere. References [] A. Bahri, Critical Point at Infinity in Some Variational Problems, Pitman Res. Notes Math. Ser., vol. 8, Longman Sci. Tech., Harlow, 989. [] A. Bahri, J.M. Coron, On a nonlinear elliptic equation involving the critical Sobolev exponent: The effect of topology of the domain, Comm. Pure Appl. Math [3] A. Bahri, Y.Y. Li, O. Rey, On a variational problem with lack of compactness: The topological effect of the critical points at infinity, Calc. Var. Partial Differential Equations [4] T. Bartsch, Critical point theory on partially ordered Hilbert spaces, J. Funct. Anal

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