Infinitely many positive solutions for nonlinear fractional Schr\"{o ...

INFINITELY MANY POSITIVE SOLUTIONS FOR NONLINEAR ¨ FRACTIONAL SCHRODINGER EQUATIONS

arXiv:1402.1902v1 [math.AP] 9 Feb 2014

WEI LONG, SHUANGJIE PENG AND JING YANG Abstract. We consider the following nonlinear fractional Schrödinger equation (−∆)s u + u = K(|x|)up , u > 0 in RN , N +2s where K(|x|) is a positive radial function, N ≥ 2, 0 < s < 1, 1 < p < N −2s . Under some asymptotic assumptions on K(x) at infinity, we show that this problem has infinitely many non-radial positive solutions, whose energy can be made arbitrarily large. Key words : fractional Laplacian; nonlinear Schrödinger equation; reduction method. AMS Subject Classifications: 35J20, 35J60

1. Introduction In this paper, we consider the following nonlinear fractional Schrödinger equation (−∆)s u + u = K(x)up , u > 0 in RN , u ∈ H s (RN )

(1.1)

with dimension N ≥ 2, where K(x) is a positive continuous potential, 0 < s < 1, 1 < p < and 2∗ (s) − 1, 2∗ (s) = N2N −2s n o |u(x) − u(y)| 2 N N H s (RN ) := u ∈ L2 (RN ) : ∈ L (R × R ) . N |x − y| 2 +s

The fractional Laplacian of a function f : RN → R is expressed by the formular Z Z f (x) − f (y) f (x) − f (y) s (−∆) f (x) = CN,s P.V. dy = CN,s lim dy, N +2s N +2s ε→0 RN |x − y| RN \Bε (x) |x − y|

(1.2)

where CN,s is some normalization constant (See Sect. 2). Problem (1.1) arises from looking for standing waves Ψ(t, x) = exp(iEt)u(x) for the following nonlinear Schrödinger equations ∂Ψ i = (−∆)s Ψ + (1 + E)Ψ − K(x)|Ψ|p−1Ψ, (t, x) ∈ R+ × RN , (1.3) ∂t where i is the imaginary unit. This equation is of particular interest in fractional quantum mechanics for the study of particles on stochastic fields modelled by Lévy processes. A path integral over the Lévy flights paths and a fractional Schrödinger equation of fractional quantum mechanics are formulated by Laskin [20] from the idea of Feynman and Hibbs’s path integrals (see also [21]). The Lévy processes occur widely in physics, chemistry and biology. The stable Lévy processes that give rise to equations with the fractional Laplacians have recently attracted 1

2

WEI LONG, SHUANGJIE PENG AND JING YANG

much research interest, and there are a lot of results in the literature on the existence of such solutions, e.g., [1, 3–5, 8, 9, 14, 19, 22, 27, 31] and the references therein. A partner problem of (1.1) is the following scalar field equation (−∆)s u + V (x)u = |u|p−1u, in RN , u ∈ H s (RN ).

(1.4)

inf V (x) < lim V (x), (or sup K(x) > lim K(x)),

(1.5)

In the sequel, we will assume that V, K is bounded, and V (x) ≥ V0 > 0, K(x) ≥ K0 > 0. It is known, but not completely trivial, that (−∆)s reduces to the standard Laplacian −∆ as s → 1. When s = 1, the classical nonlinear Schrödinger equation has been extensively studied in the last thirty years. If x∈RN

|x|→∞

x∈RN

|x|→∞

then, using the concentration compactness principle [23, 24], one can show that (1.1) and (1.4) has a least energy solution. See for example [13, 23, 24]. But if (1.5) does not hold, (1.1) or (1.4) may not have a least energy solution. So, in this case, one naturally needs to find solutions with higher energy. Recently, Cerami et al. [6] showed that problem (1.4) with s = 1 has infinitely many sign-changing solutions if V (x) goes to its limit at infinity from below at a suitable rate. In [29], Wei and Yan gave a surprising result which says that (1.1) or (1.4) with s = 1 and V (x) or K(x) being radial has solutions with large number of bumps near infinity and the energy of this solutions can be very large. This kind of results was generalized by Ao and Wei in [2] very recently to the case in which V (x) or K(x) does not have any symmetry assumption. For more results on (1.1) and (1.4) with s = 1, we can refer to [6, 7, 12, 13] and the references therein. When 0 < s < 1, Chen and Zheng [10] studied the following singularly perturbed problem ε2s (−∆)s u + V (x)u = |u|p−1u, in RN . (1.6) 1 n They showed that when N = 1, 2, 3, ε is sufficiently small, max{ 2 , 4 } < s < 1 and V satisfies some smoothness and boundedness assumptions, equation (1.6) has a nontrivial solution uε concentrated to some single point as ε → 0. Very recently, in [11], Dávila, del Pino and Wei generalized various existence results known for (1.6) with s = 1 to the case of fractional Laplacian. For results which are not for singularly perturbed type of (1.1) and (1.4) with 0 < s < 1, the readers can refer to [3, 14, 26, 28] and the references therein. As far as we know, it seems that there is no result on the existence of multiple solutions of equation (1.1) which is not a singularly perturbed problem. The aim of this paper is to obtain infinitely many non-radial positive solutions for (1.1) whose functional energy are very large, under some assumptions for K(x) = K(|x|) > 0 near the infinity. Let N + 2s < m < N + 2s. (1.7) N + 2s + 1 We assume that 0 < K(|x|) ∈ C(RN ) satisfies the following condition at infinity a 1 (K) : K(r) = K0 − m + O( m+θ ), as r → +∞, r r for some a > 0, θ > 0. Without loss of generality, we may assume that K0 = 1.

¨ ON NONLINEAR FRACTIONAL SCHRODINGER EQUATIONS

3

Our main result in this paper can be stated as follows Theorem 1.1. Suppose that N ≥ 2, 0 < s < 1, 1 < p < 2∗ (s) − 1. If K(r) satisfies (K), then problem (1.1) has infinitely many non-radial positive solutions. Remark 1.2. The radial symmetry can be replaced by the following weaker symmetry assumption: after suitably rotating the coordinated system, (i) K(x) = K(x′ , x′′ ) = K(|x′ |, |x3 |, · · · , |xN |), where x = (x′ , x′′ ) ∈ R2 × RN −2 , 1 (ii) K(x) = K0 − |x|am + O( |x|m+θ ) as |x| → +∞, where a > 0, θ > 0 and K0 > 0 are some constants. Remark 1.3. Using the same argument, we can prove that if 1 a V (x) = V (x′ , x′′ ) = V (|x′ |, |x3 |, · · · , |xN |) = V0 + m + O( m+θ ), as |x| → ∞, |x| |x| for some constants V0 > 0, a > 0, and θ > 0. Then problem (1.4) has infinitely many positive non-radial solutions. Before we close this introduction, let us outline the main idea in the proof of Theorem 1.1. We will use the unique ground state U of (−∆)s u + u = up , u > 0, x ∈ RN , u(0) = max u(x) RN

(1.8)

to build up the approximate solutions for (1.1). It is well known that when s = 1, the ground state solution of (1.8) decays exponentially at infinity. But from [15, 16], we see 1 that when s ∈ (0, 1), the unique ground solution of (1.8) decays like |x|N+2s when |x| → ∞. Let any integer k > 0, define 2(i − 1)π 2(i − 1)π , r sin , 0 , i = 1, · · · , k, xi = r cos k k N+2s

N+2s

where 0 is the zero vector in RN −2 , r ∈ [r0 k N+2s−m , r1 k N+2s−m ] for some r1 > r0 > 0. Set x = (x′ , x′′ ), x′ ∈ R2 , x′′ ∈ RN −2 . Define n H = u :u ∈ H s (RN ), u is even in xi , i = 2, · · · , N, u(r cos θ, r sin θ, x′′ ) = u(r cos(θ +

Write Ur (x) =

o 2jπ 2jπ ), r sin(θ + ), x′′ ), j = 1, · · · , k − 1 . k k

k X

Uxi (x),

i=1

where Uxi (x) = U(x − xi ). We will prove Theorem 1.1 by proving the following result Theorem 1.4. Under the assumption of Theorem (1.1), there is an integer k0 > 0, such that for any integer k ≥ k0 , (1.1) has a solution uk of the form uk = Ur (x) + ωr

4

WEI LONG, SHUANGJIE PENG AND JING YANG N+2s

N+2s

where ωr ∈ H , r ∈ [r0 k N+2s−m , r1 k N+2s−m ] and as k → +∞, Z Z |ωr (x) − ωr (y)|2 + ωr2 → 0. N +2s |x − y| 2N N R R The idea of our proof is inspired by that of [29] where infinitely many positive non-radial solutions to a nonlinear Schrödinger equations (1.3) with s = 1 are obtained when the potential approaches to a positive constant algebraically at infinity. We will use the wellknown Lyapunov-Schmidt reduction scheme to transfer our problem to a maximization problem of a one-dimensional function in a suitable range. Compared with the operator −∆, which is local, the operator (−∆)s with 0 < s < 1 on RN is nonlocal. So it is expected that the standard techniques for −∆ do not work directly. In particular, when we try to find spike solutions for (1.1) with 0 < s < 1, (−∆)s may kill bumps by averaging on the whole RN . For example, the ground state for (1.1) with 0 < s < 1 decays algebraically at infinity, which is a contrast to the fact that the ground state for −∆ decays exponentially at infinity. This kind of property requires us to establish some new basic estimates and give a precise estimate on the energy of the approximate solutions. This paper is organized as follows. In Sect.2, we will give some preliminary properties related to the fractional Laplacian operator. In Sect.3, we will establish some preliminary estimate. We will carry out a reduction procedure and study the reduced one dimensional problem to prove Theorems 1.4 in Sect.4. In Appendix, some basic estimates and an energy expansion for the functional corresponding to problem (1.1) will be established. 2. Basic Theory on Fractional Laplacian Operator In this section, we recall some properties of the fractional order Sobolev space and the ground state solution U of the limit equation (1.8). Let 0 < s < 1. Various definitions of the fractional Laplacian (−∆)s f (x) of a function f defined in RN are available, depending on its regularity and growth properties. It can be defined as a pseudo-differential operator \s f (ξ) = |ξ|2s fb(ξ), (−∆)

where b is Fourier transform. When f have some sufficiently regular, the fractional Laplacian of a function f : RN → R is expressed by the formular Z Z f (x) − f (y) f (x) − f (y) s (−∆) f (x) = CN,s P.V. dy = CN,s lim dy, N +2s ε→0 RN \B (x) |x − y|N +2s RN |x − y| ε . This integral makes sense directly when s < 12 and where CN,s = π −(2s+N/2) Γ(N/2+s) Γ(−s) f ∈ C 0,α (RN ) with α > 2s, or if f ∈ C 1,α (RN ), 1 + 2α > 2s. It is well known that (−∆)s on RN with 0 < s < 1 is a nonlocal operator. In the remarkable work of Caffarelli and Silvestre [4], this nonlocal operator was expressed as a generalized Dirichlet-Neumann map for a certain elliptic boundary value problem with nonlocal differential operator defined +1 on the upper half-space RN := {(x, y) : x ∈ RN , y > 0}. That is, for a function +


5

f : RN → R, we consider the extension u(x, y) : RN × [0, +∞) → R that satisfies the equation u(x, 0) = f (x) (2.1) 1 − 2s uy + uyy = 0. y The equation (2.2) can also be written as ∆x u +

(2.2)

div(y 1−2s ∇u) = 0,

(2.3)

which is clearly the Euler-Lagarnge equation for the functional Z J(u) = |∇u|2 y 1−2s dxdy. y>0

Then, it follows from [4] that C(−∆)s f = lim+ −y 1−2s uy = y→0

u(x, y) − u(x, 0) 1 lim+ . 2s y→0 y 2s

When s ∈ (0, 1), the space H s (RN ) = W s,2(RN ) is defined by o n |u(x) − u(y)| 2 N N s N 2 N ∈ L (R × R ) H (R ) = u ∈ L (R ) : N |x − y| 2 +s Z n o 2 N = u ∈ L (R ) : (1 + |ξ|2s )|b u(ξ)|2 < +∞ RN

and the norm is

kuks := kukH s (RN ) = Here the term

Z

Z

Z

Z

RN

RN

|u(x) − u(y)|2 dxdy + |x − y|N +2s

Z

21 |u| dx . 2

RN

21 |u(x) − u(y)|2 dxdy [u]s := N +2s RN RN |x − y| is the so-called Gagliardo (semi) norm of u. The following identity yields the relation between the fractional operator (−∆)s and the fractional Laplacian Sobolev space H s (RN ), Z 12 s 2s 2 [u]H s (RN ) = C |ξ| |b u(ξ)| dξ = Ck(−∆) 2 ukL2 (RN ) RN

for a suitable positive constant C depending only s and N. Clearly, k · kH s (RN ) is a Hilbertian norm induced by the inner product Z Z Z (u(x) − u(y))(v(x) − v(y)) hu, viH s(RN ) = dxdy + u(x)v(x)dx |x − y|N +2s RN RN RN = hu, vis + hu, viL2(RN ) . On the Sobolev inequality and the compactness of embedding, one has

6


Theorem 2.1. [25] The following imbeddings are continuous: , if N > 2s, (1) H s (RN ) ֒→ Lq (RN ), 2 ≤ q ≤ N2N −2s s N q N (2)H (R ) ֒→ L (R ), 2 ≤ q ≤ +∞, if N = 2s, (3)H s(RN ) ֒→ Cbj (RN ), if N < 2(s − j). Moreover, for any R > 0 and p ∈ [1, 2∗ (s)) the embedding H s (BR ) ֒→֒→ Lp (BR ) is compact, where n o Cbj (RN ) = u ∈ C j (RN ) : D K u is bounded on RN for |K| ≤ j .

Now, we recall some known results for the limit equation (1.8). If s = 1, the uniqueness and non-degeneracy of the ground state U for (1.8) is due to [18]. In the celebrated paper [15], Frank and Lenzemann proved the uniqueness of ground state solution U(x) = U(|x|) ≥ 0 for N = 1, 0 < s < 1, 1 < p < 2∗ (s) − 1. Very recently, Frank, Lenzemann and Silvestre [16] obtained the non-degeneracy of ground state solutions for (1.8) in arbitrary dimension N ≥ 1 and any admissible exponent 1 < p < 2∗ (s) − 1. For convenience, we summarize the properties of the ground state U of (1.8) which can be found in [15, 16]. Theorem 2.2. Let N ≥ 1, s ∈ (0, 1) and 1 < p < 2∗ (s) − 1. Then the following hold. (i) (Uniqueness) The ground state solution U ∈ H s (RN ) for equation (1.8) is unique. (ii)(Symmetry, regularity and decay) U(x) is radial, positive and strictly decreasing in N |x|. Moreover, the function U belongs to H 2s+1(R ) ∩ C ∞ (RN ) and satisfies C1 C2 ≤ U(x) ≤ f or x ∈ RN , N +2s 1 + |x| 1 + |x|N +2s with some constants C2 ≥ C1 > 0. (iii) (Non-degeneracy) The linearized operator L0 = (−∆)s +1−p|U|p−1 is non-degenerate, i.e., its kernel is given by kerL0 = span{∂x1 U, ∂x2 U, · · · ∂xN U}. By Lemma C.2 of [16], it holds that, for j = 1, · · · , N, ∂xj U has the following decay estimate, C |∂xj U| ≤ . 1 + |x|N +2s 1 when |x| → +∞. FortuFrom Theorem 2.2, the ground bound state solution like |x|N+2s nately, this polynomial decay is enough for us in the estimates of our proof. 3. Some Preliminaries Let

∂Uxi , i = 1, · · · , k, ∂r , r sin 2(i−1)π , 0) and where xi = (r cos 2(i−1)π k k Zi =

N+2s

N+2s

r ∈ [r0 k N+2s−m , r1 k N+2s−m ]


for some r1 > r0 > 0. Define

n

E= v∈H : The norm of H s (RN ) is defined by: kvks = where

Z

p

k Z X i=1

RN

7

o Z v = 0 . Uxp−1 i i

v ∈ H s (RN ),

hv, vi,

hv1 , v2 i = hv1 , v2 is + v1 v2 RN Z Z Z (u(x) − u(y))(v(x) − v(y)) = dxdy + u(x)v(x)dx. |x − y|N +2s RN RN RN

The variational functional corresponding to (1.1) is Z Z 1 1 1 2 u − K(x)|u|p+1. I(u) = hu, uis + 2 2 RN p + 1 RN

Let

J(ϕ) = I(Ur + ϕ) = I

k X i=1

Expand J(ϕ) as follows:

Uxi + ϕ , ϕ ∈ E.

1 J(ϕ) = J(0) + l(ϕ) + hL(ϕ), ϕi + R(ϕ), 2 where l(ϕ) =

Z

k X

RN i=1

Uxpi ϕ

−

hL(ϕ), ϕi = hϕ, ϕis + and 1 R(ϕ) = − p+1

Z

K(x) (Ur + ϕ) RN

p+1

Z

−

Z

K(x)

RN

RN

k X i=1

Uxi

ϕ ∈ E, p

(3.1)

ϕ,

(ϕ2 − pK(x)Urp−1 ϕ2 )

Urp+1

− (p +

1)Urp ϕ

1 p−1 2 − (p + 1)pUr ϕ . 2

In order to find a critical point for J(ϕ), we need to discuss each term in the expansion (3.1). Lemma 3.1. There is a constant C > 0 independent of k such that kR′ (ϕ)k ≤ Ckϕksmin{p,2} , kR′′ (ϕ)k ≤ Ckϕksmin{p−1,1} .

8


Proof. By direct calculation, we know that Z ′ hR (ϕ), ψi = − K(x) (Ur + ϕ)p − Urp − pUrp−1 ϕ ψ, RN

′′

hR (ϕ)(ψ, ξ)i = −p

Z

RN

K(x) (Ur + ϕ)p−1 − Urp−1 ψξ.

Firstly, we deal with the case p > 2. Since Z p p+1 Z p+1 ′ p−2 2 p−2 2 p |hR (ϕ), ψi| ≤ C Ur |ϕ| |ψ| ≤ C (Ur |ϕ| ) kψks , RN

Rn

we find

′

kR (ϕ)k ≤ C

Z

Rn

(Urp−2 |ϕ|2)

p+1 p

p p+1

.

On the other hand, it follows from Lemma A.2 that Ur is bounded. Since 2 < we obtain p Z p+1 2(p+1) ′ kR (ϕ)k ≤ C |ϕ| p ≤ Ckϕk2s .

2(p+1) p

< p+1,

Rn

For the estimate of kR′′ (ϕ)k, we have Z ′′ |R (ϕ)(ψ, ξ)| ≤ C Urp−2 |ϕ||ψ||ξ| N ZR Z ≤ C |ϕ||ψ||ξ| ≤ C RN

RN

|ϕ|

≤ Ckϕks kψks kξks .

3

31 Z

RN

|ψ|

3

31 Z

RN

|ξ|

3

13

So, kR′′ (ϕ)k ≤ Ckϕks .

Using the same argument, for the case 1 < p ≤ 2, we also can obtain that Z ′ hR (ϕ), ψi ≤ C ϕp ψ ≤ Ckϕkps kψks , RN

hR′ (ϕ)(ψ, ξ)i ≤ C

Z

RN

ϕp−1ψξ ≤ Ckϕkp−1 s kψks kξks .

4. The Finite-Dimensional reduction and proof of the main results In this section, we intend to prove the main theorem by the Lyapunov-Schmidt reduction. Associated to the quadratic form L(ϕ), we define L to be a bounded linear map from E to E, such that Z hLv1 , v2 i = hv1 , v2 is + v1 v2 − pK(x)Urp−1 v1 v2 , v1 , v2 ∈ E. RN


9

In this paper, we always assume 1 1 h B (N + 2s) N+2s B (N + 2s) N+2s i N+2s−m N+2s−m 0 0 N+2s−m r ∈ Sk := k k N+2s−m , , −α +α B1 m B1 m (4.1) where α > 0 is a small constant, B0 and B1 are defined in Proposition A.3. Next, we show the invertibility of L in E. Proposition 4.1. There exists an integer k0 > 0, such that for k ≥ k0 , there is a constant ρ > 0 independent of k, satisfying that for any r ∈ Sk , kLuk ≥ ρkuks ,

u ∈ E.

Proof. We argue by contradiction. Suppose that there are n → +∞, rk ∈ Sk , and un ∈ E, such that kLun k = o(1)kun ks , kun k2s = k. Recall n x′ xi πo ′ ′′ 2 N −2 Ωi = x = (x , x ) ∈ R × R : h ′ , i i ≥ cos , i = 1, 2, · · · , k. |x | |x | k By symmetry, we have Z Z Z (un (x) − un (y))(ϕ(x) − ϕ(y)) p−1 u ϕ − pK(x)U u ϕ + n n rk |x − y|N +2s Ω1 Ω1 RN (4.2) 1 1 = hLun , ϕi = o( √ )kϕks , ϕ ∈ E. k k In particular, Z Z Z |un (x) − un (y)|2 2 p−1 2 |u | − pK(x)U |u | = o(1) + n n rk |x − y|N +2s Ω1 RN Ω1 and Z Z Z |un (x) − un (y)|2 |un |2 = 1. + N +2s |x − y| N Ω1 Ω1 R m

Set u en (x) = un (x − x1 ). Then for any R > 0, since dist(x1 , ∂Ω1 ) = r sin πk ≥ Ck N+2s−m → +∞, BR (x1 ) ⊂ Ω1 , i.e. Z Z Z |un (x) − un (y)|2 + |un |2 ≤ 1. N +2s |x − y| 1 1 N BR (x ) BR (x ) R One can obtain

Z |e un (x) − u en (y)|2 + |e un |2 ≤ 1. N +2s |x − y| BR (0) BR (0) RN s N So we suppose that there is a u ∈ H (R ), such that as n → +∞, and

Z

Z

u en ⇀ u,

u en → u,

s in Hloc (RN )

in L2loc (RN ).

10


Since u en is even in xj , j = 2, · · · , N, it is easy to see that u is even in xj , j = 2, · · · , N. On the other hand, from Z RN

we get

Z

U p−1

RN

So, u satisfies

Z

RN

Now, we claim that u satisfies Indeed, we set

Uxp−1 1 Z1 un = 0,

∂U u en = 0. ∂x1

U p−1

∂U u = 0. ∂x1

(4.3)

(−∆)s u + u − pU p−1 u = 0 in RN . Z n s N e E = ϕ : ϕ ∈ H (R ),

RN

U p−1

(4.4)

o ∂U ϕ=0 . ∂x1

e be any function, satisfying that ϕ is even in For any R > 0, let ϕ ∈ C0∞ (BR (0)) ∩ E xj , j = 2, · · · , N. Then ϕ1 (x) = ϕ(x − x1 ) ∈ C0∞ (BR (x1 )). We may identify ϕ1 (x) as elements in E by redefining the values outside Ω1 with the symmetry. By using (4.2) and Lemma A.2, we find Z Z Z (u(x) − u(y))(ϕ(x) − ϕ(y)) p−1 + uϕ − pU uϕ = 0. (4.5) |x − y|N +2s RN RN RN On the other hand, since u is even in xj , j = 2, · · · , N, (4.5) holds for any ϕ ∈ C0∞ (BR (0))∩ e By the density of C ∞ (RN ) in H s (RN ), it is easy to show that E. 0 Z Z Z (u(x) − u(y))(ϕ(x) − ϕ(y)) p−1 e + uϕ − pU uϕ = 0, ∀ ϕ ∈ E. (4.6) N +2s |x − y| N N N R R R ∂U We know ϕ = ∂x is a solution of (4.6), thus (4.6) is true for any ϕ ∈ H s (RN ). One see 1 that u = 0 because u is even in yi , i = 2, · · · , N and (4.4). As a result, Z u2n = on (1), ∀ R > 0, BR (x1 )

where on (1) → 0 as n → +∞. Now, using the Lemma A.2, we obtained that for any 1 < η ≤ N +2s, there is a constant C > 0, such that C Urk (x) ≤ , x ∈ Ω1 . (1 + |x − x1 |)N +2s−η Thus, Z Z |un (x) − un (y)|2 2 p−1 2 on (1) = u − pK(x)U u + n rk n |x − y|N +2s Ω1 ×RN Ω1


=

Z

Z

|un (x) − un (y)|2 + |x − y|N +2s Ω1 RN Z Z + +C

Z

Ω1

11

u2n

1 2 u 1 N +2s−η n Ω1 \B 1 R (x1 ) (1 + |x − x |) B 1 R (x1 ) 2 Z Z2 Z 1 |un (x) − un (y)|2 2 ≥ u + n + on (1) + oR (1), 2 Ω1 RN |x − y|N +2s Ω1

which is impossible for large R. As a result, we get a contradiction.

Proposition 4.2. There is an integer k0 > 0, such that for each k ≥ k0 , there is a C 1 map from Sk to H : ω = ω(r), r = |x1 |, satisfying ω ∈ E, and ′ J (ω) = 0. E

Moreover, there exists a constant C > 0 independent of k such that k N+2s +τ 1 1 2 kωks ≤ Ck 2 + m +τ , r r2 where τ > 0 is a small constant.

(4.7)

Proof. We will use the contraction theorem to prove it. By the following Lemma 4.3, l(ω) is a bounded linear functional in E. We know by Reisz representation theorem that there is an lk ∈ E, such that l(ω) = hlk , ωi. So, finding a critical point for J(ω) is equivalent to solving lk + Lω + R′ (ω) = 0.

(4.8)

By Proposition 4.1, L is invertible. Thus, (4.8) is equivalent to Set

ω = A(ω) := −L−1 (lk + R′ (ω)).

1 n k N+2s+τ 1 k2 o 2 ¯ 2 Sk := ω ∈ E : kωks ≤ k + m+τ . r r 2 ¯ We shall verify that A is a contraction mapping from Sk to itself. In fact, on one hand, for any ω ∈ S¯k , by Lemmas 4.3 and 3.1, we obtain

kA(ω)k ≤ C(klk + kR′ (ω)k)

≤ Cklk + Ckωksmin{p,2} 1 1 1 k N+2s 1 k N+2s+τ +τ k 2 min{p,2} k2 2 2 2 2 ≤ C k + m+τ + m +τ + C k r r r2 r 2 1 N+2s+τ 1 k k2 2 ≤ k2 + m+τ . r r 2

12


On the other hand, for any ω1 , ω2 ∈ S¯k , kA(ω1 ) − A(ω2 )k = kL−1 R′ (ω1 ) − L−1 R′ (ω2 )k ≤ CkR′ (ω1 ) − R′ (ω2 )k ≤ CkR′′ (θω1 + (1 − θ)ω2 )kkω1 − ω2 ks ( C(kω1 kp−1 + kω2 kp−1 s s )kω1 − ω2 ks , ≤ C(kω1 ks + kω2 ks )kω1 − ω2 ks , 1 ≤ kω1 − ω2 ks . 2

if 1 < p < 2 if p ≥ 2

Then the result follows from the contraction mapping theorem. The estimate (4.7) follows Lemma 4.3. The claim that ω(r) is continuously differentiable in r can be verified by using the same argument employed to proof Lemma 4.4 in [11]. Lemma 4.3. There is a constant C > 0 and a small constant τ > 0, which are independent of k, such that 1 1 k N+2s +τ k2 2 2 + m +τ , klk k ≤ C k r r2

provided k ≥ k0 for some integer k0 > 0. Proof. Z |l(ϕ)| = ≤C

From m >

k X

RN i=1

Z

RN

N +2s , N +2s+1

Uxpi ϕ

k X

− K(x)

Uxi

i=1

p

−

k X i=1

k X i=1

we deduce that

p Uxi ϕ

Z p Uxi |ϕ| +

RN

k X Uxi )p ϕ . (K(x) − 1)(

(4.9)

i=1

p N + 2s p 1 −1 N + 2s − m . < N + 2s < (N + 2s)(p − 1) + − N − m 2 p+1 p+1 2 Hence, we can choose σ ∈ (0,

N +2s ) 2

such that

N + 2s p 1 m p + 1 (N + 2s)(p − 1) + − σ > N, − < σ. p 2 p+1 2 N + 2s − m

(4.10)


13

Using the fact Uxi ≤ Ux1 , (x ∈ Ω1 ) and Lemma A.2, we obtain Z X Z X k k k k X X p p p Uxi ) − Uxpi |ϕ| Uxi ) − Uxi |ϕ| = k ( ( RN

≤Ck

≤Ck ≤Ck

i=1

Z

Ω1

Z

Z p

=Ck p+1

k X

Uxp−1 1

i=2

Ω1

Ω1

i=2

|ϕ|

p+1

1

(1 + |x − x1 |)(N +2s)(p−1)+ Z k N+2s +σ 2

Ω1

i=1

r

i=1

Uxi |ϕ|

k p Z p+1 p+1 X p p−1 Uxi Ux1

r k N+2s +τ 1 2

≤Ck 2

Ω1

i=1

N+2s −σ 2

1

1 p+1

k

N+2s +σ 2

r

(1 + |x − x1 |)(N +2s)(p−1)+

Ω1

p+1 p

N+2s −σ 2

p p+1

(4.11)

Z

Ω1

p p+1 p+1 p

|ϕ|p+1

1 p+1

kϕks

kϕks ,

for some small constant τ > 0, where we have used (4.10). On the other hand, we have Z Z k X p p ϕ = k (K(x) − 1) U ϕ (K(x) − 1)U x1 xi RN

≤Ck

RN

i=1

Z

p+1 p

Uxp+1 1

p Z p+1

p+1

1 p+1

|K(x) − 1| |ϕ| RN Z p Z p+1 p+1 p+1 p +C ≤Ckkϕks |K(x) − 1| Ux1 RN

B r (x1 ) 2

k k ≤Ckϕks m + (N +2s)p− p N p+1 r r 1 k2 ≤C m +τ kϕks , r2

RN \B r (x1 ) 2

Uxp+1 1

p p+1

for some small constant τ > 0, where the last inequality is due to the assumption m < N + 2s. Inserting (4.11)–(4.12) into (4.9), we can complete the proof.

(4.12)

N +2s N +2s+1

<

Now, we are ready to prove our main theorem. Let ω = ω(r) be the map obtained in Proposition 4.2. Define F (r) = I(Ur + ω), ∀ r ∈ Sk . It follows from Lemma 6.1 in [11] that if r is a critical point of F (r), then Ur + ω is a solution of (1.1).

14


Proof of Theorem 1.4. It follows from Propositions 4.2 and A.3 that F (r) = I(Ur ) + O(klkkωk + kωk2) k N +2s+2τ 1 1 B0 k N +2s B1 + m+2τ = k A − N +2s + m + O m+σ + O r r r r r N +2s B0 k B1 1 = k A − N +2s + m + O m+σ . r r r Define 1 B0 k N +2s B1 F1 (r) := − N +2s + m + O m+σ . r r r We consider the following maximization problem

(4.13)

max F1 (r). r∈Sk

Suppose that rˆ is a maximizer, we will prove that rˆ is an interior point of Sk . We can check that the function B0 k N +2s B1 G(r) = − N +2s + m r r has a maximum point 1 B (N + 2s) N+2s−m N+2s 0 re = k N+2s−m B1 m and B0 k N +2s B1 m = m . N +2s re re N + 2s By direct computation, we deduce that 1 B0 k N +2s B1 F1 (e r) = − N +2s + m + O m+σ re re re 1 B1 m = m 1− + O m+σ r˜ N + 2s re N+2s N + s (N+2s)m (N+2s)m B1N+2s−m 1 = − 1 k − N+2s−m + O(k − N+2s−m −σ ). m N+2s N +2s 2m B0N+2s−m ( m ) N+2s−m On the other hand, we find 1 B (N + 2s) N+2s−m N+2s 0 N+2s−m F1 −α k B1 m (N+2s)m B0 k N +2s B1 − N+2s−m −σ = − ) + + O(k 2 m(N+2s) m (N+2s) N+2s N +2s B0 +2s B0 N+2s−m k N+2s−m N+2s−m k N+2s−m ( − α) ( Nm − α) m B1 B 1

N+2s N+2s−m

=

B1

m

N +2s B0N+2s−m ( m < F1 (e r)

N + 2s (N+2s)m (N+2s)m B1 − α − 1 k − N+2s−m + O(k − N+2s−2m −σ ) N+2s m B0 − α B1 ) N+2s−m 1

B0


15

and similarly 1 B (N + 2s) N+2s−m N+2s 0 F1 k N+2s−m +α B1 m N+2s N + 2s (N+2s)m (N+s)m 1 B1N+2s−m B1 = +α − 1 k − N+2s−m + O(k − N+2s−2m −σ ) m N+2s B N +2s m B0 B0N+2s−m ( m + α B01 ) N+2s−m < F1 (e r),

N+2s

where we have used the fact that the function f (x) = x− N+2s−m (x−1) attains its maximum +2s +2s +2s if x ∈ [ N m − α, N m + α]. at x0 = N m The above estimates implies that rˆ is indeed an interior point of Sk . Thus urˆ = Urˆ + ωrˆ is a solution of (1.1). At last, we claim that urˆ > 0. Indeed, since kωrˆks → 0 as k → ∞, noticing the fact (see [11] for example) that Z Z 2 2 1−2s |∇˜ urˆ| y dxdy + ur2ˆdx, kurˆks = RN+1 +

RN

where u˜rˆ(x, y) is the s-harmonic extension of urˆ satisfying u˜rˆ(x, 0) = urˆ(x), we can use the standard argument to verify that (urˆ)− = 0 and hence urˆ ≥ 0. Since urˆ solves (2.3), we conclude by using the strong maximum principle that urˆ > 0. Appendix A. Energy expansion In this section, we will give the energy expansion for the approximate solutions. Recall 2(i − 1)π 2(i − 1)π i x = r cos , r sin , 0 , i = 1, · · · , k, k k n πo x′ xi , i = 1, 2, · · · , k, Ωi = x = (x′ , x′′ ) ∈ R2 × RN −2 : h ′ , i i ≥ cos |x | |x | k and Z Z Z 1 |u(x) − u(y)|2 1 1 2 I(u) = u − K(x)|u|p+1 . + N +2s 2 R2N |x − y| 2 RN p + 1 RN Firstly, we will introduce Lemma B.1 ([30]). Lemma A.1. For any constant 0 < σ ≤ min{α, β}, there is a constant C > 0, such that C 1 1 1 1 ≤ + , (1 + |y − xi |)α (1 + |y − xj |)β |xi − xj |σ (1 + |y − xi |)α+β−σ (1 + |y − xj |)α+β−σ

where α, β ≥ 1 are two constants.

Then, we have the following basic estimate:

16


Lemma A.2. For any x ∈ Ω1 , and η ∈ (1, N + 2s], there is a constant C > 0, such that k X i=2

kη kη 1 ≤ C . (1 + |x − x1 |)N +2s−η |x1 |η |x1 |η

Uxi ≤ C

Proof. The proof of this lemma is similar to Lemma A.1 in [29], we sketch the proof below for the sake of completeness. For any x ∈ Ω1 , we have for i 6= 1, |x − xi | ≥ |x − x1 |, ∀ x ∈ Ω1 ,

which gives |x − xi | ≥ 21 |xi − x1 | if |x − x1 | ≥ 12 |xi − x1 |. On the other hand, if |x − x1 | ≤ 1 i |x − x1 |, then 2 1 |x − xi | ≥ |xi − x1 | − |x − x1 | ≥ |xi − x1 |. 2 So, we find 1 |x − xi | ≥ |xi − x1 |, ∀ x ∈ Ω1 . 2 Thus, k X i=2

Uxi =

k X i=2

=C

k

X C C = i N +2s 1 + |x − x | (1 + |x − xi |)N +2s i=2

k X i=2

1 1 (1 + |x − xi |)η (1 + |x − xi |)N +2s−η

k X 1 1 . ≤C 1 N +2s−η 1 (1 + |x − x |) |x − xi |η i=2

Since

|xi − x1 | = 2|x1 | sin

(i − 1)π , i = 2, · · · , k, k

we have k X i=2

k X 1 1 1 = i 1 η 1 η (i−1)π |x − x | (2|x |) i=2 (sin )η k  P k2 1 1  +  (2|x1 |)η i=2 (i−1)π (sin k )η = P[ k2 ]  1 1  , 1 η (i−1)π (2|x |)

But

0 < c′ ≤

i=2 (sin

sin (i−1)π k (i−1)π k

k

1 , (2|x1 |)η

)η

k ≤ C ′′ , j = 2, · · · , [ ]. 2

if k is even, if k is odd.


17

So, there is a constant B > 0, such that k X i=2

Thus, we obtain k X i=2

1 Bk η k = + O( ). |xi − x1 |η |x1 |η |x1 |η

Uxi ≤ C

kη kη 1 ≤ C . (1 + |x − x1 |)N +2s−η |x1 |η |x1 |η

Proposition A.3. There is a small constant τ > 0, such that

k 1 k N +2s+τ ˜0 1X B B1 I(Ur ) =k A − + + O m+τ + 2 j=2 |x1 − xj |N +2s r m r r

+

k X j=2

B0 |x1 − xj |N +2s+τ

1 B0 k N +2s B1 =k A − N +2s + m + O m+τ , r r r R 1 1 p where A = ( 2 − p ) RN U , and B0 , B1 > 0 are positive constants.

Proof. Using the symmetry,

hUr , Ur is + hUr , Ur iL2 (RN ) =

k Z k X X i=1 j=1

=k

Z

RN

=k

Z

Uxpj Uxi

RN

Uxp+1 1

U

j=2

p+1

+k

RN

It follows from Lemma A.1 that Z k Z X p Ux1 Uxi = C i=2

≤C =

RN

k X i=2

k X i=2

RN

1 1 |x − xi |N +2s

Z

+

k Z X

RN

k Z X

RN

i=2

Uxp1 Uxj

Uxp1 Uxi .

p X 1 1 1 N +2s 1 + |x − x | 1 + |x − xi |N +2s i=2

RN

k

X 1 1 + O (1 + |x − x1 |)(N +2s)p |x1 − xi |N +2s+τ i=2 k

X 1 C1 + O . 1 − xi |N +2s+τ |x1 − xi |N +2s |x i=2 k

(A.1)

18


However, k Z X

RN

i=2

Uxp1 Uxi

Z

=

p X 1 1 1 N +2s 1 + |x − x | 1 + |x − xi |N +2s i=2

RN

k

k Z X

≥

B |x1 −xi | (x1 )

i=2

+

k Z X i=2

B |x1 −xi | (xi ) 2

p 1 1 1 N +2s 1 + |x − x | 1 + |x − xi |N +2s

p 1 1 1 + |x − x1 |N +2s 1 + |x − xi |N +2s

X 1 C2 . + O |x1 − xi |N +2s |x1 − xi |N +2s+τ i=2 k

k X

≥

2

i=2

Hence, there exists B0′ (which maybe depend on k) in [C2 , C1 ], where C1 and C2 are independent of k, such that k Z X i=2

RN

Uxp1 Uxi

=

k X i=2

X 1 B0′ +O . |x1 − xi |N +2s |x1 − xi |N +2s+τ i=2 k

Now, by symmetry, we see Z

RN

K(x)Urp+1

= k

Z

Ω1

+k

K(x)Uxp+1 1

+ k(p + 1)

Z

K(x)

Ω1

k X

Uxp1 Uxi

i=2

 k Z p+1 X  p+1  2  2 i U O ) ( U , if 1 < p < 2,  x x1   Ω1 i=2  Z      O

Ω1

Uxp−1 1 (

k X i=2

Uxi ) , 2

if p ≥ 2.

For x ∈ Ω1 , we have |x − xi | ≥ 12 |xi − x1 |. By Lemma A.1, k X i=2

Uxi ≤ C ≤C ≤C

k X i=2

1 1 1 N +2s−κ (1 + |x − x |) (1 + |x − xi |)κ

k X

1 1 1 + |x1 − xi |N +2s−κ (1 + |x − x1 |)κ (1 + |x − xi |)κ

i=2

|x1

i=2

k X

1 1 , i N +2s−κ −x | (1 + |x − x1 |)κ

(A.2)


19

where κ > 0 satisfies min{ p+1 (N + 2s − κ), 2(N + 2s − κ)} > N + 2s. Hence, we get 2 Z

p+1 2 x1

U

Ω1

(1 + |x − x1 |)

k X i=2

k X

≤ C

p+1 2

1

Ω1

= C

Uxi )

i=2

Z

≤ C

(

k X

i=2

(N+2s)(p+1) 2

1 |xi − x1 |N +2s−κ 1 |xi − x1 |N +2s−κ

p+1 2

k X i=2

Z

1 |xi − x1 |N +2s−κ

1 (1 + |x − x1 |)

p+1 κ 2

1

(1 + |x − x1 |) k N +2s+τ ≤C r Ω1

p+1 2

p+1 2

(N+2s)(p+1) +κ 2

and Z

Ω1

Uxp−1 1 (

k X

Uxi )2

i=2

Z

k X 2 1 1 1 ≤ C 1 (N +2s)(p−1) i 1 N +2s−κ |x − x | (1 + |x − x1 |)2κ Ω1 (1 + |x − x |) i=2 k X 2 Z 1 1 = C i 1 N +2s−κ 1 (N +2s)(p−1)+2κ |x − x | Ω1 (1 + |x − x |) i=2

= C

k X i=2

2 k N +2s+τ 1 . ≤ C |xi − x1 |N +2s−κ r

On the other hand, Z

Ω1

K(x)Uxp1

k X

Uxi =

Z

Ω1

i=2

k X

Uxp1

Uxi +

Z

Ω1

i=2

(K(x) −

1)Uxp1

k X

Uxi .

i=2

But, from Lemma A.1 and (A.2), Z

Ω1

=

Z

RN

=

Z

RN

Uxp1

k X

Uxi =

i=2

Uxp1

k X

Z

RN

Uxi + O

i=2

Uxp1

k X i=2

Uxi + O

Uxp1

k X i=2

k σ Z r

k σ r

Uxi −

RN \Ω1

k X i=2

Z

RN \Ω1

Uxp−σ 1

Uxp1

k X

k X

Uxi

i=2

Uxi

i=2

1 i |x − x1 |N +2s

Z

RN \Ω1

p−σ Uxp−σ + U 1 xi

20

WEI LONG, SHUANGJIE PENG AND JING YANG k X

=

i=2

k N +2s+τ B0′ , + O |x1 − xi |N +2s r

where σ > 0 satisfies p − σ > 1. Moreover, similarly, Z k X p Uxi |K(x) − 1|Ux1 Ω1

= ≤

Z

Z

|K(x) −

RN

B r (x1 ) 2

C ≤ m r

Hence,

i=2

Z

1|Uxp1

k X i=2

|K(x) −

1|Uxp1

Z

Uxi −

k X

|K(x) − 1|Uxp1

RN \Ω1

Uxi + C

i=2

Z

RN \B r (x1 )

Uxp1

k X

Uxi

i=2

k X

Uxi + O

i=2

2

k N +2s+τ r

Z k 1 X 1 p−σ p−σ Uxi + O σ + U U 1 xi r i=2 |xi − x1 |N +2s RN \B r (x1 ) x RN i=2 2 k N +2s+τ Uxp1

k X

+O r k N +2s+τ 1 = O + m+τ . r r Z

Finally,

=

Ω1

K(x)Uxp1

Z

ZΩ1

Z

RN

=

Z

= =

ZR

2

N

RN

k X j=2

k N +2s+τ 1 B0′ + O . + |x1 − xi |N +2s r r m+τ

K(x)Uxp+1 1 K(x)Uxp+1 1

−

Z

RN \B 2πr (x1 )

K(x)Uxp+1 +O 1

B r (x1 )

Z

Uxi =

i=2

RN

=

k X

K(x)Uxp+1 1

Z

+

Z

K(x)Uxp+1 1

+

k

RN \B

2πr k

(x1 )

RN \B r (x1 ) 2

Z

K(x)Uxp+1 1

K(x)Uxp+1 1

Ω1 \B 2πr (x1 )

+O

K(x)Uxp+1 1

k

Z

RN \B 2πr (x1 )

k (N +2s)(p+1)−N 1 B′ U p+1 − m1 + O m+τ + O r r r ′ 1 k N +2s+τ B U p+1 − m1 + O m+τ + r r r

k

K(x)Uxp+1 1


since (N + 2s)(p + 1) − N > N + 2s. So, we have proved Z k Z X p+1 K(x)Ur = k U p+1 + RN

RN

j=2

21

1 k N +2s+τ B0′ B1′ . (A.3) − +O + |x1 − xi |N +2s r m r m+τ r

Now, inserting (A.1)–(A.3) into I(Ur ), we complete the proof.

Acknowledgements: S. Peng was partially supported by the fund from NSFC(11125101). W. Long was partially supported by the NSF of Jiangxi Province (20132BAB211004). J. Yang was partially supported by the the excellent doctorial dissertation cultivation grant from Central China Normal University (2013YBZD15). References [1] L. Abdelouhab, J. L. Bona, M. Felland, J.-C. Saut, Nonlocal models for nonlinear, dispersive waves, Phys. D, 40 (1989) 360–392. [2] W. Ao, J. Wei, Infinitely many positive solutions for nonlinear equations with non-symmetric potentials, preprint. [3] X. Cabré, J. Tan, Positive solutions of nonlinear problems involving the square root of the Laplacian, Adv. Math., 224 (2010) 2052–2093. [4] L. Caffarelli, L. Silvestre, An extension problem related to the fractional Laplacian, Comm. Partial Differential Equations, 32(7–9) (2007) 1245–1260. [5] A. Capella, J. Dávila, L. Dupaigne and Y. Sire. Regularity of radial extremal solutions for some non-local semilinear equations, Comm. Partial Differential Equations, 36 (2011) 1353–1384. [6] G. Cerami, G. Devillanova, S. Solimini, Infinitely many bound states for some nonlinear scalar field equations, Calc. Var. Partial Differential Equations, 23 (2005) 139–168. [7] G. Cerami, D. Passaseo, Infinitely many positive solutions to some scalar field equation with nonsymmetric coefficients, to appear in Comm. Pure Appl. Math.. [8] S.-M. Chang, S. Gustafson, K. Nakanishi, T.-P. Tsai, Spectra of linearized operators for NLS solitary waves, SIAM. J. Math. Anal., 39 (2007/08) 1070–1111. [9] W. Chen, C. Li, B. Ou, Classification of solutions for an integral equation, Comm. Pure Appl. Math., 59 (2006) 330–343. [10] G. Chen, Y. Zhang, Concentration phenomenon for fractionsl nonlinear Schrödinger equations, arXiv: 1305.4426. [11] J. Dávila, M. Del Pino, J. Wei, Concentrating standing waves for fractional nonlinear Schrödinger equation, J. Differerntial Equations, 256 (2014) 858–892. [12] G. Devillanova, S. Solimini, Min-max solutions to some scalar field equations, preprint. [13] W. Ding, W.M. Ni, On the existence of positive entire solutions of a semilinear elliptic equation, Arch. Ration. Mech. Anal., 91 (1986) 283–308. [14] P. Felmer, A. Quass, J. Tan, Positive solutions of nonlinear Schrödinger equation with the fractional Laplacian, Proc. Roy. Soc. Edinburgh Sect., A 142(6) (2012) 1237–1262. [15] Rupert Frank, Enno Lenzmann, Uniqueness and nondegeneracy of ground states for (−∆)s Q + Q − Qα+1 = 0 in R, Acta Math., 210 (2013) 261–318. [16] Rupert Frank, Enno Lenzmann, and Luis Silvestre, Uniqueness of radial solutions for the fractional Laplacian, arXiv preprint arXiv:1302.2652 2013. [17] A. Elgart, B. Schlein, Mean field dynamics of boson stars, Comm. Pure Appl. Math., 60 (2007) 500–545.

22


[18] M. K. Kwong, Uniqueness of positive solutions of ∆u − u + up = 0 in Rn , Arch. Ration. Mech. Anal., 105(3) (1989)243–266. [19] A. J. Majda, D. W. McLaughlin, E. G. Tabak, A one-dimensional model for dispersive wave turbulence, J. Nonlinear Sci., 7 (1997) 9–44. [20] N. Laskin, Fractional quantum mechanics and Lévy path integrals, Phys. Lett. A 268 (2000) 29–305. [21] N. Laskin, Fractional Schrödinger equation, Phys. Rev., E 66 (2002) 31–35. [22] E. H. Lieb, H.-T. Yau, The Chandrasekhar theory of stellar collapse as the limit of quantum mechanics, Comm. Math. Phys., 112 (1987) 147–174. [23] P.L. Lions, The concentration-compactness principle in the calculus of variations. The locally compact case. I. Ann. Inst. H. Poincaré Anal. Non Lineaire, 1 (1984) 109–145 . [24] P.L. Lions, The concentration-compactness principle in the calculus of variations. The locally compact case. II. Ann. Inst. H. Poincaré Anal. Non Lineaire, 1 (1984) 223–283. [25] E. Di Nezza, G. Palatucci, E. Valdinoci, Hitchhiker’s guide to the fractional Sobolev spaces, Bull. Sci. Math., 136 (2012) 521–573. [26] S. Secchi, Ground state solutions for nonlinear fractional Schr?dinger equations in RN , J. Math. Phys., 54 (2013), 031501, 17 pp. [27] Y. Sire, E. Valdinoci, Fractional Laplacian phase transitions and boundary reactions: a geometric inequality and a symmetry result, J. Funct. Anal., 256 (2009) 1842–1864. [28] J. Tan, The Brezis-Nirenberg type problem involving the square root of the Laplacian, Calc. Var. Partial Differential Equations, 42 (2011), 21-41. [29] J. Wei, S. Yan, Infinite many positive solutions for the nonlinear Schr¨ odinger equation in Rn , Calc. Var. Partial Differential Equations, 37 (2010) 423–439. [30] J. Wei, S. Yan, Infinite many positive solutions for the prescribed scalar curvature problem on SN , J. Funct. Anal., 258 (2010) 3048–3081. [31] M. I. Weinstein, Existence and dynamic stability of solitary wave solutions of equations arising in long wave propagation, Comm. Partial Differential Equations, 12(1987) 1133–1173. School of Mathematics and Statistics, Central China Normal University, Wuhan, 430079, P. R. China College of Mathematics and Information Science, Jiangxi Normal University, Nanchang, Jiangxi 330022, P. R. China E-mail address: [email protected] School of Mathematics and Statistics, Central China Normal University, Wuhan, 430079, P. R. China E-mail address: [email protected] School of Mathematics and Statistics, Central China Normal University, Wuhan, 430079, P. R. China E-mail address: [email protected] School of Mathematics and Statistics, Central China Normal University, Wuhan, 430079, P. R. China