INFINITELY MANY POSITIVE SOLUTIONS OF FRACTIONAL NONLINEAR SCHR ¨o DINGER EQUATIONS WITH NON-SYMMETRIC POTENTIALS

. We consider the fractional nonlinear Schr¨ o dinger equation ( − ∆) s u + V ( x ) u = u p in R N , u → 0 as | x | → + ∞ , where V ( x ) is a uniformly positive potential and p > 1 . Assuming that as | x | → + ∞ , and p,m,σ,s satisfy certain conditions, we prove the existence of inﬁnitely many positive solutions for N = 2. For s = 1, this corresponds to the multiplicity result given by Del Pino, Wei, and Yao [24] for the classical nonlinear Schr¨odinger equation.

1. Introduction. In this paper, we consider the fractional nonlinear Schrödinger equation (−∆) s u + V (x)u − u p = 0 in R 2 , u → 0 as |x| → +∞, (1.1) where (−∆) s , 0 < s < 1, denotes the fractional Laplace operator, V (x) is a nonnegative potential, and 1 < p < 1+s 1−s . We are interested in the existence of infinitely many spike solutions to (1.1). The natural space on which to look for solutions of (1.1) is the space H 2s (R 2 ) of all functions u ∈ L 2 (R 2 ) such that − ∆u + V (x)u = u p in R 2 , u(x) → 0 as |x| → +∞. then one can show that (1.3) has a least energy (ground state) solution by using the concentration compactness principle (cf. [37,38]). However, if (1.4) does not hold, then problem (1.3) may not admit a least energy solution, and one has to look for higher energy level solutions. Results in this direction are presented in [6,7,10], where a positive solution has been found using variational methods under a suitable decay condition on V at infinity. Let us now consider the semi-classical limit case: − ε 2 ∆u + W (x)u − u p = 0 in R N , u > 0, u ∈ H 1 (R N ), (1.5) where ε > 0 is a small parameter. Naturally, problem (1.5) is equivalent to (1.3) for V (x) = W (εx). It is known that as ε goes to zero, highly concentrated solutions can be found near critical points of the potential W [2,11,12,19,20,21,22,27,31,42,48] or near higher dimensional stationary sets of other auxiliary potentials [3,23,39]. The number of solutions of (1.5) may depend on the number or types of critical points of W (x). By assuming that V = V (|x|) is radially symmetric, Wei and Yan [50] proved that problem (1.3) admits infinitely many positive non-radial solutions if there exist

INFINITELY MANY SOLUTIONS OF FRACTIONAL SCHRÖDINGER EQUATIONS 5563
constants V ∞ > 0, a > 0, m > 1 and σ > 0 such that (1. 6) The proof given in [50] relies heavily on the radial symmetry of the potential V . Inspired by this result, Wei-Yan proposed the following conjecture: Then, problem (1.3) admits infinitely many non-radial positive solutions, whose energy can be made arbitrarily large.
Regarding the existence of multiple spike solutions for more general non-symmetric potentials, in [13] Cerami, Passaseo, and Solimini proved the following existence result: Theorem B. Let the following assumptions hold: For further results regarding the existence of (1.3) with non-symmetric potentials, we refer the reader to [5,14,15,16] and references therein.
Remark 1.1. It is not difficult to see that we can choose s sufficiently close to 1, p sufficiently large, m sufficiently close to 2 + 2s, σ sufficiently large, and µ > 1 sufficiently close to 1 such that (1.10) holds. Therefore, there exist parameters m, σ, p, and s satisfying Assumption 2. For the condition (1.9) and (1.10), we would like to make some more comments. With the help of (1.9), we can obtain the asymptotic behavior of the ground state w with its derivative in a nice form, see (2.12) in Remark 2.8; while we set the condition (1.10) to ensure the difference between the location of the spike and the vertex of the polygon is small, and the approximate solutions is good enough such that the error is small enough, see (5.25) in Proposition 5.5. Here the restriction s+3 s+2 < p is not necessary. However if we drop this assumption, then instead we would get a more complicate form in (1.10). To simplify our statement, we assume (1.9) holds. We believe that both these two conditions are for technical reasons, and can be dropped, but in that case, one need more delicate analysis of the approximate solutions. On the other hand, we can also obtain similar results for N ≥ 3 under certain conditions on m, σ, s, and p if we assume further that For the sake of the simplicity of the statement, we only consider the case with N = 2 in this paper.
Our main result in this paper is stated in the following theorem.  [24]. For s = 1, the spikes decays exponentially. As a result, each spike only interacts with neighboring spikes. Meanwhile, for 0 < s < 1 the spikes decay algebraically and each spike interacts with all the other spikes, which makes the reduced problem more complicated than in the s = 1 case. A complicated matrix T is present in the reduced problem (see Lemma 5.2), and we must precisely obtain the decay rate for this matrix. Unlike in the case that s = 1, all of the entries in the matrix are nonzero for 0 < s < 1, and we must carefully estimate the eigenvalues and determine the exact decay rate. This is a new result, and is the main contribution of this paper (see the final section). We believe that our technique can be employed in the construction of solutions for fractional Laplacian equation or the equation with a critical exponent in future work.
Before we end the introduction, we want to say a few words on the circulant matrices, which play an important role in this paper. It also appeared as an essential point in the recent work [24,40]. As we will see in Theorem 3.2, the constructed solution concentrates on the vertex of the regular polygon, which equally distributed on the circle. Then in the leading order of the reduced problem, the circulant matrices appear, see section 7 for the details.
Throughout this paper, we will employ the following notation and conventions: • For quantities G K and H K , we write G K ∼ H K to indicate that there exists a positive constant C such that 1 C ≤ G K H K ≤ C for sufficiently large K. Furthermore, G K = O(H K ) means that G K H K is uniformly bounded as K tends to infinity, and G K = o(H K ) denotes that G K H K → 0 as K → ∞. • For simplicity, the letter C denotes various generic constants that are independent of K. This is allowed to vary from line to line, as well as within the same formula. • We shall employ the notation |y| = y 2 for the Euclidean norm in various Euclidean spaces R N when no confusion can arise, and we always denote the inner product of a and b in R N by a · b. • The transpose of a matrix A shall be denoted by A T . This paper is organized as follows. Some preliminary facts and estimates are explained in Section 2. In Section 3, we describe the procedure for our construction and describe the main ideas of each step. In Sections 4-6, we shall prove each of the steps outlined in Section 3, and then complete the proof of Theorem 1.2. We omit certain technical results in the final section.

Preliminaries.
In this section, we study the fractional Laplacian operator and the ground state solution w of the following equation: (2.1) Let 0 < s < 1. Various definitions exist for the fractional Laplacian (−∆) s φ of a function φ defined in R N , depending on its regularity and growth properties.

WEIWEI AO, JUNCHENG WEI AND WEN YANG
For φ ∈ H 2s (R N ), the standard definition is given via the Fourier transform .
When φ is additionally assumed to be sufficiently regular, we obtain the direct representation for a suitable constant d s,N , where the integral is to be understood in a principal value sense. This makes sense as a direct integral when s < 1 2 and φ ∈ C 0,α (R N ) with α > 2s, or if φ ∈ C 1,α (R N ) with 1 + α > 2s. In the latter case, we can desingularize the representative integral in the following form: Another useful (local) representation, found by Caffarelli and Silverstre [9], is given via the following boundary value problem in the half space R N +1 Here,φ is the s−harmonic extension of φ, explicitly given as a convolution integral with the s−Poisson kernel p s (x, y), and C N,s satisfies R N p(x, y) = 1. Then, under suitable regularity conditions, (−∆) s φ is the Dirichlet-to-Neumann map for this problem, namely, The characterizations (2.2)-(2.4) are all equivalent, for instance, in Schwartz's space of rapidly decreasing smooth functions. For m > 0 and g ∈ L 2 (R N ), let us consider now the equation In terms of the Fourier transform, for φ ∈ L 2 this problem reads (|ξ| 2s + m)φ =ĝ, and it admits a unique solution φ ∈ H 2s (R N ) given by the convolution where G(ξ) = 1 |ξ| 2s + m .
Using the characterization (2.4) written in a weak form, φ can be characterized by φ(x) =φ(x, 0) in a trace sense, whereφ ∈ H is the unique solution of Here, H is the Hilbert space of functions ϕ ∈ H 1 loc (R N +1 + ) such that or equivalently the closure of the set of all functions in C ∞ c (R N +1 + ) under this norm. A useful fact for our purpose is the equivalence of the representations (2.5) and (2.6) for g ∈ L 2 (R N ).
Lemma 2.1. Let g ∈ L 2 (R N ). Then, the unique solutionφ ∈ H for the problem (2.6) is given by the s−harmonic extension of the function φ = T m [g] = G * g.
For the proof, we refer the reader to Lemma 2.1 in [18]. Let us recall the main properties of the fundamental solution G(x) in the representation (2.5), which are stated, for instance, in [26] and [29].
We have that G is radially symmetric and positive. That is , • |x||∇G(x)| ≤ C |x| N +2s for all |x| ≥ 1. In order to consider an a priori estimate involving the fractional Laplacian operator, we require the following Lemmas (see [18]): Lemma 2.2. Let 0 ≤ µ < N + 2s. Then, there exists a positive constant C such that Then. the following holds. If φ = T m [g], then there exists C > 0 such that
Lemma 2.4. Let ϕ ∈ H 2s be the solution of with a bounded potential W . If inf x∈R N W (x) =: m > 0, g ≥ 0, then ϕ ≥ 0 in R N .
The next lemma provides an a priori estimate for a solution φ ∈ L 2 (R N ) ∩ L ∞ (R N ) of (2.8).

5568
WEIWEI AO, JUNCHENG WEI AND WEN YANG Lemma 2.5. Let W be a continuous function, and assume that for k points q 1 , q 2 , · · · , q k there exist R > 0 and B = ∪ k j=1 B R (q j ) such that inf Then, given any number N 2 < µ < N + 2s, there exists a uniform positive constant C = C(µ, R) independent of k such that for any ϕ ∈ H 2s (R N ) ∩ L ∞ (R N ) and g satisfying (2.8) with the following estimate is valid: , and χ B is the characteristic function on B. We observe that From Lemma 2.2 with 0 < µ < N + 2s, the positive solution ϕ 0 to the problem withg ∈ L 2 . By using Lemma 2.4, we obtain that ϕ ≤φ. By a similar argument for −ϕ, we obtain that |ϕ| ≤φ. Then, it holds that The desired estimate follows immediately.
From the above Lemma, we can immediately obtain the following corollary. Then, we have that φ ∈ L ∞ (R N ), and it satisfies where C = C(µ) is independent of k.
A useful fact is that if f, g ∈ L 2 (R N ) and W = T m (f ), Z = T m (g), then the following holds: where we have used the fact that the Green kernel G is radially symmetric.
Next, we recall some basic and useful properties regarding the ground state solution w of (2.1).
3. Description of the construction. In this section, we shall briefly describe the solutions to be constructed later, and will describe the main ideas of the construction.
First, without loss of generality we can assume by suitable scaling that V ∞ = 1. Following the developments in [50], we will use the loss of compactness to construct solutions. More precisely, we will construct solutions with large numbers of spikes whose inter-distances and distances from the origin are sufficiently large.
By the asymptotic behavior of V at infinity, the basic building block is the ground state solution w of (2.1). The solutions we construct will consist of small perturbations of sums of copies of w, centered at some carefully chosen points in R 2 .
Let K ∈ N + be the number of spikes, whose locations are given by Q j ∈ R 2 , j = 1, 2, · · · , K. We define (3.1) In order to further describe the configuration space of the Q j 's, we define Here, α is a parameter representing the degeneracy resulting from rotations, and R is a positive constant to be determined later. Observe that each point Q 0 j depends on α. Thus, we write Q 0 j = Q 0 j (α). The number of spikes K and the radius R are related by the so-called balancing condition: where Γ = am 2 R 2 w 2 (x)dx > 0, and Ψ is the interaction function defined by Here, e can be any unit vector in R 2 . For Ψ(t), we obtain the following expansion: For large t, the following expansion holds for Ψ(s): Proof. By the definition of Ψ(t) and the use of a Taylor expansion, we have that (1)).

INFINITELY MANY SOLUTIONS OF FRACTIONAL SCHRÖDINGER EQUATIONS 5571
Thus, we have proved the Lemma.
From the above balancing condition (3.2) and Lemma 3.1, we can obtain that Because we require that 1 < m < 2 + 2s, we have as a consequence of (3.6) that Next, we define a small neighborhood of Q 0 = (Q 0 1 , · · · , Q 0 K ) on (R 2 ) K and introduce an additional parameter. Let f j , g j ∈ R, j = 1, 2, · · · , K. Then, we define where n j = (cos θ j , sin θ j ), and t j = (− sin θ j , cos θ j ). Note that f j and g j measure the displacement in the normal and tangential directions, respectively. Writing Q j = Q j (α), n j = n j (α) and t j = t j (α), we note the following trivial but important fact: We can now introduce an additional parameter q and define a suitable norm. Denote q = (f 1 , · · · , f K , g 1 , · · · , g K ) T ∈ R 2K . With this notation, we can define the configuration space of the Q j 's by For any (Q 1 , Q 2 , · · · , Q K ) ∈ Λ K , an easy computation shows that for j = 1, 2, · · · , K, We will prove Theorem 1.2 by demonstrating the following result.
Theorem 3.2. Under the assumption of Theorem 1.2, there exists a positive integer K 0 such that for all integers K ≥ K 0 there exist α ∈ [0, 2π) and (Q 1 , Q 2 , · · · , Q K ) ∈ Λ K such that the problem (1.1) has two solutions of the form To prove Theorem 3.2, it is sufficient to show that for sufficiently large K there exist parameters α and q such that U + φ is a genuine solution for a small perturbation φ. To achieve this, we will adopt the techniques for the singularly perturbed problem. Unlike for the problem (1.5), there is no apparent parameter in this case . As stated in Theorem 3.2, we adopt the number of spikes as the ε−type parameter.
As mentioned in Remark 1.3 in the introduction, we will employ intermediate Lyapunov-Schmidt reduction to solve this. We will solve (3.11) through the following two steps.
Step 1. Solving the projected problem. Let α ∈ R and q satisfy (3.9). We shall first solve a projected version of (3.11). More precisely, we look for a function φ and some multiplierβ ∈ R 2K such that where the vector field Z Qj is defined by (3.13) By direct computation, we have that This constitutes the first step in the Lyapunov-Schmidt reduction. It is performed in Section 4 using an a priori estimate and the contraction mapping theorem. A required condition in this step is the non-degeneracy of w. It is worth noting that the function φ and the multiplierβ found in Step 1 depend on the parameters α and q.
Step 2. Solving the reduced problem. From Step 1, we know thatβ is small. However, it is not easy to solveβ(α, q) = 0 directly, because the linear part of the expansion ofβ in q is degenerate.
More precisely, let us writeβ whereT q is the linear part and Φ(α, q) denotes the remaining term. As we will see in Section 5,T q does not depend on α, and there is a unique vector (up to scalar) such thatT q 0 = 0. By applying Lyapunov-Schmidt reduction again (called the secondary Lyapunov-Schmidt reduction), the step of solving the reduced problemβ(α, q) = 0 can be divided into two steps. To write the projected problem ofβ = 0 in a proper norm, note that where q ⊥ = (− g, f ) for q = ( f , g). Hence, we define With this notation,β Clearly, the multiplier β depends on the parameters α, q and γ. Thus, we write β = β(α, q, γ).
Step 2.A. Solving β(α, q, γ) = 0 by adjusting γ and q. In this step, for each α ∈ R we will determine parameters (γ, q) such that This can be viewed as the step of solving the projected problem in the secondary Lyapunov-Schmidt reduction. To achieve this, we will make use of condition (3.2). This step is performed in Section 5 by using various integral estimates and the contraction mapping theorem. A key condition in this step is the invertibility of a 2K × 2K matrix, the proof of which is given in the final section. As mentioned before, the analysis of this large matrix is the key contribution of this paper, where the properties of circulant matrices and the Fourier series of the Bernoulli and Euler polynomials are used. After Step 2.A is completed, we denote the unique solution of (3.16) by (γ(α), q(α)). Then, the original problem (1.1) is reduced to the problem γ(α) = 0 in one dimension.
Step 2.B. Solving γ(α) = 0 by choosing α. In this final step, we want to prove that there exists an α such that γ(α) = 0. As a result, the function u = U + φ is a genuine solution of (1.1).
This constitutes the second step of solving the reduced problem in the secondary Lyapunov-Schmidt reduction. To achieve this step, note that by Step 2.A the function φ = φ(x, α, q(α)) determined in Step 1 solves the following problem: where all of the quantities depending implicitly on (α, q) take values at (α, q(α)).
To solve γ(α) = 0, we first apply the so-called variational reduction technique to show that finding a solution of the equation γ(α) = 0 is equivalent to finding the critical point of the energy function F (α) = E(U + φ). Second, by using (3.8) it is not difficult to see that F (α) = 0 is 2π periodic in α. Hence, it has at least two critical points. More details for this step are presented in Section 6. In the following three sections, we shall complete Step 1, Step 2.A, and Step 2.B, respectively. 4. Lyapunov-Schmidt reduction. This section is devoted to completing the first step in the procedure of our construction.
Before stating the main result, we first introduce some notation. We define the weighted norm as follows: where Q j is defined in Section 3 and µ > 0 will be chosen later. In the following, we assume that (Q 1 , · · · , Q K ) ∈ Λ k , i.e., the parameter q satisfies (3.9). We first claim that To prove (4.2), it suffices to show that Indeed, for any x ∈ R 2 , suppose without loss of generality that Q 1 is the point such that where we have used that Here, [x] denotes the integer part of x. By using the relation K = O(R 1− m 2s+2 ), we can obtain that Then, B * is a Banach space with the norm h * . To demonstrate its completeness, suppose that {h n } is a Cauchy sequence in B * . By (4.2), {h n } is also a Cauchy sequence in L ∞ (R 2 ). Hence, h n converges to a function h ∞ in L ∞ (R 2 ). It is easy to see that for any ε, there exists n 0 ∈ N such that By letting k → ∞, we obtain that which implies that h n − h ∞ * → 0 as n → +∞.
In the following, we always assume that 1 < µ < 2 + 2s. Now, we can state our main result for this section. Proposition 4.1. Suppose that V (x) satisfies (1.8) for constants V ∞ > 0, α ∈ R, m ≥ 1, and σ > 0, given 1 < µ < 2 + 2s. Then, there exists a positive integer K 0 such that for all K ≥ K 0 , α ∈ R, and q satisfying (3.9), there exists a unique function φ ∈ H 2s (R 2 ) ∩ B K and a unique multiplier c ∈ R 2K such that Here, C is a positive constant that is independent of K. Moreover, (α, q) → φ(x; α, q) is of class C 1 , and The proof of Proposition 4.1 is somewhat standard, and can be divided into two steps: (i) Studying the invertibility of the linear operator.
Let M denotes a 2K × 2K matrix defined by for some constant C that is independent of K.
Proof. To prove the existence, it is sufficient to prove the a priori estimate (4.6). Suppose that |c j | = c . Then, by the definition we have that For the entries M ji , a simple computation gives that and Hence, by (4.7)-(4.9) we have for sufficiently large K that from which the desired result follows.
Before we give the proof of Proposition 4.1, we first study the following linearized problem.

Proposition 4.3.
Under the assumption of Proposition 4.1, given 1 < µ < 2 + 2s, there exists a positive integer K 0 such that for all K ≥ K 0 ,, α ∈ R, q satisfying (3.9), and h ∈ B * , there exists a unique function φ ∈ H 2s (R 2 ) ∩ B * and a unique multiplier c ∈ R 2K such that (4.10) Moreover, we have that for some constant C that is independent of K.
Proof. By following the same process as in the proof of Proposition 4.1 in [18], it suffices to prove the a priori estimate (4.11).
To prove (4.11), we first multiply the equation (4.10) by ∂u ∂q and integrate over where M is a 2K × 2K matrix as defined in (4.5).
Using integration by parts, we obtain that Then, observe that We begin by studying the first term on the right hand side of (4.12). For the sake of the simplicity of our argument, we shall discuss the first term on the right hand side of (4.12) for i = 1. The other terms can be treated in a similar manner. From Section 3, we have that ρ(x) is uniformly bounded. Then, Next, we consider the second term on the right hand side of (4.12). We divide R 2 into K parts: where

INFINITELY MANY SOLUTIONS OF FRACTIONAL SCHRÖDINGER EQUATIONS 5577
When 1 < p ≤ 2, it holds that , and for = 1 we have that Then, Therefore, we obtain that For the case that p > 2, we have that in Ω 1 , in Ω , 2 ≤ ≤ K.
As a consequence, we have that Therefore, we can conclude that On the other hand, it is easy to see that (4.14) By using (4.13), (4.14), and Lemma 4.2, we obtain that Now, we prove the a priori estimate (4.11). We argue by contradiction. Suppose that there exist h K and φ K solving (4.10) for h = h K , R = R K , with h K * → 0 and φ K * = 1 as K → ∞. For simplicity, we drop the K in the subscript.
From the conditions on the potential V, it is obvious that inf R 2 V > 0. On the other hand, from the equation satisfied by φ, we find that for any x ∈ Ω 1 \ B r (Q 1 ), which leads to inf Accordingly, from Lemma 2.5 and (4.13), it holds that Br(qj )) + o(1), from which we may assume that, up to a subsequence, (4.16) Let us setφ(x) = φ(x + Q 1 ). Then,φ satisfies For any point x in an arbitrary compact set of R 2 , we have that It is easy to see that V (x + Q 1 ) → 1, For the last term in (4.18), as 1 < p ≤ 2 it holds that , while for p > 2 we have that .

WEIWEI AO, JUNCHENG WEI AND WEN YANG
Hence,g → 0 uniformly on any compact set of R 2 as K → ∞. Meanwhile, by considering where β = min{1, 2s}. Hence, up to a subsequence, we may assume thatφ → φ 0 uniformly on any compact set. It is easy to see thatφ 0 satisfies . Furthermore, by using the fact that µ > 1 we have that which means that φ 0 ∈ L 2 (R 2 ). Then, the non-degeneracy result in [29] together with the orthogonality condition implies that φ 0 ≡ 0, which contradicts (4.16). Hence, have we proved (4.11). Once we obtain (4.11), we can follow the same steps as in [18] to obtain a solution (4.10). Then, the lemma will be proved.
Before we give the complete proof of Proposition 4.1, we will estimate the error.
Lemma 4.4. Given (Q 1 , Q 2 , · · · , Q K ) ∈ Λ K , for any fixed µ ∈ (1, 2 + 2s) there exists a constant C (independent of K) such that Proof. By the definition of E, we have that We simply assume that x ∈ Ω 1 in the following proof, because the other parts can be treated similarly. We first consider E 1 . If |x| ≥ |Q1| 2 , then and in this region we have that

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Meanwhile, for |x| ≤ R 2 , we have that Hence, In the case that µ ≤ (2s + 2)(p − 1), it holds that Otherwise, if µ > (2 + 2s)(p − 1) then It is easy to deduce that Thus, we obtain the desired result by combining the above estimates. Now, we are in the position to give the proof of Proposition 4.1.
Proof of Proposition 4.1. We write the problem (4.4) as a fixed point problem: where δ > 0 is a small number to be determined later. Let φ ∈ F. Then for 1 < p ≤ 2 we have that

By Proposition 4.3 and Lemma 4.4, we have that
Then, we can choose C(δ + δ p−1 ) ≤ 1 2 and K sufficiently large such that On the other hand, for any φ i ∈ H 2s , i = 1, 2, we have that where ξ lies between φ 1 and φ 2 . For 1 < p ≤ 2, it holds that N (t) ≤ Ct p−1 ≤ C(|φ 1 | p−1 + |φ 2 | p−1 ), which tells us that Provided that δ is small enough. For p > 2, N (t) ≤ C(W p−2 |t| + |t| p−1 ), from which we can deduce that Provided that δ is small enough. Thus, we obtain that A is a contraction mapping, and the problem (4.4) admits a unique solution φ. Clearly, Lemma 4.4 implies that Now, we will consider the differentiability of φ(x; α, q) as function of (α, q).

Consider the map
of class C 1 given by: is equivalent to T (α, q, φ, c) = 0. By the above argument, we know that given α ∈ R and q satisfying (3.9), there exists a unique local solution (φ(α, q), c(α, q)). In the following, we write (φ, c) = (φ(α, q), c(α, q)) for simplicity. We claim that the linear operator is invertible for large K. Then, the C 1 − regularity of (α, q) → (φ, c) follows from the implicit function theorem. Indeed, we have that Next, we study the dependence of φ on (α, q). Assume that we have two solutions corresponding to two sets of parameters. Let one of these be denoted by corresponding to the parameters α and q, and the other bẙ corresponding to the parametersα andq. Observe that φ is L 2 − orthogonal to ∇ q U , whileφ is L 2 − orthogonal to∇ q U . To compareφ and φ, we first choose a vector ω such thatφ ω =φ + ω∇ q U satisfies the same orthogonality condition as φ. Moreover, by the equation forφ, the functionφ ω satisfies the equation By taking the difference with the equation satisfied by φ, we obtain that For j = 1, 2, · · · , K, we have that Assuming that (R|α − α| + q − q ∞ ) ≤ 1 2 , we then have that Hence, by Lemma 3.1 we have that On the other hand, from the definition ofφ ω we have that Hence, Therefore, we have completed the proof of Proposition 4.1.

A further reduction process. The main purpose of this section is to achieve
Step 2.A. We will define The equation (4.4) then becomes Note that φ does not depend on γ, but β depends on the parameters α, q, and γ. We write β = β(α, q, γ).
In this section, we will solve β = 0 for each α ∈ R by adjusting q and γ. By multiplying (5.2) by ∂U ∂q and integrating over R 2 , we obtain that Because the matrix M is invertible, solving β = 0 is equivalent to solving In the following subsection, we will compute the projections of the error and the terms involving φ.

5.1.
Projections. We first compute R 2 E ∂U ∂q dx. We begin with the following lemma.
Proof. By the definition of the error E, we know that We first estimate I 1 : (V (x) − 1)w j ∇w k dx = I 11 + I 12 .
where in the last equality we use the following identities: For the term I 12 , we have that where we have used |Q k+j −Q k | = 2 sin jπ K in the last step. Thus, we have that where τ 1 = min m + 3, m + σ, 2m + 1, 2m + m s + 1 , 4s + 2 + m 2s + 2 . (5.4)

INFINITELY MANY SOLUTIONS OF FRACTIONAL SCHRÖDINGER EQUATIONS 5587
Next, we consider I 2 .
By the definition of the interaction function Ψ, we have that For the term I 22 , we divide our discussion into two cases. If 1 < p < 2, we write For the first term on the right hand side of (5.6), we have that where we have used that (1 + x) p ≤ 1 + px + Cx p for small x, and C is a constant that depends on p only. Similarly, Therefore, we have for 1 < p < 2 that and similarly to (5.7) we have that In summary, we have that By combining the above estimates, we complete the proof of Lemma 5.1. Now, we can analyze R 2 E ∂U ∂q dx. Before we start, we define . (5.10) By considering to the asymptotic behavior of Ψ with its derivative, it is not difficult to see that d j = −(2s + 3) + o(1).
The matrix C: where The matrix D: where By noting that n j = cos θ j n 1 + sin θ j t 1 and t j = − sin θ j n 1 + cos θ j t 1 , we can further write that By combining the above estimates with Lemma 5.1 and using the balance relation we get that where Π 3 (α, q) is a smooth vector valued function that defines the remainder term appearing in (5.17) and (5.18) and is uniformly bounded as R → ∞.
Next, we compute the projections involving φ.

5.2.
The invertibility of T . In this subsection, we study the linear problem T q = b and obtain the following result, the proof of which is deferred to the Appendix.
Lemma 5.4. There exists R 0 > 0 such that for R > R 0 and every b ∈ R 2K , there exists a unique vector q ∈ R 2K and a unique constant γ ∈ R such that where q 1 is defined by (5.27) below. Moreover, there exists a positive constant C > 0 that is independent of R such that 3. Reduction to one dimension. Now, we state the main result of the section.
Lemma 5.7. The following expansion holds: where Π i (α, q) are smooth vector valued functions that are uniformly bounded as K → ∞. where η = τ 5 − m − 5 2 − m 4s+4 and Π(α, q) is a smooth vector valued function that is uniformly bounded as K → ∞. Then, by Lemma 5.4, for q ∞ < 1 we have that On the other hand, which shows that F is a contraction map. By the Banach fixed point theorem, we obtain the existence of a solution φ to (5.23).
Next, we study the dependence of q on α. Assume that we have two solutions corresponding to two sets of parameters. Let one of these be denoted by q)], corresponding to α, and the other denoted bȳ q)], corresponding toᾱ. Assume that R|ᾱ − α| < 1 2 . Then, by direct computation and Lemma 5.4, we have that Hence, we have proved the Proposition.
6. Proof of Theorem 1.1: Variational reduction. In this section, we complete the proof of Theorem 1.1. To solve γ(α) = 0, we will apply variational reduction. To do this, we first introduce some notation. Let α ∈ R, and let φ = φ(α, q(α)) be the function given by Proposition 5.5. Then, we define the following energy functional: Both U and φ are 2π periodic in α. Hence, by Proposition 5.5 the reduced energy functional F (α) has the following property.
Lemma 6.1. The function F (α) is of class C 1 , and satisfies F (α + 2π) = F (α) for every α ∈ R. Proof. Assume thatũ is the unique s-harmonic extension of u = U + φ. Then, the well-known computation by Caffarelli and Silvestre [9] shows that By Proposition 5.5, for large K and every α ∈ R, u = U + φ satisfies the equation Thus, we obtain that Hence, we find that F (α) = 0 if and only if γ(α) = 0. Thus, we have proved the Lemma.

INFINITELY MANY SOLUTIONS OF FRACTIONAL SCHRÖDINGER EQUATIONS 5595
and P is the K × K matrix defined by Next, we analyze the matrix T . We have the following result.
Lemma 7.1. There exists K 0 > 0 such that for all K > K 0 and every b ∈ R 2K , there exists a unique q ∈ R 2K and a unique γ ∈ R such that Moreover, there exists a positive constant C, which is independent of K, such that Furthermore, the number of zero (resp. negative, positive) eigenvalues of T is given by 1 (resp. K − 1, K).
Proof. Recall that T is given by (5.11): and the entries A, B, C, D are given by (5.12)-(5.15). Thus, for = 0, · · · , K − 1 we have that the -th eigenvalues of the matrices A, B, C, D are given by the following: and Now, define P = P 0 0 P , (7.12) where P is defined in (7.5). A simple algebraic manipulation gives that where 7.14) and D X denotes the diagonal matrix of the K × K circulant matrix X.
When = 0, we note that λ B,0 = λ C,0 = λ D,0 = 0. As a consequence, Λ 1,0 = 0, and When n is even, up to a normalization constant, P n and Q n are related to the Fourier series of the Bernoulli polynomial B n (x), and when n is odd P n and Q n are related to the Fourier series of the Euler polynomial E n (x). Let Using P i (x), we can write g(x) as g(x) = P 2s+4 (0) − P 2s+4 (x).

INFINITELY MANY SOLUTIONS OF FRACTIONAL SCHRÖDINGER EQUATIONS 5597
With these functions in hand, and using the balancing condition (5.20), we set Then, we have the following: (1) . (1), we can employ a Taylor expansion on g, g , and g to obtain that We note that λ A, λ D, − λ B, λ C, < 0. Now, assume that K ≥ c for some positive constant c. In this case, because g(x) > 0 if x > 0, in the expressions of λ A, to λ D, , the leading order terms in λ A, , λ B, , λ C, , λ D, are given by the terms containing g( 2π K ). Thus, we have that λ A, + λ D, =(2s + 2)f R K √ 2π (1 + o(1)) By combining the above results, for ≥ 1 we obtain that and From the above analysis, we know that the number of zero (resp. negative, positive) eigenvalues of T is 1 (resp. K − 1, K). The eigenvector corresponding to the zero eigenvalue is q 0 = (0, · · · , 0, 1, · · · , 1). (7.24) Moreover, for the solution (q, γ) to (7.6), we have the following estimate: Proof of Lemma 5.4. To prove Lemma 5.4, it suffices to verify the a priori estimate (5.22). Let γ = − b · q 0 q 1 · q 0 . By Lemma 7.1, we have that Because q 1 · q 0 = M (Rq 0 + q ⊥ ) · q 0 = c 0 KR(1 + o(1)), Thus, we can obtain that