Doubling the equatorial for the prescribed scalar curvature problem on SN\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathbb {S}}}^N$$\end{document}

We consider the prescribed scalar curvature problem on SN\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ {{\mathbb {S}}}^N $$\end{document}ΔSNv-N(N-2)2v+K~(y)vN+2N-2=0onSN,v>0inSN,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Delta _{{{\mathbb {S}}}^N} v-\frac{N(N-2)}{2} v+{\tilde{K}}(y) v^{\frac{N+2}{N-2}}=0 \quad \text{ on } \ {{\mathbb {S}}}^N, \qquad v >0 \quad {\quad \hbox {in } }{{\mathbb {S}}}^N, \end{aligned}$$\end{document}under the assumptions that the scalar curvature K~\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\tilde{K}}$$\end{document} is rotationally symmetric, and has a positive local maximum point between the poles. We prove the existence of infinitely many non-radial positive solutions, whose energy can be made arbitrarily large. These solutions are invariant under some non-trivial sub-group of O(3) obtained doubling the equatorial. We use the finite dimensional Lyapunov–Schmidt reduction method.


Introduction
Given the N -th sphere (S N , g) equipped with the standard metric g and a fixed smooth functionK, the prescribed scalar curvature problem on S N consists in understanding whether it is possible to find another metricg in the conformal class of g, such that the scalar curvature ofg isK. For some positive function v : S N → R, and a related conformal change of the metric g = v 4 N −2 g, the scalar curvature with respect tog is given by where Δ S N is the Laplace-Beltrami operator on S N . Thus the prescribed scalar curvature problem on S N can be addressed by studying the solvability of the Testing the Eq. (1.1) against v and integrating on S N , we get that a necessary condition for the solvability of this problem is thatK(y) must be positive somewhere. There are other obstructions for the existence of solutions, which are said to be of topological type. For instance, a solution v must satisfy the following Kazdan-Warner type condition (see [15]): This condition is a direct consequence of Theorem 5.17 in [16], where Kazdan and Warner proved that given a positive solution v to for any spherical harmonics F of degree 1. Taking a = N +2 N −2 , H =K and F = y in (1.3), we can obtain condition (1.2). The problem of determining whichK(y) admits a solution has been the object of several studies in the past years. We refer the readers to [2][3][4][6][7][8]10,14,15,19,30], and the references therein.
For N = 3, Li [17] showed problem (1.4) has infinitely many solutions provided that K(y) is bounded below, and periodic in one of its variables, and the set {x | K(x) = max y∈R 3 K(y)} is not empty and contains at least one bounded connected component.
If K has the form K(y) = 1 + h(y), namely it is a perturbation of the constant 1, Cao et al. [5] proved the existence of multiple solutions.
If K(y) has a sequence of strictly local maximum points moving to infinity, Yan [32] constructed infinitely many solutions.
Wei and Yan [31] showed that problem (1.4) has infinitely many solutions provided K is radially symmetric K(y) = K(r), r = |y|, and has a local maximum around a given r 0 > 0. More precisely, they ask that there are r 0 , c 0 > 0 and m ∈ [2, N − 2) such that K(s) = K(r 0 ) − c 0 |s − r 0 | m + O |s − r 0 | m+σ , s ∈ (r 0 − δ, r 0 + δ), where σ, δ are small positive constants. In order to briefly discuss the main results in [31], we will recall the expression of Aubin-Talenti bubbles. It is well known (see [29]) that all solutions to the following problem Δu + u 2 * −1 = 0, u > 0 in R N , (1.5) are given by 4 . The solutions in [31] are obtained by gluing together a large number of Aubin-Talenti bubbles, which looks likẽ whereΛ is a positive constant and the points x j are distributed along the vertices of a regular polygon of k edges in the (y 1 , y 2 )-plane, with |x j | → r 0 as k → ∞: x j = r cos 2(j − 1)π k ,r sin 2(j − 1)π k , 0, . . . , 0 , j = 1, . . . , k, Under a weaker symmetry condition for K(y) = K(|y |, y ) with y = (|y |, y ) ∈ R 2 × R N −2 , Peng et al. [27] constructed infinitely many bubbling solutions, which concentrate at the saddle points of the potential K(y). Guo and Li [11] admitted infinitely many solutions for problems (1.4) with polyharmonic operators. For fractional case, we refer to [13,23].
The study of other aspects of problem (1.4), such as radial symmetry of their solutions, uniqueness of solutions, Liouville type theorem, a priori estimates, and bubbling analysis, have been the object of investigation of several researchers. We refer the readers to the papers [1,9,18,[20][21][22]25,26,32] and the references therein.
Recently, Guo et al. [12] investigated the spectral property of the linearized problem associated to (1.4) around the solutionũ k found in [31]. They proved a non-degeneracy result for such operator by using a refined version of local Pohozaev identities. As an application of this non-degeneracy result, they built new type of solutions by gluing another large number of bubbles, whose centers lie near the circle |y| = r 0 in the (y 3 , y 4 )-plane.
All these results concern solutions made by gluing together Aubin-Talenti bubbles with centres distributed along the vertices of one or more planar polygons, thus of two-dimensional nature. The purpose of this paper is to present a different type of solution to (1.4) with a more complex concentration structure, which cannot be reduced to a two-dimensional one.
To present our result, we assume that K is radially symmetric and satisfies the following condition (H): There are r 0 and c 0 > 0 such that There is a slight difference between our assumptions on K(s) and the ones in [31]. We will comment on this issue later. Without loss of generality, we assume r 0 = 1, K(1) = 1. For any integer k, we denote and set u(y) = r − N −2 2 v |y| r . Then the problem (1.4) can be rewritten, in terms of u, as Here 0 is the zero vector in R N −3 and h, r are positive parameters. We shall construct a family of solutions to problem (1.8) which are small perturbations of W r,h,Λ . More precisely, the Aubin-Talenti bubbles are now centred at points lying on the top and the bottom circles of a cylinder and this configuration is now invariant under a non-trivial sub-group of O(3) rather than O (2).
Throughout of the present paper, we assume N ≥ 5 and (r, h, Λ) ∈ S k , where with Λ 0 , B being the constants in (3.7), (3.10) andσ a fixed small number, independent of k.
Since h → 0 as k → ∞, then the two circles where the points x j and x j are distributed become closer to each other as k increases.
In this paper, we shall prove that for any k large enough, problem (1.8) admits a family of solutions u k with the approximate form (1.11) Moreover, these solutions are polygonal symmetry in the (y 1 , y 2 )-plane, even in the y 3 direction and radially symmetric in the variables y 4 , . . . , y N . Our solutions are thus different from the ones obtained in [31] and have strong analogies with the doubling construction of the entire finite energy sign-changing solutions for the Yamabe equation in [24]. Define the symmetric Sobolev space: where θ = arctan y2 y1 . Let us define the following norms which capture the decay property of functions and for some 1 small. The main results of this paper are the following: with r as in (1.7). Let us sketch the proof of Theorem 1.1. The first step in our argument is to find φ so that u = W r,h,Λ + φ solves the auxiliary problem for some constants c for = 1, 2, 3. In (1.16), the functions Z j and Z j are given by for j = 1, . . . , k. Moreover, the function φ belongs to the set E given by From the linear theory developed in Sect. 2, problem (1.16) can be solved by means of the contraction mapping theorem. More precisely, we prove that, for any (r, h, Λ) ∈ S k there exist φ = φ r,h,Λ ∈ E and constants c , = 1, 2, 3 which solve the auxiliary problem (1.16). After the correction φ has been found, we shall choose (r, h, Λ) ∈ S k so that the multipliers c = 0 ( = 1, 2, 3) in (1.16). As a consequence, we can derive the results as in Theorem 1.1.
Equation (1.8) is the Euler-Lagrange equation associated to the energy functional where A i for i = 1, 2, 3 and B j for j = 4, 5 are constants. We denote for some constantB. However, we now find that the term O  We need to expand the full energy F (r, h, Λ) = I(W r,h,Λ + φ r,h,Λ ). We need a strong control on the size of φ r,h,Λ in order not to destroy the critical point structure of F 1 (r, h, Λ) and to ensure the qualitative properties of the solutions as stated in Theorem 1.1. This is another delicate step of our construction, where we make full use of the assumption (H) on K.
Structure of the paper. The remaining part of this paper is devoted to the proof of Theorem 1.1, which will be organized as follows: denote certain constants and σ, τ, σ j to denote some small constants or functions. We also note that δ ij is Kronecker delta function: Furthermore, we also employ the common notation by writing O(f (r, h)), o(f (r, h)) for the functions which satisfy
We can know that We consider the following linearized problem (2.2) belong to the null space of the linearized problem associated to (1.5) around an Aubin-Talenti bubble, namely they solve It is known [28] that these functions span the set of the solutions to (2.3). This fact will be used in the following crucial lemma which concerns the linearized problem (2.1).
If f k * * tends to zero as k tends to infinity, so does φ k * .
Proof. We prove the Lemma by contradiction. Suppose that there exists a sequence of (r k , h k , Λ k ) ∈ S k , and for φ k satisfies (2. Without loss of generality, we can assume that φ k * = 1. For convenience, we drop the subscript k. From (2.1), we know that For the first term M 1 , we make use of Lemma B.5, so that For the second term M 2 , we make use of Lemma B.4, so that In order to estimate the term M 3 , we will first give the estimates of Z 1j and Z 1j Combining estimates (A.26) and Lemma B.4, we have Next, we will give the estimates of c , = 1, 2, 3. Multiply both sides of (2.1) by Z q1 , q = 1, 2, 3, then we obtain that The discussion on the left side of (2.5) may be more tricky, in fact, we have Using the property of K(s), similar to the proof of Lemma B.5, we can get For J 2 , it is easy to derive that Then, we get On the other hand, there holds Note that for some constantc q > 0. Then we can get (2.7) Combining this fact and φ * = 1, we have the following claim: Claim 1: There exist some positive constantsR, δ 1 such that Since φ ∈ H s , we assume that l = 1. By using local elliptic estimates and (2.7), we can get, up to subsequence,φ(y) = φ(y − x 1 ) converge uniformly in any compact set to a solution Since φ is even in y d , d = 2, 4, . . . , N, we know that u is also even in y d , d = 2, 4, . . . , N. Then we know that u must be a linear combination of the functions

From the assumptions
By taking limit, we have So we have u = 0. This is a contradiction to (2.8).
For the linearized problem (2.1), we have the following existence, uniqueness results. Furthermore, we can give the estimates of φ and c , = 1, 2, 3.
Proof. Recall the definition of E as in (1.17), we can rewrite problem (2.1) in the form in the sense of distribution. Furthermore, by using Riesz's representation theorem, Eq. (2.10) can be rewritten in the operational form where I is identity operator and T k is a compact operator. Fredholm's alternative yields that problem (2.11) is uniquely solvable for anyf when the homogeneous equation has only the trivial solution. Moreover, problem (2.12) can be rewritten as following (2.13) Suppose that (2.13) has nontrivial solution φ k and satisfies φ k * = 1. From Lemma 2.1, we know φ k * tends to zero as k → +∞, which is a contradiction. Thus problem (2.12) (or (2.13)) only has trivial solution. So we can get unique solvability for problem (2.1). Using Lemma 2.1, the estimates (2.9) can be proved by a standard method.
We can rewrite problem (1.16) as following where Next, we will use the Contraction Mapping Principle to show that problem (2.14) has a unique solution in the set that φ * is small enough. Before that, we will give the estimate of N(φ) and l k .
Proof. The proof is similar to that of Lemma 2.4 in [31]. Here we omit it.
We next give the estimate of l k .
where 1 is small constant given in (1.14).
Proof. We can rewrite l k as Assume that y ∈ Ω + 1 , then we get NoDEA Doubling the equatorial for the prescribed scalar curvature Thus, we have We first consider the case N = 5. It is easy to get that Then Then combining (2.16) and (2.17), we can get For S 12 , we can rewrite it as following Similarly to (2.16), we can obtain For N ≥ 6 and the same α 1 as in (2.18), it is easy to derive that where we have used the fact hr > C r k . Thus, we can obtain that Next, we consider S 13 . For y ∈ Ω + 1 , Thus we have (2.20) (2.21) We now consider the estimate of S 2 . For y ∈ Ω + 1 , we have As a result, we get As a result, Since y ∈ Ω + 1 , then for j = 2, . . . , k, there holds Therefore, it is easy to derive that (2.23) Combining (2.22) with (2.23), we obtain . Thus, we can rewrite (2.21) as Therefore, we showed (2.15).
The solvability theory for the projected problem (2.14) can be provided in the following: Proposition 2.5. Suppose that K(|y|) satisfies (H) and N ≥ 5, (r, h, Λ) ∈ S k . There exists an integer k 0 large enough, such that for all k ≥ k 0 problem (2.14) has a unique solution φ k which satisfies , for = 1, 2, 3.

(2.25)
Proof. We first denote From Proposition 2.2, we know that problem (2.14) is equivalent to the following fixed point problem where L k is the linear bounded operator defined in Proposition 2.2.
So the operator A maps from B to B. Furthermore, we can show that A is a contraction mapping. In fact, for any φ 1 , φ 2 ∈ B, we have Since N(φ) has a power-like behavior with power greater than one, then we can easily get A direct application of the contraction mapping principle yields that problem (2.14) has a unique solution φ ∈ B. The estimates for c , = 1, 2, 3 can be got easily from (2.6). We will give the expression of F (r, h, Λ). We first note that we employ the notation C(r, Λ) to denote functions which are independent of h and uniformly bounded.

Proof of
We have the following expansion as k → ∞ Proof. The proof of Proposition 3.2 is similar to that of Proposition 3.1 in [31]. We omit it here.
Next, we will give the expansions of ∂F (r,h,Λ) ∂Λ and ∂F (r,h,Λ) ∂h . Proposition 3.3. Suppose that K(|y|) satisfies (H) and N ≥ 5, (r, h, Λ) ∈ S k . We have the following expansion for k → ∞ Proof. The proof of this proposition can be found in [31]. We omit it here.

Proposition 3.4. Suppose that K(|y|) satisfies (H) and N ≥ 5, (r, h, Λ) ∈ S k .
We have the following expansion where we used the estimates (2.24)-(2.25) and the inequalities On the other hand, we have (3.5) For the second term in (3.5), using the decay property of K(|y|) and orthogonality of φ r,h,Λ , we can show this term is small. In fact, we have According to the expression of W r,h,Λ , we can obtain that And it's easy to show that .

Rewritten the expansion of the energy functional.
Then it solves and Let h be a solution of forθ is a small constant such thatθ ≤ σ 100 . In fact, S k is a subset of S k . We will find a critical point of F (r, h, Λ) in S k .
A direct Taylor expansion gives that Since G(h), G (h) are independent of h, r, Λ, for simplicity, in the following, we will denote Then combining (3.11), (3.12), (3.13), we can get Therefore, we get where We now rewrite Then we can express F (r, h, Λ) as And similarly, we have and from (A.28), by using some calculations, we have (3.17) and where η 1 > 0 small. We also define the energy level set We consider the following gradient flow system The next proposition would play an important role in the proof of Theorem 1.1. Proof. There are three positions that the flow tends to leave S k : , it is easy to derive that ♠ On the other hand, we claim that it's impossible for the flow r(t), h(t), Λ(t) leaves S k when it lies in position 2. If 1 − h −1 h = 1 kθ , then from (3.16) and (3.17), we have , there exists a constant C 2 such that Hence the flow r(t), h(t), Λ(t) does not leave S k when |Λ − Λ 0 | = 1 k 3θ 2 .
Combining above results, we conclude that the flow would not leave S k before it reachF t 1 . Now we give the proof of Theorem 1.1. We claim that c is a critical value ofF (r, h, Λ) and can be achieved by some (r, h, Λ) ∈ S k . By the minimax theory, we need to show that Using the results in Proposition 3.6 we can prove (i) and (ii) easily.
Finally, for every k large enough, we get the critical point (r k , h k , Λ k ) of F (r, h, Λ). and Proof. In fact, for 1 2 < c 3 ≤ c 4 ≤ 1, we have Without loss of generality, we can assume k is even. It is easy to derive that Direct computations show that ). Using symmetry of function sin x, we can easily show Thus we proved (A.1).
Similarly, we can obtain Combining above calculations, we can obtain that where B 2 and σ 2 are defined in (A.3), (A.4).
and 0 is constant small enough.
First, we have It is easy to check that Standard calculation implies that NoDEA Doubling the equatorial for the prescribed scalar curvature From (A.7)-(A.10), we get It's easy to get Combining (A.6), (A.11) and (A.12), we can get Similarly, we can get Lemma A.3. Suppose that K(|y|) satisfies (H) and N ≥ 5, (r, h, Λ) ∈ S k . We have the expansion for k → ∞ where C(r, Λ) denotes function independent of h and should be order of O(1), and 0 is constant can be chosen small enough.
Proof. Recalling the definition of I(u) as in (1.18), then we obtain that According to the expression of W r,h,Λ , we have For I 2 , using the symmetry of function W r,h,Λ , we have with 0 > 0 can be chosen small enough. Then we can get For I 21 , we can rewrite it as following Furthermore, recalling |x 1 | = r and using the symmetry property, we have where e 1 = (1, 0, . . . , 0). We get here C(r, Λ) denote functions which are independent of h and can be absorbed in O (1). Similarly, we can also have the following expression Then, we can obtain that Finally, we consider I 22 For I 222 , it is easy to derive that Moreover, we know that Then we get Next, we consider the term I 223 . In fact, we have And for y ∈ Ω + 1 and | |y| ). Then we can get easily where δ 1 is small constant. If |y − x 1 | ≤ δ1r k , it is easy to derive for some small δ 2 . Therefore, NoDEA Doubling the equatorial for the prescribed scalar curvature Thus we can get Combining Lemma A.1-A.3, we can get the following Proposition which gives the expression of I(W r,h,Λ ).
Proposition A.4. Suppose that K(|y|) satisfies (H) and N ≥ 5, (r, h, Λ) ∈ S k . Then we have thus provided with 0 , σ small enough, we can get k r Since m ≥ 2, we can check that   Proof. The proof of this proposition is standard and the reader can refer to [31] for details.  Proof. Recall (A. 26) We know that ∂I(W r,h,Λ ) ∂h (A.28) Then by some tedious but straightforward analysis, we can get for some 0 small enough. In fact, we know that k k Proof. For y ∈ Ω + 1 and j = 2, . . . , k, we have Proof. The proof of Lemma B.2 is similar to Lemma B.1. We omit the details for concise.