Boundary bubbling solutions for a supercritical Neumann problem with mixed nonlinearities

Accepted Manuscript Boundary bubbling solutions for a supercritical Neumann problem with mixed nonlinearities

Wenjing Chen

PII: DOI: Reference:

S0022-247X(17)30442-0 http://dx.doi.org/10.1016/j.jmaa.2017.05.005 YJMAA 21366

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Journal of Mathematical Analysis and Applications

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28 February 2017

Please cite this article in press as: W. Chen, Boundary bubbling solutions for a supercritical Neumann problem with mixed nonlinearities, J. Math. Anal. Appl. (2017), http://dx.doi.org/10.1016/j.jmaa.2017.05.005

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BOUNDARY BUBBLING SOLUTIONS FOR A SUPERCRITICAL NEUMANN PROBLEM WITH MIXED NONLINEARITIES WENJING CHEN

Abstract. In this paper, we investigate the existence of boundary bubbling solutions for a supercritical Neumann problem with mixed nonlinearities under a suitable assumption on the mean curvature of the boundary ∂Ω, the behavior of the solutions is as a tower of k bubbles when the exponent goes to Sobolev critical exponent, and the blow-up point is a critical point of the mean curvature.

MR(2010) Subject Classification: 35B33, 35J25 , 35J61. Keywords: Boundary bubbling solutions; Neumann boundary problem; Lyapunov-Schmidt reduction.

1. Introduction In this paper, we are concerned with the following Neumann boundary problem ⎧ −Δu + u = up + uq in Ω, ⎪ ⎪ ⎨ u>0 in Ω, (1.1) ⎪ ⎪ ⎩ ∂u on ∂Ω, ∂ν = 0 where Ω is a smooth bounded domain in RN , N ≥ 4, 1 < q < p, and ν is the outward normal on ∂Ω. If p = q, problem (1.1) reduces to the following problem ⎧ p in Ω, ⎪ ⎪ −Δu + u = u ⎨ u>0 in Ω, (1.2) ⎪ ⎪ ⎩ ∂u on ∂Ω, ∂ν = 0 N +2 where Ω is a smooth bounded domain. If 1 < p < N −2 , (1.2) is subcritical problem with the Neumann boundary condition. It was first studied by Lin, Ni and Takagi [21]. Many other existence results were obtained, see [17], [18], [22], [23] and references therein. N +2 If p = N −2 , (1.2) becomes the critical Neumann boundary problem, the lack of compactness of Sobolev embedding makes it harder to apply variational arguments. In order to find nontrivial solutions, one has to deal with the lack of compactness. In [33], Wang obtained the existence of non-constant least energy solution N +2 of (1.2) for p = N −2 . Many results have been obtained for the critical case, we refer to [7], [20], [29], [30] and references therein. N +2 If p is supercritical in (1.2), namely p > N −2 , Sobolev embedding no longer holds, so that variational construction of solutions becomes difficult. However, del Pino-Musso-Pistoia investigated the nearly supercritical case in [14]. More precisely, they considered ⎧ N +2 ⎨ −Δu + u = u N −2 + in Ω, (1.3) u>0 in Ω, ⎩ ∂u = 0 on ∂Ω, ∂ν

The research has been supported by NSFC (No. 11501468, No. 11501469) and Chongqing Research Program of Basic Research and Frontier Technology cstc2016jcyjA0323. 1

2

WENJING CHEN

where  > 0. Given a non-degenerate critical point ζ0 of the mean curvature H on the boundary of Ω (or, more generally, a situation of topologically non-trivial critical point) with positive critical value, that is, H(ζ0 ) > 0, a solution u exhibiting boundary bubbling around such a point as  → 0 exists u (z) ≈ Uδ ,ζ (z),

with

δ ≈ ,

ζ → ζ0 ,

where Uδ ,ζ satisfies the problem N +2

−Δu = u N −2 ,

u > 0 in RN .

It is well known that the only positive solutions of above equation are given by (1.4)

Uδ,ζ (z) = αN

δ (δ 2

N −2 2

+ |z −

ζ|2 )

N −2 2

,

αN = [N (N − 2)]

N −2 4

, δ > 0, ζ ∈ RN .

See [31]. In [4], it has been found that if N ≥ 4, the exponent in (1.3) approaches the critical exponent from below, then single-bubbling solutions exist with maximum points located on the boundary, near critical points of the mean curvature with negative value. Wang in [32] obtained the existence of arbitrarily many solutions which blow up around some points on the boundary for (1.3) when  goes to zero from below or above. Let us mention some contributions to the question of existence for problem with mixed nonlinearities. Cao, Noussair and Yan [5], [6] studied the existence of multi-peaked solutions to the following Neumann problem ⎧ N +2 ⎪ −Δu + λu = u N −2 + auq in Ω, ⎪ ⎨ (1.5) u>0 in Ω, ⎪ ⎪ ⎩ ∂u on ∂Ω, ∂ν = 0 where Ω is a smooth bounded domain in RN , N ≥ 4, λ > 0 is a large constant, a > 0 is a fixed constant, N +2 ν is the outward normal on ∂Ω, and 1 < q < N −2 . In [5], the authors established that if a = 0, problem (1.5) has no interior peak solution, while under the symmetric assumption that Ω is invariant under some group action of Γ ⊂ O(N ) (the orthogonal group in RN ), satisfying that the orbit of each point except the origin is at least three, then problem (1.5) with a > 0 has an interior single peak solution as λ → ∞. Moreover, the authors in [6] introduced a class of symmetric domains for which problem (1.5) with a > 0 has all of the three types of peaked solution: interior peaked, boundary peaked and interior-boundary peaked. Furthermore, Peng [27] constructed a unique single-peak solution to (1.5) with a > 0 and the domain Ω is assumed to satisfy some symmetry conditions. Next, we recall some results for problems with mixed nonlinearities in RN . Consider the following problem  −Δu + u = up + λuq , u > 0 in RN , (1.6) u(z) → 0 as |z| → ∞. N +2 Existence of solutions to problem (1.6) was proved by Berestycki and Lions [2] for 1 < q < p < N −2 , while N +2 non-existence follows from the Pohozaev identity [28] for p, q ≥ N . Let us mention some contributions −2 to the question of existence for (1.6) when one exponent is subcritical and the other one is critical or supercritical. Alves, de Morais Filho and Souto [1] proved that there exists a nontrivial classical solution N +2 to problem (1.6) for all λ > 0 , 1 < q < N −2 and N ≥ 4. Moreover, Ferrero and Gazzola [15] proved that N +2 ¯ > 0, such that if λ > λ, ¯ then (1.6) has at least one solution, while for for q < N −2 ≤ p, there exists λ N +2 ¯ q < N −2 < p, there exists 0 < λ < λ such that if λ < λ, then there is no solution. Recently, in [9], we N +2 N +2 investigated multiplicity results for (1.6) when p is supercritical and goes to N −2 , 1 < q < N −2 .

BOUNDARY BUBBLING SOLUTIONS

3

Motivated by above works, the aim of this paper is to consider the existence of bubble tower solutions for the following problem ⎧ −Δu + u = up+ + uq in Ω, ⎪ ⎪ ⎨ u>0 in Ω, (1.7) ⎪ ⎪ ⎩ ∂u on ∂Ω, ∂ν = 0 N +2 N where p = N −2 , Ω is a smooth bounded domain in R , N ≥ 4,  is a positive parameter, and q satisfies N 1 < q ≤ N −2 , ν denotes the outward normal on ∂Ω.

Before stating our result, let us recall that for any C 1 function ϕ on M , a critical point x0 ∈ M is called to be C 1 -stable if there is a small neighborhood Λ of x0 in M such that ∇ϕ(x) = 0 for some x ∈ Λ implies x = x0 and deg(∇ϕ, Λ, 0) = 0 (see [19]). Here deg denotes the Brouwer degree. It is known that any isolated local minimum point and maximum point is a C 1 -stable critical point. Moreover, it is a nondegenerate critical point if ϕ is a C 2 -function. For convenience, we define two positive constants, ˆ ˆ 1 N −2 2 |y|2 2 p+1 αN |y| U1,0 (y)dy + dy, c3 = 2N ∂RN 4 (1 + |y|2 )N −1 ∂RN + + ˆ 1 q+1 c4 = U1,0 (y)dy, q + 1 RN + where U1,0 (y) = αN (1 + |y|2 )− (1.8)

N −2 2

. Moreover, set  0, if 1 < q < NN−2 ; Iq = 1, if q = NN−2 .

Our result can be stated as follows. Theorem 1.1. Assume that N ≥ 4, q satisfies 1 < q  has a C 1 -stable critical point ζ0 ∈ ∂Ω with (1.9)

N N −2 ,

and the mean curvature H(ζ) on the boundary

c3 H(ζ0 ) + c4 Iq > 0.

Given an integer k  1, then for all sufficiently small  > 0, there is a family solutions u (x) of (1.7) of the following form k  N2−2 N −2 N −2  1 (1.10) u (z) = αN tj 2  2 +j−1 (1 + o(1)), 4(j−1) 2+ N −2 2 2 tj + |z − ζ | j=1  where o(1) → 0 uniformly in Ω \ {ζ }. Here ζ → ζ0 as  → 0, and the constants tj > 0, j = 1, 2, · · · , k, can be computed explicitly. Remark 1.2. It is worth pointing out that the condition (1.9) is used to guarantee that the location of bubbling solutions of problem (1.7) is around a C 1 -stable critical point of the mean curvature H on the boundary of Ω. When we consider the reduced functional (see (6)), in order to make that the mean curvature plays an important role, we need to assume that q satisfies 1 < q ≤ NN−2 . In fact, by our construction, ´ 1 U q+1 dx in the energy will not appear in the reduced if 1 < q < NN−2 , the expansion of the term q+1 Ω δ,ζ functional, but it appears if q = NN−2 . ´ q+1 Remark 1.3. When the dimension N ≥ 4 and q > 1, we have Ω U1,0 dx < ∞. However, for dimension N = 3, this integration is not finite. Some further technical difficulties arises. Rey and Wei in [29] studied problem (1.3) for N = 3. They obtained the existence of solutions, which blow up at an interior point of the domain. Moreover, del Pino-Musso-Pistoia in [14] proved the existence of bubble tower solutions of (1.3), 1 1 which concentrates at a point of the boundary of the domain with blow-up order  2 −j | log | 2 for N = 3 and j = 1, · · · , k. Therefore, we believe that problem (1.7) exhibits a similar phenomenon about the existence of

4

WENJING CHEN

bubble tower solutions for N = 3 with rates modified and q in the suitable range, but we will not develop this point in this paper. The proof of Theorem 1.1 relies on a very well known Lyapunov-Schmidt reduction procedure. Bubbletower solutions were found by del Pino, Dolbeault and Musso [11] for a slightly supercritical Brezis-Nirenberg problem in a ball, and after that have been studied intensively [3, 8–10, 12–14, 16, 24–26]. The paper is organized as follows: We give some preliminaries in Section 2, where we straight the boundary and describe the first approximation solution to problem (1.7). We introduce a polar coordinates around a reference point ζ ∈ ∂Ω, and then a transformation of the radial coordinate, a variation of the well known Emden-Fowler transformation after which dilations are converted into translations in a onedimensional variable. We deal with the linear theory and nonlinear problem in Section 3 and Section 4 respectively. Section 5 is devoted to study the variational reduction, we complete the proof of Theorem 1.1 in Section 6. 2. Some Preliminaries 2.1. Straighten the boundary. Fix ζ ∈ ∂Ω, we introduce the boundary deformation that straightens the boundary around ζ. After a rotation and translation of variables, we may assume that on a neighborhood of point ζ the domain can be described as the set of points z = (z  , zN ) with z  = (z1 , · · · , zN −1 ) ∈ RN −1 . We assume that the inward normal to ∂Ω at ζ is the direction of the positive zN −axis. Write ζ = (ζ  , ζN ) with ζ  = (ζ1 , · · · , ζN −1 ) ∈ RN −1 . Set B  (r) = {z  ∈ RN −1 : |z  − ζ  | < r}, and B(ζ, r) = {z ∈ RN : |z − ζ| < r}. Since ∂Ω is smooth, there exists a constant r > 0 such that ∂Ω ∩ B(ζ, r) can be represented by the graph of a smooth function g : B  (r) → R, with g(ζ) = 0, ∇g(ζ) = 0, and Ω ∩ B(ζ, r) = {(z  , zN ) ∈ B(ζ, r) : zN − ζN > g(z  − ζ  )}.

(2.1)

By the Taylor expansion, we have (2.2)

g(z  − ζ  ) =

N −1 N −1 1  ∂2g 1  (ζ)(zi − ζi )(zj − ζj ) + O(|z − ζ|3 ) = κi (zi − ζi )2 + O(|z − ζ|3 ), 2 i=1 ∂zi ∂zj 2 i=1

where κi , i = 1, · · · , N − 1, are the principle curvatures at ζ. The average of the principle curvatures of ∂Ω

N −1 at ζ is the mean curvature H(ζ) = N 1−1 i=1 κi . Moreover, on ∂Ω ∩ B(ζ, r), we write the normal derivative ν(z) as ν(z) = √ 1  2 (∇ g, −1) and the tangential derivatives are given by 1+|∇ g|

1 ∂ = ∂τi,z 1 + |∂g/∂zi |2

0, . . . , 1, . . . ,

∂g ∂zi

,

i = 1, · · · , N − 1.

2.2. The first approximation solutions. Fix ζ ∈ ∂Ω, we recall that the function Uδ,ζ is defined in (1.4), δ,ζ be the in order to satisfy the Neumann boundary condition, we project it into H 1 (Ω). Namely, let U unique solution of the following problem N +2

(2.3)

δ,ζ = U N −2 in Ω; δ,ζ + U −ΔU δ,ζ

δ,ζ ∂U = 0 on ∂Ω. ∂ν

δ,ζ . We write We next analyze the behavior of function U (2.4)

δ,ζ = Uδ,ζ − ϕδ,ζ . U

In order to estimate ϕδ,ζ , we introduce the following function ϕ0 , which was introduced in [30]. Let ϕ0 satisfy ⎧  in RN −Δϕ0 = 0 ⎪ + = {(z , zN ) : zN > 0}; ⎪ ⎨ N −1 2 ∂ϕ0 N −2 i=1 κi zi (2.5) on RN N +; ∂zN = αN 2 ⎪ 2 2 (1+|z| ) ⎪ ⎩ ϕ0 (z) → 0 as |z| → +∞.

BOUNDARY BUBBLING SOLUTIONS

5

Using Green’s representation, (2.6)

ˆ N −1 N −2 1  yi2 1 κi dy  , ϕ0 (z) = αN N  |N −2  2 2 ωN −1 i=1 |z − y N −1 2 (1 + |y | ) R

where ωN −1 denotes the measure of the unit sphere in RN . (2.6) implies that C C C (2.7) , |∇ϕ0 (z)| ≤ , and |D2 ϕ0 | ≤ . |ϕ0 (z)| ≤ (1 + |z|)N −3 (1 + |z|)N −2 (1 + |z|)N −1 By the results of [14, Section 2] and [30, Lemma A.1], we have the following asymptotic behaviors. Lemma 2.1. (a) Assume δ > 0 small, for N ≥ 4, we have (2.8)

−3 δ,ζ (z) = Uδ,ζ (z) + δ 12 O(Uδ,ζ (z)) N N −2 , U

|z − ζ| < r0 ,

and N −2

δ 2 , |z − ζ| > r0 , |z − ζ|N −2 with r0 > 0 small but fixed number. In particular, there are positive constants α, β, such that δ,ζ (z) ≤ βUδ,ζ (z). αUδ,ζ (z) ≤ U (2.10) (2.9)

δ,ζ (z) ≤ Cδ N 2−2 G(z, ζ) ≤ C U

Moreover, δ,ζ (z) = δ −1 ∂τ Uδ,ζ (z) + O(Uδ,ζ (z)) ∂τ U

for |z − ζ| < r0 ,

δ,ζ (z) = δ −1 ∂δ Uδ,ζ (z) + O(Uδ,ζ (z)) ∂δ U

for |z − ζ| < r0 ,

δ,ζ (z)| = O(δ δ,ζ (z)| = O(δ N2−2 ). and outside the ball, we have |∂τ U ) and |∂δ U (b) For N ≥ 4, we have the expansion z − ζ  6−N 4−N (2.11) + O δ 2 | ln δ|ϑ , ϕδ,ζ (z) = δ 2 ϕ0 δ where ϑ = 1 for N = 4 and ϑ = 0 for N ≥ 5, ϕ0 satisfies (2.5). Moreover, N −2 2

N −2

δ 2 | ln δ|ϑ (2.12) , |ϕδ,ζ (z)| ≤ C (δ + |z − ζ|)N −3 where ϑ = 1 for N = 4, 5, and ϑ = 0 for N ≥ 6. We now make a transformation of problem (1.7). Let ζ be a point on ∂Ω. We consider spherical coordinates y = y(ρ, θ) centered at ζ given by y−ζ (2.13) . ρ = |y − ζ|, and θ = |y − ζ| Define the transformation 2  p − 1 p−1  2 v(x, θ) = T (u(x, θ)) = e−x u ζ + e− N −2 x θ . 2 We denote D by the ζ−dependent subset of S = R × S N −1 where the variables (x, θ) vary. Using this transformation, finding a radial solution u(z) to problem (1.7) corresponds to that of solving the following problem ⎧ L (v) = α ex v p+ + βN e−(p−q)x v q in D; ⎪ ⎪ ⎨ 0 v>0 in D; (2.14) ⎪ ⎪   ⎩ p−1 ∂v x ∇θ v · ν θ + ∂x ν + vν x = 0 on ∂D, 2 where (2.15)

L0 (v) = −

 p − 1 2  p − 1 2 −(p−1)x ΔS N −1 v − v  + v + e v 2 2

6

WENJING CHEN

is the transformed operator associated to −Δ + Id, and 2  p − 1 − p−1  p − 1  2(p−q) p−1 , βN = . α = 2 2 Here and in what follows,  = that

∂ ∂x

and ΔS N −1 denotes the Laplace-Beltrami operator on S N −1 . We observe T (Uδ,ζ )(x, θ) = W (x − ξ),

where δ = e

− N 2−2 ξ

and

 4N N4−2  − N2−2 4 e−x 1 + e− N −2 x , N −2 which is the unique solution of the problem ⎧ W  − W + W p = 0 in (−∞, +∞); ⎪ ⎪ ⎨ W  (0) = 0; (2.16) ⎪ ⎪ ⎩ W > 0, W (x) → 0 as x → ±∞. W (x) =

Let us define

δ,ζ ), Vξ,ζ = T (U

2

δ = e− N −2 ξ .

Then v = Vξ,ζ solves the problem  L0 (v) = W (x − ξ)p in D; (2.17)  p−1  ∂v x θ x ∇θ v · ν + ∂x ν + vν = 0 on ∂D. 2 We write Vξ,ζ = W (x − ξ) + Πξ,ζ .

(2.18)

By the previous transformation and as a consequence of Lemma 2.1, we have the following estimates. Lemma 2.2. For N  4, there are positive constants C, C1 , C2 such that C1 W (x − ξ) ≤ Vξ,ζ (x, θ) ≤ C2 W (x − ξ),

(2.19) and

|Πξ,ζ | ≤ Ce−

(2.20)

2 min{x,ξ} N −2

W (x − ξ).

Let η0 > 0 be a small but fixed number. Given an integer number k, let tj , for j = 1, 2, · · · , k, be positive numbers and satisfy 1 η 0 < tj < . (2.21) η0 For 1 < q ≤

N N −2 ,

set δj = 

(2.22)

2(j−1) N −2 +1

j = 1, 2, · · · , k.

tj ,

Observe that 2 δj+1 tj+1 =  N −2 , δj tj

j = 1, · · · , k − 1.

Then let us define k points ξ1 , ξ2 , · · · , ξk in R, satisfying 2

δj = e− N −2 ξj , That is, (2.23)



ξ1 = − N 2−2 log  −

N −2 2

ξj − ξj−1 = − log  −

j = 1, · · · , k.

log t1 ,

N −2 2

log

tj tj−1 ,

j = 2, · · · , k.

BOUNDARY BUBBLING SOLUTIONS

7

Set (2.24)

Wj = W (x − ξj ),

Πj = Πξj ,ζ ,

Vj = Wj + Πj ,

V =

k 

Vj .

j=1

Looking for a solution of (1.7) of the form (2.25)

u(x) = U (x) + ψ(x)

with

U (x) =

k 

δ ,ζ (x), U j

j=1

corresponds to find a solution of (2.14) of the form v = V + φ, where V is given by (2.24) and φ = T (ψ) is a small term. Thus problem (2.14) becomes ⎧ L (φ) = N (φ) + E in D; ⎪ ⎪ ⎨ φ>0 in D; (2.26) ⎪ ⎪ ⎩  p−1  ∂φ ∇θ φ · ν θ + ∂x ν x + φν x = 0 on ∂D, 2 where L (φ) =L0 (φ) − α (p + )ex V p+−1 φ − qβN e−(p−q)x V q−1 φ,   N (φ) =α ex (V + φ)p+ − V p+ − (p + )V p+−1 φ   + βN e−(p−q)x (V + φ)q − V q − qV q−1 φ , and E = α ex V p+ − L0 (V ) + βN e−(p−q)x V q = α ex V p+ −

k 

Wjp + βN e−(p−q)x V q .

j=1

3. The linear theory In this section, we solve problem (2.26), in order to do this, we first consider the following problem: given ζ ∈ ∂Ω and points ξ = (ξ1 , · · · , ξk ), finding a function φ such that for certain constants cij , i = 1, 2, · · · , N , j = 1, 2, · · · , k, ⎧ N

k

⎪ ⎪ L (φ) = N (φ) + E + cij Vjp−1 Zij in D; ⎪ ⎨ i=1 j=1   p−1 (3.1) x x ∇θ φ · ν θ + ∂φ on ∂D, ⎪ 2 ∂x ν + φν = 0 ⎪ ⎪ ⎩´ p−1 V Zij φ = 0 for all i = 1, · · · , N, j = 1, · · · , k, D j ∂ where Vj is given in (2.24), ZN j = ∂x Vj (x, θ), and Zij = δj T (∂τi Uδj ,ζ ). for i = 1, · · · , N − 1, j = 1, 2, · · · , k. To solve (3.1), it is important to understand its linear part, thus we consider the following problem: given a function h, finding φ such that ⎧ N

k

p−1 ⎪ ⎪ in D; ⎪ ⎨ L (φ) = h + i=1 j=1 cij Vj Zij   p−1 (3.2) x x ∇θ φ · ν θ + ∂φ on ∂D, ⎪ 2 ∂x ν + φν = 0 ⎪ ⎪ ⎩´ V p−1 Zij φ = 0 for all i = 1, · · · , N, j = 1, · · · , k, D j

for certain constants cij . To solve problem (3.2), we now analyze invertibility properties of the operator L under the orthogonality conditions. We introduce the following norm which depends on the given points ξ1 , · · · , ξk and small fixed number 0 < σ < min{1, q − 1, 3q−p 2 }. For a function g defined on D, we set (3.3)

g ∗ = sup

(x,θ)∈D

k  j=1

e−σ|x−ξj |

−1

|g(x, θ)|.

8

WENJING CHEN

In the following, for any vector space X of functions u(x, θ) defined on D, we shall denote by Xs the subspace of functions in X which are even with respect to the first N − 1 variables of θ. The next result is solvability of problem (3.2). Proposition 3.1. There exist positive numbers 0 and C such that if the points 0 < ξ1 < ξ2 < · · · < ξk satisfy (2.23), then for all 0 <  < 0 and all functions h ∈ Cs0,α (D) with h ∗ < +∞, problem (3.2) has a unique solution φ = T (h). Moreover,

φ ∗ ≤ C h ∗

(3.4)

and

|cij | ≤ C h ∗ .

Observe that L (φ) = L1, (φ) − qβN e−(p−q)x V q−1 φ, where L1, (φ) = L0 (φ) − α (p + )ex V p+−1 φ, and L0 is given by (2.15). To prove Proposition 3.1, we first analyze invertibility properties of the operator L1, . We have the following result, whose proof can be seen in [14]. Lemma 3.2. Under the assumptions of Proposition 3.1, suppose φ and h1 satisfy the following problem ⎧ k N

⎪ x p+−1 ⎪ L (φ) := L (φ) − α (p + )e V φ = h + cij Vjp−1 Zij in D; 1, 0  1 ⎪ ⎨ i=1 j=1   p−1 (3.5) x x ∇θ φ · ν θ + ∂φ on ∂D, ⎪ 2 ∂x ν + φν = 0 ⎪ ⎪ ⎩´ p−1 V Zij φ = 0 for all i = 1, · · · , N, j = 1, · · · , k. D j Then

φ ∗ ≤ C h1 ∗

and

|cij | ≤ C h1 ∗

for some positive constant C. Proof of Proposition 3.1. Suppose φ and h satisfy problem (3.2), then φ and h1 satisfies problem (3.5) with h1 = h + qβN e−(p−q)x V q−1 φ. From Lemma 3.2, we then have 

φ ∗ ≤ C h ∗ + e−(p−q)x V q−1 φ ∗ , and

 |cij | ≤ C h ∗ + e−(p−q)x V q−1 φ ∗ .

In order to establish (3.4), it is sufficient to show that

e−(p−q)x V q−1 φ ∗ ≤ o(1) φ ∗ .

(3.6) Indeed, (3.7)

e−(p−q)x V q−1 φ ∗ ≤ sup

(x,θ)∈D

k 

e−σ|x−ξj |

 −1   −(p−q)x q−1  V φ . e

j=1

If x ≤ ξ1 , then we have e

−(p−q)x

V

q−1

≤

k 

e−(p−q)x e−(q−1)|x−ξj | ≤ Ce(2q−p−1)x e−(q−1)ξ1

j=1

  ≤C max e−(p−q)ξ1 , e−(q−1)ξ1 . If x ≥ ξ1 , then we obtain e−(p−q)x V q−1 ≤

k  j=1

e−(p−q)x e−(q−1)|x−ξj | ≤ Ce−(p−q)x ≤ Ce−(p−q)ξ1 .

BOUNDARY BUBBLING SOLUTIONS

It follows that

9

 

e−(p−q)x V q−1 φ ∗ ≤ C max e−(p−q)ξ1 , e−(q−1)ξ1 φ ∗ = o(1) φ ∗ .

Thus the estimate (3.6) holds. We now prove the existence and uniqueness of solution to (3.2). Consider the Hilbert space   ˆ p−1 1 H = φ ∈ Hs (D) : Vj Zij φ = 0, ∀ i = 1, · · · , N, j = 1, 2, · · · , k. D

with inner product

 p − 1 2 ˆ

ˆ

∇φ∇ψ + φψ. 2 D D Then problem (3.2) is equivalent to find φ ∈ H such that ˆ   2 2  4 α (p + )ex V p+−1 φ + qβN e−(p−q)x V q−1 φ + φ, ψ = (3.8) e− N −2 x φ + h ψdx, N −2 D φ, ψ =

∇θ φ∇θ ψ +

for all ψ ∈ H. By the Riesz representation theorem, (3.8) is equivalent to solve (3.9)

˜ φ = K(φ) + h

˜ ∈ H depending linearly on h, and where K : H → H being a compact operator. Fredholm’s with h alternative yields there is a unique solution to problem (3.9) for any h provided that (3.10)

φ = K(φ)

has only the zero solution in H. (3.10) is equivalent to problem (3.2) with h = 0. If h = 0, estimate (3.4) implies that φ = 0. We complete the proof of Proposition 3.1. Now we study the differentiability of the operator T with respect to ξ = (ξ1 , · · · , ξk ) and ζ. Consider the Banach space C∗ of all continuous functions f for which f ∗ < +∞ and are symmetric with respect to θ1 , . . . , θN −1 , endowed with this norm. The following result holds. Proposition 3.3. Under the assumptions of Proposition 3.1, the map (ξ, ζ) → T is of class C 2 (C∗ ). Moreover, there is a constant C > 0 such that

∂ξ T (h) ∗ ≤ C h ∗ ,

2

∂ξξ T (h) ∗ ≤ C h ∗ ,

∂ζ T (h) ∗ ≤ C h ∗ ,

2

∂ζζ T (h) ∗ ≤ C h ∗ ,

and

uniformly on the vectors ξ which satisfy (2.23). Proof. The argument is similar to the proof in [14] and some references therein. So we omit the proof here. 4. Nonlinear Problem In this section, our purpose is to study nonlinear problem. We first have the validity of the following result. Lemma 4.1. We have (4.1) and (4.2) for φ ∗ ≤ 1.

 min{p,2}

N (φ) ∗ ≤ C φ ∗ + φ q∗ ,  min{p−1,1}

∂φ N (φ) ∗ ≤ C φ ∗ , + φ q−1 ∗

10

WENJING CHEN

Proof. By the fundamental theorem of calculus and the definition of ∗ , we have  ˆ 1 k  −1     −σ|x−ξj | x  p+−1 p+−1

N (φ) ∗ ≤α (p + ) sup (V + tφ) φ dt e e  −V x∈D

+ qβN sup

k 

x∈D

0

j=1

e

−σ|x−ξj |

−1

e

−(p−q)x

1



(V + tφ)

q−1

0

j=1

Using the following inequality,



m

ˆ   

m

||a + b| − |a| | ≤ C

−V

q−1

|a|m−1 |b| + |b|m

if m ≥ 1;

min{|a|m−1 |b|, |b|m }

if 0 < m < 1.



  φ dt := N1 + N2 .

If p ≥ 2 and for φ ∗ ≤ 1, we have k k  −1  −1 N1 ≤C sup e−σ|x−ξj | ex V p+−2 |φ|2 + C sup e−σ|x−ξj | ex |φ|p+ x∈D

≤C φ 2∗

x∈D

j=1

+

C φ p+ ∗

j=1

C φ 2∗ .

≤

Similarly, if 1 < p < 2, we find that N1 ≤ C φ p∗ . Thus we get min{p,2}

N1 ≤ C φ ∗

.

Moreover, by similar computations as N1 , we can conclude that min{q,2}

N2 ≤ C φ ∗

= φ q∗

for 1 < q ≤ NN−2 . Thus we get (4.1). We differentiate N (φ) with respect to φ, we have     ∂φ N (φ) = α (p + )ex (V + φ)p+−1 − V p+−1 + βN qe−(p−q)x (V + φ)q−1 − V q−1 . By a similar argument as N (φ) ∗ , (4.2) holds. Lemma 4.2. Let 0 < ξ1 < ξ2 < · · · < ξk satisfy (2.23). Then there exist τ ∈ ( 12 , 1) and a constant C > 0, such that

E ∗ ≤ Cτ , and

∂ζ E ∗ ≤ Cτ ,

∂ξi E ∗ ≤ Cτ , i = 1, 2, · · · , k.

Proof. We have E =α e

x



V

p+

−V

p



+ (α e

x

p



p

− 1)V + V −

k 

p Wj

j=1

+

k  

p Wj

−

j=1

(4.3)

k 

Wjp + βN e−(p−q)x V q

j=1

:=E1 + E2 + E3 + E4 + E5 .

We next estimate Ei , for i = 1, · · · , 5, respectively. Estimate of E1 :   ˆ 1 k    V p+t log V dt ≤ C e−σ|x−ξj | . |E1 | = α ex 0

j=1

Estimate of E2 : By the Taylor expansion, we have k 2  p − 1 − p−1   ˆ 1    x |E2 | =  e − 1 V p  = x etx dt + O()ex V p ≤ C| log | e−σ|x−ξj | . 2 0 j=1

BOUNDARY BUBBLING SOLUTIONS

11

Estimate of E3 : Since k k   p     Wj  ≤ CV p−1 |Πj (x)|. |E3 | = V p − j=1

j=1

Thanks to Lemma 2.2, if 0 < x ≤ ξ1 , |E3 | ≤ CV p−1

k 

2

e−|x−ξj | e− N −2 min{x,ξj } ≤ C

j=1

If x ≥ ξ1 , for 0 < σ < min{1, q − 1, |E3 | ≤ CV p−1

k 

k 

e−σ|x−ξj | .

j=1

3q−p 2 },

we have 2

2

e−|x−ξj | e− N −2 min{x,ξj } ≤ CV p−1 e− N −2 ξ1 ≤ C

j=1

k 

e−σ|x−ξj | .

j=1

Therefore, for (x, θ) ∈ D, we get |E3 | ≤ C

k 

e−σ|x−ξj | .

j=1

Estimate of E4 : If 0 < x ≤

ξ1 +ξ2 2 ,

we have

k k      p     W (x − ξj ) − W (x − ξ1 )p  +  W (x − ξj )p  |E4 | ≤ j=1 k 

≤p

j=2

W (x − ξj )

k p−1 

j=1 k 

=p

W (x − ξj ) +

j=2

W (x − ξj )

W (x − ξj )p

j=2

k p−1−ϑ0  

j=1

k 

W (x − ξj )

k ϑ 0 

j=1

W (x − ξj ) +

k 

j=2

W (x − ξj )p

j=2

with a positive number ϑ0 , satisfying 0 < ϑ0 < p − 1 − σ. Note that k 

W (x − ξj )

k ϑ0 

j=1

W (x − ξj ) ≤ C

1+ϑ0 2

.

j=2

Moreover, k 

W (x − ξj )p ≤ C

p−σ 2

j=2

k 

e−σ|x−ξj | .

j=1

Thus for 0 < ϑ0 < p − 1 − σ, we find that |E4 | ≤ C

1+ϑ0 2

k 

e−σ|x−ξj | ,

for 0 < x ≤

j=1

Similarly, for

ξl−1 +ξl 2

≤x≤

ξl +ξl+1 2

with l = 2, · · · , k − 1, and x ≥ |E4 | ≤ C

1+ϑ0 2

k 

ξ1 + ξ2 . 2

ξk−1 +ξk , 2

we deduce that

e−σ|x−ξj | .

j=1

Therefore for (x, θ) ∈ D, we have |E4 | ≤ C

1+ϑ0 2

k  j=1

e−σ|x−ξj | ,

with

0 < ϑ0 < p − 1 − σ.

12

WENJING CHEN

Estimate of E5 : the estimate of E5 is similar to the previous ones and we get q−σ

|E5 | ≤ C max{,  p−q }

k 

e−σ|x−ξj | .

j=1

From (4.3) and the previous estimates, for 0 < ϑ0 < p − 1 − σ with 0 < σ < min{1, q − 1, 3q−p 2 } small but fixed, we find that there exists τ ∈ ( 12 , 1) such that

E ∗ ≤ Cτ .

Differentiating E with respect to ζ and ξi (i = 1, 2, · · · , k), let us denote ∂s by the differentiate with respect to ζ and ξi , i = 1, 2, · · · , k, we have x

∂s E =α (p + )e V

p+−1

∂s V − p

k 

W (x − ξj )p−1 ∂s W (x − ξj ) + βN qe−(p−q)x V q−1 ∂s V.

j=1

The proof of estimate ∂s E ∗ is similar to that of E ∗ . Proposition 4.3. Assume that 0 < ξ1 < ξ2 < · · · < ξk satisfy (2.23). Then there exists C > 0 such that for  > 0 small enough, there exists a unique solution φ = φ(ξ, ζ) to problem (3.1) with

φ ∗ ≤ Cτ , for some τ ∈ ( 12 , 1) satisfying Lemma 4.2. Moreover, the map (ξ, ζ) → φ(ξ, ζ) is of class C 1 for the · ∗ norm, and

∂ζ φ ∗ ≤ Cτ ,

∂ξj φ ∗ ≤ Cτ , j = 1, · · · , k.

Proof. Problem (3.1) is equivalent to solve a fixed point problem φ = T (N (φ) + E) := A (φ). We will show that the operator A is a contraction map in a proper region. Set Fγ = {φ ∈ Cs (D) : φ ∗ ≤ γτ }, where γ > 0 will be chosen later. For φ ∈ Fγ , by Lemmas 4.1 and 4.2, we get



A (φ) ∗ = T (N (φ) + E) ∗ ≤ C N (φ) ∗ + E ∗ ≤ C γ min{p,2} min{p−1,1}τ + γ q (q−1)τ + 1 τ .

Then we have A (φ) ∈ Fγ for φ ∈ Fγ by choosing γ large enough but fixed. Moreover, for φ1 , φ2 ∈ Fγ , writing ˆ 1 N (φ1 ) − N (φ2 ) = N  (φ2 + t(φ1 − φ2 ))dt(φ1 − φ2 ). 0

By Proposition 3.1 and using (4.2), we find

A (φ1 ) − A (φ2 ) ∗ ≤ C N (φ1 ) − N (φ2 ) ∗ min{p−1,1}  q−1 

φ1 − φ2 ∗ ≤ Cκ φ1 − φ2 ∗ ≤C max φi ∗ + max φi ∗ i=1,2

i=1,2

with some κ > 0. This implies that A is a contraction map from Fγ to Fγ . Thus A has a unique fixed point in Fγ . Now we consider the differentiability of (ξ, ζ) → φ(ξ, ζ). We write B((ξ, ζ), φ) := φ − T (N (φ) + E). First we observe that B((ξ, ζ), φ) = 0. Moreover, ∂φ B((ξ, ζ), φ)[ψ] = ψ − T (ψ(∂φ (N (φ)))) ≡ ψ + M (ψ), where M (ψ) = −T (ψ(∂φ (N (φ)))). By a direct calculation, we get

M (ψ) ∗ ≤ C ψ(∂φ (N (φ))) ∗ ≤ Cκ ψ ∗ .

BOUNDARY BUBBLING SOLUTIONS

13

So for  > 0 small enough, the operator ∂φ B((ξ, ζ), φ) is invertible with uniformly bounded inverse in · ∗ . It also depends continuously on its parameters. We denote ∂s by the differentiate with respect to ζ and ξj , j ∈ {1, 2, · · · , k}, we have ∂s B((ξ, ζ), φ) = −(∂s T )(N (φ) + E) − T ((∂s N )((ξ, ζ), φ) + ∂s E), where all these expressions depend continuously on their parameters. The implicit function theorem yields that φ(ξ, ζ) is a C 1 function into C∗ . Moreover, ∂s φ = −(∂φ B((ξ, ζ), φ))−1 [∂s B((ξ, ζ), φ)] so that

∂s φ ∗ ≤ C ( N (φ) ∗ + E ∗ + (∂s N )((ξ, ζ), φ) ∗ + ∂s E ∗ ) ≤ Cτ . 5. The finite-dimensional variational reduction According to the results in the previous section, our problem has been reduced to find points (ξ, ζ) with ξ = (ξ1 , ξ2 , · · · , ξk ), such that for all i = 1, · · · , N, j = 1, · · · , k.

k  If (5.1) holds, then v = V + φ is a solution to (2.14), and u = j=1 U μj + ψ is the solution to problem (1.7) with ψ = T −1 (φ). Define the function I : (R+ )k × ∂Ω → R as (5.1)

cij (ξ, ζ) = 0

I (ξ, ζ) := I (V + φ), where V is defined by (2.24) and I is the energy functional of (2.14) defined by ˆ ˆ ˆ 1  p − 1 2 1 1  p − 1 2 2  2 |∇θ v| + (v + v) + e−(p−1)x v 2 I (v) = 2 2 2 D 2 2 D D ˆ ˆ 1 1 α βN − ex |v|p++1 dx − e−(p−q)x |v|q+1 dx. p++1 q+1 D D Let us write (5.2)

1 J (u) = 2

ˆ

1 (|∇u| + u )dz − p + 1+ Ω 2

2

ˆ Ω

|u|

p+1+

1 dz − q+1

ˆ Ω

|u|q+1 dz,

which is the energy functional to problem (1.7). Then we have the identity (5.3)

I (v) = αN J (u),

v = T (u),

αN > 0.

We have the following result. Lemma 5.1. The function V + φ is a solution to (2.14) if and only if (ξ, ζ) is a critical point of I (ξ, ζ), where φ = φ(ξ, ζ) is given by Proposition 4.3. Proof. Let us denote ∂s by the differentiate with respect to ζ and ξj , j ∈ {1, 2, · · · , k}, we have ∂s I (ξ, ζ) =∂s (I (V + φ)) = DI (V + φ)[∂s V + ∂s φ]

ˆ ˆ k k N  N    = cij Vjp−1 Zij [∂s V + ∂s φ] = cij Vjp−1 Zij ∂s V dx + o(1) , i=1 j=1

D

i=1 j=1

D

where o(1) → 0 as  → 0 uniformly for the norm · ∗ . We note that the above relations define an almost diagonal homogeneous linear equation system for the cij . Thus (ξ, ζ) = (ξ1 , · · · , ξk , ζ) is the critical point of I if and only if cij = 0 for all i = 1, · · · , N, j = 1, 2, · · · , k. Lemma 5.2. The following expansion holds I (ξ, ζ) = I (V ) + o() 1

as  → 0, where o() is uniform in the C -sense on the vectors ξ satisfying (2.23).

14

WENJING CHEN

Proof. By the fact that DI (V + φ)[φ] = 0 and using the Taylor expansion, we have ˆ 1 D2 I (V + tφ)[φ2 ]tdt I (ξ, ζ) − I (V ) = I (V + φ) − I (V ) = ˆ

tdt

=

(N (φ) + E)φdx + (p + )α

D 1

0

ˆ

+ βN q

0 1

ˆ

ˆ

1

ˆ tdt

0

e

−(p−q)x



V

q−1

D

ˆ

tdt D

0

− (V + tφ)

q−1

  ex V p+−1 − (V + tφ)p+−1 φ2 dx



φ2 dx.

Since φ ∗ ≤ Cτ and E ∗ ≤ Cτ with τ > 12 , we get I (ξ, ζ) − I (V ) = O(2τ ) = o() uniformly on the points ξ which satisfy (2.23). Moreover, let us recall that ∂s denotes the differentiate with respect to ζ and ξj , j = 1, 2, · · · , k, we have ˆ 1ˆ ∂s [(N (φ) + E)φ]tdxdt ∂s (I (ξ, ζ) − I (V )) = 0

D

ˆ

+ α (p + ) ˆ + βN q

1

ˆ

1

tdt D

0

ˆ

tdt 0

D τ

ex ∂s



e−(p−q)x ∂s

  V p+−1 − (V + tφ)p+−1 φ2 dx



  V q−1 − (V + tφ)q−1 φ2 dx.

By the fact that ∂s φ ∗ ≤ C and ∂s E ∗ ≤ C with τ > 12 , we deduce that τ

∂s (I (ξ, ζ) − I (V )) = O(2τ ) = o(). Now we consider the energy functional of problem (1.7), which is defined by (5.2). We have the following result. Proposition 5.3. Assume that (2.21) and (2.22) hold, and 1 < q ≤ expansion: (5.4)

then we have the following

J (U ) = c0 − c1  + c2  log  − ϕ(ζ, t1 , · · · , tk ) + o(),

where 

(5.5)

N N −2 ,



ϕ(ζ, t1 , · · · , tk ) = c3 H(ζ) + c4 Iq t1 − c5

k  j=1

log tj + c6

k−1  j=1

tj+1 tj

N2−2 ,

and as  → 0, o() is uniform in the C 1 -sense on the tj ’s satisfying (2.21), and ˆ k p+1 U1,0 (y)dy, c0 = N RN +

ˆ  1 ˆ  1 p+1 p+1 U1,0 (y) ln(U1,0 (y))dy − U (y)dy , c1 = k 1,0 p + 1 RN (p + 1)2 RN + + k ˆ (N − 2)2   2(j − 1) p+1 +1 c2 = U1,0 (y)dy, N 4N N − 2 R + j=1 ˆ ˆ 1 |y|2 p+1 2 N −2 c3 = |y|2 U1,0 (y)dy + αN dy, 2N ∂RN 4 (1 + |y|2 )N −1 ∂RN + + ˆ 1 q+1 c4 = U1,0 (y)dy, q + 1 RN + ˆ (N − 2)2 p+1 U1,0 (y)dy, c5 = 4N RN +

BOUNDARY BUBBLING SOLUTIONS p+1 c6 = αN

ˆ

1

RN +

1

(1 + |y|2 )

N +2 2

|y|N −2

15

dy.

Proof. We rewrite

ˆ ˆ ˆ   1 1 1 2 2 p+1+ J (U ) = |∇U | + U dz − |U | dz − |U |q+1 dz 2 Ω p+1+ Ω q+1 Ω ˆ ˆ ˆ   1 1 1 |∇U |2 + U 2 dz − |U |p+1 dz − |U |q+1 dz = 2 Ω p+1 Ω q+1 Ω      I II ˆ ˆ 1 1 p+1+ + |U |p+1 dz − |U | dz p+1 Ω p+1+ Ω    III

(5.6)

:=I + II + III.

We estimate each term in (5.6), and obtain the following claims: N −2 ˆ ˆ k−1  δj+1 2 1 1 k p+1 p+1 U1,0 (y)dy − αN dy I= N +2 N 2 N δj (1 + |y| ) 2 |y| −2 j=1 RN +

− δ1 H(ζ)



2 αN

N −2 4

RN +

ˆ ∂RN +

1 |y|2 dy + (1 + |y|2 )N −1 2N

ˆ ∂RN +

p+1 U1,0 |y|2 dy + o(),

(5.7) N −2 2

II = −δ1

(5.8)

N +2 (N −2 −q )

1 q+1

ˆ RN +

q+1 U1,0 (y)dy + o() =



− c4 Iq t1 + o(1) ,

and ˆ k  (N − 2)2 p+1 U1,0 (y)dy ln δj 4N RN + j=1   ˆ ˆ 1 1 p+1 p+1 −k U1,0 (y) ln(U1,0 (y))dy − U1,0 (y)dy  + o(). p + 1 RN (p + 1)2 RN + +

III =

(5.9)

From (5.7)-(5.9), then we get (5.4). Proof of (5.7): Recall that U =

k

δ ,ζ , we set U j = U δ ,ζ , Uj = Uδ ,ζ , and ϕj = ϕδ ,ζ . Using the fact that U j satisfies U j j j j

j=1

(2.3), we find ! ˆ ˆ k  1 1 p p+1  I= U U Uj dz − dz 2 Ω j p+1 Ω j j=1 1 − p+1 (5.10)

ˆ    k Ω

j=1

j U

p+1

−

k  j=1

 p+1 − (p + 1) U j



 ˆ  p p   − Up U j dz Ui Uj dz − U i i

i>j

=I1 + I2 + I3 .

We estimate I1 , I2 , I3 respectively. For I1 , by Lemma 2.1 (b), we note that ! ˆ ˆ k  1 1 p+1 p I1 = U dz + U ϕj dz + o(). N Ω j 2 Ω j j=1

i>j

Ω

16

WENJING CHEN

Since ˆ Ω

Ujp+1 (z)dz

ˆ

Ujp+1 (z)dz

=

ˆ

Ω∩B(ζ,η)

Ω\B(ζ,η)

ˆ

p+1 U1,0 (y)dy + O(δjN )

= Ω−ζ η δj ∩B(0, δj

ˆ

ˆ

0

p+1  U1,0 (y , yN )dy  dyN

j

ˆ

= RN +

ˆ (5.11)

)

η δj

= |y  |≤ δη

= RN +

Ujp+1 (z)dz

+

p+1  U1,0 (y , yN )dy

1 δj

|y  |≤ δη

p+1  U1,0 (y , 0)dy 

ˆj ∂RN +

g(δj y  )

0

j

1 − δj H(ζ) 2

p+1 U1,0 (y)dy

− |y  |≤ δη

ˆ −

ˆ

ˆ



p+1  U1,0 (y , yN )dy  dyN + O(δjN )

g(δj y  ) δj

+ O(δjN )

p+1 U1,0 |y|2 dy + O(δjN ).

On the other hand, it follows from (2.5) and (2.11) that ˆ ˆ p p Uj ϕj dz =δj U1,0 (y)ϕ0 (y)dy + O(δj2 | ln δj |ϑ ) Ω

ˆ =δj

Ω−ζ δj

RN +

(−ΔU1,0 (y)ϕ0 (y) + U1,0 (y)Δϕ0 (y)) dy + o(δj )

ˆ

∂ϕ0 (y) U1,0 (y)dy + o(δj ) ∂yN ˆ N −1 N −2  yi2 κi = − δj αN N U1,0 (y)dy + o(δj ) 2 (1 + |y|2 ) 2 ∂RN + i=1 ˆ |y|2 2 N −2 dy + o(δj ). = − δj H(ζ)αN 2 (1 + |y|2 )N −1 ∂RN +

= − δj

(5.12)

∂RN +

From (5.11)-(5.12), we find that ˆ ˆ ˆ  |y|2 k 1 p+1 p+1 2 N −2 U1,0 (y)dy − αN dy + U1,0 |y|2 dy H(ζ)δ1 + o(). (5.13) I1 = 2 N −1 N 4 (1 + |y| ) 2N ∂RN ∂RN + + RN +

Estimate for I3 , we have (5.14)

−I3 =

ˆ  i>j

Ω

ˆ p − Up U j dz = O ϕi Uip−1 Uj = o(). U i i Ω

Estimate for I2 . Given η > 0 small but fixed, let δ1 , δ2 , · · · δk be given by (2.22). Set δ0 = As in [26], let us define the annuals     Al := z ∈ Ω : δl δl+1 < |z − ζ| < δl δl−1 , l = 1, · · · , k. (5.15)

η2 δ1

and δk+1 = 0.

We shall decompose the integral on Ω ∩ B(ζ, η) in the sum of Al , that is, we have Ω ∩ B(ζ, η) = ∪kl=1 Al . We write ˆ   k   p   k  j p+1 −  p+1 − (p + 1)  U  U U U −(p + 1)I2 = j i j dz Ω

j=1

j=1

i>j

BOUNDARY BUBBLING SOLUTIONS k ˆ     k

=

l=1 A

j U

p+1

−

j=1

l

ˆ

 p+1 − (p + 1) U j

j=1

k  

+

k 

17

j U

p+1

  pU  U i j dz

i>j

−

k 

j=1

Ω\B(ζ,η)



 p+1 − (p + 1) U j

j=1



  pU j dz. U i

i>j

By similar computations in [26], we can get that I2 = −

(5.16)

k−1  j=1

δj+1 δj

N2−2

p+1 αN

ˆ

RN +

1

1

|y|2 )

(1 +

|y|N −2

N +2 2

dy + o().

Therefore, from (5.13), (5.14) and (5.16), we obtain claim (5.7). Proof of (5.8): In fact, one has

−(q + 1)II =

k ˆ  Al

l=1

+



j U

q+1

j=1,j=l

k ˆ  Al

l=1

k 

l + U

k ˆ 

 q+1 + (q + 1) U l

k 

 q+1 − (q + 1)U q −U l l k 

qU  U l j +

Al j=1,j=l

l=1

ˆ

j U



j=1,j=l k 

Ω\B(ζ,η)

j U

q+1

j=1

:=II1 + II2 + II3 + II4 . By the mean value theorem, for some t ∈ [0, 1], we have k

II1 =

q(q + 1)  2

ˆ Al

l=1

k 

≤C

ˆ Al

j,l=1,j=l



k 

l + t U

j U

k q−1  

j=1,j=l

ˆ

k 

Ulq−1 Uj2 + C

i,j,l=1, i,j=l

j U

2

j=1,j=l

Al

Uiq−1 Uj2 .

H¨older inequality gives that ˆ

k 

Al

j,l=1,j=l

Ulq−1 Uj2 dz

ˆ

k 

=

j,l=1,j=l k 

≤

(5.17)

q−1 q

Al

(Ulq−1 Uj

ˆ Al

j,l=1,j=l

Ulq Uj

q+1 q

)Uj

q−1 ˆ q Al

Ujq+1

q1 ,

and k 

(5.18)

ˆ

i,j,l=1, i,j=l

Al

Uiq−1 Uj2 dz

ˆ

k 

≤

i,j,l=1, i,j=l

Al

Uiq+1

q−1 ˆ q+1 Al

Ujq+1

2

q+1

.

If j > l, then ˆ

Ulq Uj dz

=

Al

(5.19)

=

δj δl

N2−2

√

− N 2−2 q+

δl

ˆ

q+1 αN

δl δl+1 ≤|z−ζ|≤

N +2  2

q+1 αN

N −2 2

N −2 2 q

δj

δl √

ˆ RN +

δl δl−1

(δl2 + |z − ζ|2 ) 1

(1 + |z|2 )

|z|

(δj2 + |z − ζ|2 )

 dz + o(1) . N −2 1

N −2 2 q

N −2 2 q

N −2 2

dz

18

WENJING CHEN

If j < l, then ˆ

Ulq Uj dz

=

Al

≤

δl δj

δl δj

N2−2

− N −2 q+ N2+2 q+1 δl 2 αN 

N2−2

− N 2−2 q+ N2+2

δl

ˆ

1 

δl+1 δl

≤|z|≤

(1 + δl−1 δl

ˆ

q+1 αN 

1 N −2 2 q

1 

δl+1 δl

|z|2 )

≤|z|≤

δl−1 δl

(1 + |z|2 )

N −2 2 q

(1 +

N −2 ( δδjl )2 |z|2 ) 2

dz.

(5.20) For i = l, we have ˆ (5.21)

Uiq+1 dz ≤

⎧ N ⎨ ( δδl ) 2

− N −2 q+ N2+2 Cδi 2

if i ≤ l − 1 < l;

i

⎩(

Al

δi2

δl δl−1 )

N −2 2 q−1

if i ≥ l + 1 > l.

From (5.17)-(5.21) and (2.22), we get II1 = o(). Moreover, =

II2

k ˆ  Al

l=1

=

k 

Ulq+1 +

N −2 2

N +2 (N −2 −q )

l=1

=

ˆ

 =

(5.22)

N +2 (N −2 −q )

δ1

I q t1

Al

l=1

δl

N −2 2

k ˆ 

RN +

ˆ

RN +

ˆ RN +

 q+1 − U q+1 ) (U l l

q+1 U1,0 (y)dy + o()

q+1 U1,0 (y)dy + o()

q+1 U1,0 (y)dy + o(1) ,

where Iq is defined in (1.8). From (5.19) and (5.20), we have (5.23)

II3 ≤ C

k ˆ  l=1

k 

Al j=1,j=l

qU  U l j ≤C

k ˆ  l=1

k 

Al j=1,j=l

Ulq Uj = o().

Finally, ˆ (5.24)

II4 =

Ω\B(ζ,η)

⎛ ⎞q+1 k k ˆ    ⎝ ⎠ Uj ≤C j=1

j=1

Ω\B(ζ,η)

Ujq+1 dz = o().

Thus from (5.22)-(5.24), we derive (5.8). Proof of (5.9): We have ˆ ˆ 1 1 p+1 p+1+ III = |U | dz − |U | dz p+1 Ω p+1+ Ω ˆ  ˆ 1 1 1 p+1 p+1 − |U | dz −  + o() |U | (1 +  ln |U | + o()) dz = p+1 Ω p + 1 (p + 1)2 Ω ˆ  1 ˆ  1 p+1 =− |U |p+1 ln U dz − |U | dz + o(). p+1 Ω (p + 1)2 Ω

dz

BOUNDARY BUBBLING SOLUTIONS

We write

ˆ Ω

|U |

p+1

ln U dz =

k ˆ  Aj

j=1

19

k k  p+1     i  i dz j + U U ln U  Uj + i=j

ˆ + Ω\∪k j=1 Aj

i=j

k K  p+1     i  i dz. j + ln U U U Uj + i=j

i=j

We use the annulus defined in (5.15), and by a similar way as [26], we can get k ˆ k k  p+1      i  i dz j + U U ln U Uj + j=1

= − Moreover, ˆ Ω\∪k j=1 Aj

Aj

N −2 2

ˆ RN +

i=j

i=j

p+1 U1,0 (y)dy

k 

ln δj + k

j=1

ˆ RN +

p+1 U1,0 (y) ln(U1,0 (y))dy + o().

& k ˆ k k  p+1          ln Uj + Ui  Ui dz = O Uj + i=j

Ω\B(ζ,η)

j=1

i=j

Ujp+1

ln

k 

'

Uj dz

= o().

j=1

Using (5.11), we thus obtain (5.9). Therefore from (5.7)-(5.9), then (5.4) holds in the C 0 sense. By similar computations, we can get C 1 estimate, we omit it here. 6. Proof of the main result Now we are ready to prove our main result. Proof of Theorem 1.1 Thanks to Lemma 5.1, we know that u=

k 

μ + ψ U j

with ψ = T −1 (φ)

j=1

is a solution to problem (1.7) if and only if (ξ, ζ) is a critical point of I (ξ, ζ), where the existence of φ is guaranteed by Proposition 4.3. From Lemma 5.2, Proposition 5.3 and (5.3), we know that finding a critical point of I (ξ, ζ) is equivalent to find that of I (ξ, ζ), which is defined by ! 1 1  c0 − c1  + c2  log  − I (ξ, ζ) = ϕ(ζ, t1 , · · · , tk ) + o(1). I (ξ, ζ) :=  αN where o(1) is uniform in the C 1 sense as  → 0, and 



ϕ(ζ, t1 , · · · , tk ) = c3 H(ζ) + c4 Iq t1 − c5

k  j=1

log tj + c6

k−1  j=1

tj+1 tj

N2−2 ,

where constants ci , i = 0, 1, · · · , 6 are given in Proposition 5.3. We change variables by setting s = Θ(t), where  t t tk 2 3 Θ(t) = t1 , , , . . . , . t1 t2 tk−1 We then get k k k−1    N −2  2 (k − j + 1) log sj + c6 sj+1 := ϕ˜1 − ϕ˜j , ϕ(ζ, s1 , · · · , sk ) = c3 H(ζ) + c4 Iq s1 − c5 j=1

with

 ϕ˜1 = c3 H(ζ) + c4 Iq s1 − c5 k log s1 ,

j=1

j=2

N −2 2

and ϕ˜j = c5 (k − j + 1) log sj − c6 sj

.

20

WENJING CHEN

By our assumption, we get the existence of a C 1 stable critical point ζ0 of the mean curvature H(ζ) on the boundary ∂Ω, with c3 H(ζ0 ) + c4 Iq > 0. Since s¯1 = satisfies ∂s1 ϕ˜1 = 0, and

s¯j =

c5 k c3 H(ζ0 ) + c4 Iq

2c5 (k − j + 1) (N − 2)c6

N2−2 ,

j = 2, · · · , k,

is the critical point of ϕ˜j . Then (ζ0 , t∗ ) := (ζ0 , s¯1 , s¯2 s¯1 , s¯3 s¯2 s¯1 , · · · , s¯k × · · · × s¯2 s¯1 ) is a nondegenerate critical point of ϕ(ζ, t1 , · · · , tk ). It follows that the local degree deg(∇ϕ(ζ, t), O, 0) is well defined and is nonzero, here O is an arbitrarily small neighborhood of (ζ0 , t∗ ). Hence for  > 0 small ¯ 0) = 0. Thus by standard properties of the Brouwer degree, we then enough, we have that deg(∇I (ξ, ζ), O, get the existence of a family of critical points (ζ , t ) of I (ξ, ζ) with ζ → ζ0 and t → t∗ as  → 0. The formula (1.10) follows from our construction. We complete the proof. References [1] C. O. Alves, D. C. De Morais Filho and M. A. S. Souto, Radially symmetric solutions for a class of critical exponent elliptic problems in RN , Electron. J. Differential Equations 07 (1996), 12 p. [2] H. Berstick and P. L. Lions, Nonlinear scalar field equations, I. Existence of a ground state, Arch. Ration. Mech. Anal. 82 (1983), 313-345. [3] J. Campos, Bubble-tower phenomena in a semilinear elliptic equation with mixed Sobolev growth, Nonlinear Anal. 68 (2008), 1382-1397. [4] D. Cao and T. Kupper, On the existence of multipeaked solutions to a semilinear Neumann problem, Duke Math. J. 97 (1999), 261–300. [5] D. Cao, E.S. Noussair and S. Yan, Existence and nonexistence of interior-peaked solution for a nonlinear Neumann problem, Pacific J. Math., 200 (2001) 19-41. [6] D. Cao, E.S. Noussair and S. Yan, Symmetric solutions for a Neumann problem involving critical exponent, Proc. Roy. Soc. Edinburgh Sect. A, 131 (2001) 1039-1064. [7] D. Cao and S. Yan, Infinitely many solutions for an elliptic Neumann problem involving critical Sobolev growth, J. Differential Equations 251 (2011), 1389-1414. [8] W. Chen and I. Guerra, Multiplicity of solutions to nearly critical elliptic equation in the bounded domain of R3 , J. Math. Anal. Appl. 424 (2015), 179-200. [9] W. Chen, J. D´ avila and I. Guerra, Bubble tower solutions for supercritical elliptic problem in RN , Ann. Sc. Norm. Super. Pisa Cl. Sci. (5), Vol. XV (2016), 85-116. [10] J. D´ avila, M. Del Pino and I. Guerra, Non-uniqueness of positive ground states of nonlinear schr¨ odinger equations, Proc. London Math. Soc. (3) 106 (2013), 318-344. [11] M. del Pino, J. Dolbeault and M. Musso, Bubble-tower radial solutions in the slightly supercritical Brezis-Nirenberg problem, J. Differential Equations 193 (2003), 280-306. [12] M. del Pino, J. Dolbeault and M. Musso, The Brezis-Nirenberg problem near criticality in dimension 3, J. Math. Pures Appl. 83 (2004), 1405-1456. [13] M. del Pino and I. Guerra, Ground states of a prescribed mean curvature equation, J. Differential Equations 241 (2007), 112-129. [14] M. del Pino, M. Musso and A. Pistoia, Super-critical boundary bubbling in a semilinear Neumann problem, Ann. Inst. H. Poincar´ e Anal. Non Lin´ eaire 22 (2005), 45-82. [15] A. Ferrero and F. Gazzola, On subcriticality assumptions for the existence of ground states of quasilinear elliptic equations, Adv. Differential Equations 8 (2003), 1081-1106. [16] Y. Ge, M. Musso and A. Pistoia, Sign changing tower of bubbles for an elliptic problem at the critical exponent in pierced non-symmetric domains, Comm. Partial Differential Equations 35 (2010), 1419-1457. [17] C. Gui and J. Wei, Multiple interior peak solutions for some singularly perturbed Neumann problems, J. Differential Equations 158 (1999), 1-27. [18] C. Gui and J. Wei, Multiple boundary peak solutions for some singularly perturbed Neumann problems, Ann. Inst. H. Poincar´e Anal. Non Lin´ eaire 17 (2000), 47-82. [19] Y. Y. Li, On a singularly perturbed elliptic equation, Adv. Differential Equations 2 (1997), 955-980. [20] C. S. Lin, L. P. Wang and J. C. Wei, Bubble accumulations in an elliptic Neumann problem with critical Sobolev exponent, Calc. Var. Partial Differential Equations 30 (2007), 153-182.

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21

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