View
0
Download
0
Category
Preview:
Citation preview
J. Math. Anal. Appl. 448 (2017) 1079–1119
Contents lists available at ScienceDirect
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: tangzw@bnu.edu.cn (Z. Tang).
1 The first author was supported by National Science Foundation of China (11571040).
http://dx.doi.org/10.1016/j.jmaa.2016.11.0440022-247X/© 2016 Elsevier Inc. All rights reserved.
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).
εZ. 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
1082 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
‖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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1083
{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
1084 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
⎧⎪⎪⎨⎪⎪⎩
−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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1085
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
1086 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
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)
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1087
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
1088 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
∫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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1089
‖(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. �
1090 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1091
∫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)
for some C > 0 which is independent of ε and K.
1092 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
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)).
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1093
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,
1094 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
∂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. �
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1095
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
].
1096 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1097
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
1098 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1099
≤ 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 ε1100 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
∫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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1101
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
1102 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
∫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
M−12 ‖(u, v)‖L2(Ωε)×L2(Ωε) + ‖h‖L2(Ωε)×L2(Ωε)
). (3.8)
Similarly,
|bij | ≤ C(ε
M−12 ‖(u, v)‖L2(Ωε)×L2(Ωε) + ‖h‖L2(Ωε)×L2(Ωε)
). (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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1103
⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩
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)
1104 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
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,
Sε,2(u + Uε,P, v + V ε,Q) =∑ij
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)
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1105
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
for some C > 0 which is independent of ε and K.
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
1106 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
‖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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1107
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). �
1108 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1109
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
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
Sε,1(u + Uε,P, v + V ε,Q) =∑ij
aijXij in Ωε,
Sε,2(u + Uε,P, v + V ε,Q) =∑ij
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,
1110 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1111
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. �
1112 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1113
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
1114 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
+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
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1115
+∫
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
1116 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
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 \ΩεZ. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1117
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
1118 Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119
∫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.
Z. Tang, L. Wang / J. Math. Anal. Appl. 448 (2017) 1079–1119 1119
[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.
Recommended