New multiplicity results in prescribing Q-curvature on standard spheres

Mohamed Ben Ayed; Khalil El Mehdi

doi:10.1515/ans-2023-0135

Article Open Access

New multiplicity results in prescribing Q-curvature on standard spheres

Mohamed Ben Ayed and Khalil El Mehdi

Published/Copyright: April 24, 2024

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From the journal Advanced Nonlinear Studies Volume 24 Issue 3

Abstract

In this paper, we study the problem of prescribing Q-Curvature on higher dimensional standard spheres. The problem consists in finding the right assumptions on a function K so that it is the Q-Curvature of a metric conformal to the standard one on the sphere. Using some pinching condition, we track the change in topology that occurs when crossing a critical level (or a virtually critical level if it is a critical point at infinity) and then compute a certain Euler-Poincaré index which allows us to prove the existence of many solutions. The locations of the levels sets of these solutions are determined in a very precise manner. These type of multiplicity results are new and are proved without any assumption of symmetry or periodicity on the function K.

Keywords: partial differential equations; analysis on manifolds; Q curvature problem

AMS subject classification: 35A15; 58J05; 58E05

1 Introduction and main results

In [1], Branson introduced a conformally operator defined on n-dimensional Riemannian manifolds with n ≥ 5 which generalizes the Paneitz operator defined on 4-manifolds [2]. The conformal Branson operator is defined on a smooth compact Riemannian n-manifold (M, g ₀), n ≥ 5, by the following formula:

P g 0 n u = Δ g 0 2 u − div g 0 ( α n R g 0 g 0 + β n Ric g 0 ) d u + n − 4 2 Q u ,

where

α n = ( n − 2 ) 2 + 4 2 ( n − 1 ) ( n − 2 ) , β n = − 4 n − 2

Q ≔ Q g 0 n = − 1 2 ( n − 1 ) Δ g 0 R g 0 + n 3 − 4 n 2 + 16 n − 16 8 ( n − 1 ) 2 ( n − 2 ) 2 R g 0 2 − 2 ( n − 2 ) 2 | Ric g 0 | 2

with R g 0 denotes the scalar curvature of (M, g ₀) and Ric g 0 denotes the Ricci curvature of (M, g ₀).

Notice that, if g = u ^4/(n−4) g ₀ is a metric conformal to g ₀, where u is a smooth positive function, then for all ψ ∈ C ^∞(M), the conformal Branson operator satisfies the following covariance property:

P g 0 n ( u ψ ) = u ( n + 4 ) / ( n − 4 ) P g n ( ψ ) .

Taking ψ ≡ 1, we then obtain

(1) P g 0 n ( u ) = n − 4 2 Q g n u ( n + 4 ) / ( n − 4 ) .

In view of equation (1), a natural question is whether it is possible to prescribe the Q curvature, that is: given a function h : M → R , does there exist a metric g, in the conformal class of g ₀, whose Q-curvature is equal to h, that is equivalent to solving the following problem

(2) P g 0 n ( u ) = n − 4 2 h u ( n + 4 ) / ( n − 4 ) , u > 0 in M .

The Q-Curvature problem has attracted a lot of attention in the last half century. For more background information on Q-Curvature problem, we refer to the lecture notes [3], [4] and the survey articles of Chang [5] and Chang-Yang [6]. More recently, there are more and more interests in Q-Curvature problem. See [6], [7], [8], [9], [10], [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27] and references therein.

In this paper, we deal with the case of standard sphere S n endowed with its standard metric. Thus, we are looking for a solution u of the following nonlinear problem involving the critical Sobolev exponent:

( P K ) P u ≔ Δ 2 u − c n Δ u + d n u = K u n + 4 n − 4 , u > 0 on S n ,

where c n = 1 2 ( n 2 − 2 n − 4 ) , d n = n − 4 16 n ( n 2 − 4 ) and where K is a given function defined on S n .

The requirement about the positivity of the function u is necessary for the metric g to be Riemannian.

Problem ( P K ) has a variational structure in the sense that its solutions correspond to the positive critical points of the associated Euler-Lagrange functional J _K defined by:

(3) J K ( u ) = ∫ S n K | u | 2 n / ( n − 4 ) ( 4 − n ) / n

defined on the unit sphere Σ of H 2 ( S n ) endowed with the following norm:

‖ u ‖ 2 = ⟨ u , u ⟩ P = ∫ S n P u ⋅ u = ∫ S n ∣ Δ u ∣ 2 + c n ∫ S n ∣ ∇ u ∣ 2 + d n ∫ S n u 2 .

Studying problem ( P K ) by using variational methods is quite difficult since the functional J _K does not satisfy the Palais-Smale condition. This means that there exist a sequence ( u k ) k along which J _K(u _k) is bounded, ∇J _K(u _k) goes to zero and ( u k ) k does not converge. In addition, due to topological obstructions of Kazdan-Warner type (see [13] and [28]), problem ( P K ) does not have solutions for every function K.

In dimensions n = 5, 6, we studied problem ( P K ) by using the tools of the critical point at infinity theory and an Euler-Poincaré counting formula, see [8], [9]. For the case of higher dimensions, problem ( P K ) becomes difficult due to the complexity of the lack of compactness. Indeed every k – tuple of critical points of K with negative Laplacian gives rise a critical point at infinity and under the assumption that there are no solutions the above mentioned index counting formulae are always equal to one. Furthermore every solution gives rise to a mixed critical point at infinity corresponding to this solution plus a sum of Dirac functions, See [16]. Recently, inspired by the results obtained by Malchiodi and Mayer [29] for the conformal Laplacian equation, we have proved some existence results under pinching conditions [10].

In this paper, we consider the case of higher dimensional spheres and our aim is to give sufficient conditions on the function K such that equation ( P K ) has multiple solutions. Our approach is based on the theory of the critical points at infinity introduced by Bahri [30].

This theory deals with the variational problems with lack of compactness, that is, when the associated Euler-Lagrange functionals do not satisfy the Palais-Smale condition. In fact, the Palais-Smale condition is not the best tool to study variational problems. Indeed, if we try to detect the critical points of a function by deforming one of its level sets onto the other by using the gradient flow, two types of obstruction can occur: a stop at a critical point or the emergence of a critical point at infinity, that is, an orbit of the gradient along which the gradient goes to zero, the functional remains bounded, and the orbit does not converge. The second case is forbidden by the Palais-Smale condition. Furthermore, the Palais-Smale condition also excludes situations which do not represent any danger from a variational viewpoint [30]. It therefore appears that the important thing is whether or not the Palais-Smale condition is satisfied along the flow lines, that is, whether or not critical points at infinity exist. Thus, if we look for critical points of a functional by studying the difference of topology between its level sets, we have to determine the critical points at infinity in order to compute their contribution to the topological changes of the level sets. In the present paper, we make a finer study of the topological contribution of the critical points at infinity which allows us to prove the existence of several solutions of the problem ( P K ) having their energies close to the same level set of the functional J _K. To state our results, we need to introduce some notation.

For ɛ > 0, we set

K ε = K : S n → R / 0 < K ∈ C 3 ( S n ) and K max K min − 1 < ε ,

where K max ≔ max S n K and K min ≔ min S n K .

We introduce the following non degeneracy conditions:

The critical points y’s of K are non degenerate and satisfy ΔK(y) ≠ 0.
The critical points of J _K are non degenerate.
m ≔ # K ∞ ≥ 2 , where K ∞ ≔ { y ∈ S n : ∇ K ( y ) = 0 and Δ K ( y ) < 0 } and #A denotes the cardinal of A.

In the first part of this paper, we consider the case where the Euler-Poincaré index formula related to the topological contribution of critical points of J _K is different from 1, that is, B ₁ ≠ 1 where

B 1 ≔ ∑ y ∈ K ∞ ( − 1 ) ι ( y ) where ι ( y ) = n − morse index ( K , y ) .

Taking N an arbitrary integer, our aim is to prove that, for ɛ small and 1 ≤ p ≤ N, problem ( P K ) has solutions whose energies are close to ( p S n ) 4 / n , where S _n is defined by (49). Assume that K ∞ = { y 1 , … , y m } with y ₁ is a maximum point of K. Without loss of generality, we can assume that y ₁, …, y _m are ordered so that we are in one of the following four cases:

Case 1. ι(y _j) is even for all j.
Case 2. ι(y _j) is odd for all j ≥ 2.
Case 3. For some l ≥ 1 such that 2l ≤ m − 2, we have

ι ( y 2 j ) is odd and ι ( y 2 j + 1 ) is even ∀ j ∈ { 1 , … , ℓ } and ι ( y j ) is even ∀ 2 ℓ + 2 ≤ j ≤ m .

Case 4. For some ℓ ≥ 1 such that 2ℓ ≤ m − 2, we have

ι ( y 2 j ) is odd and ι ( y 2 j + 1 ) is even ∀ j ∈ { 1 , … , ℓ } and ι ( y j ) is odd ∀ 2 ℓ + 2 ≤ j ≤ m .

Note that, in cases 3 and 4, we have used the fact that B ₁ ≠ 1 which implies that m cannot be equal to 2ℓ + 1 and therefore: m ≥ 2ℓ + 2.

In order to give the precise number of solutions obtained by our argument as well as their levels, we will present our results for each of the above cases. Under the above notation, our first multiplicity results read as follows:

Theorem 1.1.

for any p ∈ {1, 2, …, N}, ( P K ) has at least (m − 1) solutions u _1,p, …, u _m−1,p satisfying J K ( u i , p ) ∈ ( p S n ( 1 − η ) ) 4 / n , ( p S n ( 1 + η ) ) 4 / n for any i ∈ {1, …, m − 1}, for some small positive constant η. In particular ( P K ) has at least N(m − 1) solutions. (Here and in the sequel S _n denotes the constant defined by (49) ).

Theorem 1.2.

Let n ≥ 9 and N ∈ N be an integer. There exists ɛ _N > 0 such that, if 0 < ɛ < ɛ _N and K ∈ K ε satisfying (H1), (H2), (H3), ∑ y ∈ K ∞ ( − 1 ) ι ( y ) ≠ 1 and the assumption stated in Case 3 or Case 4, then there exist at least N solutions u ₁, …, u _N of ( P K ) satisfying J K ( u p ) ∈ ( p S n ( 1 − η ) ) 4 / n , ( p S n ( 1 + η ) ) 4 / n for each p ∈ {1, 2, …, N}, for some small positive constant η.

Remark 1.3.

If the cardinal of K ∞ is even then we have: ∑ y ∈ K ∞ ( − 1 ) ι ( y ) ≠ 1 .

In the second part of this paper, we deal with the case where the above mentioned index counting formula is equal to 1, that is ∑ y ∈ K ∞ ( − 1 ) ι ( y ) = 1 . Note that, in this case, # K ∞ has to be odd, say 2r + 1 with r + 1 points having ι(y) even and r points having ι(y) odd. Furthermore, since we have assumed that # K ∞ ≥ 2 , we see that r ≥ 1. In this part, we fail to prove the existence of solutions at each level set ( p S n ) 4 / n with 1 ≤ p ≤ N. However, we are able to show the existence of solutions of ( P K ) whose energies are close to the level ( 2 p S n ) 4 / n for 1 ≤ p ≤ [ N 2 ] . More precisely, we have:

Theorem 1.4.

Let n ≥ 9 and N ∈ N be an integer. There exists ɛ _N > 0 such that, if 0 < ɛ < ɛ _N and K ∈ K ε satisfying (H1), (H2), (H3) and ∑ y ∈ K ∞ ( − 1 ) ι ( y ) = 1 , then for any p ∈ { 1,2 , … , [ N 2 ] } , ( P K ) has at least (m − 1)/2 solutions u _1,p, …, u _(m−1)/2,p satisfying J K ( u i , p ) ∈ ( 2 p S n ( 1 − η ) ) 4 / n , ( 2 p S n ( 1 + η ) ) 4 / n for any i ∈ {1, …, (m − 1)/2}, for some small positive constant η. In particular, ( P K ) has at least [N/2](m − 1)/2 solutions.

The idea of the proof of Theorems 1.1, 1.2 and 1.4 is as follows: taking an arbitrary integer N ∈ N and using the Pinching condition introduced in the definition of K ε , we first prove that, for ɛ small, the Euler-Poincaré counting index formula related to the topological contribution of the critical points and the critical points at infinity at the first level set of the associated variational functional is equal to 1 while the above mentioned index counting formula related to the relative topological contribution of these critical points and critical points at infinity at each level set going from 2 to N is equal to zero. Second, we compute the above mentioned index counting formula related to the topological contribution of the critical points at infinity at the first level and then two cases may occur:

If the above mentioned index counting formula is different from 1 (that means B ₁ ≠ 1), we deduce the existence of a solution ω whose energy is close to the first level. Since we have assumed that n ≥ 9, this solution gives rise critical points at infinity (y _i, ω) (see Proposition 2.6 below) at the 2nd level with y i ∈ K ∞ . Thus, we have at this level two types of critical points at infinity: Some of them are of type (y _i, y _j) with i ≠ j, and others are of type (y _i, ω) y i , y j ∈ K ∞ . But by computing the Euler-Poincaré counting index formula related to the contribution of these critical points at infinity we see that it is different from 0. This shows the existence of a solution whose energy is close to the second level. This argument is repeated up to the level N, which proves the existence of at least N solutions.
If B ₁ = 1, we cannot conclude the existence of solutions with energies close to the first level. But using assumption (H3), we prove that this mentioned index counting formula is different from 0 at all even levels. This gives the existence of solutions at all even levels.

The rest of the paper is organized as follows: In Section 2, we set up the analytical framework of the associated variational problem, recall some preliminaries and we give the characterization of the critical points at infinity. Lastly, we prove our multiplicity results in Section 3.

2 Preliminaries

This section is devoted to set up the analytical setting of the associated variational problem, to recall the lack of compactness and to give the characterization of non converging orbits of the gradient flow, the so called critical points at infinity. Note that ( P K ) has a variational structure. Namely, the positive critical points of the functional J _K are in one to one correspondence with the solutions of ( P K ) . The functional J _K fails to satisfy the Palais-Smale condition. To recall this failure, we introduce the following notation.

For a ∈ S n and λ > 0, let

δ ̃ ( a , λ ) ( x ) = β n λ ( n − 4 ) / 2 2 + ( λ 2 − 1 ) ( 1 − cos ⁡ d ( x , a ) ) ( n − 4 ) / 2 ,

where d is the geodesic distance on ( S n , g ) and β _n = [(n − 4)(n − 2)n(n + 2)]^(n−4)/8.

It is well known that these functions are the solutions of the problem ([31])

P u = u ( n + 4 ) / ( n − 4 ) , u > 0 , in S n .

Let ω be a non degenerate positive critical point of J _K, q ∈ N and ν > 0, we set

V ( q , ν , ω ) = u ∈ Σ + ∣ ∃ a 1 , … , a q ∈ S n , ∃ λ 1 , … , λ q > ν − 1 , ∃ α 0 , α 1 , … , α q > 0 with ∥ u − ∑ i = 1 q α i δ ̃ ( a i , λ i ) − α 0 ω ∥ < ν , ε i j < ν ∀ i ≠ j , J K ( u ) n n − 4 α i 8 n − 4 K ( a i ) − 1 < ν ∀ i ≥ 1 , | α 0 J K ( u ) n / 8 − 1 | < ν ,

where

Σ + ≔ { u ∈ Σ : u > 0 } and ε i j = λ i λ j + λ j λ i + λ i λ j 2 ( 1 − cos ⁡ d ( a i , a j ) ) ( 4 − n ) / 2 .

We also set

(4) V ( q , ν , ω ) ≔ ( α , a , λ , h ) ∈ R q + 1 × ( S n ) q × ( ν − 1 , ∞ ) q × T ω ( W u ( ω ) ) : c ̄ − 1 ≤ α i ≤ c ̄ , ∀ i , ‖ h ‖ ≤ c ̄ ′ and ε i j ≤ ν ,

where c ̄ and c ̄ ′ are two positive constants and T _ω(W _u(ω)) denotes the tangent space at ω of the unstable manifold W _u(ω).

According to [21], we know that positive non converging Palais-Smale sequences have to enter some V(q, ν, ω). To parametrize V(q, ν, ω), we follow the corresponding statement in [32]. Denoting by ω a non degenerate positive critical point of J _K, we are going to prove the following result:

Proposition 2.1.

For any q ∈ N , there is ν _q > 0 such that if 0 < ν ≤ ν _q and u ∈ V(q, ν, ω), then the following minimization problem

(5) min u − ∑ i = 1 q α i δ ̃ ( a i , λ i ) − α 0 ( ω + h ) , α i > 0 , λ i > ν − 1 , a i ∈ S n , h ∈ T ω ( W u ( ω ) )

has a unique solution (α ₀, α, λ, a, h) = (α ₀, α ₁, …, α _q, λ ₁, …, λ _q, a ₁, …, a _q, h). In particular, u can be written as follows:

u = ∑ i = 1 q α i δ ̃ ( a i , λ i ) + α 0 ( ω + h ) + v ,

where v ∈ H 2 ( S n ) ∩ T ω ( W s ( ω ) ) and satisfies (W ₀). Here T _ω(W _s(ω)) and T _ω(W _u(ω)) denote the tangent spaces at ω of the stable and unstable manifolds of ω respectively, and (W ₀) is defined by:

( W 0 ) : ⟨ v , ψ i ⟩ P = 0 for i = 1 , … , q and for each ψ i = δ ̃ ( a i , λ i ) , ∂ δ ̃ ( a i , λ i ) / ∂ λ i , ∂ δ ̃ ( a i , λ i ) / ∂ ( a i ) j , j = 1 , … , n , where ( a i ) 1 , … , ( a i ) n are some system of coordinates on S n near a i , ⟨ v , h 1 ⟩ P = 0 ∀ h 1 ∈ T ω ( W u ( ω ) ) ⟨ v , ω ⟩ P = 0 .

In addition, when ω = 0, the above result holds with α ₀ = 0 and h = 0.

To prove Proposition 2.1, we need to prove three preparation lemmas.

Lemma 2.2.

Let ν _k > 0 be a sequence with lim ν _k = 0 and let ( α k , a k , λ k , h k ) ∈ V ( q , ν k , ω ) and ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) ∈ V ( q , ν k , ω ) be such that

(6) lim k → ∞ ‖ φ ( α k , a k , λ k , h k ) − φ ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) ‖ = 0

(7) where φ ( α , a , λ , h ) = ∑ i = 1 q α i δ ̃ a i , λ i + α 0 ( ω + h ) .

Then (modulo permutation on ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) ),

(8) lim k → ∞ λ i k λ ̃ i k = 1 ∀ i ∈ { 1 , … , q } ,

(9) lim k → ∞ λ i k d a i k , a ̃ i k = 0 ∀ i ∈ { 1 , … , q } ,

(10) lim k → ∞ | α i k − α ̃ i k | = 0 ∀ i ∈ { 0,1 , … , q } ,

(11) lim k → ∞ ‖ h k − h ̃ k ‖ = 0 .

Proof.

For sake of simplicity, we will often omit the index k. We start by proving the following claim:

Claim: For each i ∈ {1, …, q}, there exists j(i) ∈ {1, …, q} such that

(12) λ i k λ ̃ j ( i ) k + λ ̃ j ( i ) k λ i k + λ i k λ ̃ j ( i ) k 1 2 ( 1 − cos ( d a i k , a ̃ j k ) ) is bounded.

The proof will be by contradiction. Let i ∈ {1, …, q} and assume that

λ i k λ ̃ j k + λ ̃ j k λ i k + λ i k λ ̃ j k 1 2 ( 1 − cos ( d a i k , a ̃ j k ) ) → ∞ ∀ j ∈ { 1 , … , q } .

Combining this information with the fact that ( α k , a k , λ k , h k ) ∈ V ( q , ν k , ω ) and ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) ∈ V ( q , ν k , ω ) , we deduce that

(13) lim k → ∞ 〈 δ ̃ a i k , λ i k , δ ̃ a j k , λ j k 〉 P = 0 ∀ j ≠ i and lim k → ∞ 〈 δ ̃ a i k , λ i k , δ ̃ a ̃ j k , λ ̃ j k 〉 P = 0 ∀ 1 ≤ j ≤ q .

Furthermore, easy computations imply that

(14) lim k → ∞ ⟨ δ ̃ a i , λ i , ω ⟩ P = 0 and lim k → ∞ ⟨ δ ̃ a i , λ i , ψ ⟩ P = 0 ∀ ψ ∈ T ω ( W u ( ω ) ) .

Now, using (6), (13) and (14), we get

o k ( 1 ) = ‖ α i δ ̃ a i , λ i + ( φ ( α , a , λ , h ) − α i δ ̃ a i , λ i ) − φ ( α ̃ , a ̃ , λ ̃ , h ̃ ) ‖ 2 = α i 2 ‖ δ ̃ a i , λ i ‖ 2 + ‖ ( φ ( α , a , λ , h ) − α i δ ̃ a i , λ i ) − φ ( α ̃ , a ̃ , λ ̃ , h ̃ ) ‖ 2 + o k ( 1 )

which gives a contradiction, where o _k(1) is some quantity satisfying lim_k→∞ o _k(1) = 0.

Hence our claim follows.

In the following, we will take j(i) = i (modulo permutation on ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) ). Note that, since ( α k , a k , λ k , h k ) ∈ V ( q , ν k , ω ) and ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) ∈ V ( q , ν k , ω ) , we deduce that

lim k → ∞ λ i k λ ̃ j k + λ ̃ j k λ i k + λ i k λ ̃ j k 1 2 1 − cos d a i k , a ̃ j k = ∞ ∀ j ≠ i

and therefore

⟨ δ ̃ a i , λ i , δ ̃ a j , λ j ⟩ P = o k ( 1 ) , ⟨ δ ̃ a ̃ i , λ ̃ i , δ ̃ a ̃ j , λ ̃ j ⟩ P = o k ( 1 ) , ⟨ δ ̃ a i , λ i , δ ̃ a ̃ j , λ ̃ j ⟩ P = o k ( 1 ) ∀ j ≠ i .

Observe that (6) implies that

o k ( 1 ) = ‖ ∑ i = 1 q ( α i δ ̃ a i , λ i − α ̃ i δ ̃ a ̃ i , λ ̃ i ) + ( α 0 − α ̃ 0 ) ω + ( α 0 h − α ̃ 0 h ̃ ) ‖ 2 = ∑ i = 1 q ‖ α i δ ̃ a i , λ i − α ̃ i δ ̃ a ̃ i , λ ̃ i ‖ 2 + ( α 0 − α ̃ 0 ) 2 ‖ ω ‖ 2 + ‖ α 0 h − α ̃ 0 h ̃ ‖ 2 + o k ( 1 )

which implies that

(15) lim k → ∞ | α 0 − α ̃ 0 | = 0 ; lim k → ∞ ‖ α i δ ̃ a i , λ i − α ̃ i δ ̃ a ̃ i , λ ̃ i ‖ = 0 ∀ i and lim k → ∞ ‖ α 0 h − α ̃ 0 h ̃ ‖ = 0 .

Note that, from the first and the last informations, we derive that lim ‖ h − h ̃ ‖ = 0 which implies (11). To prove (10) for i ≥ 1, we write

o k ( 1 ) = ‖ α i δ ̃ a i , λ i − α ̃ i δ ̃ a ̃ i , λ ̃ i ‖ 2 = α i 2 ‖ δ ̃ a i , λ i ‖ 2 + α ̃ i 2 ‖ δ ̃ a ̃ i , λ ̃ i ‖ 2 − 2 α i α ̃ i 〈 δ ̃ a i , λ i , δ ̃ a ̃ i , λ ̃ i 〉 P = α i 2 S n + α ̃ i 2 S n − 2 α i α ̃ i 〈 δ ̃ a i , λ i , δ ̃ a ̃ i , λ ̃ i 〉 ≥ ( α i − α ̃ i ) 2 S n ,

which completes the proof of (10) (for i = 0, the proof is done in (15)).

Now, observe that

o k ( 1 ) = ‖ α i δ ̃ a i , λ i − α ̃ i δ ̃ a ̃ i , λ ̃ i ‖ ≥ α ̃ i ‖ δ ̃ a i , λ i − δ ̃ a ̃ i , λ ̃ i ‖ − | α i − α ̃ i | ‖ δ ̃ a i , λ i ‖ .

Using (10) for i ≥ 1 and (15), we derive that

(16) lim k → ∞ ‖ δ ̃ a i , λ i − δ ̃ a ̃ i , λ ̃ i ‖ = 0 .

Notice that, from (12), we deduce that d ( a i , a ̃ i ) = o ( 1 ) . Furthermore, observe that, using Lemma 2.4 with the stereographic projection Π − a i and using the notation of Lemma 2.4, we obtain

〈 δ ̃ a i , λ i , δ ̃ a ̃ i , λ ̃ i 〉 P = ∫ R n δ 0 , λ i n + 4 n − 4 δ a ̃ ̂ i , λ ̃ ̂ i = β n 2 n n − 4 ∫ R n λ i n + 4 2 1 + λ i 2 | x | 2 n + 4 2 λ ̃ ̂ i n − 4 2 1 + λ ̃ ̂ i 2 | x − a ̃ ̂ i | 2 n − 4 2 d x = β n 2 n n − 4 ∫ R n 1 ( 1 + | y | 2 ) n + 4 2 μ i n − 4 2 1 + μ i 2 | x − b i | 2 n − 4 2 d x = 〈 δ 0,1 , δ b i , μ i 〉 ≔ ∫ R n Δ δ 0,1 Δ δ b i , μ i ,

(by using the change of variables: y = λ _i x) where b i ≔ λ i a ̃ ̂ i and μ i ≔ λ ̃ ̂ i / λ i . This implies that

(17) ‖ δ ̃ a i , λ i − δ ̃ a ̃ i , λ ̃ i ‖ 2 = ‖ δ 0,1 − δ b i , μ i ‖ 2 ≔ ∫ R n | Δ ( δ 0,1 − δ b i , μ i ) | 2 .

Finally, (16) and (17) imply that ( δ b i , μ i ) converges to δ _0,1 which means that

lim k → ∞ b i = 0 and lim k → ∞ μ i = 1 .

Notice that, from (12), it follows that d ( a i , a ̃ i ) = o ( 1 ) and therefore cos ⁡ d ( a ̃ i , − a i ) = − 1 + o ( 1 ) . Thus, from the precise value of λ ̃ ̂ i (see Lemma 2.4 below) it follows that λ ̃ ̂ i / λ ̃ i = 1 + o ( 1 ) . Thus (8) follows. Concerning (9), using the precise value of a ̃ ̂ i given in Lemma 2.4, it follows that a ̃ ̂ i = P R n a ̃ i ( 1 + o ( 1 ) ) and therefore | a ̃ ̂ i | = d ( a i , a ̃ i ) ( 1 + o ( 1 ) ) . Thus (9) follows from the fact that |b _i| = o(1). This ends the proof of Lemma 2.2. ■

Lemma 2.3.

Let a, a ̃ , λ and λ ̃ be such that ∧ ≔ ( λ ̃ / λ ) − 1 and ξ ≔ λ ( a ̃ − a ) are small. Then it holds

(18) δ ̃ a ̃ , λ ̃ = δ ̃ a , λ + ∧ λ ∂ δ ̃ a , λ ∂ λ + ξ 1 λ ∂ δ ̃ a , λ ∂ a + O ( ∧ 2 + | ξ | 2 ) δ ̃ a , λ ,

(19) ⟨ δ ̃ a ̃ , λ ̃ − δ ̃ a , λ , δ ̃ a , λ ⟩ P = O ( ∧ 2 + | ξ | 2 ) .

Proof.

First, we remark that

(20) | y − x | 2 = 2 ( 1 − cos ( d ( y , x ) ) ) for each y , x ∈ S n ,

where |y − x| is the Euclidian norm in R n + 1 . Furthermore, by our notation, we have: λ ̃ = ( ∧ + 1 ) λ and therefore it holds

2 + ( λ ̃ 2 − 1 ) ( 1 − cos ( d ( y , a ̃ ) ) ) = 2 + 1 2 ( ( ∧ + 1 ) 2 λ 2 − 1 ) | y − a ̃ | 2 = 2 + 1 2 [ ( λ 2 − 1 ) + 2 ∧ λ 2 + ∧ 2 λ 2 ] | y − a | 2 + | a − a ̃ | 2 + 2 ⟨ y − a , a − a ̃ ⟩ = 2 + 1 2 ( λ 2 − 1 ) | y − a | 2 ︸ A 1 + 1 A ∧ λ 2 | y − a | 2 + 1 A ( λ 2 − 1 ) ⟨ y − a , a − a ̃ ⟩ + O ( ∧ 2 + | ξ | 2 ) = 2 + ( λ 2 − 1 ) ( 1 − cos ( d ( y , a ) ) ) 1 + ∧ A λ 2 | y − a | 2 + λ 2 − 1 A ⟨ y − a , a − a ̃ ⟩ + O ( ∧ 2 + | ξ | 2 )

which implies that (since ∧ and ξ are small)

δ ̃ a ̃ , λ ̃ ( y ) ≔ β n λ ̃ ( n − 4 ) / 2 [ 2 + ( λ ̃ 2 − 1 ) ( 1 − cos ( d ( y , a ̃ ) ) ) ] ( n − 4 ) / 2 = δ ̃ a , λ ( y ) ( 1 + ∧ ) n − 4 2 1 + 1 A ∧ λ 2 | y − a | 2 + 1 A ( λ 2 − 1 ) 〈 y − a , a − a ̃ 〉 + O ( ∧ 2 + | ξ | 2 ) 4 − n 2 = δ ̃ a , λ ( y ) 1 + n − 4 2 ∧ − n − 4 2 A ∧ λ 2 | y − a | 2 − n − 4 2 A ( λ 2 − 1 ) 〈 y − a , a − a ̃ 〉 + O ( ∧ 2 + | ξ | 2 ) .

The proof of (18) follows from the computations of ∂ δ ̃ a , λ / ∂ λ and ∂ δ ̃ a , λ / ∂ a .

Concerning the second claim (19), it follows immediately from (18) and the following result

‖ δ ̃ a , λ ‖ 2 = S n ; 〈 δ ̃ a , λ , ∂ δ ̃ a , λ ∂ λ 〉 P = 0 and 〈 δ ̃ a , λ , ∂ δ ̃ a , λ ∂ a 〉 P = 0 .

Hence the proof of the lemma is completed. ■

In the following lemma, we need to precise the corresponding function to δ ̃ a , λ , for a ∈ S n and λ > 0 by using the stereographic projection. Precisely, for a ∈ S n , let Π − a : R n → S n be the stereographic projection. Without loss of the generality, by taking a rotation of S n , we can assume that a = (0, …, 0, − 1) and therefore Π_−a is defined by

(21) Π − a ( x 1 , … , x n ) = 2 1 + | x | 2 x 1 , … , x n , | x | 2 − 1 2 .

It is well known that (see Eq. (A2) of [33])

(22) | J a c ( Π − a ) | = 2 1 + | x | 2 n .

Let us define the map ι defined by

(23) for u ∈ H 2 ( S n ) , ι ( u ) ( x ) ≔ 2 1 + | x | 2 n − 4 2 u ( Π − a ( x ) ) , x ∈ R n .

Then, we have:

Lemma 2.4.

Let a ∈ S n and Π_−a be defined by (21) . We have

For each λ > 0, it holds
ι ( δ ̃ a , λ ) = δ 0 , λ ≔ β n λ ( n − 4 ) / 2 ( 1 + λ 2 | . | 2 ) ( n − 4 ) / 2 .
For b ∈ S n and μ > 0, it holds
ι ( δ ̃ b , μ ) = δ b ̂ , μ ̂ with b ̂ ≔ μ 2 − 1 1 + μ 2 − ( μ 2 − 1 ) cos ⁡ d ( b , − a ) P R n ( b ) , μ ̂ ≔ 1 2 μ 1 + μ 2 − ( μ 2 − 1 ) cos ⁡ d ( b , − a ) ,
where P R n ( b ) is the projection of b on R n .
Furthermore, we have
⟨ δ ̃ a , λ , δ ̃ b , μ ⟩ P = ∫ S n δ ̃ a , λ ( n + 4 ) / ( n − 4 ) δ ̃ b , μ = ∫ R n δ 0 , λ ( n + 4 ) / ( n − 4 ) δ b ̂ , μ ̂ .

Proof.

We start by proving (1). Observe that, since we used −a for the stereographic projection, it follows that Π_−a(0) = a. By using (23), for each x ∈ R n , we get

(24) ι ( δ ̃ a , λ ) ( x ) = 2 1 + | x | 2 n − 4 2 δ ̃ a , λ ( Π − a ( x ) ) = 2 1 + | x | 2 n − 4 2 β n λ ( n − 4 ) / 2 2 + ( λ 2 − 1 ) ( 1 − cos ⁡ d ( a , Π − a ( x ) ) ) n − 4 2 .

Note that

cos ⁡ d ( a , Π − a ( x ) ) = ⟨ a , Π − a ( x ) ⟩ = 1 − | x | 2 1 + | x | 2 ,

where ⟨., .⟩ denotes the scalar product in R n .

Therefore, we obtain

2 + ( λ 2 − 1 ) ( 1 − cos ⁡ d ( a , Π − a ( x ) ) ) = 2 + ( λ 2 − 1 ) 2 | x | 2 1 + | x | 2 = 2 1 + | x | 2 ( 1 + λ 2 | x | 2 ) .

Putting this estimate in (24), we obtain the proof of (1).

Concerning the proof of (2), we know that ι ( δ ̃ b , μ ) is equal to δ b ̂ , μ ̂ and we need to precise the values of the new parameters. Observe that, expanding the two functions when |x| is very large, we get

(25) δ b ̂ , μ ̂ ( x ) = β n μ ̂ ( n − 4 ) / 2 ( 1 + μ ̂ 2 [ | x | 2 + | b ̂ | 2 − 2 〈 x , b ̂ 〉 ] ) ( n − 4 ) / 2

(26) = β n μ ̂ ( n − 4 ) / 2 | x | n − 4 1 + ( n − 4 ) 〈 x , b ̂ 〉 | x | 2 + O 1 | x | 2 .

On the other hand, (21) implies that

Π − a ( x ) = − a + 2 x | x | 2 + O 1 | x | 2 , for | x | very large

and therefore, we get

cos ( d ( b , Π − a ( x ) ) ) = ⟨ b , Π − a ( x ) ⟩ = ⟨ b , − a ⟩ + 2 ⟨ b , x ⟩ | x | 2 + O 1 | x | 2

which implies that

2 + ( μ 2 − 1 ) ( 1 − cos ⁡ d ( b , Π − a ( x ) ) ) = 1 + μ 2 − ( μ 2 − 1 ) ⟨ b , − a ⟩ − 2 ( μ 2 − 1 ) ⟨ b , x ⟩ | x | 2 + O 1 | x | 2 = 1 + μ 2 − ( μ 2 − 1 ) ⟨ b , − a ⟩ 1 − 2 μ 2 − 1 1 + μ 2 − ( μ 2 − 1 ) ⟨ b , − a ⟩ ⟨ b , x ⟩ | x | 2 + O 1 | x | 2 .

Putting this expansion in the definition of ι ( δ ̃ b , μ ) , for |x| very large, we obtain

(27) ι ( δ ̃ b , μ ) ( x ) = 2 | x | 2 n − 4 2 1 + O 1 | x | 2 β n μ ( n − 4 ) / 2 [ 1 + μ 2 − ( μ 2 − 1 ) ⟨ b , − a ⟩ ] n − 4 2 × 1 + ( n − 4 ) ( μ 2 − 1 ) 1 + μ 2 − ( μ 2 − 1 ) ⟨ b , − a ⟩ ⟨ b , x ⟩ | x | 2 + O 1 | x | 2 .

Using (26) and (27), by identification, we get

μ ̂ = 1 2 μ 1 + μ 2 − ( μ 2 − 1 ) cos ⁡ d ( b , − a ) and b ̂ = μ 2 − 1 1 + μ 2 − ( μ 2 − 1 ) cos ⁡ d ( b , − a ) P R n ( b ) ,

where P R n ( b ) is the projection of b on R n . Hence the proof of (2) is completed.

Finally, (3) follows from (22), (23) and Claims (1) and (2) by using the change of variables y = Π_−a(x). This ends the proof of the lemma. ■

Now, we are ready to prove Proposition 2.1.

Proof of Proposition 2.1.

From the definition of V(q, ν, ω), for u ∈ V(q, ν, ω), there exists ( β , b , μ , h 0 ) ∈ V ( q , ν , ω ) satisfying the assumptions stated in V ( q , ν , ω ) such that

(28) ‖ u − φ ( β , b , μ , h 0 ) ‖ < ν ,

where φ is defined in (7). Note that the variables α, a, h in the minimizing problem (5) are in compact sets. However λ ∈ ( ν − 1 , ∞ ) q . But, by using Lemma 2.2 and (28), it follows that λ _i has to be close to μ _i for each i ∈ {1, …, q}. Therefore, it has to be in a compact set. Thus the problem (5) has a minimizing solution ( α ̄ , a ̄ , λ ̄ , h ̄ ) .

It remains to prove that this solution is unique, for ν small. For this aim, we argue by contradiction. Assume that there exist ν _k > 0 (with ν _k → 0) and u _k ∈ V(q, ν _k, ω) such that the problem (5) is achieved by two different solutions

(29) ( α k , a k , λ k , h k ) ≠ ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) .

Denoting by v _k ≔ u _k − φ(α ^k, a ^k, λ ^k, h ^k) and v ̃ k ≔ u k − φ ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) , it is easy to see that the following hold

(30) 0 = 〈 v k , ψ 〉 P = 〈 v ̃ k , ψ 〉 P ∀ ψ ∈ T ω ( W u ( ω ) ) ,

(31) 0 = 〈 v k , ω 〉 P = 〈 v ̃ k , ω 〉 P ,

(32) 0 = 〈 v k , δ ̃ a i k , λ i k 〉 P = 〈 v ̃ k , δ ̃ a ̃ i k , λ ̃ i k 〉 P ∀ i ∈ { 1 , … , q } ,

(33) 0 = 〈 v k , ∂ δ ̃ a i k , λ i k / ∂ λ i k 〉 P = 〈 v ̃ k , ∂ δ ̃ a ̃ i k , λ ̃ i k / ∂ λ ̃ i k 〉 P ∀ i ∈ { 1 , … , q } ,

(34) 0 = 〈 v k , ∂ δ ̃ a i k , λ i k / ∂ a i k j 〉 P = 〈 v ̃ k , ∂ δ ̃ a ̃ i k , λ ̃ i k / ∂ a ̃ i k j 〉 P ∀ i ∈ { 1 , … , q } ∀ j ∈ { 1 , … , n } .

Observe that, since ‖v _k‖ → 0 and ‖ v ̃ k ‖ → 0 , we derive that

‖ φ ( α k , a k , λ k , h k ) − φ ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) ‖ = ‖ v k − v ̃ k ‖ → 0

and therefore, by using Lemma 2.2, we derive that (up permutation on ( α ̃ k , a ̃ k , λ ̃ k , h ̃ k ) ),

(35) λ i k λ ̃ i k = 1 + o k ( 1 ) ; λ i d a i k , a ̃ i k = o k ( 1 ) ; | α i k − α ̃ i k | = o k ( 1 ) ∀ i ; ‖ h k − h ̃ k ‖ = o k ( 1 ) .

In the following, we will denote by

(36) ∧ i , k ≔ λ i k λ ̃ i k − 1 ; ξ i , k ≔ λ i k a i k − a ̃ i k ; ζ i , k ≔ | α i k − α ̃ i k | ∀ i ; θ k ≔ α 0 k h k − α ̃ 0 k h ̃ k .

Notice that, (35) implies that the new variables are small and we need to prove that these variables, for k large, are zero. In the following, we will omit the index k.

From (32), for each i ∈ {1, …, q}, we derive that

(37) ⟨ v ̃ , δ ̃ a i , λ i − δ ̃ a ̃ i , λ ̃ i ⟩ P = ⟨ v ̃ , δ ̃ a i , λ i ⟩ P = ⟨ u − φ ( α ̃ , a ̃ , λ ̃ , h ̃ ) , δ ̃ a i , λ i ⟩ P = ⟨ φ ( α , a , λ , h ) − φ ( α ̃ , a ̃ , λ ̃ , h ̃ ) , δ ̃ a i , λ i ⟩ P = ∑ j = 1 q ⟨ α j δ ̃ a j , λ j − α ̃ j δ ̃ a ̃ j , λ ̃ j , δ ̃ a i , λ i ⟩ P + ( α 0 − α ̃ 0 ) ⟨ ω , δ ̃ a i , λ i ⟩ P + ⟨ α 0 h − α ̃ 0 h ̃ , δ ̃ a i , λ i ⟩ P .

Observe that, the right hand side can be estimated as

(38) ( α 0 − α ̃ 0 ) ⟨ ω , δ ̃ a i , λ i ⟩ P = o ( | ζ 0 | ) , ⟨ α 0 h − α ̃ 0 h ̃ , δ ̃ a i , λ i ⟩ P = ∫ S n ( P δ ̃ a i , λ i ) θ = O ‖ θ ‖ ∞ ∫ S n | P δ ̃ a i , λ i | = o ( ‖ θ ‖ ) , ⟨ α j δ ̃ a j , λ j − α ̃ j δ ̃ a ̃ j , λ ̃ j , δ ̃ a i , λ i ⟩ P = ( α j − α ̃ j ) ⟨ δ ̃ a j , λ j , δ ̃ a i , λ i ⟩ P + α ̃ j ⟨ δ ̃ a j , λ j − δ ̃ a ̃ j , λ ̃ j , δ ̃ a i , λ i ⟩ P = o ( | ζ j | + | ∧ j | + | ξ j | ) , for j ≠ i , ⟨ α i δ ̃ a i , λ i − α ̃ i δ ̃ a ̃ i , λ ̃ i , δ ̃ a i , λ i ⟩ P = ( α i − α ̃ i ) ‖ δ ̃ a i , λ i ‖ 2 + α ̃ i ⟨ δ ̃ a i , λ i − δ ̃ a ̃ i , λ ̃ i , δ ̃ a i , λ i ⟩ P = S n ζ i + o ( | ∧ i | + | ξ i | ) ,

(by using Lemma 2.3 and the fact that ‖θ‖_∞ ≤ c‖θ‖ since the dimension of W _u(ω) is finite). Furthermore, using Lemma 2.3, the left hand side of (37) can be estimated as

(39) ⟨ v ̃ , δ ̃ a i , λ i − δ ̃ a ̃ i , λ ̃ i ⟩ P = o ( ‖ δ ̃ a i , λ i − δ ̃ a ̃ i , λ ̃ i ‖ ) , ‖ δ ̃ a i , λ i − δ ̃ a ̃ i , λ ̃ i ‖ 2 = ⟨ δ ̃ a j , λ j − δ ̃ a ̃ j , λ ̃ j , δ ̃ a i , λ i ⟩ P − ⟨ δ ̃ a j , λ j − δ ̃ a ̃ j , λ ̃ j , δ ̃ a ̃ j , λ ̃ j ⟩ P = O | ∧ i | 2 + | ξ i | 2 ,

Combining the previous estimates, we obtain

(40) ζ i = o | ζ 0 | + ∑ ( | ζ j | + | ∧ j | + | ξ j | ) for each i ∈ { 1 , … , q } .

Now, we will focus on ζ ₀. As in the proof of (37), using (31), we get

(41) 0 = ⟨ v ̃ − v , ω ⟩ P = ⟨ φ ( α , a , λ , h ) − φ ( α ̃ , a ̃ , λ ̃ , h ̃ ) , ω ⟩ P = ∑ j = 1 q ⟨ α j δ ̃ a j , λ j − α ̃ j δ ̃ a ̃ j , λ ̃ j , ω ⟩ P + ( α 0 − α ̃ 0 ) ⟨ ω , ω ⟩ P + ⟨ α 0 h − α ̃ 0 h ̃ , ω ⟩ P = ∑ j = 1 q ( α j − α ̃ j ) ⟨ δ ̃ a j , λ j , ω ⟩ P + α ̃ j ⟨ δ ̃ a j , λ j − δ ̃ a ̃ j , λ ̃ j , ω ⟩ P + ‖ ω ‖ 2 ( α 0 − α ̃ 0 ) ,

(since ⟨ h , ω ⟩ P = ⟨ h ̃ , ω ⟩ P = 0 ). Now, observe that: ⟨ δ a j , λ j , ω ⟩ P = o ( 1 ) and using Lemma 2.3, we have

⟨ δ ̃ a j , λ j − δ ̃ a ̃ j , λ ̃ j , ω ⟩ P = ∫ S n ( P ω ) ( δ ̃ a j , λ j − δ ̃ a ̃ j , λ ̃ j ) = O ( | ∧ j | + | ξ j | ) ∫ S n δ ̃ a j , λ j = o ( | ∧ j | + | ξ j | ) .

Thus (41) implies that

(42) ζ 0 = o | ζ 0 | + ∑ ( | ζ j | + | ∧ j | + | ξ j | ) .

Concerning the variable θ k ≔ α 0 h − α ̃ 0 h ̃ , as in the estimate (41), using (30) and the fact that ⟨ ω , θ ⟩ P = 0 , we get

(43) 0 = ⟨ v ̃ − v , θ k ⟩ P = ⟨ φ ( α , a , λ , h ) − φ ( α ̃ , a ̃ , λ ̃ , h ̃ ) , θ k ⟩ P = ∑ j = 1 q ( α j − α ̃ j ) ⟨ δ ̃ a j , λ j , θ k ⟩ P + α ̃ j ⟨ δ ̃ a j , λ j − δ ̃ a ̃ j , λ ̃ j , θ k ⟩ P + ⟨ α 0 h − α ̃ 0 h ̃ , θ k ⟩ P .

The first integral is computed in (38) and the last one is exactly ‖θ _k‖². For the second one, using Lemma 2.3, it holds

(44) ⟨ δ ̃ a j , λ j − δ ̃ a ̃ j , λ ̃ j , θ k ⟩ P = ∧ j ⟨ λ j ∂ δ ̃ a j , λ j ∂ λ j , θ k ⟩ P + ξ j ⟨ 1 λ j ∂ δ ̃ a j , λ j ∂ a j , θ k ⟩ P + O | ∧ j | 2 + | ξ j | 2 ‖ θ k ‖ = o ∧ j 2 + | ξ j | 2 + ‖ θ k ‖ 2 .

Combining (43) with (38) and (44), we derive that

(45) ‖ θ k ‖ = ∑ j = 1 q o | ∧ j | + | ξ j | + ‖ θ k ‖ .

Now, we will focus on the variables ∧_j’s. As in the computations done in (37), and using (33), we get

(46) 〈 v ̃ , λ i ∂ δ ̃ a i , λ i ∂ λ i − λ ̃ i ∂ δ ̃ a ̃ i , λ ̃ i ∂ λ ̃ i 〉 P = 〈 φ ( α , a , λ , h ) − φ ( α ̃ , a ̃ , λ ̃ , h ̃ ) , λ i ∂ δ ̃ a i , λ i ∂ λ i 〉 P = ∑ j = 1 q ( α j − α ̃ j ) 〈 δ ̃ a j , λ j , λ i ∂ δ ̃ a i , λ i ∂ λ i 〉 P + α ̃ j 〈 δ ̃ a j , λ j − δ ̃ a ̃ j , λ ̃ j , λ i ∂ δ ̃ a i , λ i ∂ λ i 〉 P + ( α 0 − α ̃ 0 ) 〈 ω , λ i ∂ δ ̃ a i , λ i ∂ λ i 〉 P + 〈 α 0 h − α ̃ 0 h ̃ , λ i ∂ δ ̃ a i , λ i ∂ λ i 〉 P .

Observe that

〈 δ ̃ a i , λ i , λ i ∂ δ ̃ a i , λ i ∂ λ i 〉 P = 0 , 〈 δ ̃ a j , λ j , λ i ∂ δ ̃ a i , λ i ∂ λ i 〉 P = o ( 1 ) for j ≠ i , 〈 ω , λ i ∂ δ ̃ a i , λ i ∂ λ i 〉 P = o ( 1 ) , 〈 α 0 h − α ̃ 0 h ̃ , λ i ∂ δ ̃ a i , λ i ∂ λ i 〉 P = o ( ‖ θ k ‖ ) ,

(the proof of the last term can be done as the proof of (38)). It remains to estimate the second term in (46). Using Lemma 2.3, we have

⟨ δ ̃ a ̃ j , λ ̃ j − δ ̃ a j , λ j , λ i ∂ δ ̃ a i , λ i ∂ λ i ⟩ P = ∧ j ⟨ λ j ∂ δ ̃ a j , λ j ∂ λ j , λ i ∂ δ ̃ a i , λ i ∂ λ i ⟩ P + ξ j ⟨ 1 λ j ∂ δ ̃ a j , λ j ∂ a j , λ i ∂ δ ̃ a i , λ i ∂ λ i ⟩ P + O ∧ j 2 + | ξ j | 2 = ∧ i ‖ λ i ∂ δ ̃ a i , λ i / ∂ λ i ‖ 2 + o ( | ∧ i | + | ξ i | ) if j = i , o ( | ∧ j | + | ξ j | ) if j ≠ i .

Combining the previous estimates, we derive that

(47) | ∧ i | = ∑ j = 1 q o | ∧ j | + | ξ j | + ‖ θ k ‖ for each i ∈ { 1 , … , q } .

Finally, it remains to focus on the variables ξ _j’s. Following the proof of (47) (step by step by taking the derivative with respect to a unstead of the derivative with respect to λ), we derive that

(48) | ξ i | = ∑ j = 1 q o | ∧ j | + | ξ j | + ‖ θ k ‖ for each i ∈ { 1 , … , q } .

Now, combining equations (40), (42), (45), (47) and (48), we derive that all the variables are zero, that is

ζ i = 0 ∀ i ∈ { 0,1 , … , q } ; ∧ j = 0 ; ξ j = 0 ∀ j ∈ { 1 , … , q } and θ k = 0

which gives a contradiction with (29). Hence the proof of the lemma is completed. ■

Next, as in [34], we define the critical points at infinity of J _K as follows:

Definition 2.5.

A critical point at infinity of J _K in Σ⁺ ≔ {u ∈ Σ: u > 0} is defined as a limit of the flow-line u(s) of the equation ∂ u ∂ s = − ∇ J K ( u ) , u ₀ ∈ Σ⁺ such that, for s large, u(s) remains in V(q, ν(s), ω), where q ∈ N , ν(s) tends to zero as s tends to ∞ and ω is zero or a positive critical point of J _K. By Proposition 2.1, u(s) can be written as

u ( s ) = ∑ i = 1 q α i ( s ) δ ̃ ( a i ( s ) , λ i ( s ) ) + α 0 ( s ) ( ω + h ( s ) ) + v ( s ) .

Denoting a _i = lim_s→+∞ a _i(s), ( a 1 , … , a q , ω ) ∞ is called a critical point at infinity of J _K. When ω ≠ 0, we call ( a 1 , … , a q , ω ) ∞ a mixed type of critical point at infinity of J _K.

Note that, on the contrary to the case where n ≤ 7, when n ≥ 9, any solution ω of ( P K ) and a collection of critical points y _i of K having ΔK(y _i) < 0 create a critical point at infinity of J _K. For n = 8, a balance phenomenon appears; that is, the self-interaction of the functions δ ̃ a i , λ i and the scalar product ⟨ δ ̃ a i , λ i , ω ⟩ are of the same size. For n ≥ 9, we have the following characterization of the critical points at infinity.

Proposition 2.6.

Let n ≥ 9. Then, the only critical points at infinity of J _K correspond to

( y i 1 , … , y i q , ω ) ∞ and ( y i 1 , … , y i q ) ∞ ,

where { y i 1 , … , y i q } are any collections of q critical points of K in S n satisfying Δ K ( y i k ) < 0 and ω is a non-degenerate positive critical point of J _K.

The Morse index of these critical points (i.e. the number of the decreasing directions for J _K ) are equal to

i ∞ ( y i 1 , … , y i q , ω ) ≔ index ( ω ) + q + ∑ k = 1 q ( n − morse index ( K , y i k ) ) , i ∞ ( y i 1 , … , y i q ) ≔ q − 1 + ∑ k = 1 q ( n − morse index ( K , y i k ) ) .

Furthermore, the levels of these critical points at infinity are given by

C ∞ ( y i 1 , … , y i q ) ≔ S n 4 / n ∑ k = 1 q 1 K ( y i k ) ( n − 4 ) / 4 4 / n C ∞ ( y i 1 , … , y i q , ω ) ≔ S n ∑ k = 1 q 1 K ( y i k ) ( n − 4 ) / 4 + ‖ ω ‖ 2 4 / n ,

where S _n is defined by

(49) S n = ∫ R n δ ( 0,1 ) 2 n / ( n − 4 ) .

Proof.

For ω = 0, the result is extracted from [10], (see Proposition 3.2). For ω ≠ 0, the characterization of the critical points at infinity and their Morse index are proved in Proposition 3.3 of [16]. We notice that in [16], the author assumed that K has only two critical points with ΔK < 0. However its proof is done in the general case. Concerning the level C ∞ ( y i 1 , … , y i q , ω ) , using Proposition 5.1 of [16] and the fact that u ≔ ∑ i = 1 q α i δ ̃ i + α 0 ( ω + h ) + v ∈ V ( q , ν , ω ) with a _i close to a critical point y _i for each i, we have

J K ( u ) = S n ∑ α i 2 + α 0 2 ‖ ω ‖ 2 S n ∑ α i 2 n / ( n − 4 ) K ( a i ) + α 0 2 n / ( n − 4 ) ‖ ω ‖ 2 ( n − 4 ) / n ( 1 + o ( 1 ) ) = S n ∑ 1 K ( y i ) ( n − 4 ) / 4 + ‖ ω ‖ 2 4 / n ( 1 + o ( 1 ) ) .

This achieves the proof of the proposition. ■

3 Proof of the theorems

This section is devoted to the proof of our results. Their proofs are basically based on some deformation arguments. To see the effect of the pinching condition in K ε on the level sets of the functional J _K, we start by proving the following proposition which is crucial in the sequel.

Proposition 3.1.

Let N ∈ N and η be a positive constant satisfying η < 2/(2N + 1). There exists ɛ _N,η > 0 such that for

0 < ε < min ε N , η , N + 1 N 4 / ( n − 2 ) 1 − η 1 + η 4 / ( n − 2 ) − 1

and K ∈ K ε we have, J K C ≔ u ∈ Σ + : J K ( u ) ≤ C is contractible for any

C ∈ k 4 / n S n 4 / n ( 1 + η ) 4 / n , ( k + 1 ) 4 / n S n 4 / n ( 1 − η ) 4 / n

and for each k ∈ {1, …, N}.

Proof.

Notice that, in S n , the only positive solutions of ( P 1 ) (that is ( P K ) with K ≡ 1) are the functions δ ̃ ( a , λ ) with a ∈ S n and λ > 0 [31]. Furthermore, it follows from the characterization of the lack of compactness for J ₁ (see [21]), each Palais-Smale sequence ( u q ) q (under the level ( N + 1 ) 4 / n S n 4 / n ) has to be close to a sum of δ ̃ i and therefore J ₁(u _q) will be close to k 4 / n × S n 4 / n for some k ≤ N. Hence |∇J ₁(u)| ≥ 2C _N,η for some positive constant C _N,η and for each u satisfying

J 1 ( u ) ∈ ∪ 1 ≤ k ≤ N k 4 / n S n 4 / n ( 1 + η ) 4 / n , ( k + 1 ) 4 / n S n 4 / n ( 1 − η ) 4 / n .

Now, using the fact that ∇J _K → ∇J ₁ as ɛ → 0, we see that there exists ɛ _N,η such that for 0 < ɛ ≤ ɛ _N,η and K ∈ K ε , we have

(50) | ∇ J K ( u ) | ≥ C N , η > 0

for each u satisfying

(51) J K ( u ) ∈ k 4 / n S n 4 / n ( 1 + η ) 4 / n , ( k + 1 ) 4 / n S n 4 / n ( 1 − η ) 4 / n with k ∈ { 1 , … , N } .

Hence, for k ∈ {1, …, N} and letting A ̲ ≔ [ k S n ( 1 + η ) ] 4 / n , we see that (50) implies that

(52) J K C retracts by deformation onto J K A ̲ ∀ C ∈ k S n ( 1 + η ) 4 n , ( k + 1 ) S n ( 1 − η ) 4 n .

Observe that, easy computations imply that

A ̄ ≔ A ̲ ( K max / K min ) ( n − 4 ) / n < [ ( k + 1 ) S n ( 1 − η ) ] 4 / n .

Thus, for each u ∈ J K A ̄ \ J K A ̲ , we have J K ( u ) ∈ k S n ( 1 + η ) 4 n , ( k + 1 ) S n ( 1 − η ) 4 n and therefore, by using (50), we see that J _K does not have any critical point nor critical point at infinity in J K A ̄ \ J K A ̲ . Thus, using Proposition 4.1 of [10], we derive that J K C is contractible for each C ∈ [ A ̲ , A ̄ ] , in particular J K A ̲ is contractible. Then the result follows from (52). ■

Next, we are going to study the contribution of solutions in the Euler-Poincaré characteristic formulae. We notice that, from (50) and (51), under the level ( ( N + 1 ) S n ) 4 / n , each positive critical point ω of J _K (if it exists) has to satisfy

J K ( ω ) ∈ ∪ k = 1 N ( k S n ( 1 − η ) ) 4 / n , ( k S n ( 1 + η ) ) 4 / n .

In the sequel, if J K ( ω ) ∈ ( k S n ( 1 − η ) ) 4 / n , ( k S n ( 1 + η ) ) 4 / n , we say that the level of ω is close to ( k S n ) 4 / n . Now, to proceed further, we need to introduce some notation. Let m be the cardinal of the set K ∞ defined in (H3) and for y ∈ K ∞ , we denote by

ι ( y ) = n − morse index ( K , y ) .

Suppose that K ∞ = { y 1 , … , y m } with y ₁ is a maximum point of K.

In the sequel, denoting by ω _k a positive critical point of J _K having an energy close to ( k S n ) 4 / n , we will use the following notation:

(53) S p ; S ≥ k ∞ , p and S ≥ k ; k ∞ , p

to denote

S p : the set of positive critical points ω _p of J _K,
S ≥ k ∞ , p : the critical points at infinity ω _∞,p having the form ( y i 1 , … , y i p ) ∞ with k ≤ i ₁ < … < i _p or ( y i 1 , … , y i r , ω p − r ) ∞ with 1 ≤ r < p and k ≤ i ₁ < … < i _r,
S ≥ k ; k ∞ , p : the subset of S ≥ k ∞ , p satisfying i ₁ = k.

We remark that, if k > m, then S ≥ k ∞ , p = ∅ . We also need to introduce the following notation:

μ p ≔ ∑ ω p ∈ S p ( − 1 ) m ( ω p ) ; μ ≥ k ∞ , p ≔ ∑ ω ∞ , p ∈ S ≥ k ∞ , p ( − 1 ) m ( ω ∞ , p ) and μ ≥ k ; k ∞ , p ≔ ∑ ω ∞ , p ∈ S ≥ k ; k ∞ , p ( − 1 ) m ( ω ∞ , p )

where m(ω) denotes the Morse index of ω.

We notice that, if k > m ≔ # K ∞ , we have μ ≥ k ∞ , p = 0 for each p ≥ 1 (since it does not exist critical point y _k in K ∞ with k > m).

Remark 3.2.

The number of positive critical points of J _K (hence the number of solutions of ( P K ) ) having an energy close to ( p S n ) 4 / n is at least equal to |μ _p|.
Since Σ⁺ is invariant under the gradient flow [19], the solutions obtained by our argument are positive.
The variational solutions of ( P K ) are regular [35].

In the next two lemmas, we provide some useful information on μ _p, μ ≥ k ∞ , p and μ ≥ k ; k ∞ , p .

Lemma 3.3.

Under the above notation, for each k ≥ 1, it holds

(54) μ ≥ k ∞ , p = μ ≥ k ; k ∞ , p + μ ≥ k + 1 ∞ , p and μ ≥ k ; k ∞ , p = − ( − 1 ) ι ( y k ) μ ≥ k + 1 ∞ , p − 1 + μ p − 1 .

Proof.

First, we remark that, for ω ∞ , p ∈ S ≥ k ∞ , p , two cases may occur:

either y _k does not appear in the configuration of ω _∞,p. In this case, ω ∞ , p ∈ S ≥ k + 1 ∞ , p .
or y _k appears in the configuration of ω _∞,p. In this case, ω ∞ , p ∈ S ≥ k ; k ∞ , p .

This implies the first claim of the lemma.

Second, the above second situation can be divided in three subcases:

ω ∞ , p = ( y k , ω p − 1 ) ∞ and in this case, by Proposition 2.6, we have
( − 1 ) morse index ( ω ∞ , p ) = − ( − 1 ) ι ( y k ) ( − 1 ) morse index ( ω p − 1 ) .
ω ∞ , p = ( y k , y i 2 , … , y i p ) ∞ with k < i ₂ < … < i _p. Note that ω ∞ , p − 1 ≔ ( y i 2 , … , y i p ) ∞ ∈ S ≥ k + 1 ∞ , p − 1 and Proposition 2.6 implies that
( − 1 ) morse index ( J K , ω ∞ , p ) = − ( − 1 ) ι ( y k ) ( − 1 ) morse index ( J K , ω ∞ , p − 1 ) .
ω ∞ , p = ( y k , y i 2 , … , y i r , ω p − r ) ∞ with k < i ₂ < … < i _r and 2 ≤ r < p. Note that ω ∞ , p − 1 ≔ ( y i 2 , … , y i r , ω p − r ) ∞ ∈ S ≥ k + 1 ∞ , p − 1 and Proposition 2.6 implies that
( − 1 ) morse index ( J K , ω ∞ , p ) = − ( − 1 ) ι ( y k ) ( − 1 ) morse index ( J K , ω ∞ , p − 1 ) .

Hence the proof of the second claim of (54) is complete. ■

Lemma 3.4.

With the above notation, we have

(55) ( i ) μ ≥ 1 ∞ , 1 + μ 1 = 1 and μ ≥ 1 ∞ , p + μ p = 0 ∀ p ≥ 2 .

(56) ( i i ) μ ≥ 2 ∞ , p + μ p = 0 ∀ p ≥ 1 .

Proof.

The proof is based on the deformation of the level sets of J _K in Σ⁺ ≔ {u ∈ Σ: u > 0}. Taking u 0 ∈ J K [ p S n ( 1 + η ) ] 4 / n ≔ u ∈ Σ + : J K ( u ) ≤ [ p S n ( 1 + η ) ] 4 / n , we consider the flow

∂ u ∂ s = − ∇ J K ( u ( s ) ) = − 2 J K ( u ) u − J K ( u ) n n − 4 P − 1 K | u | 8 n − 4 u u ( 0 ) = u 0 .

Notice that Σ⁺ is invariant under the above flow [19]. Now, if we use the above flow in order to deform the level set J K [ p S n ( 1 + η ) ] 4 / n onto the level set J K [ p S n ( 1 − η ) ] 4 / n , we see that three cases may occur:

there exists s ₁ > such that u ( s 1 , u 0 ) ∈ J K [ p S n ( 1 − η ) ] 4 / n .
a stop at a critical point of J _K and since Σ⁺ is invariant under the flow, such a critical point is positive.
the emergence of a critical point at infinity, that is, for s large, the flow line u(s, u ₀) enters and remains in V(q, ν(s), ω), where q ∈ N , ν(s) tends to zero as s tends to ∞ and ω is zero or a positive critical point of J _K.

Thus, the Euler-Poincaré theorem implies that

χ ( J K [ p S n ( 1 + η ) ] 4 / n ) = χ ( J K [ p S n ( 1 − η ) ] 4 / n ) + μ p + μ ≥ 1 ∞ , p for p ≥ 2 , χ ( J K S n 4 / n ( 1 + η ) 4 / n ) = μ 1 + μ ≥ 1 ∞ , 1 .

Furthermore, Proposition 3.1 implies that J K [ p S n ( 1 + η ) ] 4 / n and J K [ p S n ( 1 − η ) ] 4 / n are contractible sets. Therefore, Claim (i) follows.

Now, using (54) and (55), we derive that

(57) ( − 1 ) ι ( y 1 ) + μ ≥ 2 ∞ , 1 + μ 1 = 1 and − ( − 1 ) ι ( y 1 ) μ ≥ 2 ∞ , p − 1 + μ p − 1 + μ ≥ 2 ∞ , p + μ p = 0 ∀ p ≥ 2 .

Since we have assumed that y ₁ is a maximum point of K, it has to satisfy ΔK(y ₁) < 0 and ι(y ₁) = 0 is even. This implies that the first equation in (57) is equivalent to μ ≥ 2 ∞ , 1 + μ 1 = 0 . However, denoting by x p ≔ μ ≥ 2 ∞ , p + μ p , we see that the second equation in (57) is equivalent to x _p = x _p−1 for each p ≥ 2. Thus, we obtain Claim (56) which completes the proof of the lemma. ■

In the next remark, we explain the importance of the assumption (H3) in our argument.

Remark 3.5.

If K ∞ is reduced to a point, we have μ ≥ 2 ∞ , p = 0 for any p ≥ 1. This implies that μ _p = 0 for any p ≥ 1. Hence we cannot deduce the existence of solutions of ( P K ) .

In the sequel, we assume that m ≔ # K ∞ ≥ 2 . Next we are going to prove the following crucial Lemmas.

Lemma 3.6.

Assume that ι(y _j) is even for each j ≥ 2. Then it holds that

(58) μ ≥ k ∞ , p + μ p < 0 ∀ p ≥ 1 ∀ 3 ≤ k ≤ m + 1 .

Furthermore, we have

μ p ≤ − ( m − 1 ) ∀ p ≥ 1 .

Proof.

Using (54) with k = 2, (56) becomes

1 + μ ≥ 3 ∞ , 1 + μ 1 = 0 and − μ ≥ 3 ∞ , p − 1 + μ p − 1 + μ ≥ 3 ∞ , p + μ p = 0 ∀ p ≥ 2

which implies that the sequence μ ≥ 3 ∞ , p + μ p p ≥ 1 is constant. Thus we get

(59) μ ≥ 3 ∞ , p + μ p = − 1 ∀ p ≥ 1 .

Therefore (58) holds for k = 3. By induction, assume that (58) is true for k, i.e.:

(60) μ ≥ k ∞ , p + μ p < 0 ∀ p ≥ 1 .

Arguing as before, using (54) and (60), it holds:

1 + μ ≥ k + 1 ∞ , 1 + μ 1 < 0 and − μ ≥ k + 1 ∞ , p − 1 + μ p − 1 + μ ≥ k + 1 ∞ , p + μ p < 0 ∀ p ≥ 2 ,

which implies that

μ ≥ k + 1 ∞ , 1 + μ 1 < − 1 and μ ≥ k + 1 ∞ , p + μ p < μ ≥ k + 1 ∞ , 1 + μ 1 ∀ p ≥ 2 .

Hence (58) is true for k + 1 and therefore (58) follows for each 3 ≤ k ≤ m + 1.

Now using (58), we see that

μ ≥ k ∞ , p = μ ≥ k + 1 ∞ , p + μ ≥ k ; k ∞ , p = μ ≥ k + 1 ∞ , p − μ ≥ k + 1 ∞ , p − 1 + μ p − 1 > μ ≥ k + 1 ∞ , p .

Thus we derive that

μ p = μ ≥ m + 1 ∞ , p + μ p < μ ≥ m ∞ , p + μ p < … < μ ≥ 2 ∞ , p + μ p = 0

which implies that

μ p ≤ − ( m − 1 ) .

■

Lemma 3.7.

Assume that ι(y ₂) is odd and let ℓ ₀ ≥ 1 be such that ι(y _2j) is odd and ι(y _2j+1) is even for each 1 ≤ j ≤ ℓ ₀. Then it holds:

μ ≥ 2 j + 2 ∞ , 1 + μ 1 = 0 ; μ ≥ 2 j + 2 ∞ , 2 p + μ 2 p < 0 and μ ≥ 2 j + 2 ∞ , 2 p + 1 + μ 2 p + 1 = 0 ∀ p ≥ 1 ; ∀ 1 ≤ j ≤ ℓ 0 .

Proof.

We will prove the lemma by induction on j. Using (54) and (56) and the fact that ι(y ₂) is odd, we derive that

(61) − 1 + μ ≥ 3 ∞ , 1 + μ 1 = 0 and μ ≥ 3 ∞ , p − 1 + μ p − 1 + μ ≥ 3 ∞ , p + μ p = 0 ∀ p ≥ 2 .

Denoting by x p ≔ μ ≥ 3 ∞ , p + μ p , the second equation in (61) implies that x _p = −x _p−1 for all p ≥ 2. Therefore we have x _p = (−1)^p+1 x ₁. Thus we deduce that

(62) μ ≥ 3 ∞ , p + μ p = ( − 1 ) p + 1 ∀ p ≥ 1 .

Now, using (54), (62) and the fact that ι(y ₃) is even, we deduce that

(63) 1 + μ ≥ 4 ∞ , 1 + μ 1 = 1 and − μ ≥ 4 ∞ , p − 1 + μ p − 1 + μ ≥ 4 ∞ , p + μ p = ( − 1 ) p + 1 ∀ p ≥ 2 .

Denoting by z p ≔ μ ≥ 4 ∞ , p + μ p and using (63), we see that

z k = ∑ j = 2 k ( z j − z j − 1 ) = ∑ j = 2 k ( − 1 ) j + 1 = 0 if k is odd − 1 if k is even .

Thus we obtain

μ ≥ 4 ∞ , 1 + μ 1 = 0 ; μ ≥ 4 ∞ , 2 p + μ 2 p = − 1 and μ ≥ 4 ∞ , 2 p + 1 + μ 2 p + 1 = 0 ∀ p ≥ 1 .

Hence, we obtain the desired result for j = 1.

Now, we will assume that the result holds for j − 1 ≤ ℓ ₀ − 1 and we need to prove it for j. Note that ι(y _2j) is odd and ι(y _2j+1) is even. Then we get

(64) μ ≥ 2 j ∞ , 1 + μ 1 = 0 ; μ ≥ 2 j ∞ , 2 p + μ 2 p < 0 and μ ≥ 2 j ∞ , 2 p + 1 + μ 2 p + 1 = 0 ∀ p ≥ 1 .

Using (54), (64) and the fact that ι(y _2j) is odd, we obtain

− 1 + μ ≥ 2 j + 1 ∞ , 1 + μ 1 = 0 ; μ ≥ 2 j + 1 ∞ , 2 p − 1 + μ 2 p − 1 + μ ≥ 2 j + 1 ∞ , 2 p + μ 2 p < 0 ;

and

μ ≥ 2 j + 1 ∞ , 2 p + μ 2 p + μ ≥ 2 j + 1 ∞ , 2 p + 1 + μ 2 p + 1 = 0 ∀ p ≥ 1 .

Denoting by r p ≔ μ ≥ 2 j + 1 ∞ , p + μ p , we see that the previous equations imply that

r 1 = 1 , r 2 p < − r 2 p − 1 and r 2 p = − r 2 p + 1 ∀ p ≥ 1 .

This implies that r _2p < r _2p−2 which gives that ( r 2 p ) p ≥ 1 is a decreasing sequence. Thus

r 2 p < r 2 < − r 1 = − 1 .

Hence we obtain

(65) μ ≥ 2 j + 1 ∞ , 1 + μ 1 = 1 and μ ≥ 2 j + 1 ∞ , 2 p + 1 + μ 2 p + 1 = − μ ≥ 2 j + 1 ∞ , 2 p + μ 2 p > 0 ∀ p ≥ 1 .

Now we need to repeat this argument using y _2j+1 which satisfies ι(y _2j+1) is even.

Using (54), (65) and the fact that ι(y _2j+1) is even, we obtain 1 + μ ≥ 2 j + 2 ∞ , 1 + μ 1 = 1 which implies that μ ≥ 2 j + 2 ∞ , 1 + μ 1 = 0 . Furthermore, using (54) and (65), we get:

0 = μ ≥ 2 j + 1 ∞ , 2 p + 1 + μ 2 p + 1 + μ ≥ 2 j + 1 ∞ , 2 p + μ 2 p = μ ≥ 2 j + 2 ∞ , 2 p + 1 + μ 2 p + 1 − μ ≥ 2 j + 2 ∞ , 2 p − 1 + μ 2 p − 1 ∀ p ≥ 1 ,

which implies that the sequence μ ≥ 2 j + 2 ∞ , 2 p + 1 + μ 2 p + 1 is constant and it is equal to μ ≥ 2 j + 2 ∞ , 1 + μ 1 which is 0.

For the other indices, combining (54) and (65), we obtain

0 > μ ≥ 2 j + 1 ∞ , 2 p + μ 2 p = − μ ≥ 2 j + 2 ∞ , 2 p − 1 + μ 2 p − 1 + μ ≥ 2 j + 2 ∞ , 2 p + μ 2 p = μ ≥ 2 j + 2 ∞ , 2 p + μ 2 p ∀ p ≥ 1 .

Hence the result is proved for j and therefore the proof of the lemma is complete. ■

Lemma 3.8.

Assume that ι(y ₂) is odd and let k ₀ ≥ 3 be such that ι ( y k 0 ) is even and μ ≥ k 0 ∞ , p + μ p ≤ 0 for each p ≥ 1. Then it holds:

μ ≥ k 0 + 1 ∞ , p + μ p < 0 ∀ p ≥ 1 .

Proof.

For p = 1, since μ ≥ k 0 ∞ , 1 + μ 1 ≤ 0 and ι ( y k 0 ) is even, using (54), we get that 0 ≥ 1 + μ ≥ k 0 + 1 ∞ , 1 + μ 1 which implies the result for p = 1.

Now using again (54), for p ≥ 2, we obtain that

0 ≥ μ ≥ k 0 ∞ , p + μ p = − μ ≥ k 0 + 1 ∞ , p − 1 + μ p − 1 + μ ≥ k 0 + 1 ∞ , p + μ p

which implies that the sequence μ ≥ k 0 + 1 ∞ , p + μ p p ≥ 1 is a decreasing sequence. Thus we get μ ≥ k 0 + 1 ∞ , p + μ p ≤ μ ≥ k 0 + 1 ∞ , 1 + μ 1 ≤ − 1 . Hence the proof of the lemma is complete. ■

Lemma 3.9.

Assume that ι(y ₂) is odd and let k ₀ ≥ 2 be such that ι ( y k 0 ) is odd and

μ ≥ k 0 ∞ , 1 + μ 1 ≥ 0 ; μ ≥ k 0 ∞ , 2 p + μ 2 p ≤ 0 and μ ≥ k 0 ∞ , 2 p + 1 + μ 2 p + 1 ≥ 0 ∀ p ≥ 1 .

Then it holds:

μ ≥ k 0 + 1 ∞ , 1 + μ 1 > 0 ; μ ≥ k 0 + 1 ∞ , 2 p + μ 2 p < 0 and μ ≥ k 0 + 1 ∞ , 2 p + 1 + μ 2 p + 1 > 0 ∀ p ≥ 1 .

Furthermore, if ι(y _j) is odd for each j ≥ 2, it holds that

μ 2 p ≤ − ( m − 1 ) ∀ 1 ≤ p ≤ [ N / 2 ] ; μ 2 p − 1 ≥ ( m − 1 ) ∀ 1 ≤ p ≤ [ ( N + 1 ) / 2 ] .

Proof.

The first claim follows immediately from the fact that ι ( y k 0 ) is odd and μ ≥ k 0 ∞ , 1 = ( − 1 ) ι ( y k 0 ) + μ ≥ k 0 + 1 ∞ , 1 . For the other claims, using (54), it holds:

0 ≥ ∑ j = 2 p ( − 1 ) j μ ≥ k 0 ∞ , j + μ j = ∑ j = 2 p ( − 1 ) j μ ≥ k 0 + 1 ∞ , j − 1 + μ j − 1 + μ ≥ k 0 + 1 ∞ , j + μ j = ( − 1 ) p μ ≥ k 0 + 1 ∞ , p + μ p + μ ≥ k 0 + 1 ∞ , 1 + μ 1 ,

which implies that ( − 1 ) p μ ≥ k 0 + 1 ∞ , p + μ p < 0 . The first claim of the lemma is thereby proved.

Concerning the second claim, by using the first one, we see that

μ ≥ k 0 ∞ , 2 p + μ 2 p = μ ≥ k 0 + 1 ∞ , 2 p − 1 + μ 2 p − 1 + μ ≥ k 0 + 1 ∞ , 2 p + μ 2 p > μ ≥ k 0 + 1 ∞ , 2 p + μ 2 p ∀ p ≥ 1

which implies that

μ 2 p = μ ≥ m + 1 ∞ , 2 p + μ 2 p < … < μ ≥ 2 ∞ , 2 p + μ 2 p = 0 .

Therefore, we deduce that

μ 2 p ≤ − ( m − 1 ) ∀ 1 ≤ p ≤ [ N / 2 ] .

In the same way we prove that

μ 2 p − 1 ≥ ( m − 1 ) ∀ 1 ≤ p ≤ [ ( N + 1 ) / 2 ] .

Thus the proof of the lemma is complete. ■

Now, we are ready to prove our results. We start by considering the case where the Euler-Poincaré index formula is equal to 1.

Proof of Theorem 1.4.

We assume that B 1 ≔ ∑ y ∈ K ∞ ( − 1 ) ι ( y ) = 1 . In this case, # K ∞ has to be odd, say 2r + 1 with r + 1 points having ι(y) even and r points having ι(y) odd. Furthermore, since we have assumed that # K ∞ ≥ 2 , we deduce that r ≥ 1. In addition, since # K ∞ = 2 r + 1 , we derive that μ ≥ 2 r + 2 ∞ , p = 0 for each p ≥ 1.

Now, we order the points y _k’s of K ∞ such that ι(y ₁), ι(y ₃), …., ι(y _2r+1) are even and ι(y ₂), ι(y ₄), …., ι(y _2r) are odd. Applying (54), we deduce that

μ ≥ 2 j ∞ , 2 p + μ 2 p = μ ≥ 2 j + 1 ∞ , 2 p − 1 + μ 2 p − 1 + μ ≥ 2 j + 1 ∞ , 2 p + μ 2 p = − μ ≥ 2 j + 2 ∞ , 2 p − 2 + μ 2 p − 2 + μ ≥ 2 j + 2 ∞ , 2 p + μ 2 p

Using Lemma 3.7, we obtain

μ ≥ 2 j ∞ , 2 p + μ 2 p > μ ≥ 2 j + 2 ∞ , 2 p + μ 2 p

which implies that

μ 2 p = μ ≥ 2 r + 2 ∞ , 2 p + μ 2 p < μ ≥ 2 r ∞ , 2 p + μ 2 p < … < μ ≥ 2 ∞ , 2 p + μ 2 p = 0 .

Therefore we obtain

μ 2 p ≤ − r ≔ − ( # K ∞ − 1 ) / 2 .

In addition, applying Lemma 3.7 with j = l ₀ = r, we deduce that

μ 2 p + 1 = 0 ∀ p ≥ 1 .

Hence the proof of the theorem is completed. ■

Next, we consider the case where B 1 ≔ ∑ y ∈ K ∞ ( − 1 ) ι ( y ) ≠ 1 . More precisely, our aim is to prove Theorems 1.1 and 1.2.

Proof of Theorem 1.1.

We assume that the assumption stated in Case 1 or Case 2 is satisfied, that is, we assume that y ₁, …, y _m are ordered so that we are in one of the following two cases:

Case 1. ι(y _j) is even for all j.

Case 2. ι(y _j) is odd for all j ≥ 2.

In Case 1, i.e. all the points y _k’s of K ∞ satisfy: ι(y _k) is even. Using Lemma 3.6, we derive that μ _p ≤ −(m − 1) for each p ≥ 1 and therefore, at each level ( p S n ) 4 / n with p ≤ N, there exist at least (m − 1) solutions of ( P K ) .
In Case 2, i.e. we assume that ι(y _k) is odd for each k ≥ 2. Using Lemma 3.9, we derive that, at each level ( p S n ) 4 / n with p ≤ N, ( P K ) has at least (m − 1) solutions.

This completes the proof of the theorem. ■

Proof of Theorem 1.2.

We assume that the assumption stated in Case 3 or Case 4 is satisfied, that is, we assume that y ₁, …, y _m are ordered so that we are in one of the following two cases:

Case 3. For some l ≥ 1 such that 2l ≤ m − 2, we have

ι ( y 2 j ) is odd and ι ( y 2 j + 1 ) is even ∀ j ∈ { 1 , … , ℓ } and ι ( y j ) is even ∀ 2 ℓ + 2 ≤ j ≤ m .

Case 4. For some ℓ ≥ 1 such that 2ℓ ≤ m − 2, we have

ι ( y 2 j ) is odd and ι ( y 2 j + 1 ) is even ∀ j ∈ { 1 , … , ℓ } and ι ( y j ) is odd ∀ 2 ℓ + 2 ≤ j ≤ m .

Note that, since B ₁ ≠ 1, m cannot be equal to 2ℓ + 1 and therefore: m ≥ 2ℓ + 2.

In Case 3, we know that ι(y _2k+1) is even and ι(y _2k) is odd for each k ≤ ℓ and ι(y _j) is even for each j ≥ 2ℓ + 2. Applying Lemma 3.7 for ℓ ₀ = ℓ, we have
(66) μ ≥ 2 ℓ + 2 ∞ , 1 + μ 1 = 0 ; μ ≥ 2 ℓ + 2 ∞ , 2 p + μ 2 p < 0 and μ ≥ 2 ℓ + 2 ∞ , 2 p + 1 + μ 2 p + 1 = 0 ∀ p ≥ 1 .

Now we notice that (66) implies that the assumptions of Lemma 3.8 are satisfied for k ₀ = 2ℓ + 2 and therefore we deduce that μ ≥ 2 ℓ + 3 ∞ , p + μ p < 0 for each p ≥ 1. Again, the assumptions of Lemma 3.8 are satisfied for k ₀ = 2ℓ + 3. Hence by induction we deduce that
μ ≥ m + 1 ∞ , p + μ p = μ p < 0 for each p ≥ 1 .

Thus the existence of at least one solution of ( P K ) at each level ( p S n ) 4 / n with p ≤ N.
In Case 4, we know that ι(y _2k+1) is even and ι(y _2k) is odd for each k ≤ ℓ and ι(y _j) is odd for each j ≥ 2ℓ + 2. Clearly, in this case, assumptions of Lemma 3.7 are satisfied for ℓ ₀ = ℓ. Thus, we see that (66) holds in this case. Hence, applying Lemma 3.9 with k ₀ = 2ℓ + 2, we get
μ ≥ 2 ℓ + 3 ∞ , 1 + μ 1 > 0 ; μ ≥ 2 ℓ + 3 ∞ , 2 p + μ 2 p < 0 and μ ≥ 2 ℓ + 3 ∞ , 2 p + 1 + μ 2 p + 1 > 0 ∀ p ≥ 1 .
Observe that the assumptions of Lemma 3.9 are again satisfied for k ₀ = 2ℓ + 3, then by induction, we get that μ _2p−1 > 0 and μ _2p < 0 for each p ≥ 1. Hence the existence of at least one solution of ( P K ) at each level ( p S n ) 4 / n with p ≤ N.

The proof of the theorem is thereby completed. ■

Corresponding author: Khalil El Mehdi, Department of Mathematics, College of Science, Qassim University, Buraydah 51452, Saudi Arabia; and Faculté des Sciences et Techniques, Université de Nouakchott, Nouakchott, Mauritania, E-mail: K.Jiyid@qu.edu.sa

Acknowledgments

The Researchers would like to thank the Deanship of Graduate Studies and Scientific Research at Qassim University for financial support (QU-APC-2024-9/1).

Research ethics: Not applicable.
Author contributions: The authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Competing interests: The authors state no conflicts of interest.
Research funding: This research received no external funding.
Data availability: Not applicable.

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Received: 2023-03-16

Accepted: 2024-03-26

Published Online: 2024-04-24

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Keywords for this article

partial differential equations; analysis on manifolds; Q curvature problem

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