Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 568245, 11 pages http://dx.doi.org/10.1155/2013/568245 Research Article Existence of Conformal Metrics with Prescribed Q-Curvature Mohammed Ali Al-Ghamdi,1 Hichem Chtioui,1 and Afef Rigane2 1 2 Department of Mathematics, King Abdulaziz University, P.O. Box 80230, Jeddah, Saudi Arabia Department of Mathematics, Faculty of Sciences of Sfax, Road Soukra, Sfax, Tunisia Correspondence should be addressed to Hichem Chtioui; hichem.chtioui@fss.rnu.tn Received 13 October 2012; Revised 5 February 2013; Accepted 8 February 2013 Academic Editor: Svatoslav StaneΜk Copyright © 2013 Mohammed Ali Al-Ghamdi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We consider the problem of existence of conformal metrics with prescribed Q-curvature on standard sphere ππ , π ≥ 5. Under the assumption that the order of flatness at critical points of prescribed Q-curvature function πΎ(π₯) is π½ ∈ ]1, π − 4], we give precise estimates on the losses of the compactness, and we prove new existence and multiplicity results through an Euler-Hopf type formula. 1. Introduction and Main Result The Paneitz operator on π-Riemannian manifold (π, π0 ) is a fourth-order differential operator which arises in conformal geometry and satisfies a certain covariance property (see [1]). For π ≥ 5 it is defined by πππ0 π’ = Δ2π0 π’ − divπ0 (ππ ππ0 π0 + ππ Ricπ0 ) ππ’ + π−4 π ππ0 π’, 2 (1) where (π − 2)2 + 4 ππ = , 2 (π − 1) (π − 2) πππ0 = − − 4 ππ = − , π−2 1 π3 − 4π2 + 16π − 16 2 Δ π0 ππ0 + ππ0 2 (π − 1) 8(π − 1)2 (π − 2)2 (2) 2 σ΅¨σ΅¨ σ΅¨2 σ΅¨Ricπ0 σ΅¨σ΅¨σ΅¨ . 2 σ΅¨σ΅¨ σ΅¨ (π − 2) Such a πππ0 is a fourth-order invariant called Q-curvature or Paneitz curvature. See [2, 3] for details about the properties of πππ0 . If π = π’4/(π−4) π0 is a conformal metric to π0 , where π’ is a smooth positive function, then the conformal covariance property of the Paneitz operator reads as follows: ∀π ∈ πΆ∞ (π) : πππ0 (π’π) = π’(π+4)/(π−4) πππ (π) . (3) If one prescribes the Q-curvature for the metric π by a function πΎ, this leads to the following equation: πππ0 (π’) = π’(π+4)/(π−4) πππ (1) = π − 4 (π+4)/(π−4) . πΎπ’ 2 (4) The literature for the existence of solutions of the prescribed Q-curvature problem on compact manifolds is considerably bigger. In [4–7], existence results for the constant Qcurvature problem in 4-dimensional manifolds are given. On manifolds of dimension greater than 4, existence results were given for Einstein manifolds in [3]. On the sphere ππ , we refer to results of [8–14] and the references therein. In this paper we continue to study the problem of prescribing Q-curvature on the standard sphere (ππ , π0 ), π ≥ 5. The Paneitz operator in this case is coercive on the Sobolev space π»22 (ππ ) and has the following expression: P (π’) := πππ0 (π’) = Δ2π0 π’ − ππ Δ π0 π’ + ππ π’, (5) 2 Abstract and Applied Analysis where 1 2 π−4 (6) ππ = (π − 2π − 4) , π (π2 − 4) . 2 16 According to problem (4), the problem can be formulated as follows. Given a smooth function πΎ on ππ , we look for solutions of π − 4 (π+4)/(π−4) , P (π’) = πΎπ’ 2 (7) ππ = π’>0 on ππ . The special case where the manifold is a sphere endowed with its standard metric deserves particular attention. Indeed due to Kazdan-Warner type obstructions, see [3], conditions have to be imposed on the function πΎ. To state our result we set up the following conditions and notation. (π)π½ Assume that πΎ : ππ → R, π ≥ 5 be a πΆ1 positive function such that for any critical point π¦ of πΎ, there exists some real number π½ = π½(π¦) such that in some geodesic normal coordinate system centered at π¦ (through the exponential map), we have and πΊ is the Green function for the operator P(π’) on ππ . Here π₯1 is the first component of π₯ in some geodesic normal coordinate system. Let π(ππ ) be the least eigenvalue of π(ππ ). It was first pointed out by Bahri [15] that when the interaction between the different bubbles is of the same order as the self interaction, the least eigenvalue of some matrices like (11) plays a fundamental role in the existence of solutions to problems like (7). Regarding problem (7), such kind of phenomenon appears under (π)π½ condition in Kπ−4 ; see [11]. (π΄ 1 ) Assume that π(ππ ) =ΜΈ 0 for each distinct point π¦π1 , . . . , π¦ππ ∈ K+ ∩ Kπ−4 . Now, we introduce the following set: π C∞ := {ππ = (π¦π1 , . . . , π¦ππ ) ∈ (K+ ) , π ≥ 1, s.t. π¦π =ΜΈ π¦π ∀π =ΜΈ π, and if {π¦π1 , . . . , π¦ππ } ∩ Kπ−4 =ΜΈ 0, we denote by π¦π1 , . . . , π¦ππ all elements of π σ΅¨ σ΅¨π½ πΎ (π₯) = πΎ (π¦) + ∑ ππ σ΅¨σ΅¨σ΅¨(π₯ − π¦)π σ΅¨σ΅¨σ΅¨ + π (π₯ − π¦) , {π¦π1 , . . . , π¦ππ } with π½ (π¦ππ ) = π − 4 (8) π=1 where ππ = ππ (π¦) =ΜΈ 0, for all π = 1, . . . , π, ∑ππ=1 ππ =ΜΈ 0 and [π½] ∑π =0 |∇π π (π₯ − π¦)||π₯ − π¦|−π½+π = π(1) as π₯ tends to π¦. Here π ∇ denotes all possible derivatives of order π and [π½] is the integer part of π½. For each critical point π¦ of πΎ, we denote Μπ (π¦) =: β― {ππ , π = 1, . . . , π, such that ππ < 0} . (9) Let, for each π = 1, . . . , π then π (π¦π1 , . . . , π¦ππ ) > 0} . The main result of this paper is the following. Theorem 1. Assume that πΎ satisfies (π΄ 1 ) and (π)π½ , with 1 < π½ ≤ π − 4. K = {π¦ ∈ ππ , ∇π0 πΎ (π¦) = 0} K = {π¦ ∈ K, − ∑ ππ > 0} , (10) π=1 For each π-tuple, π ≥ 1 of distinct points ππ := (π¦π1 , . . . , π¦ππ ) such that for all π = 1, . . . , π, π¦ππ ∈ K+ ∩ Kπ−4 , we define a π × π symmetric matrix π(ππ ) = (πππ ) by π π − 4 − ∑π=1 ππ (π¦ππ ) Μπ1 , π/4 π πΎ(π¦ ) −πΊ (π¦ππ , π¦ππ ) (π−4)/8 [πΎ (π¦ππ ) πΎ (π¦ππ )] (11) , where π1 = ∫ Rπ Μπ1 = ππ₯ (1 + |π₯|2 ) π02π/(π−4) (π+4)/2 , σ΅¨σ΅¨ σ΅¨σ΅¨π−4 σ΅¨σ΅¨π₯1 σ΅¨σ΅¨ ∫ π ππ₯, Rπ (1 + |π₯|2 ) ∑ π Μ (−1)π−1+∑π=1 π−π(π¦π ) − 1 =ΜΈ 0. (15) Then (7) has at least one solution. Moreover if (π − 4)/2 < π½ ≤ π − 4, for generic πΎ one has β―π ≥ |π| , (16) where π denotes the set of solutions to (7). ππ πππ = 2(π−4)/2 π1 π= (π¦1 ,...,π¦π )∈C∞ Kπ−4 = {π¦ ∈ K, π½ = π½ (π¦) = π − 4} . πππ = (14) If π + (13) (12) Our argument uses a careful analysis of the lack of compactness of the Euler Lagrange functional π½ associated with problem (7). Namely, we study the noncompact orbits of the gradient flow of π½ the so-called critical points at infinity following the terminology of Bahri [15]. These critical points at infinity can be treated as usual critical points once a Morse lemma at infinity is performed from which we can derive just as in the classical Morse theory the difference of topology induced by these noncompact orbits and compute their Morse index. Such a Morse lemma at infinity is obtained through the construction of suitable pseudogradient for which the Palais-Smale condition is satisfied along the decreasing flow lines, as long as these flow lines do Abstract and Applied Analysis 3 not enter the neighborhood of a finite number π¦1 , . . . , π¦π of critical points of πΎ such that (π¦1 , . . . , π¦π ) ∈ C∞ . Similar Morse lemma at infinity has been established for the problem (7) on the sphere ππ , π ≥ 5, under the hypothesis that the order of flatness at critical points of πΎ is π½ ∈ [π−4, π[; see [11]. The rest of this paper is organized as follows. In Section 2, we set up the variational problem and we recall the expansion of the gradient of the associated Euler-Lagrange functional near infinity. In Section 3, we characterize the critical points at infinity of the associated variational problem. Section 4 is devoted to the proof of the main result Theorem 1, while we give in Section 5 a more general statement than Theorem 1. 2. General Framework and Some Known Facts 2.1. Variational Structure and Lack of Compactness. Our problem (7) enjoys a variational structure. Indeed, solutions to (7) correspond to positive critical points of the functional π½ (π’) = ∫ππ Pπ’π’πVπ0 2π/(π−4) (∫ππ πΎ|π’| πVπ0 ) (π−4)/2π , (17) defined on = [(π π /π π ) + (π π /π π ) + (π π π π /2)(1 − Here, πππ cos π(ππ , ππ ))]−(π−4)/2 . For π€ a solution to (7) we also define π(π, π, π€) as { π (π, π, π€) = { { π’ ∈ Σ/∃ πΌ0 > 0 s.t. π’ − πΌ0 π€ ∈ π (π, π) , σ΅¨σ΅¨ 8/(π−4) σ΅¨ σ΅¨σ΅¨πΌ0 π½(π’)π/(π−4) − 1σ΅¨σ΅¨σ΅¨σ΅¨ < π. σ΅¨ (22) If π’ is a function in π(π, π, π€), one can find an optimal representation, following the ideas introduced in Proposition 5.2 of [15] and [16, pages 348–350]. Namely, we have the following. Proposition 2. For any π ∈ N∗ , there is ππ > 0 such that if π ≤ ππ and π’ ∈ π(π, π, π€), then the following minimization problem σ΅©σ΅© σ΅©σ΅© π σ΅© σ΅©σ΅© σ΅©σ΅©π’ − ∑πΌπ πΏ(π ,π ) − πΌ0 (π€ + β)σ΅©σ΅©σ΅© min σ΅©σ΅© σ΅© π π (23) σ΅© σ΅©σ΅© πΌπ >0,π π >0,ππ ∈ππ , σ΅© π=1 σ΅© β∈ππ€ (ππ’ (π€)). has a unique solution (πΌ, π, π, β), up to a permutation. In particular, we can write π’ as follows: Σ = {π’ ∈ π»22 π 2 (π ) , βπ’β := ∫ Pπ’π’πVπ0 = 1} . (18) ππ However, it is delicate from a variational viewpoint because the functional π½ does not satisfy the Palais-Smale condition. This means that there exist sequences along which π½ is bounded, its gradient goes to zero and which not converge. The analysis of sequences failing Palais-Smale condition can be analyzed along the ideas introduced in [16, pages 325 and 334]. In order to describe such a characterization in our case, we need to introduce some notations. For π ∈ ππ and π > 0, let 1 πΏ(π,π) (π₯) = π0 (π−4)/2 2 × π(π−4)/2 (π−4)/2 (1 + ((π2 − 1) /2) (1 − cos (π (π₯, π)))) , (19) π where π is the geodesic distance on (π , π) and π0 is chosen so that πΏ(π,π) is the family of solutions of the problem P (π’) = π’ (π+4)/(π−4) , π π’ > 0, on π (20) (see [17]). We define now the set of potential critical points at infinity associated with the function π½. For π > 0 and π ∈ N∗ , let us define π (π, π) π’ ∈ Σ/∃ π1 , . . . , ππ ∈ ππ , ∃π 1 , . . . , π π > π−1 , { { { σ΅©σ΅© σ΅©σ΅© π { { σ΅© σ΅©σ΅© σ΅© = { ∃πΌ1 , . . . , πΌπ > 0 with σ΅©σ΅©π’ − ∑πΌπ πΏ(ππ ,π π ) σ΅©σ΅©σ΅©σ΅© < π, { σ΅©σ΅© σ΅© { σ΅©σ΅© π=1 { σ΅© {σ΅¨ σ΅¨ 8/(π−4) π/(π−4) σ΅¨ πΌπ πΎ (ππ ) − 1σ΅¨σ΅¨σ΅¨σ΅¨ < π ∀π, πππ < π ∀π =ΜΈ π. { σ΅¨σ΅¨σ΅¨π½(π’) (21) π π’ = ∑πΌπ πΏ(ππ ,π π ) + πΌ0 (π€ + β) + V, (24) π=1 where V belongs to π»22 (ππ ) ∩ ππ€ (ππ (π€)) and it satisfies (π0 ), ππ€ (ππ’ (π€)) and ππ€ (ππ (π€)) are the tangent spaces at π€ of the unstable and stable manifolds of π€ for a decreasing pseudogradient of π½, and (π0 ) is the following: β¨V, πβ© = 0 for π ∈ {πΏπ , ππΏπ ππΏπ , , π = 1, . . . , π} ππ π πππ β¨V, π€β© = 0, β¨V, ββ© = 0 (π0 ) ∀β ∈ ππ€ ππ’ (π€) . Here, πΏπ = πΏ(ππ ,π π ) and β¨⋅, ⋅β© denotes the scalar product defined on π»22 (ππ ) by β¨π’, Vβ© = ∫ Δ π0 π’Δ π0 VπVπ0 + ππ ∫ ∇π0 π’∇π0 VπVπ0 ππ ππ (25) + ππ ∫ π’VπVπ0 . ππ Notice that Proposition 2 is also true if we take π€ = 0 and, therefore, β = 0 and π’ in π(π, π). We are ready now to state the characterization of the sequences failing the Palais-Smale condition. For technical reasons, we introduce the following subsets of Σ. Let Σ+ = {π’ ∈ Σ, π’ ≥ 0} and let for π positive very small, σ΅© σ΅©8/(π−4) ππ (Σ+ ) = {π’ ∈ Σ, π½(π’)(2π−4)/(π−4) π2π½(π’) σ΅©σ΅©σ΅©π’− σ΅©σ΅©σ΅©πΏ2π/(π−4) < π} , (26) where π’− = max(0, −π’). Using the idea introduced in [16, 18], see also [19], we have the following proposition. 4 Abstract and Applied Analysis Proposition 3. Let (π’π ) be a sequence in ππ (Σ+ ) such that π½(π’π ) is bounded and ππ½(π’π ) goes to zero. Then there exists an integer π ∈ N∗ , a sequence (ππΎ ) > 0, ππ tends to zero, and an extracted subsequence of π’π ’s, again denoted (π’π ), such that π’π ∈ π(π, ππ , π€) where π€ is zero or a solution to (7). Now arguing as in [16] (pages 326, 327, and 334), we have the following Morse lemma which completely gets rid of the V-contributions and shows that it can be neglected with respect to the concentration phenomenon. Proposition 4. There is a C1 -map which to each (πΌπ , ππ , π π , β) π such that ∑π=1 πΌπ πΏ(ππ ,π π ) + πΌ0 (π€ + β) belongs to π(π, π, π€) associates V = V(πΌ, π, π, β) such that V is unique and satisfies π π½ (∑πΌπ πΏ(ππ ,π π ) + πΌ0 (π€ + β) + V) π=1 (27) π = min {π½ (∑πΌπ πΏ(ππ ,π π ) + πΌ0 (π€ + β) + V)} . V∈(π0 ) π=1 Moreover, there exists a change of variables V − V → π such that π π½ (∑πΌπ πΏ(ππ ,π π ) + πΌ0 (π€ + β) + V) positive function tending to zero when π → +∞. Using Proposition 2, π’(π ) can be written as π π’ (π ) = ∑ πΌπ (π ) πΏ(ππ (π ),π π (π )) + πΌ0 (π ) (π€ + β (π )) + V (π ) . (31) π=1 Μ π := limπ → +∞ ππ (π ), we Μ π := limπ → +∞ πΌπ (π ), π¦ Denoting πΌ denote by π Μ 0 π€ or (Μ Μ π , π€) Μ π πΏ(Μπ¦π ,∞) + πΌ π¦1 , ..., π¦ ∑πΌ ∞ π=1 such a critical point at infinity. If π€ =ΜΈ 0, it is called of π€-type or mixed type. Notice that ππ (Σ+ ) remains invariant under the flow generated by (−ππ½) (see Lemma 5.1 of [20]; see also Lemma 4.1 of [18]). 2.2. Expansion of the Gradient of the Functional. In this subsection, we recall the expansion of the gradient of the functional π½ in π(π, π), π ≥ 1. π π=1 (28) π = π½ (∑πΌπ πΏ(ππ ,π π ) + πΌ0 (π€ + β) + V) + βπβ2 . Proposition 7 (see [11]). For any π’ = ∑π=1 πΌπ πΏπ in π(π, π), the following expansion holds (i) π=1 The following proposition gives precise estimates of V. β¨ππ½ (π’) , π π ππππ ππΏπ β© = − 2π2 π½ (π’) ∑ πΌπ π π ππ π ππ π π =ΜΈ π (33) π Proposition 5 (see [11, Lemma 3.1]). Let π’ = ∑π=1 πΌπ πΏπ + πΌ0 (π€ + β) ∈ π(π, π, π€) and let V be defined in Proposition 4. One has the following estimates: there exists π > 0 independent of π’ such that the following holds: π βVβ ≤ π∑ [ π=1 1 ππ/2 π (π+4)/2π , ππ π ≥ 12 ππ π < 12. (29) At the end of this subsection, we give the following definition. Definition 6 (see [16, 18]). A critical point at infinity of π½ on ππ (Σ+ ) is a limit of a flow line π’(π ) of the equation ππ’ = −ππ½ (π’ (π )) , ππ 1 + π (∑ πππ ) + π ( ) , ππ π =ΜΈ π where π2 = π02π/(π−4) ∫Rπ ππ¦/(1 + |π¦|2 )(π+4)/2 . (ii) If ππ ∈ π΅(π¦ππ , π), π¦ππ ∈ K and π is a positive constant small enough, one has (π+4)/2π σ΅¨ σ΅¨σ΅¨ σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ (log π π ) + π½+σ΅¨ + ] ππ π(π+4)/2 ππ π 1 −1 ) ∑ π(π+4)/2(π−4) (log πππ { { {π =ΜΈ π ππ + π{ −1 (π−4)/π { { ∑ πππ (log πππ ) , {π = ΜΈ π (32) (30) π’ (0) = π’0 , such that π’(π ) remains in π(π, π(π ), π€) for π ≥ π 0 . Here π€ is either zero or a solution to (7) and π(π ) is some β¨ππ½ (π’) , π π ππΏπ β© ππ π ππππ π − 4 2π/(π−4) πΌ + π½ π = 2π½ (π’) [−π2 ∑ πΌπ π π π0 ππ π 2π πΎ (ππ ) [ π =ΜΈ π × ∑ππ=1 ππ π½ ππ ∫ Rπ signe(π₯π + π π (ππ − π¦ππ )π ) (34) σ΅¨σ΅¨π½−1 σ΅¨ × σ΅¨σ΅¨σ΅¨σ΅¨π₯π + π π (ππ − π¦ππ )π σ΅¨σ΅¨σ΅¨ × π₯π π ππ₯ (1 + |π₯|2 ) π +π (∑ πππ + ∑ π =ΜΈ π 1 π½ π=1 π π )] . ] Abstract and Applied Analysis 5 (iii) Furthermore if π π |ππ − π¦ππ | < πΏ, for πΏ very small, one then has β¨ππ½ (π’) , π π ππΏπ β© ππ π π ππππ πΌπ ∑π=1 ππ π−4 − π2 ∑ πΌπ π π = 2π½ (π’) × [ π½π3 π½ 2π ππ π πΎ (ππ ) π π =ΜΈ π π [ π +π (∑ πππ + ∑ 1 π½ π=1 π π π =ΜΈ π )] , ] enter in the neighborhood of finite number of critical points π¦π , π = 1, . . . , π of πΎ such that (π¦1 , . . . , π¦π ) ∈ C∞ . Now we introduce the following main result. Theorem 9. Assume that πΎ satisfies (π΄ 1 ) and (π)π½ , 1 < π½ ≤ π − 4. Let π½ := max{π½(π¦)/π¦ ∈ K}. For π ≥ 1, there exists a pseudogradient π in π(π, π) so that the following holds. There exists a constant π > 0 independent of π’ = π ∑π=1 πΌπ πΏ(ππ π π ) ∈ π(π, π) such that (π) σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ + + ∑ πππ ) . ∑ π½ ππ π=1 π π π=1 π =ΜΈ π (35) π ≤ −π (∑ where π3 = π02π/(π−4) ∫ππ |π₯1 |π½ /(1 + |π₯|2 )π ππ₯. π Proposition 8 (see [11]). Letting π’ = ∑π=1 πΌπ πΏπ ∈ π(π, π), one has (i) = −π5 (π½ (π’))2(π−2)/(π−4) πΌπ(π+4)/(π−4) 1 π π =ΜΈ π π (ππ) ∇πΎ (ππ ) ππ (36) σ΅¨σ΅¨ ππ σ΅¨σ΅¨ 1 σ΅¨σ΅¨ ππ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨) + π (∑ πππ + ) , π σ΅¨σ΅¨σ΅¨ πππ σ΅¨σ΅¨σ΅¨ π π =ΜΈ π 1 β¨ππ½ (π’ + V) , π (π’) + πV (π (π’))β© π (πΌπ , ππ , π π ) (38) σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ + + ∑ πππ ) . ∑ π½ ππ π=1 π π π=1 π =ΜΈ π π ≤ −π (∑ 1 ππΏπ β© β¨ππ½ (π’) , π π πππ + π (∑ β¨ππ½ (π’) , π (π’)β© 1 Furthermore |π| is bounded and the only case where the maximum of the π π ’s is not bounded is when ππ ∈ π΅(π¦ππ , π) with π¦ππ ∈ K, for all π = 1, . . . , π, (π¦π1 , . . . , π¦ππ ) ∈ C∞ . We will prove Theorem 9 at the end of the section. Now we state two results which deal with two specific cases of Theorem 9. Let where π5 = ∫Rπ ππ¦/(1 + |π¦|2 )π . (ii) If ππ ∈ π΅(π¦ππ , π), π¦ππ ∈ K, one has π π1 (π, π) = {π’ = ∑πΌπ πΏπ ∈ π (π, π) s.t., ππ ∈ π΅ (π¦ππ , π) , π=1 1 ππΏπ β¨ππ½ (π’) , β© π π π(ππ )π = −2 (π − 4) π02π/(π−4) πΌπ(π+4)/(π−4) (π½ (π’))2(π−2)/(π−4) σ΅¨ σ΅¨π½ × ∫ ππ σ΅¨σ΅¨σ΅¨σ΅¨π₯π + π π (ππ − π¦ππ )π σ΅¨σ΅¨σ΅¨σ΅¨ Rπ π₯π 2 π+1 (1 + |π₯| ) 1 π½ ππ ππ¦ σ΅¨ σ΅¨ 1 1 σ΅¨σ΅¨ ππππ σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨) , + π (∑ πππ ) + π (∑ π½ ) + π (∑ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ π =ΜΈ π π=1 π π π =ΜΈ π π π σ΅¨σ΅¨ πππ σ΅¨σ΅¨ (37) π where π = 1, . . . , π and (ππ )π is the πth component of ππ in some geodesic normal coordinates system. 3. Characterization of the Critical Points at Infinity This section is devoted to the characterization of the critical points at infinity in π(π, π), π ≥ 1, under π½-flatness condition with 1 < π½ ≤ π − 4. This characterization is obtained through the construction of a suitable pseudogradient at infinity for which the Palais-Smale condition is satisfied along the decreasing flow-lines as long as these flow lines do not π¦ππ ∈ K \ Kπ−4 ∀π = 1, . . . , π} , π π2 (π, π) = {π’ = ∑πΌπ πΏπ ∈ π (π, π) s.t., ππ ∈ π΅ (π¦ππ , π) , π=1 π¦ππ ∈ Kπ−4 ∀π = 1, . . . , π} . (39) We then have the following. Proposition 10 (see [11], Proposition 3.7). For π ≥ 1 there exists a pseudogradient π2 in π2 (π, π) such that for all π’ = π ∑π=1 πΌπ πΏπ ∈ π2 (π, π), one has π β¨ππ½ (π’) , π2 (π’)β© ≤ −π (∑ 1 π−4 π=1 π π σ΅¨ π σ΅¨σ΅¨ σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ + ∑ πππ + ∑ σ΅¨ ), ππ π =ΜΈ π π=1 (40) where π is a positive constant independent of π’. Furthermore, one has |π2 | which is bounded and the only case where the maximum of π π ’s is not bounded is when ππ ∈ π΅(π¦ππ , π), π¦ππ ∈ K+ , for all π = 1, . . . , π, with π(π¦π1 , . . . , π¦ππ ) > 0. 6 Abstract and Applied Analysis Proposition 11. For π ≥ 1, there exists a pseudogradient π1 in π1 (π, π) so that the following holds. π There exists π > 0 independent of π’ = ∑π=1 πΌπ πΏπ ∈ π1 (π, π) such that σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ + π + ∑ ). ∑ ππ π½ ππ π=1 π π π =ΜΈ π π=1 (41) π β¨ππ½ (π’) , π1 (π’)β© ≤ −π (∑ 1 Now, we introduce the following two lemmas. π Lemma 12. Let π’ = ∑π=1 πΌπ πΏ(ππ ,π π ) ∈ π(π, π), such that ππ ∈ π΅(π¦ππ , π), π¦ππ ∈ K, for all π = 1, . . . , π. One then has β¨ππ½ (π’) , ππ (π’)β© = − 2π2 π½ (π’) ∑ πΌπ πΌπ π =ΜΈ π ππ (π’) = πΌπ π π ππΏ(ππ ,π π ) ππ π σ΅¨σ΅¨ σ΅¨π½ σ΅¨σ΅¨π₯π + π π (ππ − π¦ππ ) σ΅¨σ΅¨σ΅¨ π 1 ππΏ(ππ ,π π ) σ΅¨ πσ΅¨ ππ = πΌπ ∑ ∫ π π π(ππ )π Rπ π (1 + π σ΅¨σ΅¨σ΅¨(π − π¦ ) σ΅¨σ΅¨σ΅¨)π½π −1 π=1 π π σ΅¨σ΅¨ π ππ π σ΅¨σ΅¨ π₯π × ππ₯. π+1 (1 + |π₯|2 ) (42) π½ ) 1 ), where ππ is defined in (45). Proof. Using Proposition 7, we have β¨ππ½ (π’) , ππ (π’)β© ππππ ππ π + 2 π − 4 2π/(π−4) πΌπ π½ π0 2π πΎ (ππ ) σ΅¨π½−1 σ΅¨ ×∫ signe (π₯π + π π (ππ − π¦ππ )π ) σ΅¨σ΅¨σ΅¨σ΅¨π₯π + π π (ππ − π¦ππ )π σ΅¨σ΅¨σ΅¨σ΅¨ Rπ π₯π π 1 ππ₯ + π (∑ πππ ) + π ( ∑ π½ ) . 2 π π (1 + |π₯| ) π =ΜΈ π π=1 π π (47) Observe that for π ∈ {1, . . . , π}, if π π |(ππ − π¦ππ )π | > π1 /√π, we have ∫ signe (π₯π + π π (ππ − π¦ππ )π ) σ΅¨σ΅¨ σ΅¨π½ σ΅¨σ΅¨π₯π + π π (ππ − π¦ππ ) σ΅¨σ΅¨σ΅¨ π π₯π σ΅¨ πσ΅¨ ππ₯ ∫ π+1 Rπ (1 + |π₯|2 ) Rπ (43) σ΅¨ σ΅¨ π½π −1 = π (signe π π (ππ − π¦ππ )π ) (π π σ΅¨σ΅¨σ΅¨σ΅¨(ππ − π¦ππ )π σ΅¨σ΅¨σ΅¨σ΅¨) σ΅¨σ΅¨ σ΅¨π½ −1 σ΅¨σ΅¨π₯π + π π (ππ − π¦ππ ) σ΅¨σ΅¨σ΅¨ π π₯π σ΅¨ πσ΅¨ × ππ₯ π (1 + |π₯|2 ) (48) = π signe (π π (ππ − π¦ππ )π ) × (1 + π (1)) . σ΅¨ σ΅¨ π½π −2 × (π π σ΅¨σ΅¨σ΅¨σ΅¨(ππ − π¦ππ )π σ΅¨σ΅¨σ΅¨σ΅¨) (1 + π (1)) , Hence our claim follows. Next, we will say that π ∈ πΏ2 1 ππ π (46) × We claim that ππ is bounded. Indeed, the claim is trivial if π π |ππ −π¦ππ | ≤ π1 . If π π |ππ −π¦π | > π1 , for any π, 1 ≤ π ≤ π, such π that π π |(ππ − π¦ππ )π | > π1 /√π, we have the following estimate: + π( π½π π=1 π π π =ΜΈ π π =ΜΈ π π π ∈ πΏ1 π + π (∑ πππ ) + π (∑ = −2π2 π½ (π’) ∑ πΌπ πΌπ π π , ππ π σ΅¨σ΅¨ σ΅¨π½ −2 σ΅¨σ΅¨(ππ − π¦π ) σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨ π ππ σ΅¨ [ ] σ΅¨ + [π ( ) , if π ∈ πΏ 2 ] π2π [ ] Furthermore |π1 | is bounded and the only case where the maximum of the π π ’s is not bounded is when ππ ∈ π΅(π¦ππ , π) with π¦ππ ∈ K+ , for all π = 1, . . . , π, and π¦i =ΜΈ π¦π for all π =ΜΈ π. Observe that in π1 (π, π) the interaction of two bubbles is negligible with respect to the self-interaction. Similar phenomena occur for the scalar curvature problem, see [21], so the proof of Proposition 11 is similar to the corresponding statement in [21]. Before giving the proof of Theorem 9, we state the following notations extracted from [11]. Let π1 be a fixed positive constant large enough and let π π’ = ∑π=1 πΌπ πΏ(ππ ,π π ) ∈ π(π, π) such that ππ ∈ π΅(ππ , π), π¦ππ ∈ K, for all π = 1, . . . , π. For any index π, 1 ≤ π ≤ π, we define the following vector fields: ππππ σ΅¨ σ΅¨ if π π σ΅¨σ΅¨σ΅¨σ΅¨ππ − π¦ππ σ΅¨σ΅¨σ΅¨σ΅¨ ≤ π1 , σ΅¨ σ΅¨ if π π σ΅¨σ΅¨σ΅¨σ΅¨ππ − π¦ππ σ΅¨σ΅¨σ΅¨σ΅¨ > π1 , (44) σ΅¨σ΅¨ σ΅¨π½ −1 σ΅¨σ΅¨π₯π + π π (ππ − π¦ππ ) σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨σ΅¨σ΅¨π₯π σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ πσ΅¨ ππ₯ = π (1) . ∫ π Rπ (1 + |π₯|2 ) and we will denote by ππ the index satisfying σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨(ππ − π¦π ) σ΅¨σ΅¨σ΅¨ = max σ΅¨σ΅¨σ΅¨σ΅¨(ππ − π¦π ) σ΅¨σ΅¨σ΅¨σ΅¨ . σ΅¨σ΅¨ π ππ σ΅¨ π πσ΅¨ σ΅¨ 1≤π≤π σ΅¨ It easy to see that if π ∈ πΏ 2 , then π π |(ππ − π¦ππ )ππ | > π1 /√π. taking π1 large enough. If not, we have (49) (45) Using the fact that ππ defined in (45) satisfies π π |(ππ − π¦ππ )ππ | > π1 /√π, if π ∈ πΏ 2 , Lemma 12 follows. Abstract and Applied Analysis 7 π Lemma 13. For π’ = ∑π=1 πΌπ πΏ(ππ ,π π ) ∈ π(π, π), such that ππ ∈ π΅(π¦ππ , π), π¦ππ ∈ K, for all π = 1, . . . , π, one has π Subset 1. We consider here the case of π’ = ∑π=1 πΌπ πΏ(ππ π π ) = ∑π∈πΌ1 πΌπ πΏ(ππ π π ) + ∑π∈πΌ2 πΌπ πΏ(ππ π π ) such that β¨ππ½ (π’) , ππ (π’)β© 1 π =ΜΈ π π π ≤ π (∑ σ΅¨σ΅¨ ππ σ΅¨σ΅¨ 1 σ΅¨σ΅¨ ππ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ πππ σ΅¨σ΅¨σ΅¨) + π [( ππ½π ) , if π ∈ πΏ 1 ] σ΅¨ σ΅¨ π πΌ1 =ΜΈ 0, πΌ2 =ΜΈ 0, σ΅¨σ΅¨ σ΅¨π½ −1 σ΅¨σ΅¨(ππ − π¦π ) σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨ π ππ σ΅¨ 1 [ ] (50) σ΅¨ + [−π ( π½ + ) , if π ∈ πΏ 2 ] ππ ππ π [ ] π + π (∑ remains (using Propositions 10 and 11) to focus attention on the two following subsets of π(π, π). 1 π½π π=1 π π ∑ πΌπ πΏ(ππ π π ) ∈ π1 π∈πΌ1 ∑ πΌπ πΏ(ππ π π ) ∈ π2 π∈πΌ2 (β―πΌ1 , π) , (52) (β―πΌ2 , π) . Without loss of generality, we can assume that π1 ≤ ⋅ ⋅ ⋅ ≤ ππ. ), (53) We distinguish three cases. where ππ is defined in (45). Case 1. π’1 := ∑π∈πΌ1 πΌπ πΏ(ππ π π ) ∉ π11 (β―πΌ1 , π) = {π’ = β―πΌ ∑π=11 πΌπ πΏ(ππ ,π π ) , ππ ∈ π΅(π¦ππ , π), π¦ππ ∈ K+ for all π = 1, . . . , β―πΌ1 and π¦ππ =ΜΈ π¦ππ for all π =ΜΈ π}. Μ1 be the pseudogradient on π(π, π) defined by Let π Μ1 (π’) = π1 (π’1 ), where π1 is the vector filed defined by π Proposition 11 in π1 (β―πΌ1 , π). Note that if π’1 ∉ π11 (β―πΌ1 , π), then the pseudo-gradient π1 (π’1 ) does not increase the maximum of the π π ’s, π ∈ πΌ1 . Using Proposition 11, we have Proof. Using Proposition 8, we have β¨ππ½ (π’) , ππ (π’)β© = −2 (π − 4) π02π/(π−4) πΌπ2π/(π−4) 1 π π π½π σ΅¨σ΅¨ σ΅¨π½ σ΅¨σ΅¨π₯π + π π (ππ − π¦ππ ) σ΅¨σ΅¨σ΅¨ π σ΅¨ πσ΅¨ × ∑ (∫ ππ σ΅¨ σ΅¨ (π½π −1)/2 Rπ π=1 (1 + π π σ΅¨σ΅¨σ΅¨σ΅¨(ππ − π¦ππ )π σ΅¨σ΅¨σ΅¨σ΅¨) π × 1 + π (∑ π π =ΜΈ π π π₯π Μ1 (π’)β© β¨ππ½ (π’) , π 2 2 π+1 (1 + |π₯| ) π½π π∈πΌ1 π π ππ₯) π σ΅¨σ΅¨ ππ σ΅¨σ΅¨ 1 σ΅¨σ΅¨ ππ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨) + π ( ∑ ) π½π σ΅¨σ΅¨σ΅¨ πππ σ΅¨σ΅¨σ΅¨ π=1 π π 1 + π (∑ π π =ΜΈ π π π+1 (1 + |π₯|2 ) σ΅¨σ΅¨ σ΅¨ σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ ) ∑ πππ + ∑ σ΅¨ ππ π =ΜΈ π,π,π∈πΌ1 π∈πΌ1 (54) π∈πΌ1 ,π∈πΌ2 (51) 2 × + + π ( ∑ πππ ) . σ΅¨σ΅¨ σ΅¨π½ σ΅¨σ΅¨π₯π + π π (ππ − π¦π ) σ΅¨σ΅¨σ΅¨ π σ΅¨ π 1 ππ σ΅¨σ΅¨ σ΅¨ ≤ −π π½ (∫ πππ σ΅¨ σ΅¨ σ΅¨ σ΅¨ (π½π −1)/2 Rπ ππ i (1 + π π σ΅¨σ΅¨σ΅¨(ππ − π¦ππ )π σ΅¨σ΅¨σ΅¨) σ΅¨ πσ΅¨ π₯ππ 1 ≤ −π (∑ ππ₯) π σ΅¨σ΅¨ ππ σ΅¨σ΅¨ 1 σ΅¨σ΅¨ ππ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨) + π ( ∑ ). π½π σ΅¨σ΅¨σ΅¨ πππ σ΅¨σ΅¨σ΅¨ π=1 π π An easy calculation yields πππ = π ( 1 (π π π π ) (π−4)/2 ) = π( 1 π½ ππ π ) + π( 1 π½ πππ ), (55) Fix π0 ∈ πΌ1 , we denote by 1 π½ ≥ π π0π0 } , π½1 = {π ∈ πΌ2 , s.t, ππ−4 π 2 π½2 = πΌ2 \ π½1 . (56) Using (54) and (55), we find that Μ1 (π’)β© β¨ππ½ (π’) , π σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ + + ∑ πππ ) ∑ π½π ππ π∈πΌ1 ∪π½1 π π π∈πΌ1 π =ΜΈ π∈πΌ1 Using (43) and the fact that π π |(ππ − π¦ππ )ππ | > π1 /√π, if π ∈ πΏ 2 , Lemma 13 follows. ≤ −π ( ∑ Proof of Theorem 9. In order to complete the construction of the pseudogradient π suggested in Theorem 9, it only + π (∑ π 1 1 π½π π=1 π π ). (57) 8 Abstract and Applied Analysis Let π1 > 0 be small enough; using Lemma 13, (63), and (55), we get From another part, by Lemma 12 we have β¨ππ½ (π’) , ∑ − 2π ππ (π’)β© β¨ππ½ (π’) , ∑ − 2π ππ (π’) + π1 ∑ ππ (π’)β© π∈π½1 π ≤ π ∑ 2 ππ π =ΜΈ π,π∈π½1 ππππ ππ π + π (∑ π∈π½1 1 π½π π∈π½1 π π ) (58) σ΅¨σ΅¨ σ΅¨π½ −2 σ΅¨σ΅¨(ππ − π¦π ) σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨ π ππ σ΅¨ σ΅¨ + π( ∑ ). 2 π π π∈π½1 ∩πΏ 2 ππππ ππ π ≤ − ππππ , π½π π∈π½1 π π if π π ≥ π π or π π ∼ π π ππ π + 2π π π ππππ ππ π (60) ), π∈π½1 ∩πΏ 2 π =ΜΈ π∈πΌ1 σ΅¨σ΅¨ σ΅¨ σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ πππ + ∑ σ΅¨ ) ππ π =ΜΈ π,π∈π½1 ,π∈π½1 ∪π½2 π∈πΌ1 ∪π½1 π + π (∑ ∑ 1 π½π π=1 π π ). (65) Μ := We need to add the remainder indices π ∈ π½2 . Note that π’ ∑π∈π½2 πΌπ πΏπ ∈ π2 (β―π½2 , π). Thus using Proposition 10, we apply Μ2 . We the associated vector field which we will denote by π then have the following estimate: π∈π½1 π =ΜΈ π,π∈π½1 ,π∈π½1 ∪π½2 π½π π=1 π π + ∑ πππ π½π π∈πΌ1 ∪π½1 π π + ≤ −ππππ , 1 ≤ −π ( ∑ β¨ππ½ (π’) , ∑ − 2π ππ (π’)β© ∑ 1 π∈π½1 (59) and for π ∈ π½1 and π ∈ π½2 we have π π ≤ π π , we obtain π π (ππππ /ππ π ) ≤ −ππππ . These estimates yield ≤ −π π ) + π (∑ Μ1 (π’) + π1 (∑ − 2π ππ (π’) + π1 ∑ ππ (π’))β© β¨ππ½ (π’) , π Since for π < π, we have ππππ 1 (64) and by (57) we obtain σ΅¨ σ΅¨ or σ΅¨σ΅¨σ΅¨σ΅¨ππ − ππ σ΅¨σ΅¨σ΅¨σ΅¨ ≥ πΏ0 > 0. 2π π π σ΅¨σ΅¨ σ΅¨ σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ πππ + ∑ σ΅¨ ) ππ π =ΜΈ π,π∈π½1 ,π∈π½1 ∪π½2 π∈π½1 ∑ + π (∑ Observe that using a direct calculation, we have ππ ≤ −π ( π∈π½1 ∩πΏ 2 πππ + π (∑ 1 π½π π∈π½1 π π ) σ΅¨σ΅¨ σ΅¨π½ −2 σ΅¨σ΅¨(ππ − π¦π ) σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨ π ππ σ΅¨ σ΅¨ + π( ∑ ) 2 π π π∈π½1 ∩πΏ 2 Μ2 (π’)β© β¨ππ½ (π’) , π (61) ≤ −π ( ∑ 1 π½π π∈π½2 π π σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨) + ∑ πππ + ∑ ππ π =ΜΈ π,π,π∈π½2 π∈π½2 π + π ( ∑ πππ ) + π (∑ π½π π=1 π π π∈π½2 ,π∈π½1 + π ( ∑ πππ ) . It is easy to see that for any index π ∈ πΏ 2 , we have (62) for any π ∈ πΏ 2 . σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ + + ∑ πππ ) . ∑ π½π ππ π=1 π π π=1 π =ΜΈ π (67) π β¨ππ½ (π’) , π (π’)β© ≤ −π (∑ where ππ is defined in (45) and π1 is large enough. Thus, we derive that σ΅¨σ΅¨ σ΅¨π½ −2 σ΅¨π½ −1 σ΅¨σ΅¨ σ΅¨σ΅¨(ππ − π¦π ) σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨(ππ − π¦π ) σ΅¨σ΅¨σ΅¨ π σ΅¨ σ΅¨ σ΅¨σ΅¨ π ππ σ΅¨ π ππ σ΅¨ σ΅¨ = π (σ΅¨ ), 2 ππ ππ ), since |ππ − ππ | ≥ π for π ∈ π½2 and π ∈ πΌ1 . Μ1 + π1 (π Μ2 + ∑π∈π½ −2π ππ + Let in this case π = π 1 π1 ∑π∈π½1 ∩πΏ 2 ππ ). From (65) and (66) we obtain π∈π½1 ,π∈πΌ1 σ΅¨σ΅¨ σ΅¨π½ −2 σ΅¨σ΅¨ σ΅¨σ΅¨π½π −1 σ΅¨σ΅¨(ππ − π¦π ) σ΅¨σ΅¨σ΅¨ π σ΅¨ σ΅¨ σ΅¨σ΅¨ π ππ σ΅¨ √π σ΅¨σ΅¨σ΅¨(ππ − π¦ππ )ππ σ΅¨σ΅¨σ΅¨ σ΅¨ ≤ , π1 ππ π2π 1 (66) (63) 1 Case 2. π’1 := ∑π∈πΌ1 πΌπ πΏ(ππ π π ) ∈ π11 (β―πΌ1 , π) and π’2 := β―πΌ ∑π∉πΌ2 πΌπ πΏ(ππ π π ) ∉ π21 (β―πΌ2 , π) := {π’ = ∑π=12 πΌπ πΏ(ππ ,π π ) , ππ ∈ π΅(π¦ππ , π), π¦ππ ∈ K+ , for all π = 1, . . . , β―πΌ2 and π(π¦π1 , . . . , π¦β―πΌ2 ) > 0}. Abstract and Applied Analysis 9 Since π’2 ∈ π2 (β―πΌ2 , π), by Proposition 10, we can apply the associated vector field which we will denote by π1 . We get Using the previous estimate and Lemma 13, we find that σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ + + ∑ πππ ) . ∑ π½π ππ π=1 π π π=1 π =ΜΈ π (74) π β¨ππ½ (π’) , π (π’)β© ≤ −π (∑ β¨ππ½ (π’) , π1 (π’)β© σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨ + ∑ π ) + ∑ ππ π½π ππ π∈πΌ2 π π π∈πΌ2 π =ΜΈ π,π,π∈πΌ2 1 ≤ −π (∑ (68) + π ( ∑ πππ ) . π∈πΌ2 , π∈πΌ1 Observe that π1 (π’) does not increase the maximum of the π π ’s, π ∈ πΌ2 , since π’2 ∉ π21 (β―πΌ2 , π). Fix π0 ∈ πΌ2 and let 1 π½ }, π½Μ1 = {π ∈ πΌ1 , s.t, π π π ≥ ππ−4 2 π0 π½Μ2 = πΌ1 \ π½Μ1 . Case 3. π’1 ∈ π11 (β―πΌ1 , π) and π’2 ∈ π21 (β―πΌ2 , π). Μ2 ) be the pseudo-gradient in π(π, π) defined Μ1 (resp., π Let π Μ Μ2 (π’) = π2 (π’2 )) where π1 by π1 (π’) = π1 (π’1 ) (resp., π (resp., π2 ) is the vector field defined by Proposition 11 (resp., Proposition 10) in π11 (β―πΌ1 , π) (resp., π21 (β―πΌ2 , π)) and let in this case Μ2 . Μ1 + π π=π (69) σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ + + ∑ πππ ) . ∑ π½π ππ π=1 π π π=1 π =ΜΈ π (76) π β¨ππ½ (π’) , π (π’)β© ≤ −π (∑ β¨ππ½ (π’) , π1 (π’)β© 1 π + π (∑ 1 π½π π=1 π π (70) ). Μ := We need to add the indices π, π ∈ π½Μ2 . Letting π’ Μ ∈ π1 (β―Μπ½2 , π), we can apply the ∑π∈Μπ½2 πΌπ πΏ(ππ π π ) , since π’ associated vector field given by Proposition 10. Let π2 be this vector field. By Proposition 11 we have β¨ππ½ (π’) , π2 (π’)β© ≤ −π ( ∑ 1 π½π π∈π½Μ2 π π σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨+ +∑ π π Μ π∈π½2 ∑ π =ΜΈ π, π,π∈π½Μ2 πππ ) (71) + π ( ∑ πππ ) . π∈π½Μ2 ,π∉π½Μ2 Observe that πΌ1 = π½Μ1 ∪ π½Μ2 and we are in the case where for all π =ΜΈ π ∈ πΌ1 , we have |ππ − ππ | ≥ π. Thus by (55), we get π π ( ∑ πππ ) = π (∑ π∈π½Μ2 ,π∉π½Μ2 1 π½π π=1 π π ), (72) and hence σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨ + ∑ π ) . + ∑ ππ π½π ππ π=1 π π π =ΜΈ π π∈πΌ2 ∪π½Μ2 π 1 1 Notice that in the first and second cases, the maximum of the π π ’s, 1 ≤ π ≤ π, is a bounded function and hence the Palais-Smale condition is satisfied along the flow lines of π. However in the third case all the π π ’s, 1 ≤ π ≤ π, will increase and go to +∞ along the flow lines generated by π. π Subset 2. We consider the case of π’ = ∑π=1 πΌπ πΏπ ∈ π(π, π), such that there exist ππ satisfying ππ ∉ ∪π¦∈K π΅(π¦, π). In this region, the construction of the pseudo-gradient π proceeds exactly as the proof of Theorem 3.2, of subset 2, of [11]. Finally, observe that our pseudogradient π in π(π, π) satisfies claim (i) of Theorem 9 and it is bounded, since βπ π (ππΏ(ππ,π ) /ππ π )β and β(1/π π )(ππΏ(ππ,π ) /πππ )β are bounded. π π From the definition of π, the π π ’s, 1 ≤ π ≤ π, decrease along the flow lines of π as long as these flow lines do not enter in the neighborhood of finite number of critical points π¦ππ , π = 1, . . . , π, of K such that (π¦π1 , . . . , π¦ππ ) ∈ C∞ . Now, arguing as in Appendix 2 of [16], see also Appendix B of [18], claim (ii) of Theorem 9 follows from (i) and Proposition 5. This completes the proof of Theorem 9. Corollary 14. Let π ≥ 1. The critical points at infinity of π½ in π(π, π) correspond to π (π¦π1 , . . . , π¦ππ ) ∞ := ∑ 1 π=1 πΎ(π¦π ) π πΏ , (π−4)/2 (π¦ππ ,∞) (77) where (π¦π1 , . . . , π¦ππ ) ∈ C∞ . Moreover, such a critical point at β¨ππ½ (π’) , π1 (π’) + π2 (π’)β© ≤ −π (∑ (75) Using Proposition 10, Proposition 11, and (55) we get Using (68) and (55), we get σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨∇πΎ (ππ )σ΅¨σ΅¨σ΅¨ + ∑ πππ ) ≤ −π ( ∑ π½ + ∑ π ππ π∈πΌ2 π =ΜΈ π,π,π∈πΌ2 π∈πΌ2 ∪π½Μ1 π π 1 π (73) Let in this case π = π1 + π2 + π1 ∑π∈π½Μ1 ππ (π’), π1 small enough. infinity has an index equal to π(π¦π1 , . . . , π¦ππ )∞ = π−1+∑π=1 π− Μπ(π¦). 4. Proof of Theorem 1 We prove the existence result by contradiction. Assume that π½ has no critical point in ππ (Σ+ ). It follows from Corollary 14 10 Abstract and Applied Analysis that the critical points at infinity of the associated variational problem are in one to one correspondence with the elements of C∞ defined in (13). Notice that, just like for usual critical points, it is associated with each critical point at infinity π€∞ of π½ stable and unstable manifolds ππ ∞ (π€∞ ) and ππ’∞ (π€∞ ) (see [16, pages 356-357]). These manifolds can be easily described once a finite-dimensional reduction like the one we performed in Section 3 is established. For any π€∞ = (π¦π1 , . . . , π¦ππ ) ∈ C∞ , let π(π€)∞ = π ππ (∑π=1 (π−4)/2 4/π 1/πΎ(π¦ππ ) ) denote the associated critical value. Here we choose to consider a simplified situation, σΈ , π(π€)∞ =ΜΈ π(π€σΈ )∞ , and thus order the where for any π€∞ =ΜΈ π€∞ π(π€)∞ ’s, π€∞ ∈ C∞ as π(π€1 )∞ < ⋅ ⋅ ⋅ < π(π€π0 )∞ . (78) By using a deformation lemma (see [22, proposition 7.24 and Theorem 8.2]), we know that if π(π€π−1 )∞ < π < π(π€π )∞ < π < π(π€π+1 )∞ , then π½π β π½π ∪ ππ’∞ (π€π )∞ , (79) where π½π = {π’ ∈ ππ (Σ+ ), π½(π’) ≤ π} and β denotes retracts by deformation. We apply the Euler-PoincareΜ characteristic of both sides of (79); we find that π (π½π ) = π (π½π ) + (−1)π(π€π )∞ , (80) where π(π€π )∞ denotes the index of the critical point at infinity (π€π )∞ . Let π1 < π(π€1 )∞ = min π½ (π’) < π2 < π(π€2 )∞ π’∈ππ (Σ+ ) (81) < ⋅ ⋅ ⋅ < ππ0 < π(π€π0 )∞ < ππ0 +1 . Since we have assumed that (4) has no solution, π½ππ +1 is a 0 retard by deformation of ππ (Σ+ ). Therefore π(π½ππ +1 ) = 1, since 0 ππ (Σ+ ) is a contractible set. Now using (80), we derive after recalling that π(π½π1 ) = π(0) = 0 π0 1 = ∑(−1)π(π€π )∞ . (82) π=1 Hence if (82) is violated, π½ has a critical point in ππ (Σ+ ). Now, arguing as in the proof of theorem of [18, pages 659660], we prove that such a critical point is positive. To prove the multiplicity part of the statement, we observe that it follows from Sard-Smale theorem that for generic πΎ’s, the solutions to (7) are all nondegenerate, in the sense that the associated linearized operator does not admit zero as eigenvalue. We need to introduce the following lemma extracted from [11]. Lemma 15 (see [11, Section 3.2]). Let π€ be a solution to (7). Assume that the function πΎ satisfies condition (π)π½ , with (π − 4)/2 < π½ ≤ π − 4; then for each π ∈ N∗ , there are neither critical points nor critical points at infinity in π(π, π, π€). Once the existence of mixed critical points at infinity is ruled out, it follows from the previous arguments that π0 ππ (Σ+ ) β βππ’∞ (π€π )∞ ∪ π=1 β ππ’ (π€) . (83) π€,ππ½(π€)=0 Now using the Euler-PoincareΜ theorem, we derive that π0 1 = ∑(−1)π(π€π )∞ + π=1 (−1)morse(π€) . ∑ (84) π€,ππ½(π€)=0 Hence our theorem follows. 5. A General Existence Result In this last section of this paper, we give a generalization of Theorem 1. Namely, in view of the result of Theorem 1, one may think about the situation where the degree π given in Theorem 1 is equal to zero; that is, the total sum in π is equal to 1, but a partial one is not equal to 1. A natural question arises: is it possible in this case to use such information to derive an existence result? In the following theorem we give a partial answer to this question. Theorem 16. Let πΎ be a function satisfying (π΄ 1 ) and (π)π½ , with 1 < π½ ≤ π − 4. If there exists π ∈ N such that (1) ∀π€∞ ∈ C∞ , (2) ∑ π(π€)∞ =ΜΈ π + 1 (−1)π(π€∞ ) =ΜΈ 1, (85) π€∞ ∈C∞ ,π(π€∞ )≤π then the problem (7) has at least one solution. Moreover for generic πΎ’s, if (π − 4)/2 < π½ ≤ π − 4, then the number of solutions is lower bounded by σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ π(π€∞ ) σ΅¨σ΅¨σ΅¨ ∑ (−1) σ΅¨σ΅¨ . σ΅¨σ΅¨1 − σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ π€∞ ∈C∞ ,π(π€∞ )≤π (86) Let πβ― be the maximal index over all elements of C∞ . Please observe that the integer π = πβ― satisfies (1) of Theorem 16; it follows that Theorem 1 is a corollary of Theorem 16. Proof of Theorem 16. We set ππ∞ = β π₯∞ ∈C∞ ,π(π₯)∞ ≤π ππ’∞ (π₯∞ ), (87) where ππ’∞ (π₯∞ ) is the closure of the unstable manifolds of π₯∞ . Observe that ππ∞ is a stratified set of top dimension π, which is contractible set in ππ (Σ+ ), since ππ (Σ+ ) is a contractible set. Let π denote the image of such a contraction. To prove the first part of Theorem 16, arguing by contradiction, we assume that (7) has no solution. Using the pseudo-gradient constructed in Theorem 9, we can deform π. By transversality arguments, we can assume that such a deformation avoids all critical points at infinity of index Abstract and Applied Analysis 11 greater or equal to π + 1. Note that, from assumption (1) of Theorem 16, there is no critical point at infinity with index π + 1. It follows then from a theorem of Bahri and Rabinowitz [22] that π β ππ∞ . (88) Hence from the Euler-PoincareΜ theorem, we get 1 = π (ππ∞ ) = ∑ (−1)π(π€∞ ) , π€∞ ∈πΆ∞ ,π(π€∞ )≤π (89) which is a contradiction with assumption (2) of Theorem 16. Regarding the multiplicity result, we observe that for generic πΎ’s the functional π½ admits only nondegenerate critical points. 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