Journal of Mathematical Analysis and Applicationsmath0.bnu.edu.cn/~tangzw/papers/jmaa2017.pdf · Journal of Mathematical Analysis and Applications. ... for nonlinear coupled elliptic

J. Math. Anal. Appl. 448 (2017) 1079–1119

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Journal of Mathematical Analysis and Applications

www.elsevier.com/locate/jmaa

Number of synchronized and segregated interior spike solutions

for nonlinear coupled elliptic systems

Zhongwei Tang ∗,1, Lushun WangSchool of Mathematical Sciences, Beijing Normal University, Beijing, 100875, PR China

a r t i c l e i n f o a b s t r a c t

Article history:Received 18 August 2016Available online 23 November 2016Submitted by Y. Du

Keywords:Lyapunov–Schmidt reduction methodsSynchronized and segregated solutionInterior spikesElliptic systems

In this paper, we consider the following nonlinear coupled elliptic systems

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

−ε2Δu + u = μ1u3 + βuv2 in Ω,

−ε2Δv + v = μ2v3 + βu2v in Ω,

u > 0, v > 0 in Ω,

∂u∂ν

= ∂v∂ν

= 0 on ∂Ω,

(Aε)

where ε > 0, μ1 > 0, μ2 > 0, β ∈ R, and Ω is a bounded domain with smooth boundary in R3. Due to Lyapunov–Schmidt reduction method, we proved that (Aε) has at least O( 1

ε3| ln ε| ) synchronized and segregated vector solutions for ε small enough and some β ∈ R. Moreover, for each m ∈ (0, 3) there exist synchronized and segregated vector solutions for (Aε) with energies in the order of ε3−m. Our result extends the result of Lin, Ni and Wei [20], from the Lin–Ni–Takagi problem to the nonlinear elliptic systems.

© 2016 Elsevier Inc. All rights reserved.

1. Introduction and main results

In this paper, we study the following nonlinear elliptic systems⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

−ε2Δu + u = μ1u3 + βuv2 in Ω,

−ε2Δv + v = μ2v3 + βu2v in Ω,

u > 0, v > 0 in Ω,

∂u∂ν = ∂v

∂ν = 0 on ∂Ω,

(Bε)

* Corresponding author.E-mail address: [email protected] (Z. Tang).

1 The first author was supported by National Science Foundation of China (11571040).

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1080 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119

where ε > 0, μ1 > 0, μ2 > 0, β ∈ R, Ω is a bounded domain with smooth boundary in R3.For Ω = R

N , N ≤ 3 and ε = 1, (Bε) leads to investigate the following problem⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

−Δu + u = μ1u3 + βuv2 in R

N ,

−Δv + v = μ2v3 + βu2v in R

N ,

u > 0, v > 0 in RN ,

u → 0, v → 0 as x → +∞.

(1.1)

Problem (Bε) arises in the Hartree–Fock theory for a double condensate, i.e. a binary mixture of Bose–Einstein condensates in two different hyperfine states |1〉 and |2〉 (see [9]). Physically, u and v are the corresponding condensate amplitudes. μj and β are the intraspecies and interspecies scattering lengths. The sign of scattering length β determines whether the interactions of states |1〉 and |2〉 are repulsive or attractive. When β > 0, the interactions of states |1〉 and |2〉 are attractive, the components of a vector solution tend to go along with each other, leading to synchronization. In contrast, when β < 0, the interac-tions of states |1〉 and |2〉 are repulsive, the components tend to segregate from each other, leading to phase separations.

Recently, B. Sirakov [24] discussed the whole β ∈ R and analyzed for which β the problem (1.1)assures a least energy solution and for which β the problem (1.1) has no least energy solution. In [31], Wei and Yao proved that for 0 < β < min{μ1, μ2} small and β > max{μ1, μ2}, the solution of (1.1) is unique and non-degenerate. Furthermore, Peng and Wang [23] proved there exists a decreas-ing sequence {βk} ⊂ (−√

μ1μ2, 0) with βk → −√μ1μ2, the solution of (1.1) is non-degenerate for

β ∈ (−√μ1μ2, 0) ∪ (0, min{μ1, μ2}) ∪ (max{μ1, μ2}, +∞) and β = βk for any k.

For the scalar case, there are many investigations to the following elliptic equations⎧⎪⎪⎨⎪⎪⎩

−ε2Δu + u = up in Ω,

u > 0, v > 0 in Ω,

∂u∂ν = 0 on ∂Ω,

(1.2)

where Ω is a smooth bounded domain in Rn and p is subcritical, i.e. 2 < p < 2nn−2 for n ≥ 3, and

2 < p < +∞ for n = 1, 2. It is well known that problem (1.2) has both interior spike solutions and boundary spike solutions. For single interior spike solutions, we refer to [7,28–30] and references therein. For multiple interior spikes solutions, we refer to [4,5,15] and references therein. For single boundary spike solutions, we refer to [8,19,21,22] and references therein. For multiple boundary spikes solutions, we refer to [6,13,16,17]and references therein. By Lyapunov–Schmidt reduction method, Gui and Wei [14] constructed a solution to (1.2) with both k interior spikes and l boundary spikes for any k ≥ 0, l ≥ 0, k + l > 0 and ε > 0 small enough.

We also want to mention the paper by Lin, Ni and Wei [20], where the authors showed that there exists ε0 such that for each 0 < ε < ε0 and for each integer k bounded by 1 ≤ k ≤ C(Ω,n)

εn(ln ε)n , (1.2) has a solution with k-interior spikes. Recently, Ao, Wei and Zeng [1] promoted the upper bound of the number of interior spikes to C(Ω,n)

εn which is optimal due to the fact that each spike contributes to at least O(εn) energy.For the singularly perturbed nonlinear Schrödinger equations, we want to refer the readers to the work

by Felmer, Martinez and Tanaka [10,11], where in 1-dimensional situation, they considered the following problem

ε2Δu− V (x)u + up = 0, u > 0, u ∈ H1(R). (1.3)

They constructed a solution to (1.3) with C spikes for some V (x).
ε

Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1081

In this paper, we are interested in the number of interior spikes of synchronized and segregated solutions to (Bε). Under the Dirichlet conditions, Lin and Wei [18] showed that there exist ε0 > 0 and β0 ∈ (0, √μ1μ2)such that for any β ∈ (−∞, β0) and 0 < ε < ε0, (Bε) has a least energy solution (uε, vε). Moreover, if β < 0the spikes Pε and Qε of (uε, vε) repel each other and ∂Ω repels these spikes. As a result, the locations of Pε and Qε reach a sphere packing position in Ω. On the other hand, if β > 0, Pε and Qε bound each other and the locations reach the innermost part of the Ω due to the repelling effect of ∂Ω. Under the Neumann conditions, the first author proved that for any ε > 0 the problem (Cε) has a least energy solution for β ∈ (−∞, min{μ1, μ2}) ∪(max{μ1, μ2}, +∞) and no solution for β ∈ (min{μ1, μ2}, max{μ1, μ2}). Moreover, the spikes Pε and Qε of (uε, vε) locate at the most curved part of ∂Ω if β < 0, Pε and Qε repel each other in the sense |Pε−Qε|

ε → +∞ and they may converge to the same point of ∂Ω if ∂Ω has a only most curved point. If β > 0, Pε and Qε bound each other and converge to a same point of ∂Ω. By Lyapunov–Schmidt reduction method, the first author also constructed a multiple boundary spikes solution to (Bε) for β ∈ [0, min{μ1, μ2}) ∪ (max{μ1, μ2}, +∞) and a multiple segregated peak solutions to (Bε) for −∞ < β < min{μ1, μ2} (see [25–27]).

Our main results in this paper can be stated as follows.

Theorem 1.1. Suppose μ1 > 0, μ2 > 0, there exists a decreasing sequence {βn} ⊂ (−√μ1μ2, 0) with βn →

−√μ1μ2 as n → +∞ such that for β ∈ (−√

μ1μ2, 0) ∪ (0, min{μ1, μ2}) ∪ (max{μ1, μ2}, +∞) and β = βk

for any k, (Bε) has at least O(

1ε3| ln ε|

)synchronized vector solutions. Moreover, for each m ∈ (0, 3), (Bε)

has a synchronized vector solution (uε, vε) with the energy of the order ε3−m.

Theorem 1.2. Suppose μ1 > 0, μ2 > 0, then there exists a 0 < β∗ < min{μ1, μ2} such that for any

β ∈ (−∞, β∗), (Bε) has at least O(

1ε3| ln ε|

)segregated vector solutions. Moreover, for each m ∈ (0, 3), (Bε)

has a segregated vector solution (uε, vε) with the energy of the order ε3−m.

Before approaching to the proof of Theorem 1.1 and Theorem 1.2, we introduce some notations first and then we will introduce another version of the above two main results which we will mainly focus on the proof in the following.

Without loss of generality, we may assume 0 ∈ Ω. Put x = εz, by changing variables, (Bε) is equivalent to a system as follows

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

−Δu + u = μ1u3 + βuv2 in Ωε,

−Δv + v = μ2v3 + βu2v in Ωε,

u > 0, v > 0 in Ωε,

∂u∂ν = ∂v

∂ν = 0 on ∂Ωε.

(Cε)

Let

H1N (Ωε) :=

{u ∈ H1(Ωε) : ∂u

∂ν= 0 on ∂Ωε

}

be the Hilbert space with the inner product

〈u, v〉 :=∫Ωε

(∇u∇v + uv)

and the corresponding norm


‖u‖2H1(Ωε) :=

∫Ωε

(|∇u|2 + u2).

The functional corresponding to (Cε) is defined by for any (u, v) ∈ H1N (Ωε) ×H1

N (Ωε),

Jε(u, v) = 12

∫Ωε

(|∇u|2 + |∇v|2 + u2 + v2) − 14

∫Ωε

(μ1u4 + μ2v

4 + 2βu2v2).

Let ω be the unique solution of

⎧⎪⎪⎨⎪⎪⎩

Δu− u + u3 = 0 in R3,

u > 0 in R3,

u(0) = maxR3

u(x), u(x) → 0 as |x| → +∞.

According to Gidas, Ni and Nirenberg [12], ω is radially symmetric and ω′(r) < 0 for r = |x| > 0. Moreover, the asymptotic behavior of ω can be described as follows

ω(r) = A3r−1e−r

(1 + O

(1r

)), ω′(r) = −A3r

−1e−r

(1 + O

(1r

))as r → +∞.

Put

a =

√μ2 − β

μ1μ2 − β2 , b =

√μ1 − β

μ1μ2 − β2 ,

it is easy to see that (U, V ) = (aω, bω) satisfies⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

−Δu + u = μ1u3 + βuv2 in R

3,

−Δv + v = μ2v3 + βu2v in R

3,

u > 0, v > 0 in R3,

u → 0, v → 0 as x → +∞.

(1.4)

The functional corresponding to (1.4) is defined by for any (u, v) ∈ H1(R3) ×H1(R3),

I(u, v) = 12

∫R3

(|∇u|2 + |∇v|2 + u2 + v2) − 14

∫R3

(μ1u4 + μ2v

4 + 2βu2v2).

Let ωε,P be the unique solution of

{Δu− u + ω3(· − P

ε ) = 0 in Ωε,

∂u∂ν = 0 on ∂Ωε,

(1.5)

then Uε,P = aωε,P satisfies

{Δu− u + μ1U

3(· − Pε ) + βU(· − P

ε )V 2(· − Pε ) = 0 in Ωε,

∂u∂ν = 0 on ∂Ωε,

(1.6)

and Vε,P = bωε,P satisfies


{Δu− u + μ2V

3(· − Pε ) + βU2(· − P

ε )V (· − Pε ) = 0 in Ωε,

∂u∂ν = 0 on ∂Ωε.

(1.7)

In this paper, we will use (Uε,P , Vε,P ) to build up the synchronized approximate vector solutions for (Cε).Let M > 3√

2−1 be a fixed constant, we define our configuration space as follows:

ΛK :={P = (P1, P2, . . . , PK) ∈ ΩK : ϕ(P) ≥ Mε| ln ε|

}, (1.8)

where

ΩK = Ω × Ω × . . .Ω︸︷︷︸K

, ϕ(P) = mini�=j

i,j,k=1,2,··· ,K

{|Pi − Pj |, 2dist(Pk, ∂Ω)

}.

Let

Uε,P =k∑

i=1Uε,Pi

, Vε,P =k∑

i=1Vε,Pi

.

In stead of giving the proof of Theorem 1.1 directly, in this paper, we will prove the following results which immediately implies Theorem 1.1.

Theorem 1.3. Under the assumption of Theorem 1.1, there exist ε0 > 0 such that for each 0 < ε < ε0 and any integer K satisfying

1 ≤ K ≤ KΩ

ε3| ln ε|3 ,

where KΩ is a constant depending only on Ω, problem (Cε) has a synchronized vector solution of the form

(uε, vε) = (Uε,Pε, Vε,Pε

) + (uε,Pε, vε,Pε

),

where Pε ∈ ΛK and (uε,Pε, vε,Pε

) satisfies

‖(uε,Pε, vε,Pε

)‖H2(Ωε) = o(1)

and we have the following energy estimates

Jε(uε,Pε, vε,Pε

) = KI(U, V ) + o(1)

as ε → 0.As a consequence, (Cε) has at least O

(1

ε3| ln ε|3)

synchronized vector solutions. For each m ∈ (0, 3), (Cε)has a synchronized vector solution (uε,Pε

, vε,Pε) with energy in the order of ε−m. When m = 3, the energy

of (uε,Pε, vε,Pε

) is in the order of O(

1ε3| ln ε|3

). In addition, the Morse index of (uε,Pε

, vε,Pε) is at least K.

Put

U = 1√μ1

w, V = 1√μ2

w,

then U and V satisfy respectively for i = 1, 2


⎧⎪⎪⎨⎪⎪⎩

−Δu + u = μiu3 in R

3,

u > 0 in R3,

u → 0 as x → +∞.

(1.9)

The energy functional corresponding to (1.9) is defined by for any u ∈ H1(R3),

Iμi(u) = 1

2

∫R3

(|∇u|2 + u2) − 14

∫R3

μiu4.

Let Uε,P = 1√μ1ωε,P and V ε,P = 1√

μ2ωε,P , then Uε,P satisfies

{Δu− u + μ1U

3(· − Pε ) = 0 in Ωε,

∂u∂ν = 0 on ∂Ωε,

(1.10)

and V ε,Q satisfies

{Δu− u + μ2V

3(· − Qε ) = 0 in Ωε,

∂u∂ν = 0 on ∂Ωε.

(1.11)

In this paper, we will use (Uε,P , V ε,Q) to build up the segregated approximate vector solutions for (Cε).Our configuration space can be defined as follows.Assume Ω1 and Ω2 are two nonempty smooth domain in Ω such that δ = dist(Ω1, Ω2) > 0 and M > 3√

2−1is a fixed constant. Let

ΛK :={P = (P1, P2, · · · , PK) ∈ ΩK

1 : ϕ(P) ≥ Mε| ln ε|},

ΛL :={Q = (Q1, Q2, · · · , QL) ∈ ΩL

2 : ϕ(Q) ≥ Mε| ln ε|}

(1.12)

where

ϕ(P) = mini�=j

i,j,k=1,2,··· ,K

{|Pi − Pj |, 2dist(Pk, ∂Ω)}

and

ϕ(Q) = mini�=j

i,j,k=1,2,··· ,L

{|Qi −Qj |, 2dist(Qk, ∂Ω)} .

Let

Uε,P =K∑i=1

Uε,Pi, V ε,Q =

L∑i=1

V ε,Qi.

Similarly, in this paper we will prove the following result which implies Theorem 1.2 immediately as well.

Theorem 1.4. Under the assumption of Theorem 1.2, there exist ε0 > 0 such that for each 0 < ε < ε0 and any integer pair (K, L) satisfying


1 ≤ K ≤ KΩ1

ε3| ln ε|3 , 1 ≤ L ≤ KΩ2

ε3| ln ε|3 ,

where KΩiis a constant depending only on Ωi, problem (Cε) has a segregated vector solution of the form

(uε,Pε,Qε, vε,Pε,Qε

) = (Uε,Pε, V ε,Qε

) + (uε,Pε,Qε, vε,Pε,Qε

),

where (Pε, Qε) ∈ ΛK × ΛL and (uε,Pε,Qε, vε,Pε,Qε

) satisfies

‖(uε,Pε,Qε, vε,Pε,Qε

)‖H2(RN ) = o(1)

and we have the following energy estimate

Jε(uε,Pε,Qε, vε,Pε,Qε

) = KIμ1(U) + LIμ2(V ) + o(1), as ε → 0.

As a consequence, (Cε) has at least O(

1ε3| ln ε|3

)segregated vector solutions. For each m ∈ (0, 3), (Cε) has

a segregated vector solution (uε,Pε,Qε, vε,Pε,Qε

) with energy in the order of ε−m. When m = 3, the energy

of (uε,Pε,Qε, vε,Pε,Qε

) is in the order of O(

1ε3| ln ε|3

). In addition, the Morse index of (uε,Pε,Qε

, vε,Pε,Qε) is

at least KL.

This paper is organized as following. In Section 2, we construct a synchronized vector solution with at most O(ε−3| ln ε|−3) interior spikes for ε > 0 small enough and prove Theorem 1.1 by proving Theorem 1.3. In Section 3, we construct a segregated vector solution with at most O(ε−3| ln ε|−3) interior spikes for ε > 0 small enough and prove Theorem 1.2 by proving Theorem 1.4. Finally, we will collect some technical estimates and some energy expansions in Appendix A.

2. Synchronized solutions

In this section, we consider synchronized solutions and prove Theorem 1.1 by proving Theorem 1.3. We firstly prove the existence of solutions to a linear projected problem for any P ∈ ΛK in Subsection 2.1. Secondly, we use Banach fixed point Theorem to prove the existence of solutions to a nonlinear projection problem for any P ∈ ΛK in Subsection 2.2. Thirdly, in Subsection 2.3, we study a maximum problem in P ∈ ΛK and prove the maximum problem is achieved by some interior point Pε of ΛK . Finally, we prove Theorem 1.1 by proving Theorem 1.3 in Subsection 2.4.

2.1. A linear projected problem

Let (U, V ) be the solution of problem (1.4), we denote UPi= U

(· − Pi

ε

), VPi

= V(· − pi

ε

). Take

Zij = (Zij,1, Zij,2)T =(χPi

∂UPi

∂xj, χPi

∂VPi

∂xj

)T

,

where χPi(x) = χ

(2

(M−1)| ln ε| (x− Pi

ε ))

and χ(t) is a smooth cutoff function such that

χ(t) = 1 for |t| ≤ 1, χ(t) = 0 for |t| ≥ M2

M2 − 1 and 0 ≤ χ(t) ≤ 1.

In this subsection, we will consider the following linear projected problem


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Lε(u, v) = h +k∑

i=1

3∑j=1

cijZij in Ωε,

∫Ωε

(uZij,1 + vZij,2) = 0, i = 1, 2, · · · , k, j = 1, 2, 3,

∂u

∂ν= ∂v

∂ν= 0 on ∂Ωε.

(2.1)

Here h = (h1, h2)T , Lε(u, v) = (Lε,1(u, v), Lε,2(u, v))T where

Lε,1(u, v) = Δu− u + 3μ1U2ε,Pu + βV 2

ε,Pu + 2βUε,PVε,Pv,

Lε,2(u, v) = Δv − v + 3μ2V2ε,Pv + βU2

ε,Pv + 2βUε,PVε,Pu. (2.2)

The first lemma deals with the non-degeneracy of (U, V ) which is the solution of problem (1.4).

Lemma 2.1. There exists a decreasing subsequence {βk} ⊂ (−√μ1μ2, 0) with βk → −√

μ1μ2 as k → +∞such that for β ∈ (−√

μ1μ2, 0) ∪ (0, min{μ1, μ2}) ∪ (max{μ1, μ2}, +∞) and β = βk for any k, the solution (U, V ) of problem (1.4) is non-degenerate. Namely, the solution space E of the linearized equation

{Δu− u + 3μ1U

2u + βV 2u + 2βUV v = 0 in R3,

Δv − v + 3μ2V2v + βU2v + 2βUV u = 0 in R

3

is exactly 3 dimensional, i.e.

E = span

{(∂U∂x1

∂V∂x1

),

(∂U∂x2

∂V∂x2

),

(∂U∂x3

∂V∂x3

)}.

Proof. See the proof of Proposition 2.3 in Peng and Wang [23] or Lemma 3.1 in Bartsch, Dancer and Wang [3]. �

The following lemma deals with a prior estimate for solutions to (2.1).

Lemma 2.2. Let h = (h1, h2) ∈ L2(Ωε) ×L2(Ωε), assume that ((u, v), {cij}) is a solution to (2.1), then there exists ε0 > 0 and C > 0 such that for any 0 < ε < ε0 and P ∈ ΛK , one has

‖(u, v)‖H2(Ωε)×H2(Ωε) ≤ C‖h‖L2(Ωε)×L2(Ωε). (2.3)

Proof. Multiplying Zij on both sides of the equation (2.1) and integrating over Ωε, we have

∫Ωε

Lε(u, v)Zijdx =∫Ωε

hZijdx + cij

∫Ωε

Z2ijdx. (2.4)

Note that

ΔuχPiUij + ΔvχPi

Vij

= div (χPiUij∇u + χPi

Vij∇v) − div (∇(χPiUij)u + ∇(χPi

Vij)v)

+ ΔχPi(Uij + Vij) + χPi

(ΔUiju + ΔVijv) + 2∇χPi∇(Uiju + Vijv)


where Uij = ∂UPi

∂xjand Vij = ∂VPi

∂xj. Then

∫Ωε

Lε(u, v)Zijdx = I + II +∫Ωε

[ΔχPi(Uiju + Vijv) + 2∇χPi

(∇Uiju + ∇Vijv)] ,

where

I =∫Ωε

χPiu(ΔUij − Uij + 3μ1U

2PiUij + βV 2

PiUij + 2βUPi

VPiVij

)

+∫Ωε

χPiv(ΔVij − Vij + 3μ2V

2PiVij + βU2

PiVij + 2βUPi

VPiUij

)

and

II =∫Ωε

χPiUij

[3μ1(U2ε,P − U2

Pi

)u + β

(V 2ε,P − V 2

Pi

)u + 2β (Uε,PVε,P − UPi

VPi) v]

+∫Ωε

χPiVij

[3μ2(V 2ε,P − V 2

Pi

)v + β

(U2ε,P − U2

Pi

)v + 2β (Uε,PVε,P − UPi

VPi)u].

According to Lemma 2.1, we obtain I = 0.By Lemma A.2, for x ∈ Ωε,Pi

, we have

∣∣U2ε,P − U2

Pi

∣∣ ≤ CUPiε

M2 , |Uε,PVε,P − UPi

VPi| ≤ C(UPi

+ VPi)εM

2 .

Then by Hölder’s inequality, we obtain |II| ≤ CεM2 ‖(u, v)‖L2(Ωε)×L2(Ωε).

By Hölder’s inequality again, we get∫Ωε

[ΔχPi(Uiju + Vijv) + 2∇χPi

(∇Uiju + ∇Vijv)]

≤ C

⎛⎝∫

Ωε

(|ΔχPi

(Uij + Vij) |2 + |∇χPi∇ (Uij + Vij) |2

)⎞⎠12

‖(u, v)‖L2(Ωε)×L2(Ωε)

≤ C(M + 1)| ln ε|

⎛⎜⎜⎝

(M2−1)| ln ε|2(M+1)∫

(M−1)| ln ε|2

e−2r

⎞⎟⎟⎠

12

‖(u, v)‖L2(Ωε)×L2(Ωε)

≤ CεM−1

2 ‖(u, v)‖L2(Ωε)×L2(Ωε).

Thus we have ∣∣∣∣∣∣∫Ωε

Lε(u, v)Zijdx

∣∣∣∣∣∣ ≤ CεM−1

2 ‖(u, v)‖L2(Ωε)×L2(Ωε). (2.5)

Since, by Lebesgue Dominated Convergence Theorem, we have


∫Ωε

Z2ijdx =

∫Ωε

χ2Pi

(U2ij + V 2

ij

)=∫R3

(U2ij + V 2

ij

)+ o(1), (2.6)

and by Hölder’s inequality, we have∫Ωε

hZijdx =∫Ωε

χPi(h1Uij + h2Vij)

≤(‖Uij‖L2(R3) + ‖Vij‖L2(R3)

)‖h‖L2(Ωε)×L2(Ωε). (2.7)

Then according to (2.4)–(2.7), we have

|cij | ≤ C(ε

M−12 ‖(u, v)‖L2(Ωε)×L2(Ωε) + ‖h‖L2(Ωε)×L2(Ωε)

). (2.8)

Let

(ui, vi) = χ′Pi

(u, v) and (uK+1, vK+1) =(u−

K∑i=1

ui, v −K∑i=1

vi

),

where χ′Pi

= χ (

2M(M2−1)| ln ε|

∣∣z − Pi

ε

∣∣) for i = 1, 2, · · · , K. Then (uK+1, vK+1) satisfies

⎧⎪⎪⎪⎨⎪⎪⎪⎩

Lε(uK+1, vK+1) =(

1 −K∑i=1

χ′Pi

)h − 2

K∑i=1

(∇u∇χ′Pi,∇v∇χ′

Pi)T −

K∑i=1

Δχ′Pi

(u, v)T in Ωε,

∂uK+1

∂ν= ∂vK+1

∂ν= 0 on ∂Ωε.

(2.9)

Multiplying (uK+1, vK+1) on both sides of (2.9), by Lemma 2.1, we can easily get

‖(uK+1, vK+1)‖H1(Ωε)×H1(Ωε)

≤ C

∥∥∥∥∥(

1 −K∑i=1

χ′Pi

)h

∥∥∥∥∥L2(Ωε)×L2(Ωε)

+ C

∥∥∥∥∥2K∑i=1

(∇u∇χ′

Pi,∇v∇χ′

Pi

)+

K∑i=1

Δχ′Pi

(u, v)

∥∥∥∥∥L2(Ωε)×L2(Ωε)

.

Thus by equation (2.9), we have

‖(uK+1, vK+1)‖H2(Ωε)×H2(Ωε)

≤ C(‖(ΔuK+1,ΔvK+1)‖L2(Ωε)×L2(Ωε) + ‖(uK+1, vK+1)‖L2(Ωε)×L2(Ωε)

)≤ C

∥∥∥∥∥(

1 −K∑i=1

χ′Pi

)h

∥∥∥∥∥L2(Ωε)×L2(Ωε)

+ C

∥∥∥∥∥2K∑i=1

(∇u∇χ′

Pi,∇v∇χ′

Pi

)+

K∑i=1

Δχ′Pi

(u, v)

∥∥∥∥∥L2(Ωε)×L2(Ωε)

. (2.10)

Now we complete our proof by a contradiction argument. Assume ‖hn‖L2(Ωεn )×L2(Ωεn ) → 0 as n → +∞and ‖(un, vn)‖H2(Ωε )×H2(Ωε ) = 1 for any positive integer n. According to (2.10), we have as n → +∞

n n


‖(un,K+1, vn,K+1)‖H2(Ωεn )×H2(Ωεn )

≤ C

(‖hn‖L2(Ωεn )×L2(Ωεn ) +

4M‖(un, vn)‖H1(Ωεn )×H1(Ωεn )

| ln ε|

)→ 0.

Thus there exists some i ∈ {1, 2, · · · , K} and a subsequence {εnk} ⊂ {εn} such that

‖(unk,i, vnk,i)‖H2(Ωεnk)×H2(Ωεnk

) ≥1

2K .

It is easy to see that (un,i, vn,i) is bounded in H1(Ωεn) ×H1(Ωεn). Thus (un,i, vn,i) ⇀ (u∞, v∞) weakly in H1(R3) ×H1(R3). Moreover, (u∞, v∞) satisfies⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

Δu− u + 3μ1U2u + βV 2u + 2βUV v = 0 in R

3,

Δv − v + 3μ2V2v + βU2v + 2βUV u = 0 in R

3,∫R3

(u∂U

∂xj+ v

∂v

∂xj

)= 0, j = 1, 2, 3.

By the non-degeneracy of (U, V ), we obtain (u∞, v∞) = (0, 0) which leads to a contradiction. Thus we complete the proof of this Lemma. �Proposition 2.3. Let ε0 and C be the positive number in Lemma 2.2, for any 0 < ε < ε0 and any given h = (h1, h2) ∈ L2(Ωε) ×L2(Ωε), there exists a unique solution ((u, v), {cij}) to the problem (2.1) satisfying

‖(u, v)‖H2(Ωε)×H2(Ωε) ≤ C‖h‖L2(Ωε)×L2(Ωε).

Proof. Let

H :=

⎧⎨⎩(u, v) ∈ H1

N (Ωε) ×H1N (Ωε) :

∫Ωε

(uZij,1 + vZij,2) = 0, i = 1, 2, · · · , k, j = 1, 2, 3

⎫⎬⎭

be the Hilbert space and by Riesz lemma we can rewrite the problem (2.1) as

(u, v) + K(u, v) = h in H,

where

〈K(u, v), (ψ, φ)〉 = −∫Ωε

[3μ1U

2ε,Puψ + 3μ2V

2ε,Pvφ

+ β(V 2ε,Puψ + U2

ε,Pvφ) + 2βUε,PVε,P(vψ + uφ)]

and

〈h, (ψ, φ) =∫Ωε

(h1ψ + h2φ)

for each (ψ, φ) ∈ H. According to Rellich embedding theorem, we can easily obtain that K is a compact operator on H. By Lemma 2.2 and Fredholm alternative theorem, we know that (2.1) has a unique solution for any h ∈ L2(Ωε) ×L2(Ωε). By the smooth assumption on the boundary ∂Ωε and elliptic regularity theory we finished the proof of the proposition. �


2.2. A nonlinear projected problem

In this subsection, for P ∈ Λk, we consider the following nonlinear projected problem

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

Sε(u + Uε,P, v + Vε,P) =∑ij

cijZij in Ωε,

∂u

∂ν= ∂v

∂ν= 0 on ∂Ωε,∫

Ωε

(uZij,1 + vZij,2) = 0, i = 1, 2, · · · , k, j = 1, 2, 3,

(2.11)

where Sε(u, v) := (Sε,1(u, v), Sε,2(u, v))T and

Sε,1(u, v) = Δu− u + μ1u3 + βuv2,

Sε,2(u, v) = Δv − v + μ2v3 + βu2v. (2.12)

We denote Nε(u, v) := (Nε,1(u, v), Nε,2(u, v))T , then equation (2.11) can be rewritten as

Lε(u, v) = −Sε(Uε,P, Vε,P) − Nε(u, v) +∑ij

cijZij , (2.13)

where

Nε,1(u, v) = μ1(u + Uε,P)3 − 3μ1U2ε,Pu− μ1U

3ε,P + β(u + Uε,P)(v + Vε,P)2

− 2βUε,PVε,Pv − βUε,PV2ε,P − βV 2

ε,Pu,

Nε,2(u, v) = μ2(v + Vε,P)3 − 3μ2V2ε,Pv − μ2V

3ε,P + β(u + Uε,P)2(v + Vε,P)

− 2βUε,PVε,Pu− βU2ε,PVε,P − βU2

ε,Pv. (2.14)

The following lemma gives the estimate of Sε(Uε,P, Vε,P).

Lemma 2.4. Suppose P ∈ ΛK and ε > 0 small enough, then

‖Sε(Uε,P, Vε,P)‖L2(Ωε)×L2(Ωε) ≤ CKεM

for some C > 0 which is independent of ε and K.

Proof. According to (2.11) and equation (1.6), we have

Sε,1(Uε,P, Vε,P) = ΔUε,P − Uε,P + μ1U3ε,P + βUε,PV

2ε,P

= μ1U3ε,P + βUε,PV

2ε,P − μ1

K∑i=1

U3Pi

− βK∑i=1

UPiV 2Pi

= μ1

(U3ε,P −

K∑i=1

U3Pi

)+ β

(Uε,PV

2ε,P −

K∑i=1

UPiV 2Pi

). (2.15)

Notice that


∫Ωε

(U3ε,P −

K∑i=1

U3Pi

)2

=K∑i=1

∫Ωε,Pi

⎛⎝U3

ε,P − U3Pi

−∑j �=i

U3Pj

⎞⎠2

+∫

Ωε\∪Ki=1Ωε,Pi

⎛⎝U3

ε,P − U3Pi

−∑j �=i

U3Pj

⎞⎠2

≤ 2K∑i=1

∫Ωε,Pi

(U2ε,P + Uε,PUPi

+ U2Pi

)2⎛⎝∑

j �=i

Uε,Pj− ψε,Pi

⎞⎠2

+K∑i=1

∫Ωε,Pi

⎛⎝∑

j �=i

U3Pj

⎞⎠2

+ O(ε3M−3)

≤ C

⎛⎜⎝K

∫R3

U4Pi

⎛⎝∑

j �=i

UPj

⎞⎠2

−K∑i=1

∫R3\Ωε,Pi

U4Pi

⎛⎝∑

j �=i

UPj

⎞⎠2⎞⎟⎠

+ C

⎛⎜⎝K

∫R3

U4Piψ2ε,Pi

−K∑i=1

∫R3\Ωε,Pi

U4Piψ2ε,Pi

⎞⎟⎠+ O(ε3M−3) ≤ CK(K − 1)ε2M .

Similarly,

∫Ωε

(Uε,PV

2ε,P −

K∑i=1

UPiV 2Pi

)2

≤ CK(K − 1)ε2M .

Thus

‖Sε,1(Uε,P, Vε,P)‖L2(Ωε) ≤ CKεM .

Therefore,

‖Sε(Uε,P, Vε,P)‖L2(Ωε)×L2(Ωε) ≤ CKεM . �The following lemma gives the estimates of Nε(u, v).

Lemma 2.5. Suppose P ∈ ΛK and ε > 0 small enough, then

‖Nε(u, v)‖L2(Ωε)×L2(Ωε) ≤ C(‖(u, v)‖3

H2(Ωε)×H2(Ωε) + ‖(u, v)‖2H2(Ωε)×H2(Ωε)

)and

‖Nε(u1, v1) − Nε(u2, v2)‖L2(Ωε)×L2(Ωε)

≤ C(‖(u, v)‖H2(Ωε)×H2(Ωε) + ‖(u, v)‖2

H2(Ωε)×H2(Ωε)

)‖(u1, v1) − (u2, v2)‖H2(Ωε)×H2(Ωε)



Proof. A simple computation shows that

Nε,1(u, v) = μ1(u + Uε,Q)3 − 3μ1U2ε,Qu− μ1U

3ε,Q

+ β(u + Uε,Q)(v + Vε,Q)2 − 2βUε,QVε,Qv

= μ1(u3 + 3Uε,Qu2) + β(uv2 + 2Vε,Quv).

Then by Hölder’s inequality and Sobolev inequality, we have

‖Nε,1(u, v)‖L2(Ωε) ≤ C(‖(u, v)‖3

L6(Ωε)×L6(Ωε) + ‖(u, v)‖2L4(Ωε)×L4(Ωε)

)≤ C

(‖(u, v)‖3

H2(Ωε)×H2(Ωε) + ‖(u, v)‖2H2(Ωε)×H2(Ωε)

).

Thus

‖Nε(u, v)‖L2(Ωε)×L2(Ωε) ≤ C(‖(u, v)‖3

H2(Ωε)×H2(Ωε) + ‖(u, v)‖2H2(Ωε)×H2(Ωε)

).

Note that

Nε,1(u1, v1) −Nε,1(u2, v2) = μ1(u21 + u1u2 + u2

2)(u1 − u2) + 3μ1Uε,Q(u1 + u2)(u1 − u2)

+ βv21(u1 − u2) + βu2(v1 + v2)(v1 − v2)

+ 2βVε,Qv1(u1 − u2) + 2βVε,Qu2(v1 − v2).

Thus

‖Nε,1(u1, v1) −Nε,1(u2, v2)‖L2(Ωε)

≤ C(‖(u, v)‖2

H2(Ωε)×H2(Ωε) + ‖(u, v)‖H2(Ωε)×H2(Ωε)

)‖(u1, v1) − (u2, v2)‖H2(Ωε)×H2(Ωε)

and hence

‖Nε(u1, v1) − Nε(u2, v2)‖L2(Ωε)×L2(Ωε)

≤ C(‖(u, v)‖2

H2(Ωε)×H2(Ωε) + ‖(u, v)‖H2(Ωε)×H2(Ωε)

)‖(u1, v1) − (u2, v2)‖H2(Ωε)×H2(Ωε).

This completes the proof of the lemma. �Our main result in this subsection is as follows.

Proposition 2.6. There exists ε0 > 0 such that for all 0 < ε < ε0 and for any P ∈ ΛK , there exists a unique solution ((uε,P, vε,P), {cij}) to the problem (2.11). Moreover, (uε,P, vε,P) is C1 in P and the following estimate holds

‖(uε,P, vε,P)‖H2(Ωε)×H2(Ωε) ≤ CKεM

for some constants C > 0.

Proof. Let A(h) denotes the solution to (2.1), then problem (2.11) can be rewritten as

(u, v) = A(−Sε(Uε,P, Vε,P) − Nε(u, v)).


Let T(u, v) := A(−Sε(Uε,P, Vε,P) −Nε(u, v)). Then solving the equation (2.11) is equivalent to finding the fixed point of T in some suitable space.

For some r > 0 which will be given later, define

B :={(u, v) ∈ C2(Ωε) × C2(Ωε) : ‖(u, v)‖H2(Ωε)×H2(Ωε) ≤ KrεM

}. (2.16)

Thus solving the equation (2.11) is equivalent to finding the fixed point of T onto B.We claim that

T is a contraction mapping onto B for some r > 0.

In fact, by Lemma 2.4, we have

‖T(u, v)‖H2(Ωε)×H2(Ωε) ≤ C(‖Sε(Uε,Q, Vε,Q)‖L2(Ωε)×L2(Ωε) + ‖Nε(u, v)‖L2(Ωε)×L2(Ωε)

)≤ C

(KεM + ‖(u, v)‖2

H2(Ωε)×H2(Ωε) + ‖(u, v)‖3H2(Ωε)×H2(Ωε)

)≤ KrεM

for some r > 0. Thus T is a mapping onto B.Furthermore, for ε small enough and some r > 0, by Lemma 2.5, we have

‖T(u1, v1) − T(u2, v2)‖H2(Ωε)×H2(Ωε)

= ‖Nε(u1, v1) − Nε(u2, v2)‖H2(Ωε)×H2(Ωε)

≤ C(‖(u1, v1)‖H2(Ωε)×H2(Ωε) + ‖(u2, v2)‖H2(Ωε)×H2(Ωε)

)‖(u1, v1) − (u2, v2)‖H2(Ωε)×H2(Ωε)

≤ 12‖(u1, v1) − (u2, v2)‖H2(Ωε)×H2(Ωε).

Hence, T is a contraction mapping onto B.According to Banach Fixed Point Theorem, we obtain that T has a unique fixed point (uε,P, vε,P) in B.

This implies that (uε,P, vε,P) is a unique solution to (2.11) satisfies

‖(uε,P, vε,P)‖H2(Ωε)×H2(Ωε) ≤ CKεM .

Next, we prove (uε,P, vε,P) is C1 in P. Define a mapping from ΛK ×H× R3K to H× R

3K

Hε(P, (u, v), c) =

⎛⎜⎜⎜⎝

(Δ − I)−1Sε(Uε,P + u, Vε,P + v) −∑

ij cij(Δ − I)−1Zij

〈(u, v), (Δ − I)−1Z11〉· · ·〈(u, v), (Δ − I)−1Zk3〉

⎞⎟⎟⎟⎠ .

For any P ∈ ΛK , (2.11) has a unique solution ((uε,P, vε,P), cε,P), i.e.

Hε(P, (uε,P, vε,P), cε,P) = 0.

By a direct calculation,


∂Hε(P, (uε,P, vε,P), cε,P)∂((u, v), c)

[ψ,d]

=

⎛⎜⎜⎜⎝

(Δ − I)−1Lε[Uε,P + uε,P, Vε,P + vε,P](ψ) −∑

ij dij(Δ − I)−1Zij

〈ψ, (Δ − I)−1Z11〉· · ·〈ψ, (Δ − I)−1Zk3〉

⎞⎟⎟⎟⎠ ,

where

Lε,1[Uε,P + uε,P, Vε,P + vε,P](ψ) = Δψ1 − ψ1 + 3μ1(Uε,P + uε,P)2ψ1 + β(Vε,P + vε,P)2ψ1

+ 2β(Uε,P + uε,P)(Vε,P + vε,P)ψ2,

Lε,2[Uε,P + uε,P, Vε,P + vε,P](ψ) = Δψ2 − ψ2 + 3μ2(Vε,P + vε,P)2ψ2 + β(Uε,P + uε,P)2ψ2

+ 2(βUε,P + uε,P)(Vε,P + vε,P)ψ1.

Let ∂Hε(P,(uε,P,vε,P),cε,P)∂((u,v),c) [ψ, d] = 0, then (ψ, d) is a solution to

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Lε[Uε,P + uε,P, Vε,P + vε,P](ψ) =k∑

i=1

3∑j=1

dijZij in Ωε,

∫Ωε

(ψ1Zij,1 + ψ2Zij,2) = 0, i = 1, 2, · · · , k, j = 1, 2, 3,

∂ψ1

∂ν= ∂ψ2

∂ν= 0 on ∂Ωε.

Thus ψ = A(N1(ψ), N2(ψ)) where

N1(ψ) = 3μ1(U2ε,P − (Uε,P + uε,P)2)ψ1 + β(V 2

ε,P − (Vε,P + vε,P)2)ψ1

+ 2β(Uε,PVε,P − (Uε,P + uε,P)(Vε,P + vε,P))ψ2,

N2(ψ) = 3μ2(V 2ε,P − (Vε,P + vε,P)2)ψ2 + β(U2

ε,P − (Uε,P + uε,P)2)ψ2

+ 2β(Uε,PVε,P − (Uε,P + uε,P)(Vε,P + vε,P))ψ1.

By Lemma 2.2, we can easily obtain that

‖ψ‖H2(Ωε)×H2(Ωε) ≤ C‖N(ψ)‖L2(Ωε)×L2(Ωε) ≤ CKεM‖ψ‖H2(Ωε)×H2(Ωε).

Thus ψ = 0 which implies that

∂Hε(P, (uε,P, vε,P), cε,P)∂((u, v), c)

is invertible. Since

∂Hε(P, (u, v), c)∂((u, v), c) and ∂Hε(P, (u, v), c)

∂P

are continuous. Then by the implicit function theorem, we know that (uε,P, vε,P) is C1 in P. �


2.3. A maximum problem

In this subsection, we study a maximum problem.Fix P ∈ ΛK , we define a new functional from ΛK to R as follows

Mε(P) := Jε(uε,P, vε,P) = Jε(Uε,P + uε,P, Vε,P + vε,P), (2.17)

where (uε,P, vε,P) is the solution to (2.11).Our main proposition in this subsection is as follows.

Proposition 2.7. The maximum problem

Cε,K = maxP∈ΛK

Mε(P)

is achieved at some Pε ∈ intΛK , the interior part of ΛK .

Proof. It is easy to see that Mε(P) is continuous in compact set ΛK . So Cε,K is achieved by some Pε in ΛK , namely

Mε(Pε) = maxP∈ΛK

{Mε(P)}. (2.18)

For any P ∈ Λk, a simple computation shows that

Mε(P) = Jε (Uε,P + uε,P, Vε,P + vε,P)

= Jε (Uε,P, Vε,P) − 12‖ (uε,P, vε,P) ‖2

H1(Ωε)×H1(Ωε)

+∫Ωε

[∇ (Uε,P + uε,P)∇uε,P + (Uε,P + uε,P)uε,P]

+∫Ωε

[∇ (Vε,P + vε,P)∇vε,P + (Vε,P + vε,P) vε,P]

− μ1

4

∫Ωε

[(Uε,P + uε,P)4 − U4

ε,P

]− μ2

4

∫Ωε

[(Vε,P + vε,P)4 − V 4

ε,P

]

− β

2

∫Ωε

[(Uε,P + uε,P)2 (Vε,P + vε,P)2 − U2

ε,PV2ε,P

]

= Jε (Uε,P, Vε,P) − 12‖ (uε,P, vε,P) ‖2

H1(Ωε)×H1(Ωε)

− μ1

4

∫Ωε

[(Uε,P + uε,P)4 − 4 (Uε,P + uε,P)3 uε,P − U4

ε,P

]

− μ2

4

∫Ωε

[(Vε,P + vε,P)4 − 4 (Uε,P + uε,P)3 uε,P − V 4

ε,P

]

− β

2

∫Ωε

[(Uε,P + uε,P)2 (Vε,P + vε,P)2 − (Uε,P + uε,P) (Vε,P + vε,P)2 uε,P

− (Uε,P + uε,P)2 (Vε,P + vε,P) vε,P − U2ε,PV

2ε,P

].


Since ∫Ωε

∣∣∣(Uε,P + uε,P)4 − 4 (Uε,P + uε,P)3 uε,P − U4ε,P

∣∣∣=∫Ωε

∣∣6U2ε,Pu

2ε,P + 4Uε,Pu

3ε,P + u3

ε,P∣∣

≤ C(‖uε,P‖2

H1(Ωε) + ‖uε,P‖3H1(Ωε) + ‖uε,P‖4

H1(Ωε)

).

Then by Lemma A.5, we have

Mε(P) = Jε (Uε,P, Vε,P) + O(‖(uε,P, vε,P)‖2H1(Ωε)×H1(Ωε))

= KI(U, V ) − μ1a4 + μ2b

4 + 2βa2b2

2

⎛⎝ K∑

i=1B(Pi) +

∑i�=j

B(Pi, Pj)

⎞⎠

+ o(ω(M | ln ε|))

= KI(U, V ) − (μ1a4 + μ2b

4 + 2βa2b2)(γ + o(1))2

×

⎛⎝ K∑

j=1ω

(2dist(Pj , ∂Ω)

ε

)+∑i�=j

ω

(|Pi − Pj |

ε

)⎞⎠+ o(ω(M | ln ε|)).

Let KΩ(r) be the maximum number of nonoverlapping balls with radius r packed in Ω. Now we choose Ksuch that

1 ≤ K ≤ KΩ

((M + 3)ε| ln ε|

2

).

Let P̂ = (P̂1, P̂2, · · · , P̂K) be the centers of arbitrary K balls. It is easy to see that P̂ ∈ ΛK and

ω

(2dis(P̂i, ∂Ω)

ε

)≤ εM+3 and ω

(|P̂i − P̂j |

ε

)≤ εM+3

for i, j = 1, 2 · · · , K and i = j. Hence,

Mε(Pε) ≥ Mε(P̂)

= KI(U, V ) − μ1a4 + μ2b

4 + 2βa2b2

2

⎛⎝∫

R3

ω3e−y1dy + o(1)

⎞⎠

×

⎛⎝ K∑

j=1ω

(2dist(P̂j , ∂Ω)

ε

)+∑i�=j

ω

(|P̂i − P̂j |

ε

)⎞⎠+ o(ω((M + 3)| ln ε|))

≥ KI(U, V ) − (μ1a4 + μ2b

4 + 2βa2b2) (γ + o(1))KεM+3 + o(ω((M + 3)| ln ε|))= KI(U, V ) + o(ω(M | ln ε|)). (2.19)

If Pε ∈ ∂ΛK , then |Pε,i − Pε,j | = Mε| ln ε| or dist(Pε,k, ∂Ω) = Mε| ln ε| for some i, j, k. In both two cases, we have


Mε(Pε) ≤ KI(U, V ) − (μ1a4 + μ2b

4 + 2βa2b2)(γ + o(1))2 ω(M | ln ε|) + o(ω(M | ln ε|)),

which contradicts to (2.19). Thus we complete our proof. �2.4. The proof of Theorem 1.1

Instead of proving Theorem 1.1 directly, we will Theorem 1.3 and thus Theorem 1.1 comes immediately.

Proof of Theorem 1.3. By Proposition 2.6, for each P ∈ Λk, there exists (uε,P, vε,P) satisfies⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

Sε(Uε,P + uε,P, Vε,P + vε,P) =∑K

i=1∑3

j=1 cijZij in Ωε,∫Ωε

(uε,PZij,1 + vε,PZij,2) = 0,

∂uε,P

∂ν= ∂uε,P

∂ν= 0 on ∂Ωε.

(2.20)

Let Pε = (Pε,1, Pε,2, · · · , Pε,K) be the maximum point of ΛK and denote

(uε,Pε, vε,Pε

) = (Uε,Pε, Vε,Pε

) + (uε,Pε, vε,Pε

). (2.21)

By Proposition 2.7, we have ∂Mε(Pε)

∂Pij= 0, namely

0 =∫Ωε

(∇uε,Pε

∇∂uε,Pε

∂Pij+ ∇vε,Pε

∇∂vε,Pε

∂Pij

)

+∫Ωε

(uε,Pε

∂uε,Pε

∂Pij+ vε,Pε

∂vε,Pε

∂Pij

)

+∫Ωε

(μ1u

3ε,Pε

∂uε,Pε

∂Pij+ μ2v

3ε,Pε

∂vε,Pε

∂Pij

)

+∫Ωε

(βuε,Pε

v2ε,Pε

∂uε,Pε

∂Pij+ βu2

ε,Pεvε,Pε

∂vε,Pε

∂Pij

). (2.22)

By equation (2.20), we obtain the following system

K∑k=1

3∑l=1

ckl

∫Ωε

(∂(Uε,Pε

+ uε,Pε)

∂PijZkl,1 + ∂(Vε,Pε

+ vε,Pε)

∂PijZkl,2

)= 0. (2.23)

Since ∫Ωε

(uε,PεZkl,1 + vε,Pε

Zkl,2) = 0,

then

∂

∂Pij

∫(uε,Pε

Zkl,1 + vε,PεZkl,2) = 0,

Ωε


more precisely,

∫Ωε

(∂uε,Pε

∂PijZkl,1 + ∂vε,Pε

∂PijZkl,2

)= −

∫Ωε

(uε,Pε

∂Zkl,1

∂Pij+ vε,Pε

∂Zkl,2

∂Pij

).

If k = i, we have

∫Ωε

(∂(Uε,Pε

+ uε,Pε)

∂PijZkl,1 + ∂(Vε,Pε

+ vε,Pε)

∂PijZkl,2

)

=∫Ωε

(∂Uε,Pε,i

∂PijZkl,1 +

∂Vε,Pε,i

∂PijZkl,2

)= O(εM ).

If k = i and l = j, we have

∫Ωε

(∂(Uε,Pε

+ uε,Pε)

∂PijZil,1 + ∂(Vε,Pε

+ vε,Pε)

∂PijZil,2

)

=∫Ωε

(∂Uε,Pε

∂PijZil,1 +

∂Vε,Pε,i

∂PijZil,2

)

+ O

(‖(uε,Pε

, vε,Pε)‖L2(Ωε)×L2(Ωε)

∥∥∥∥(∂Zil,1

∂Pij,∂Zil,2

∂Pij

)∥∥∥∥L2(Ωε)×L2(Ωε)

)

= O(KεM ).

If k = i and l = j, we have

∫Ωε

(∂(Uε,Pε

+ uε,Pε)

∂PijZij,1 + ∂(Vε,Pε

+ vε,Pε)

∂PijZij,2

)

=∫Ωε

(∂Uε,Pε,i

∂PijZij,1 +

∂Vε,Pε,i

∂PijZij,2

)+ O(KεM )

= −∫R3

∂2

∂x2j

(U + V ) + o(1).

Thus the equation (2.23) is a diagonally dominant system, then cij = 0, i = 1, · · · , k, j = 1, 2, 3. Hence (uε,Pε

, vε,Pε) = (Uε,Pε

+ uε,Pε, Vε,Pε

+ vε,Pε) is a solution to (Cε) if uε,Pε

and vε,Pεare both positive.

Since H2(Ωε) can be continuously embedded into Cγ(Ωε), then ‖(uε,Pε, vε,Pε

)‖L∞(Ωε)×L∞(Ωε) → 0 as ε → 0. If uε(x) < 0, then 0 < Uε,Pε

(x) < −uε,P(x) which implies Uε,Pε→ 0 and Vε,Pε

= baUε,Pε

→ 0 as ε → 0. Thus uε,− = max{0, −uε} ⇒ 0 in Ωε and vε = Vε,Pε

+ vε,Pε⇒ 0 in Ωε ∩ {uε < 0} as ε → 0. Note

that ∫Ωε

(|∇uε,−|2 + u2

ε,−)

=∫Ωε

(μ1u

4ε,− + βu2

ε,−v2ε

)

≤ C‖u2ε,− + v2

ε‖L∞(Ωε∩{uε<0})

∫u2ε,−

Ωε


≤ o(1)∫Ωε

(|∇uε,−|2 + u2

ε,−),

then uε,− = 0 which leads to a contradiction. Thus uε,Pεand vε,Pε

are both nonnegative. By strong maximum principle, we obtain that uε,Pε

and vε,Pεare both positive.

Furthermore, since ‖(uε,Pε, vε,Pε

)‖L∞(Ωε)×L∞(Ωε) = O(‖(uε,Pε

, vε,Pε)‖H2(Ωε)×H2(Ωε)

), then

uε,Pε(x) = Uε,Pε

+ uε,Pε={

U(x− Pε,i

ε ) + O(εM2 ) in Ωε,Pε,i

, i = 1, 2, · · · ,K,

O(εM2 ) in Ωε \ ∪K

i=1Ωε,Pε,i,

and

vε,Pε(x) = Vε,Pε

+ vε,Pε={

V (x− Pε,i

ε ) + O(εM2 ) in Ωε,Pε,i

, i = 1, 2, · · · ,K,

O(εM2 ) in Ωε \ ∪K

i=1Ωε,Pε,i.

Thus there exists R > 1 such that the maximum points of uε,Pεand vε,Pε

locate in ∪Ki=1BR

(Pε,i

ε

).

Let P ′ε,i

ε and P ′′

ε,i

ε be the maximum points of uε and vε in BR

(Pε,i

ε

)respectively. Since uε,Pε

=

U(x− Pε,i

ε

)+ o(1) and vε,Pε

= V(x− Pε,i

ε

)+ o(1). Then

P ′ε,i−Pε,i

ε = o(1) and P ′′

ε,i−Pε,i

ε = o(1) as ε → 0.

Let ξε = uε,Pε− U

(x− P ′

ε,i

ε

)and ζε = vε,Pε

− V(x− P ′′

ε,i

ε

). Then (ξε, ζε) satisfies

⎧⎪⎨⎪⎩

Δξε − ξε + μ1

[u3ε,Pε

− U3(x− P ′

ε,i

ε

)]+ β

[uε,Pε

v2ε,Pε

− U(x− P ′

ε,i

ε

)V 2(x− P ′

ε,i

ε

)]= 0,

Δζε − ζε + μ2

[v3ε,Pε

− V 3(x− P ′′

ε,i

ε

)]+ β

[u2ε,Pε

vε,Pε− U2

(x− P ′′

ε,i

ε

)V(x− P ′′

ε,i

ε

)]= 0

in BR

(Pε,i

ε

). Thus by the standard regularity of elliptic equations, we have

‖ξε‖C1,α

(BR

2

(Pε,iε

)) → 0 and ‖ζε‖C1,α

(BR

2

(Pε,iε

)) → 0

as ε → 0 for some 0 < α < 1. Hence, uε has exactly K local maximum points P′ε,1ε ,

P ′ε,2ε , · · · , P

′ε,K

ε such that P ′

ε,i−Pε,i

ε = o(1) and vε has exactly K local maximum points P′′ε,1ε ,

P ′′ε,2ε , · · · , P

′′ε,K

ε such that P′′ε,i−Pε,i

ε = o(1)for each i = 1, 2 · · · , K as ε → 0. This also implies P

′′ε,i−P ′

ε,i

ε = o(1) for each i = 1, 2 · · · , K as ε → 0, which shows that uε,Pε

and vε,Pεare synchronized.

From the above discussion, if we take K = [ε−m], then Jε(uε,Pε, vε,Pε

) = O(ε−m). If we take K =O(

1εN | ln ε|N

), then Jε(uε,Pε

, vε,Pε) = O

(1

εN | ln ε|N).

Finally, let (U0, V0) be the principle eigenfunction for the operator L(u, v) = (L1(u, v), L2(u, v)) in H1(R3) ×H1(R3) where

L1(u, v) = Δu− u + 3μ1U2u + βV 2u + 2βUV v, L2(u, v) = Δv − v + 3μ1V

2v + βU2v + 2βUV u.

According to Lemma 2.1, we know that the principle eigenvalue λ1 of L(u, v) is positive. Set (U0,i, V0,i) =χj(x)

(U0

(x− Pε,i

), V0

(x− Pε,i

))for i = 1, 2, · · · , K. Then it is easy to see that
ε ε


∫Ωε

(|∇U0,i|2 + U20,i + |∇V0,i|2 + V 2

0,i)

−∫Ωε

(3μ1u2ε,Pε

U20,i + 3μ2v

2ε,Pε

V 20,i + βv2

ε,PεU2

0,i + βu2ε,Pε

V 20,i + 4βuε,Pε

vε,PεU0,iV0,i)

=∫Ωε

(|∇U0,i|2 + U20,i + |∇V0,i|2 + V 2

0,i)

−∫Ωε

(3μ1U2Pε,i

U20,i + 3μ2V

2Pε,i

V 20,i + βV 2

Pε,iU2

0,i + βU2Pε,i

V 20,i + 4βUPε,i

VPε,iU0,iV0,i)

+[ ∫Ωε

(3μ1U2Pε,i

U20,i + 3μ2V

2Pε,i

V 20,i + βV 2

Pε,iU2

0,i + βU2Pε,i

V 20,i + 4βUPε,i

VPε,iU0,iV0,i)

−∫Ωε

(3μ1u2ε,Pε

U20,i + 3μ2v

2ε,Pε

V 20,i + βv2

εU20,i + βu2

ε,PεV 2

0,i + 4βuε,Pεvε,Pε

U0,iV0,i)]

≤ −λ1

2

∫Ωε

(U20,i + V 2

0,i)

for i = 1, 2, · · · , K. Since (U0,i, V0,i) are nonlinear independent, then the Morse index of (uε,Pε, vε,Pε

) is at least K. �Proof of Theorem 1.1. By changing variables, Theorem 1.1 is a direct result of Theorem 1.3. �3. Segregated solutions

In this section, we consider segregated solutions and prove Theorem 1.2 by proving Theorem 1.3. We firstly prove the existence of solutions to a linear projected problem for any (P, Q) ∈ ΛK × ΛL in Subsection 3.1. Secondly, we use Banach Fixed Point Theorem to prove the existence of solutions to a nonlinear projection problem for any (P, Q) ∈ ΛK × ΛL in Subsection 3.2. Thirdly, in Subsection 3.3, we study a maximum problem in (P, Q) ∈ ΛK ×ΛL and prove the maximum problem is achieved by some interior points (Pε, Qε)of ΛK × ΛL. Finally, we prove Theorem 1.2 by proving Theorem 1.4 in Subsection 3.4.

3.1. A linear projected problem

In this subsection, we consider the following linear projected problem

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Lε,1(u, v) = h1 +K∑i=1

3∑j=1

aijXij in Ωε,

Lε,2(u, v) = h2 +L∑

s=1

3∑j=1

bsjYsj in Ωε,

∫Ωε

uXij =∫Ωε

vYsj = 0, i = 1, 2, · · · ,K, s = 1, 2, · · · , L, j = 1, 2, 3,

∂u = ∂v = 0 on ∂Ωε.

(3.1)

∂ν ∂ν


Here Xij = χPi

∂∂xj

U(x − Pi

ε ), Ysj = χQs

∂∂xj

V (x − Qs

ε ) and

Lε,1(u, v) = Δu− u + 3μ1U2ε,Pu + βV 2

ε,Qu + 2βUε,PVε,Qv,

Lε,2(u, v) = Δv − v + 3μ2V2ε,Qv + βU2

ε,Pv + 2βUε,PVε,Qu. (3.2)

The following lemma deals with a prior estimate for solutions to (3.1).

Lemma 3.1. Let h = (h1, h2) ∈ L2(Ωε) ×L2(Ωε), assume that ((u, v), (aij), (bsj)) is a solution to (3.1), then there exists ε0 > 0 and C > 0 such that for any 0 < ε < ε0 and (P, Q) ∈ ΛK × ΛL, one has

‖(u, v)‖H2(Ωε)×H2(Ωε) ≤ C‖h‖L2(Ωε)×L2(Ωε). (3.3)

Proof. Multiplying the first equation of (3.1) against Xij and integrating over Ωε, we have

∫Ωε

Lε,1(u, v)Xijdx =∫Ωε

h1Xijdx + aij

∫Ωε

X2ijdx. (3.4)

Note that

ΔuχPiU ij = div

(χPi

U ij∇u)− div

(∇(χPi

U ij)u)

+ ΔχPiU ij + χPi

ΔU iju + 2∇χPi∇U iju

where U ij = ∂UPi

∂xj. Then

∫Ωε

Lε,1(u, v)Xijdx =∫Ωε

χPiu(ΔU ij − U ij + 3μ1U

2PiU ij

)+∫Ωε

χPiUij

[3μ1(U2ε,P − U2

Pi

)u]

+∫Ωε

(ΔχPiUiju + 2∇χPi

∇Uiju) +∫Ωε

βV2ε,QuχPi

U ij

+ 2β∫Ωε

Uε,PV ε,QvχPiU ij .

By the non-degeneracy of U , we obtain that

∫Ωε

χPiu(ΔU ij − U ij + 3μ1U

2PiU ij

)= 0.

By Lemma A.2 and Hölder’s inequality, we have

∫Ωε

χPiUij

(U2ε,P − U2

Pi

)u

≤ CεM2

∫Ωε

U ijUPiu ≤ Cε

M2 ‖u‖L2(Ωε).

By Hölder’s inequality, we get


∫Ωε

(ΔχPiUiju + 2∇χPi

∇Uiju)

≤ C

⎡⎣ ∫

Ωε

(|ΔχPi

Uij |2 + |∇χPi∇Uij |2

)⎤⎦12

‖u‖L2(Ωε)

≤ C(M + 1)| ln ε|

⎛⎜⎜⎝

(M2−1)| ln ε|2(M+1)∫

(M−1)| ln ε|2

e−2r

⎞⎟⎟⎠

12

‖u‖L2(Ωε)

≤ CεM−1

2 ‖u‖L2(Ωε).

It is easy to see that∫Ωε

βV2ε,QuχPi

U ij + 2β∫Ωε

Uε,PV ε,QvχPiU ij = o(εM

2(‖u‖L2(Ωε) + ‖v‖L2(Ωε))

).

Thus we have ∣∣∣∣∣∣∫Ωε

Lε,1(u, v)Xijdx

∣∣∣∣∣∣ ≤ CεM−1

2 ‖(u, v)‖L2(Ωε)×L2(Ωε). (3.5)

Since, by Lebesgue Dominated Convergence Theorem, we have∫Ωε

X2ijdx =

∫Ωε

χ2PiU

2ij =

∫R3

U2ij + o(1), (3.6)

and by Hölder’s inequality, we have∫Ωε

h1Xijdx =∫Ωε

χPih1U ij ≤ ‖U ij‖L2(R3)‖h1‖L2(Ωε). (3.7)

Then according to (3.4)–(3.7), we have

|aij | ≤ C(ε


). (3.8)

Similarly,

|bij | ≤ C(ε


). (3.9)

Let (uPi, vPi

) = χ′Pi

(u, v), (uQs, vQs

) = χ′Qs

(u, v) and

(uc, vc) =(u−

K∑i=1

uPi−

L∑s=1

uQs, v −

K∑i=1

vPi−

L∑s=1

vQs

),

where χ′P = χ

(2M

2

∣∣z − Pi∣∣) for i = 1, 2, · · · , K. Then (uc, vc) satisfies

i (M −1)| ln ε| ε


⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Lε,1(uc, vc) =[1 −∑is

(χ′Pi

+ χ′Qs

)]h1 − 2

∑is

∇u∇(χ′Pi

+ χ′Qs

) −∑is

Δ(χ′Pi

+ χ′Qs

)u in Ωε,

Lε,2(uc, vc) =[1 −∑

is(χ′Pi

+ χ′Qs

)]h2 − 2

∑is ∇v∇(χ′

Pi+ χ′

Qs) −∑

is Δ(χ′Pi

+ χ′Qs

)v in Ωε,

∂uc

∂ν= ∂vc

∂ν= 0 on ∂Ωε.

(3.10)

Multiplying (uc, vc) on both sides of (3.10), by Lemma 2.1, we can easily get

‖(uc, vc)‖H1(Ωε)×H1(Ωε) ≤ C

∥∥∥∥∥(

1 −∑is

(χ′Pi

+ χ′Qs

))

(h1 + h2)

∥∥∥∥∥L2(Ωε)

+ C

∥∥∥∥∥2∑is

∇(u + v)∇(χ′Pi

+ χ′Qs

) +∑is

Δ(χ′Pi

+ χ′Qs

)(u + v)

∥∥∥∥∥L2(Ωε)

.

Thus by equation (3.10), we have

‖(uc, vc)‖H2(Ωε)×H2(Ωε)

≤ C(‖Δ(uc, vc)‖L2(Ωε)×H2(Ωε) + ‖(uc, vc)‖L2(Ωε)×H2(Ωε)

)≤ C

∥∥∥∥∥(

1 −∑is

(χ′Pi

+ χ′Qs

))

(h1, h2)

∥∥∥∥∥L2(Ωε)×L2(Ωε)

+ C

∥∥∥∥∥2∑is

∇(u, v)∇(χ′Pi

+ χ′Qs

) +∑is

Δ(χ′Pi

+ χ′Qs

)(u, v)

∥∥∥∥∥L2(Ωε)×L2(Ωε)

. (3.11)

Now we complete our proof by a contradiction argument. Assume ‖hn‖L2(Ωεn )×L2(Ωεn ) → 0 as n → +∞and ‖(un, vn)‖H2(Ωεn )×H2(Ωεn ) = 1 for any positive integer n. According to (3.11), we have

‖(un,c, vn,c)‖H2(Ωεn )×H2(Ωεn ) → 0

as n → +∞. Thus without loss of generality, we may assume that there exists some i ∈ {1, 2, · · · , K} and a subsequence {εnk

} ⊂ {εn} such that ‖(unk,Pi, vnk,Pi

)‖H2(Ωεnk)×H2(Ωεnk

) ≥ 12(K+L) . It is easy to see that

(un,Pi, vn,Pi

) is bounded in H1(Ωεn) ×H1(Ωεn). Thus (un,Pi, vn,Pi

) ⇀ (u∞, v∞) weakly in H1(R3) ×H1(R3). Moreover, (u∞, v∞) satisfies

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Δu− u + 3μ1U2u = 0 in R

3,

Δv − v + βU2v = 0 in R

3,∫R3

u∂U

∂xj= 0, j = 1, 2, 3.

(3.12)

By the non-degeneracy of U , we obtain u∞ = 0. Since β < min{μ1, μ2}, then multiplying v on both sides of the second equation of (3.12) and integrate over Ω, we can easily get v∞ = 0. Thus (u∞, v∞) = (0, 0)which leads to a contradiction. Thus we complete the proof of this Lemma. �Proposition 3.2. Let ε0, ρ0 and C be the positive number in Lemma 3.1, for any 0 < ε < ε0, ρ > ρ0 and any given h = (h1, h2) ∈ L2(Ωε) ×L2(Ωε), there exists a unique solution ((u, v), {cij}) to the problem (3.1)


satisfying

‖(u, v)‖H2(Ωε)×H2(Ωε) ≤ C‖h‖L2(Ωε)×L2(Ωε).

Proof. Let

H :={

(u, v) ∈ H1N (Ωε) ×H1

N (Ωε) :∫Ωε

uXij =∫Ωε

vYsj,2 = 0,

i = 1, 2, · · · ,K, s = 1, 2, · · · , L, j = 1, 2, 3}

be the Hilbert space and by Riesz lemma we can rewrite the problem (3.1) as

(u, v) + K(u, v) = h in H,

where

〈K(u, v), (ψ, φ)〉 = −∫Ωε

[3μ1U

2ε,Puψ + 3μ2V

2ε,Qvφ

+ β(V 2ε,Quψ + U

2ε,Pvφ) + 2βUε,PV ε,Q(vψ + uφ)

]

and

〈h, (ψ, φ) =∫Ωε

(h1ψ + h2φ)

for each (ψ, φ) ∈ H. According to Rellich embedding theorem, we can easily obtain that K is a compact operator on H. By Lemma 3.1 and Fredholm alternative theorem, we know that (3.1) has a unique solution for any h ∈ L2(Ωε) ×L2(Ωε). Again by the smooth assumption on the boundary ∂Ωε and elliptic regularity theory we finish the proof of the proposition. �3.2. A nonlinear projected problem

In this subsection, for (P, Q) ∈ ΛK × ΛL, we consider the following nonlinear projected problem

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Sε,1(u + Uε,P, v + V ε,Q) =∑ij

aijXij in Ωε,


bijYij in Ωε,

∂u

∂ν= ∂v

∂ν= 0 on ∂Ωε,∫

Ωε

uXij =∫Ωε

vYsj = 0, i = 1, 2, · · · ,K, s = 1, 2, · · · , L, j = 1, 2, 3,

(3.13)

where

Sε,1(u, v) = Δu− u + μ1u3 + βuv2,

Sε,2(u, v) = Δv − v + μ2v3 + βu2v. (3.14)


We denote Nε(u, v) := (Nε,1(u, v), Nε,2(u, v))T , then equation (3.13) can be rewritten as

⎧⎨⎩Lε,1(u, v) = −Sε,1(Uε,P, V ε,Q) −Nε,1(u, v) +

∑ij aijXij ,

Lε,2(u, v) = −Sε,2(Uε,P, V ε,Q) −Nε,2(u, v) +∑

sj bsjYsj ,(3.15)

where

Nε,1(u, v) = μ1(u + Uε,P)3 − 3μ1U2ε,Pu− μ1U

3ε,P + β(u + Uε,P)(v + V ε,Q)2

− 2βUε,PV ε,Qv − βUε,PV2ε,Q − βV

2ε,Qu,

Nε,2(u, v) = μ2(v + V ε,Q)3 − 3μ2V2ε,Qv − μ2V

3ε,Q + β(u + Uε,P)2(v + V ε,Q)

− 2βUε,PV ε,Qu− βU2ε,PV ε,Q − βU

2ε,Pv. (3.16)

The following lemma gives the estimate of Sε(Uε,P, Vε,Q) = (Sε,1(Uε,P, Vε,Q), Sε,2(Uε,P, Vε,Q))T .

Lemma 3.3. Suppose (P, Q) ∈ ΛK × ΛL and ε > 0 small enough, then

‖Sε(Uε,P, V ε,Q)‖L2(Ωε)×L2(Ωε) ≤ C(K + L)εM


Proof. According to (3.14) and equation (1.10), we have

Sε,1(Uε,P, V ε,P) = ΔUε,P − Uε,P + μ1U3ε,P + βUε,PV

2ε,Q

= μ1U3ε,P + βUε,PV

2ε,Q − μ1

K∑i=1

U3Pi

+ βUε,PV2ε,Q

= μ1

(U

3ε,P −

K∑i=1

U3Pi

)+ βUε,PV

2ε,Q. (3.17)

As proved in Lemma 2.4, we obtain that

∫Ωε

∣∣Sε,1(Uε,P, V ε,P)∣∣2 ≤ C

∫Ωε

(U

3ε,P −

K∑i=1

U3Pi

)2

+ C

∫Ωε

U2ε,PV

4ε,Q ≤ CK(K − 1)ε2M .

Therefore,

‖Sε(Uε,P, V ε,P)‖L2(Ωε)×L2(Ωε) ≤ C(K + L)εM . �The following lemma gives the estimates of Nε(u, v).

Lemma 3.4. Suppose (P, Q) ∈ ΛK × ΛL and ε > 0 small enough, then

‖Nε(u, v)‖L2(Ωε)×L2(Ωε) ≤ C(‖(u, v)‖3

H2(Ωε)×H2(Ωε) + ‖(u, v)‖2H2(Ωε)×H2(Ωε)

)and


‖Nε(u1, v1) − Nε(u2, v2)‖L2(Ωε)×L2(Ωε)

≤ C(‖(u, v)‖H2(Ωε)×H2(Ωε) + ‖(u, v)‖2

H2(Ωε)×H2(Ωε)

)‖(u1, v1) − (u2, v2)‖H2(Ωε)×H2(Ωε)

for some C > 0 which is independent of ε, K and L.

Proof. The proof similar to Lemma 2.5, we omit the details. �Our main result in this subsection is as follows.

Proposition 3.5. There exists ε0 > 0, ρ0 > 0 such that for all 0 < ε < ε0, ρ > ρ0 and for any (P, Q) ∈ΛK × ΛL, there exists a unique solution ((uε,P,Q, vε,P,Q), (aij), (bsj)) to the problem (3.13). Moreover, (uε,P,Q, vε,P,Q) is C1 in (P,Q), and for some constants C > 0 which is independent of K, L and ε, the following estimate holds

‖(uε,P,Q, vε,P,Q)‖H2(Ωε)×H2(Ωε) ≤ C(K + L)εM . (3.18)

Proof. Let A(h) be the solution to (3.1), T(u, v) := A(−Sε(Uε,P, V ε,Q) − Nε(u, v)), and define

B :={(u, v) ∈ H2(Ωε) ×H2(Ωε) : ‖(u, v)‖H2(Ωε)×H2(Ωε) ≤ (K + L)rεM

}(3.19)

For some r > 0. As proved in Proposition 2.6, we see that T is a contraction mapping onto B for some r > 0and thus T has a unique fixed point (uε,P,Q, vε,P,Q) in B by Banach Fixed Point Theorem. This implies that (uε,P,Q, vε,P,Q) is a unique solution to (3.13) satisfies

‖(uε,P,Q, vε,P,Q)‖H2(Ωε)×H2(Ωε) ≤ C(K + L)εM .

Next, we prove (uε,P,Q, vε,P,Q) is C1 in (P,Q). Define a mapping from ΛK×H×R3(K+L) to H×R

3(K+L)

Hε(P,Q, (u, v),a,b) =

⎛⎜⎜⎜⎜⎜⎜⎝

(Δ − I)−1Sε,1(Uε,P + u, V ε,Q + v) −∑

ij aij(Δ − I)−1Xij

(Δ − I)−1Sε,2(Uε,P + u, V ε,Q + v) −∑

ij bij(Δ − I)−1Yij

〈u, (Δ − I)−1X11〉· · ·〈v, (Δ − I)−1YL3〉

⎞⎟⎟⎟⎟⎟⎟⎠ .

For any (P,Q) ∈ ΛK × ΛL, (3.13) has a unique solution ((uε,P,Q, vε,P,Q), aε,P,Q,bε,P,Q), i.e.

Hε(P,Q, (uε,P,Q, vε,P,Q),aε,P,Q,bε,P,Q) = 0.

By direct calculation,

∂Hε(P,Q, (uε,P,Q, vε,P,Q),aε,P,Q,bε,P,Q)∂((u, v),a,b) [ψ, c,d]

=

⎛⎜⎜⎜⎜⎜⎜⎝

(Δ − I)−1Lε,1[Uε,P + uε,P,Q, V ε,Q + vε,P,Q](ψ) −∑

ij cij(Δ − I)−1Xij

(Δ − I)−1Lε,2[Uε,P + uε,P,Q, V ε,Q + vε,P,Q](ψ) −∑

ij dij(Δ − I)−1Yij

〈ψ1, (Δ − I)−1X11〉· · ·〈ψ , (Δ − I)−1Y 〉

⎞⎟⎟⎟⎟⎟⎟⎠ ,

2 L3


where

Lε,1[Uε,P + uε,P,Q, V ε,Q + vε,P,Q](ψ)

= Δψ1 − ψ1 + 3μ1(Uε,P + uε,P,Q)2ψ1 + β(V ε,P + vε,P,Q)2ψ1

+ 2β(Uε,P + uε,P,Q)(V ε,Q + vε,P,Q)ψ2,

Lε,2[Uε,P + uε,P,Q, V ε,Q + vε,P,Q](ψ)

= Δψ2 − ψ2 + 3μ2(V ε,Q + vε,P,Q)2ψ2 + β(Uε,P + uε,P,Q)2ψ2

+ 2(βUε,P + uε,P,Q)(V ε,Q + vε,P,Q)ψ1.

Let

∂Hε(P,Q, (uε,P,Q, vε,P,Q),aε,P,Q,bε,P,Q)∂((u, v), c,d) [ψ, c,d] = 0,

then (ψ, c,d) is a solution to⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Lε,1[Uε,P + uε,P,Q, V ε,Q + vε,P,Q](ψ) =K∑i=1

3∑j=1

cijXij in Ωε,

Lε,2[Uε,P + uε,P,Q, V ε,Q + vε,P,Q](ψ) =L∑

s=1

3∑j=1

dsjYsj in Ωε,

∫Ωε

ψ1Xij,1 =∫Ωε

ψ2Ysj = 0, i = 1, 2, · · · ,K, s = 1, 2, · · · , L, j = 1, 2, 3,

∂ψ1

∂ν= ∂ψ2

∂ν= 0 on ∂Ωε.

Thus ψ = A(N1(ψ), N2(ψ)) where

N1(ψ) = 3μ1(U2ε,P − (Uε,P + uε,P,Q)2)ψ1 + β(V 2

ε,Q − (V ε,Q + vε,P,Q)2)ψ1

+ 2β(Uε,PV ε,Q − (Uε,P + uε,P,Q)(V ε,Q + vε,P,Q))ψ2,

N2(ψ) = 3μ2(V2ε,Q − (V ε,Q + vε,P,Q)2)ψ2 + β(U2

ε,P − (Uε,P + uε,P,Q)2)ψ2

+ 2β(Uε,PV ε,Q − (Uε,P + uε,P,Q)(V ε,Q + vε,P,Q))ψ1.

By Lemma 3.1, we can easily obtain that

‖ψ‖H2(Ωε)×H2(Ωε) ≤ C‖N(ψ)‖L2(Ωε)×L2(Ωε) ≤ CKεM‖ψ‖H2(Ωε)×H2(Ωε).

Thus ψ = 0 which implies that

∂Hε(P,Q, (uε,P,Q, vε,P,Q),aε,P,Q,bε,P,Q)∂((u, v),a,b)

is invertible. Since

∂Hε(P, (u, v), c)∂((u, v),a,b) and ∂Hε(P,Q, (u, v),a,b)

∂P,Q

are continuous. Then by the implicit function theorem, we know that (uε,P,Q, vε,P,Q) is C1 in (P,Q). �


3.3. A maximum problem

In this subsection, we study a maximum problem.Fix (P,Q) ∈ ΛK × ΛL, we define a new functional from ΛK × ΛL to R as follows

Mε(P,Q) := Jε(uε,P,Q, vε,P,Q) = Jε(Uε,P + uε,P,Q, V ε,Q + vε,P,Q), (3.20)

where (uε,P,Q, vε,P,Q) is the solution to (3.13).Our main proposition in this subsection is as following.

Proposition 3.6. The maximum problem

Cε,K,L = max(P,Q)∈ΛK×ΛL

Mε(P,Q)

is achieved at some (Pε, Qε) ∈ int(ΛK × ΛL), the interior part of ΛK × ΛL.

Proof. It is easy to see that Mε(P,Q) is continuous in compact set ΛK × ΛL. So Cε,K is achieved by some (Pε, Qε) in ΛK × ΛL, namely

Mε(Pε,Qε) = max(P,Q)∈ΛK×ΛL

{Mε(P,Q)}. (3.21)

For any (P,Q) ∈ ΛK × ΛL, similar to the calculation in Proposition 2.7, we have

Mε(P,Q) = Jε (Uε,P, Vε,Q) + O(‖(uε,P,Q, vε,P,Q)‖2

H1(Ωε)×H1(Ωε)

). (3.22)

According to (3.22) and Lemma A.6, we have

Mε(P,Q) = KIμ1(U) + LIμ2(V ) − 12μ1

⎛⎝ K∑

i=1B(Pi) +

∑i�=j

B(Pi, Pj)

⎞⎠

− 12μ2

⎛⎝ K∑

s=1B(Qs) +

∑s �=l

B(Qs, Ql)

⎞⎠+ o(ω(M | ln ε|))

= KIμ1(U) + LIμ2 −γ + o(1)

2μ1

⎛⎝ K∑

j=1ω

(2dist(Pj , ∂Ω)

ε

)+∑i�=j

ω

(|Pi − Pj |

ε

)⎞⎠

− γ + o(1)2μ2

⎛⎝ L∑

s=1ω

(2dist(Qs, ∂Ω)

ε

)+∑s �=l

ω

(|Qs −Ql|

ε

)⎞⎠+ o(ω(M | ln ε|)).

Let KΩ(r) be the maximum number of nonoverlapping balls with radius r packed in Ω. Now we choose Kand L such that

1 ≤ K ≤ KΩ1

((M + 3)ε| ln ε|

2

)and 1 ≤ L ≤ KΩ2

((M + 3)ε| ln ε|

2

).

Let P̂ = (P̂1, P̂2, · · · , P̂K) be the centers of arbitrary K balls in Ω1 and Q̂ = (Q̂1, Q̂2, · · · , Q̂L) be the centers of arbitrary L balls in Ω2. Since (P̂, Q̂) ∈ ΛK × ΛL, then


Mε(Pε,Qε) ≥ Mε(P̂, Q̂)

= KIμ1(U) + LIμ2(V ) − γ + o(1)2μ1

⎛⎝ K∑

i=1ω

(2dist(P̂i, ∂Ω)

ε

)+∑i�=j

ω

(|P̂i − P̂j |

ε

)⎞⎠

− γ + o(1)2μ1

⎛⎝ L∑

s=1ω

(2dist(Q̂s, ∂Ω)

ε

)+∑s �=l

ω

(|Q̂s − Q̂l|

ε

)⎞⎠+ o(ω((M + 3)| ln ε|))

≥ KIμ1(U) + LIμ2(V ) − (γ + o(1))(μ2K + μ1L)μ1μ2

εM+3 + o(ω((M + 3)| ln ε|))

= KIμ1(U) + LIμ2(V ) + o(ω(M | ln ε|)). (3.23)

If (Pε, Qε) ∈ ∂(ΛK × ΛL), then

maxi,j,s,l,m,n

{|Pε,i − Pε,j |, |Qε,s −Qε,l|, dist(Pε,m, ∂Ω),dist(Qε,n, ∂Ω)} = Mε| ln ε|.

Thus

Mε(Pε,Qε) ≤ KIμ1(U) + LIμ2(V ) − (γ + o(1)) min{μ1K,μ2L}2μ1μ2

ω(M | ln ε|) + o(ω(M | ln ε|))

which contradicts to (3.23). Thus we complete our proof. �3.4. The proof of Theorem 1.2

Similarly, in this subsection we mainly focus on the proof of Theorem 1.4 and Theorem 1.2 is an direct result of Theorem 1.4.

Proof of Theorem 1.4. By Proposition 3.5, for any (P, Q) ∈ ΛK × ΛL, (uε,P,Q, vε,P,Q) satisfies

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩


aijXij in Ωε,


bijYij in Ωε,

∂u

∂ν= ∂v

∂ν= 0 on ∂Ωε,∫

Ωε

uXij =∫Ωε

vYsj = 0, i = 1, 2, · · · ,K, s = 1, 2, · · · , L, j = 1, 2, 3.

(3.24)

Let (Pε, Qε) ∈ intΛK × ΛL be the maximum point of Mε(Pε, Qε) over ΛK × ΛL and denote

(uε,P,Q, vε,P,Q) = (Uε,P , V ε,Q) + (uε,P,Q, vε,P,Q),

then

∂Mε(Pε,Qε)∂Pmn

= 0, ∂Mε(Pε,Qε)∂Qkt

= 0

for m = 1, 2 · · · , K, k = 1, 2, · · · , L andn, t = 1, 2, 3. Namely,


0 =∫Ωε

(∇uε,Pε,Qε

∇∂uε,Pε,Qε

∂Pmn+ ∇vε,Pε,Qε

∇∂vε,Pε,Qε

∂Pmn

)

+∫Ωε

(uε,Pε,Qε

∂uε,Pε,Qε

∂Pmn+ vε,Pε,Qε

∂vε,Pε,Qε

∂Pij

)

+∫Ωε

(μ1u

3ε,Pε,Qε

∂uε,Pε,Qε

∂Pmn+ μ2v

3ε,Pε,Qε

∂vε,Pε,Qε

∂Pmn

)

+∫Ωε

(βuε,Pε,Qε

v2ε,Pε,Qε

∂uε,Pε,Qε

∂Pij+ βu2

ε,Pε,Qεvε,Pε,Qε

∂vε,Pε,Qε

∂Pmn

)

and

0 =∫Ωε

(∇uε,Pε,Qε

∇∂uε,Pε,Qε

∂Qkt+ ∇vε,Pε,Qε

∇∂vε,Pε,Qε

∂Qkt

)

+∫Ωε

(uε,Pε,Qε

∂uε,Pε,Qε

∂Qkt+ vε,Pε,Qε

∂vε,Pε,Qε

∂Qkt

)

+∫Ωε

(μ1u

3ε,Pε,Qε

∂uε,Pε,Qε

∂Pmn+ μ2v

3ε,Pε,Qε

∂vε,Pε,Qε

∂Qkt

)

+∫Ωε

(βuε,Pε,Qε

v2ε,Pε,Qε

∂uε,Pε,Qε

∂Pij+ βu2

ε,Pε,Qεvε,Pε,Qε

∂vε,Pε,Qε

∂Qkt

).

By equation (3.24), we obtain the following system

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

K∑i=1

3∑j=1

aij

∫Ωε

∂uε,Pε,Qε

∂PmnXij +

L∑s=1

3∑l=1

bsl

∫Ωε

∂vε,Pε,Qε

∂PmnYsl = 0,

K∑i=1

3∑j=1

aij

∫Ωε

∂uε,Pε,Qε

∂QktXij +

L∑s=1

3∑l=1

bsl

∫Ωε

∂vε,Pε,Qε

∂QktYsl = 0

(3.25)

for m = 1, 2, · · · , K, s = 1, 2, · · · , L and n, t = 1, 2, 3. By the fourth equality in (3.24), we can easily get

∫Ωε

∂uε,Pε,Qε

∂QktXij =

∫Ωε

∂vε,Pε,Qε

∂PmnYsl = 0.

As calculated in the proof of Theorem 1.3, we know that

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫Ωε

∂uε,Pε,Qε

∂PmnXij = −

∫R3

∂2

∂x2j

U + o(1) if m = i, n = j,

∫Ωε

∂uε,Pε,Qε

∂PmnXij = O(εM ) if m = i,

∫∂uε,Pε,Qε

∂PmnXij = O(KεM ) if m = i, n = j

Ωε


and ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫Ωε

∂vε,Pε,Qε

∂QktYsl = −

∫R3

∂2

∂x2j

V + o(1) if k = s, t = l,

∫Ωε

∂vε,Pε,Qε

∂QktYsl = O(εM ) if k = s,

∫Ωε

∂vε,Pε,Qε

∂QktYsl = O(KεM ) if k = s, t = l.

Thus (3.25) is a diagonally dominant system which implies then aij = bsl = 0, for i = 1, · · · , k, s =1, 2, · · · , L and j, l = 1, 2, 3. Hence (uε,Pε,Qε

, vε,Pε,Qε) = (Uε,Pε

+ uε,Pε,Qε, V ε,Pε

+ vε,Pε,Qε) is indeed a

solution to (Cε) if uε,Pε,Qεand vε,Pε,Qε

are both positive.Since H2(Ωε) can be continuously embedded into Cγ(Ωε), then

‖(uε,Pε,Qε, vε,Pε,Qε

)‖L∞(Ωε)×L∞(Ωε) → 0

as ε → 0. If uε,Pε,Qε(x) < 0, then 0 < Uε,Pε

(x) < −uε,Pε,Qε(x) which implies Uε,Pε

→ 0 as ε → 0. Thus uε,− = max{0, −uε,Pε,Qε

} ⇒ 0 uniformly in Ωε as ε → 0. If β < 0, then∫Ωε

(|∇uε,−|2 + u2

ε,−)

=∫Ωε

(μ1u

4ε,− + βu2

ε,−v2ε

)

≤ μ1‖u2ε,−‖L∞(Ωε)

∫Ωε

u2ε,−

≤ o(1)∫Ωε

(|∇uε,−|2 + u2

ε,−),

which implies that uε,− = 0 and leads to a contradiction. Thus uε,Pε,Qε≥ 0. Similarly, vε,Pε,Qε

≥ 0. Since −Δuε,Pε,Qε

+ (1 − βv2ε,Pε,Qε

)uε,Pε,Qε= μ1u

3ε,Pε,Qε

≥ 0, then strong maximum theorem implies that uε,Pε,Qε

> 0. Similarly, vε,Pε,Qε> 0. If β > 0, then∫

Ωε

(|∇uε,−|2 + u2

ε,−)

=∫Ωε

(μ1u

4ε,− + βu2

ε,−v2ε

)

≤(μ1‖u2

ε,−‖L∞(Ωε) + β∗‖v2ε‖L∞(Ωε)

) ∫Ωε

u2ε,−

≤ (12 + o(1))

∫Ωε

(|∇uε,−|2 + u2

ε,−),

which implies uε,− = 0 and leads to a contradiction. Here vε is uniformly bounded in Ωε by the construction of vε and we can select β∗ > 0 such that β∗‖vε,Pε,Qε

‖L∞(Ωε) <12 . Thus by strong maximum principle, we

obtain uε,Pε,Qε> 0. Similarly, vε,Pε,Qε

> 0.Similar to proof of Theorem 1.3, uε,Pε,Qε

and vε,Pε,Qεare segregated. If we take K = O(ε−m) and L =

(ε−m), then Jε(uε, vε) = O(ε−m). If we take K = O(

1εN | ln ε|N

)and L = O

(1

εN | ln ε|N), then Jε(uε, vε) =

O(

1εN | ln ε|N

). In addition, the Morse index of (uε, vε) is at least KL. Thus we complete our proof of

Theorem 1.4. �


Proof of Theorem 1.2. By changing variables, the results of Theorem 1.2 follows directly from Theo-rem 1.4. �Appendix A. Some technical computations

In this section, we collect some technical computations which are needed in our arguments in above Sections.

Let P ∈ Ω, we denote

ϕε,P = ω(z − P

ε) − ωε,P (z

ε),

ψε,P = U(z − P

ε) − Uε,P (z

ε),

φε,P = V (z − P

ε) − Vε,P (z

ε),

ψε,P = U(z − P

ε) − Uε,P (z

ε),

φε,P = V (z − P

ε) − V ε,P (z

ε),

then it is easy to see that

ψε,P = aϕε,P , φε,P = bϕε,P , ψε,P = 1√μ1

ϕε,P , φε,P = 1√μ2

ϕε,P .

We first present a lemma about the properties of ϕε,P .

Lemma A.1 (See Lemma 2.1 in [20]). Let Cε ≤ d(P, ∂Ω) ≤ 10ε| ln ε|, c ≥ ρ2 , then

ϕε,P = a

(A0 + O

(1√ρ

))K

(z − P ∗

ε

)+ O

(e−2ρ) .

Let P = (P1, P2, · · · , PK) ∈ ΩK , we denote

ωε,P =K∑i=1

ωε,Pi, Uε,P =

K∑i=1

Uε,Pi, Vε,P =

K∑i=1

Vε,Pi, Uε,P =

K∑i=1

Uε,Pi, V ε,P =

K∑i=1

V ε,Pi,

then we have

Uε,P = aωε,P, Vε,P = bωε,P, Uε,P = 1√μ1

ωε,P, V ε,P = 1√μ2

ωε,P.

The following lemma gives the estimate ωε,P in Ωε.

Lemma A.2 (See Lemma 2.1 in [1]). For i = 1, 2, · · · , K, we denote

Ωε,Pi= {x ∈ Ωε : |x− Pi

ε| ≤ M | ln ε|

2 },

then


ωε,P ={

ωPi+ O(ε−M

2 ), x ∈ Ωε,Pi, i = 1, 2, · · · , k;

O(ε−M2 ), x ∈ Ωε \ ∪K

i=1Ωε,Pi.

Lemma A.3 (See Lemma 2.3 in [2]). Let f ∈ C(RN ) ∩ L∞(RN ), g ∈ C(RN ) be radially symmetric and satisfy for some α ≥ 0, β ≥ 0, γ0 ∈ R,

f(x)|x|βeα|x| → γ0 as |x| → ∞,∫RN

|g(x)|(1 + |x|β)eα|x|dx < ∞.

Then as |y| → ∞, we have

|y|βeα|y|∫RN

g(x + y)f(x)dx → γ0

∫RN

e−αx1g(x)dx.

For P = {P1, P2, · · · , PK} ∈ ΛK and i = j, we denote

Bε(Pi) = −∫Ωε

ω3Piϕε,Pi

, Bε(Pi, Pj) =∫Ωε

ω3PiωPj

.

The following lemma gives the estimates of Bε(Pi) and Bε(Pi, Pj).

Lemma A.4 (See Lemma 2.5 in [20]). For P = {P1, P2, · · · , PK} ∈ ΛK and i = j, we have

Bε(Pi) = (γ + o(1))ω(

2dis(Pi, ∂Ω)ε

)+ o(ω(M | ln ε|)),

Bε(Pi, Pj) = (γ + o(1))ω(|Pi − Pj |

ε

)+ o(ω(M | ln ε|)),

where γ =∫R3

ω3e−y1dy.

The following lemma gives the estimates on the energy expansion Jε(Uε,P, Vε,P) for P ∈ ΛK .

Lemma A.5. For any P ∈ ΛK and ε > 0 small enough, we have

Jε(Uε,P, Vε,P) = KI(U, V ) − μ1a4 + μ2b

4 + 2βa2b2

2

⎛⎝ K∑

i=1Bε(Pi) +

∑i�=j

Bε(Pi, Pj)

⎞⎠+ o(ω(M | ln ε|)).

Proof. A simple computation shows that

Jε(Uε,P, Vε,P) =K∑i=1

Jε(Uε,Pi, Vε,Pi

) +∑i�=j

12

∫Ωε

(∇Uε,Pi

∇Uε,Pj+ ∇Vε,Pi

∇Vε,Pj

)

+∫ (

Uε,PiUε,Pj

+ Vε,PiVε,Pj

)− 1

4

∫ (μ1U

4ε,P + μ2U

4ε,P + 2βU2

ε,PV2ε,P)

Ωε Ωε


+K∑i=1

14

∫Ωε

(μ1U

4ε,Pi

+ μ2U4ε,Pi

+ 2βU2ε,Pi

V 2ε,Pi

)

=K∑i=1

Jε(Uε,Pi, Vε,Pi

) − μ1

4

∫Ωε

⎛⎝U4

ε,P −K∑i=1

U4ε,Pi

− 4∑i�=j

U3PiUε,Pj

⎞⎠

− μ2

4

∫Ωε

⎛⎝V 4

ε,P −K∑i=1

V 4ε,Pi

− 4∑i�=j

V 3PiVε,Pj

⎞⎠

− β

2

∫Ωε

⎛⎝U2

ε,PV2ε,P −

K∑i=1

U2ε,Pi

V 2ε,Pi

− 2∑i�=j

UPiV 2ε,Pi

Uε,Pj− 2∑i�=j

U2PiVPi

Vε,Pj

⎞⎠

−∑i�=j

12

∫Ωε

(μ1U

3PiUε,Pj

+ μ2V3PiVε,Pj

+ βUPiV 2PiUε,Pj

+ βU2PiVPi

Vε,Pj

). (A.1)

A direct calculation shows∫Ωε

U3PiUε,Pj

=∫Ωε

U3PiUPj

−∫Ωε

U3Piψε,Pj

= a4B(Pi, Pj) + O

(ω

( |Pi − P ∗j |

ε

)),

then

∑i�=j

12

∫Ωε

(μ1U

3PiUε,Pj

+ μ2V3PiVε,Pj

+ βUPiV 2PiUε,Pj

+ βU2PiVPi

Vε,Pj

)

= μ1a4 + μ2b

4 + 2βa2b2

2∑i�=j

B(Pi, Pj) + O(Kε

√2M). (A.2)

Note that

∫Ωε

⎛⎝U4

ε,P −K∑i=1

U4ε,Pi

− 4∑i�=j

U3PiUε,Pj

⎞⎠

=K∑

k=1

∫Ωε,Pk

⎛⎝U4

ε,P −K∑i=1

U4ε,Pi

− 4∑i�=j

U3PiUε,Pj

⎞⎠

+∫

Ωε\∪Kk=1Ωε,Pj

⎛⎝U4

ε,P −K∑i=1

U4ε,Pi

− 4∑i�=j

U3PiUε,Pj

⎞⎠ .

On the other hand by an elementary computation, we have

∫Ωε,Pk

∣∣∣∣∣∣U4ε,P −

K∑i=1

U4ε,Pi

− 4∑i�=j

U3PiUε,Pj

∣∣∣∣∣∣≤∫

Ω

⎛⎝U4

ε,P − U4ε,Pk

− 4∑j �=k

U3ε,Pk

Uε,Pj

⎞⎠

ε,Pk


+∫

Ωε,Pk

⎛⎝∑

i�=k

U4ε,Pi

+∑j �=k

|U3ε,Pk

− U3Pk|Uε,Pj

+ 4∑

i�=k,i �=j

U3PiUε,Pj

⎞⎠

≤ C

∫Ωε,Pk

U2ε,Pk

⎛⎝∑

i�=k

Uε,Pi

⎞⎠2

+ C

∫Ωε,Pk

⎛⎝∑

i�=k

U4Pi

+∑j �=k

|U3ε,Pk

− U3Pk|UPj

+ 4∑

i�=k,i �=j

U3PiUPj

⎞⎠

≤ C(K + 4)∑i�=k

∫R3

U2PkU2Pidx− C

∑j �=k

∫R3

U2Pkψε,Pk

UPj≤ C(K + 5)ε2M

and

∫Ωε\∪K

k=1Ωε,Pj

∣∣∣∣∣∣U4ε,P −

K∑i=1

U4ε,Pi

− 4∑i�=j

U3PiUε,Pj

∣∣∣∣∣∣ ≤ CKε2M−3.

Then

∫Ωε

⎛⎝U4

ε,P −K∑i=1

U4ε,Pi

− 4∑i�=j

U3PiUε,Pj

⎞⎠ = O

(K(K + 5)ε2M + Kε2M−3) = o(ω(M | ln ε|)). (A.3)

Similarly, we have

∫Ωε

⎛⎝V 4

ε,P −K∑i=1

V 4ε,Pi

− 4∑i�=j

V 3PiVε,Pj

⎞⎠ = o(ω(M | ln ε|)) (A.4)

and

∫Ωε

⎛⎝U2

ε,PV2ε,P −

K∑i=1

U2ε,Pi

V 2ε,Pi

− 2∑i�=j

UPiV 2ε,Pi

Uε,Pj− 2∑i�=j

U2PiVPi

Vε,Pj

⎞⎠ = o(ω(M | ln ε|)). (A.5)

Thus by (A.1)–(A.5), we have

Jε(Uε,P, Vε,P) =K∑i=1

Jε(Uε,Pi, Vε,Pi

) − μ1a4 + μ2b

4 + 2βa2b2

2∑i�=j

B(Pi, Pj) + o(ω(M | ln ε|)). (A.6)

Note that

Jε(Uε,Pi, Vε,Pi

) = 14

∫Ωε

(μ1U

4Pi

+ μ2V4Pi

+ 2βU2PiV 2Pi

)− μ1

4

∫Ωε

(U4ε,Pi

− U4Pi

+ 2U3Piψε,Pi

)

− μ2

4

∫Ωε

(V 4ε,Pi

− V 4Pi

+ 2V 3Piφε,Pi

)

− β

2

∫Ωε

(U2ε,Pi

V 2ε,Pi

− U2PiV 2Pi

+ UPiV 2Piψε,Pi

+ U2PiVPi

φε,Pi

).

Again by a direct calculation, we have


14

∫Ωε

(μ1U

4Pi

+ μ2V4Pi

+ 2βU2PiV 2Pi

)= I(U, V ) −

∫R3\Ωε

(μ1U

4Pi

+ μ2V4Pi

+ 2βU2PiV 2Pi

)

= I(U, V ) + O(ε2M )

and

∫Ωε

(U4ε,Pi

− U4Pi

+ 2U3Piψε,Pi

)= −2

∫ωε

U3Piψε,Pi

+ O

⎛⎝ ∫

Ωε

U2Piψ2ε,Pi

⎞⎠ = 2a4B(Pi) + O(ε2M ),

then

Jε(Uε,Pi, Vε,Pi

) = I(U, V ) − μ1a4 + μ2b

4 + 2βa2b2

2 B(Pi) + O(ε2M ). (A.7)

Thus by (A.6) and (A.7), we obtain

Jε(Uε,P, Vε,P) = KI(U, V ) − μ1a4 + μ2b

4 + 2βa2b2

2

⎛⎝ K∑

i=1Bε(Pi) +

∑i�=j

Bε(Pi, Pj)

⎞⎠+ o(ω(M | ln ε|)).

Thus the proof of the lemma is finished. �The following lemma gives the estimates on the energy expansion Jε(Uε,P, Vε,Q) for (P, Q) ∈ ΛK × ΛL.

Lemma A.6. For any (P, Q) ∈ ΛK × ΛL and ε > 0 small enough, we have

Jε(Uε,P, V ε,Q) = KIμ1(U) + LIμ2(V ) − 12μ1

⎛⎝ K∑

i=1Bε(Pi) +

∑i�=j

Bε(Pi, Pj)

⎞⎠

− 12μ2

⎛⎝ L∑

s=1Bε(Qs) +

∑s �=l

Bε(Qs, Ql)

⎞⎠+ o(ω(M | ln ε|)).

Proof. A simple computation shows

Jε(Uε,P, V ε,Q) =

⎛⎝μ1

4

K∑i=1

∫Ωε

U4Pi

+ μ2

4

L∑s=1

∫Ωε

V4Pi

⎞⎠− β

2

∫Ωε

U2ε,PV

2ε,Q

− μ1

4

∫Ωε

⎛⎝U

4ε,P −

K∑i=1

U4Pi

− 2∑i�=j

U3PiUε,Pj

+ 2K∑i=1

U3Piψε,Pi

⎞⎠

− μ2

4

∫Ωε

⎛⎝V

4ε,P −

L∑s=1

V4Qs

− 2∑s �=l

V3Qs

V ε,Ql+ 2

L∑s=1

V3Qs

φε,Qs

⎞⎠ . (A.8)

Since

μ1

4

∫U

4Pi

= Iμ1(U) − μ1

4

∫3

U4Pi

= Iμ1(U) + O(ε2M ),
Ωε R \Ωε


then

μ1

4

K∑i=1

∫Ωε

U4Pi

+ μ2

4

L∑s=1

∫Ωε

V4Pi

= KIμ1(U) + O(Kε2M ). (A.9)

Notice

∫Ωε

⎛⎝U

4ε,P −

K∑i=1

U4Pi

− 2∑i�=j

U3PiUε,Pj

+ 2K∑i=1

U3Piψε,Pi

⎞⎠

=∫Ωε

⎛⎝U

4ε,P −

K∑i=1

U4ε,Pi

− 4∑i�=j

U3PiUε,Pj

⎞⎠

+K∑i=1

∫Ωε

(U

4ε,Pi

− U4Pi

+ 2U3Piψε,Pi

)+ 2∑i�=j

∫Ωε

U3PiUε,Pj

.

As calculated in Lemma A.5, we have

∫Ωε

⎛⎝U

4ε,P −

K∑i=1

U4ε,Pi

− 4∑i�=j

U3PiUε,Pj

⎞⎠ = o(ω(M | ln ε|)),

and

K∑i=1

∫Ωε

(U

4ε,Pi

− U4Pi

+ 2U3Piψε,Pi

)+ 2∑i�=j

∫Ωε

U3PiUε,Pj

= 2μ2

1

K∑i=1

B(Pi) + 2μ2

1

∑i�=j

B(Pi, Pj) + o(ω(M | ln ε|)).

Thus

∫Ωε

⎛⎝U

4ε,P −

K∑i=1

U4Pi

− 2∑i�=j

U3PiUε,Pj

+ 2K∑i=1

U3Piψε,Pi

⎞⎠

= 2μ2

1

⎛⎝ K∑

i=1B(Pi) +

∑i�=j

B(Pi, Pj)

⎞⎠+ o(ω(M | ln ε|)). (A.10)

Similarly,

∫Ωε

⎛⎝V

4ε,Q −

L∑s=1

V4Qs

− 2∑s �=l

V3Qs

V ε,Ql+ 2

L∑s=1

V3Qs

φε,Qs

⎞⎠

= 2μ2

2

⎛⎝ L∑

s=1B(Qs) +

∑i�=j

B(Qs, Ql)

⎞⎠+ o(ω(M | ln ε|)). (A.11)

On the other hand since


∫Ωε

U2ε,PV

2ε,Q ≤ CKL

∑is

∫Ωε

U2PiV Qs

) ≤ CKLε−2δε = o(ω(M | ln ε|)),

then by (A.8)–(A.11), we have

Jε(Uε,P, V ε,Q) = KIμ1(U) + LIμ2(V ) − 12μ1

⎛⎝ K∑

i=1Bε(Pi) +

∑i�=j

Bε(Pi, Pj)

⎞⎠

− 12μ2

⎛⎝ L∑

s=1Bε(Qs) +

∑s �=l

Bε(Qs, Ql)

⎞⎠+ o(ω(M | ln ε|)).

This completes the proof of lemma. �References

[1] W. Ao, J. Wei, J. Zeng, An optimal bound on the number of interior peak solutions for the Lin–Ni–Takagi problem, J. Funct. Anal. 265 (2013) 1324–1356.

[2] A. Bahri, Y. Li, On a minimax procedure for the existence of a positive solution for certain scaler field equation in Rn, Rev. Mat. Iberoam. 6 (1990) 1–15.

[3] T. Bartsch, N. Dancer, Z. Wang, A Liouville theorem, a priori bounds, and bifurcating branches of positive solutions for a nonlinear elliptic system, Calc. Var. Partial Differential Equations 37 (2010) 345–361.

[4] P. Bates, G. Fusco, Equilibria with many nuclei for the Cahn–Hilliard equation, J. Differential Equations 160 (2000) 283–356.

[5] G. Cerami, J. Wei, Multiplicity of multiple interior spike solutions for some singularly perturbed Neumann problem, Int. Math. Res. Notes 12 (1998) 601–626.

[6] E.N. Dancer, S. Yan, Multipeak solutions for a singularly perturbed Neumann problem, Pacific J. Math. 189 (1999) 241–262.

[7] M. del Pino, P. Felmer, J. Wei, On the role of the distance function for some singular perturbation problems, Comm. Partial Differential Equations 25 (2000) 155–177.

[8] M. del Pino, P. Felmer, J. Wei, On the role of mean curvature in some singularly perturbed Neumann problems, SIAM J. Math. Anal. 31 (1999) 63–79.

[9] B.D. Esry, C.H. Greene, J.P. Burker Jr., J.L. Bohn, Hartree–Fock theory for double condensates, Phys. Rev. Lett. 78 (1997) 3584–3597.

[10] P. Felmer, S. Martinez, K. Tanaka, High frequency solutions for the singularly-perturbed one dimensional nonlinear Schrödinger equation, Arch. Ration. Mech. Anal. 182 (2006) 253–268.

[11] P. Felmer, S. Martinez, K. Tanaka, On the number of positive solutions of singularly perturbed 1D nonlinear Schrödinger equations, J. Eur. Math. Soc. (JEMS) 8 (2006) 253–268.

[12] B. Gidas, W. Ni, L. Nirenberg, Symmetry of positive solutions of nonlinear elliptic equations in Rn , Adv. in Math., Suppl. Stud. 7A (1981) 369–402.

[13] C. Gui, Multi-peak solutions for a semilinear Neumann problem, Duke Math. J. 84 (1996) 739–769.[14] C. Gui, J. Wei, On multiple mixed interior and boundary peak solutions for some singularly perturbed Neumann problems,

Canad. J. Math. 52 (2000) 522–538.[15] C. Gui, J. Wei, Multiple interior peak solutions for some singularly perturbed Neumann problems, J. Differential Equations

158 (1999) 1–27.[16] C. Gui, J. Wei, M. Winter, Multiple boundary peak solutions for some singularly perturbed Neumann problems, Ann.

Inst. H. Poincaré Anal. Non Linéaire 17 (2000) 47–82.[17] Y. Li, On a singularly perturbed equation with Neumann boundary condition, Comm. Partial Differential Equations 23

(1998) 487–545.[18] T. Lin, J. Wei, Spikes in two coupled nonlinear Schrödinger equations, Ann. Henri Poincaré 22 (2005) 403–439.[19] C. Lin, W. Ni, I. Takagi, Large amplitude stationary solutions to a chemotaxis systems, J. Differential Equations 72 (1988)

1–27.[20] F. Lin, W. Ni, J. Wei, On the number of interior peak solutions for singularly perturbed Neumann problem, Comm. Pure

Appl. Math. 60 (2007) 252–281.[21] W. Ni, I. Takagi, On the shape of least-energy solution to a semilinear Neumann problem, Comm. Pure Appl. Math. 41

(1991) 819–851.[22] W. Ni, I. Takagi, Locating the peaks of least energy solutions to a semilinear Neumann problem, Duke Math. J. 70 (1993)

247–281.[23] S. Peng, Z. Wang, Segregated and synchronized vector solutions for nonlinear Schrödinger systems, Arch. Ration. Mech.

Anal. 208 (2013) 305–339.[24] B. Sirakov, Least energy solitary waves for a system of nonlinear Schrödinger equations in Rn, Comm. Math. Phys. 271

(2007) 199–221.

http://refhub.elsevier.com/S0022-247X(16)30734-X/bib61777As1

http://refhub.elsevier.com/S0022-247X(16)30734-X/bib61777As1

http://refhub.elsevier.com/S0022-247X(16)30734-X/bib626Cs1


http://refhub.elsevier.com/S0022-247X(16)30734-X/bib626477s1














http://refhub.elsevier.com/S0022-247X(16)30734-X/bib666D7431s1




http://refhub.elsevier.com/S0022-247X(16)30734-X/bib676E6Es1

http://refhub.elsevier.com/S0022-247X(16)30734-X/bib676E6Es1










http://refhub.elsevier.com/S0022-247X(16)30734-X/bib6C77s1

http://refhub.elsevier.com/S0022-247X(16)30734-X/bib6C6E74s1




http://refhub.elsevier.com/S0022-247X(16)30734-X/bib6E7431s1









[25] Z. Tang, Spike-layer solutions to singularly perturbed semilinear systems of coupled Schrödinger equations, J. Math. Anal. Appl. 377 (2011) 336–352.

[26] Z. Tang, Multi-peak solutions to coupled Schrödinger systems with Neumann boundary conditions, J. Math. Anal. Appl. 409 (2014) 684–704.

[27] Z. Tang, Segregated peak solutions of coupled Schrödinger systems with Neumann boundary conditions, Discrete Contin. Dyn. Syst. 34 (2014) 5299–5323.

[28] J. Wei, On the interior spike layer solutions of singularly perturbed semilinear Neumann problem, Tohoku Math. J. 50 (2) (1998) 159–178.

[29] J. Wei, On the interior spike layer solutions for some singular perturbation problems, Proc. Roy. Soc. Edinburgh Sect. A 128 (1998) 849–874.

[30] J. Wei, M. Winter, On the Cahn–Hilliard equations II: interior spike Layer solutions, J. Differential Equations 148 (1998) 231–267.

[31] J. Wei, W. Yao, Uniqueness of positive solutions to some coupled nonlinear Schrödinger equations, Commun. Pure Appl. Anal. 11 (2012) 1003–1011.















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