Infinitely many non-radial solutions for the prescribed curvature problem of fractional operator

We consider the following problem: 
\begin{equation*} 
\left\{\begin{array}{ll} 
(-\Delta)^s u=K(y)u^{p-1} \hbox { in } \ \mathbb{R}^N, \\ 
u>0, \ y \in \mathbb{R}^N, 
\end{array}\right.             (P) 
\end{equation*} 
where $s\in(0,\frac{1}{2})$ for $N=3$, $s\in(0,1)$ for $N\geq4$ and $p=\frac{2N}{N-2s}$ is the fractional critical Sobolev exponent. Under the condition that the function $K(y)$ has a local maximum point, we prove the existence of infinitely many non-radial solutions for the problem $(P)$.

It is well known that when K(y) = 1, the following functions , are the unique (up to the translation and scaling) solutions for the problem (see [16]): We are interested in the existence of non-radial positive solutions. Note that, in general, even for s = 1, the problem (1) does not always admit a solution. The sufficient condition on K, under which the following problem: admits a solution, has been extensively studied, see [1], [6]- [9], [14], [15], [17], [18], [22] and references therein. In this paper, following the idea from Wei and Yan [21] dealing with the problem (3), we use the techniques in dealing with the singularly perturbed elliptic problems and use the number of the bubbles solutions to the equation (2) as the parameter in the construction of non-radially symmetric solutions for problem (1).
We consider the Schwartz space S of rapidly decaying C ∞ functions on R N . For any ϕ ∈ S, we define the Fourier transformation of ϕ by: We will look for solutions of the problem (1) in the energy space: where (−∆) s 2 u L 2 (R N ) is defined by ( R N |ξ| 2s |Fu(ξ)| 2 dξ) 1 2 . We define the functional I on D s (R N ) by: Then the solutions of problem (1) correspond to the critical points of the functional I. Set µ = k N −2s N −2s−m to be the scaling parameter. Using the transformation u(y) → µ − N −2s 2 u( y µ ), problem (1) becomes: Let y = (y , y ), y ∈ R 2 , y ∈ R N −2 . Define H s = {u : u ∈ D s (R N ), u is even in y i , i = 3, . . . , N, u(rcos(θ + 2πj k ), rsin(θ + 2πj k ), y ) = u(rcosθ, rsinθ, y )}.
2. Some basic estimates. In this section, we give some basic estimates.
Lemma 2.1. For any constants 0 < σ < N − 2s, there exists a constant C > 0, such that Proof. The proof of this lemma is similar to that of Lemma B.2 in [21] and Lemma 2.2 in [12]. For the sake of completeness, we give a sketch of the proof. Without loss of generality, we set |y| ≥ 2, and let d = |y| 2 , then . Then there is a small θ > 0, such that Proof. Since for any λ > 0, we have Note that there exist two constants C 1 , C 2 > 0 such that For Define Since, for any z ∈ Ω 1 and j > 1, Similar to the proof of Lemma 2.1, we have By the symmetry, we have where we choose τ 1 < (N −2s)2s N +2s .
3. Finite dimensional reduction. In this section, we perform a finite dimensional reduction by using W r,Λ (y) as the approximation solution and considering the linearization of the problem (5) around the approximation solution W r,Λ (y). We first introduce the following norms: where max{ 6s−N 2 , N −2s−m N −2s } < τ < min{1 + η, 2s, N −2s 2 } and η > 0 is small. Let

YUXIA GUO AND JIANJUN NIE
We consider the following problem: for some numbers c l , where u, v = R N uv. Lemma 3.1. Assume that φ k solves problem (10). If h k * * → 0 as k → ∞, then φ k * → 0 as k → ∞.
Proof. We follow the idea in [21] and proceed the proof by contradiction arguments.
Without loss of generality, we can assume that φ k * ≡ 1. We have for some explicit positive constant σ N,s . For the first term J 1 , by Lemma 2.2, we have For the second term J 2 , we make use of Lemma 2.1, so that Then, we have Multiply both sides of (10) by Z 1,t , we have First of all, there exists a constant c > 0 such that On the other hand, we have For j > 1, by the symmetry, Moreover, one has If p > 3, then 2s > 1 and For l = 1, by (8), we have Note that N > 4s, 2s > 1 and τ > 6s−N 2 . For l > 1, we have Since m ≥ 2 and N < 6s, N −2s−m where σ > 0 is a small constant. Similarly, we have By the condition (K) and Lemma 2.2, we have At last, we get The case p ≤ 3 can be discussed in a similar way. By (14), (15), (16)and (17), we have . Then by (13) and φ k * = 1, there is R > 0 such that for some i. The translated version φ k = φ k (y − x i ) converges uniformly in any compact set to a solution u of the following equation: Since u is perpendicular to the kernel of this equation, u = 0. This is a contradiction to (18).
Using the same argument as in the proof of Proposition 4.1 in [10], we have the following proposition. Proposition 1. There exist k 0 > 0 and a constant C > 0, independent of k, such that for all k ≥ k 0 and all h ∈ L ∞ (R N ), problem (10) has a unique solution φ = L k (h). Besides, Now we consider the following problem: In the rest of this section, we devote ourselves to prove the following proposition by using the contraction mapping theorem.
Proposition 2. There exist k 0 > 0 and a constant C > 0, independent of k, such that for all k ≥ k 0 , where σ > 0 is a small constant.
We first rewrite (20) as where In order to use the contraction mapping theorem to prove Proposition 2, we need to estimate N (φ) and l k . In the following, we assume that φ * is small.
We only give the proof for s ∈ ( 5 6 , 1). Firstly, we consider N ≥ 6. From the symmetry, we can assume that y ∈ Ω 1 . One has, |y − x j | ≥ 1 2 |x 1 − x j | for j > 1. Thus, by (8), Using the Hölder inequality, we obtain: When N = 4, 5, we have Next, we estimate l k . Proof. We have By the symmetry, we can assume that y ∈ Ω 1 . Then |y − x j | ≥ |y − x 1 |, and Since τ < 1 + η for small η > 0, N +2s 4s ( N −2s 2 − N −2s N +2s τ ) > 1, using the Hölder inequality, we have Now, we estimate J 2 . For y ∈ Ω 1 and j > 1, we have We split the slice Ω 1 into two parts, namely: I := {y ∈ Ω 1 : |y| − µr 0 ≥ δµ} and II := {y ∈ Ω 1 : |y| − µr 0 < δµ}, where δ > 0 is the constant in (K). When y ∈ I, we have When y ∈ II, by the condition (K), we have and |y| − |x 1 | ≤ |y| − µr 0 | + |µr 0 − |x 1 | ≤ 2δµ. As a result, Combining with the obtained results, we obtain Proof of Proposition 2. Let y = (y , y ), y ∈ R 2 , y ∈ R N −2 , and letẼ be the completion of C ∞ 0 (R N ) under the norm of || · || * . Set By Proposition 1, the solution φ of (20) is equivalent to the following fixed point problem: Hence, it is sufficient to prove that the operator A is a contraction map from the set E to itself. In fact, for any φ ∈ E, by Proposition 1, Lemma 3.2 and Lemma 3.3, we have which shows that A maps E to E itself and E is invariant under A operator.
On the other hand, we have |N (t)| ≤ C|t| p−2 for p ≤ 3 and |N (t)| ≤ C(W p−3 r,Λ |t|+ |t| p−2 ) for p > 3. If p ≤ 3, ∀φ 1 , φ 2 ∈ E, by the Hölder inequality, we have The case p > 3 can be discussed in a similar way. Hence A is a contraction map. The Banach fixed point theorem tells us that there exists a unique solution φ ∈ E for the problem (20). where φ is the function obtained in Proposition 2 and I is the functional of problem (5), that is: In this section, we give the energy expansion for F (r, Λ) and ∂F (r,Λ) ∂Λ . We begin with where A and B i , i = 1, 2, 3 are positive constants, and r = |x 1 |.
Proof. By using the symmetry of slices Ω j (defined in (9)), we have The curvature term can be rewritten into three terms as the following: For the first term, we have We split the slice Ω 1 into two parts, namely, where δ > 0 is the constant in (K).
Proof. By (20), we have I (W r,Λ + φ), φ = 0 and the functional F (r, Λ) can be expanded as the following: Thus, By using the similar arguments, we have By Lemma 4.1, we get the desired result.

5.
Proof of the main theorem. By (6) and (7), there exists a constant B 4 > 0 such that We rewrite the expansion of F r,Λ and ∂Fr,Λ ∂Λ as the following: and .