Stable spike clusters on a compact two-dimensional Riemannian manifold

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Stable spike clusters on a compact two-dimensional Riemannian manifold Weiwei Ao a , Juncheng Wei b , Matthias Winter c,∗ a Wuhan University, Department of Mathematics, 430072, China b Department of Mathematics, University of British Columbia, Vancouver, B.C., V6T 1Z2, Canada c Brunel University London, Department of Mathematics, Uxbridge UB8 3PH, United Kingdom

Received 28 August 2019; accepted 4 October 2019

Abstract We consider the Gierer-Meinhardt system with small inhibitor diffusivity and very small activator diffusivity on a compact two-dimensional Riemannian manifold without boundary. We study steady state solutions which are far from spatial homogeneity. We construct two different spike clusters, each consisting of two spikes, which both approach the same nondegenerate local maximum point of the Gaussian curvature. We show that one of these spike clusters is stable, the other one is unstable. © 2019 The Authors. Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/). MSC: 35K45; 53B20 Keywords: Pattern formation; Mathematical biology; Singular perturbation; Reaction-diffusion system; Riemannian manifold

* Corresponding author.

E-mail addresses: [email protected] (W. Ao), [email protected] (J. Wei), [email protected] (M. Winter). https://doi.org/10.1016/j.jde.2019.10.005 0022-0396/© 2019 The Authors. Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/).

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1. Introduction 1.1. The problem Since the pioneering work of Turing in 1952 [39], many different reaction-diffusion system in biological modeling have been proposed and the occurrence of pattern formation has been investigated by studying what is now called Turing instability. One of the most popular models in biological pattern formation is the Gierer-Meinhardt system [14], see also [26]. In this paper, we consider the following Gierer-Meinhardt system on a compact two-dimensional Riemannian manifold (M, g) without boundary: 

ε 2 g A − A + Dg H − H

A2 H

=0

+ A2

=0

in M.

(1.1)

Throughout the paper, we assume that 0 < ε << 1, 0 < D << 1. We prove the existence and study the stability of a cluster of two spikes near a non-degenerate local maximum point p 0 of the Gaussian curvature of the manifold M. 1.2. The geometric setting Before stating the results, we first introduce the geometric setting of the problem. Let Tp M be the tangent place to M at p, and given an orthonormal basis {e1 (p), e2 (p)} of Tp M, we can obtain via the exponential map expp : Tp M → M, a natural correspondence Ep (x) = x1 e1 (p) +x2 e2 (p) → q = expp (x1 e1 (p) +x2 e2 (p)). Since M is a compact manifold, one knows that there exists a constant ig > 0 such that Xp := Ep−1 ◦ exp−1 p : Bg (p, ig ) → B(0, ig ) is a diffeomorphism for every p ∈ M. The values of this natural chart Xp are called normal coordinates about p. We now define function spaces. Set ⎧ ⎫ ⎪ ⎪  ⎨ ⎬ 2 2 L (Mε ) = u measurable function defined on Mε , u (q)dvgε < ∞ , ⎪ ⎪ ⎩ ⎭ Mε

where dvgε denotes the Riemannian measure with respect to the metric gε . We further set H 1 (Mε ) = {u ∈ L2 (Mε ), ∇gε u ∈ L2 (Mε )}. We will construct cluster solutions near a non-degenerate local maximum point of the Gaussian curvature function K(p). In the rest of the paper, we assume that there is a local maximum of K(p) is at p 0 = 0, i.e. we have

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 ∇K(0) = 0, ∇ 2 K(0) =

K11 0

0 K22

3



where K11 , K22 < 0. Let the local normal coordinates around p be x = (x1 , x2 ). Then we set χ = 1 for |x| ≤ and χ = 0 for |x| ≥

ig 2,

ig 4

and introduce χε = χ( xε ).

1.3. The main results Let w be the unique solution of the problem w − w + w 2 = 0, w > 0 in R2 , w(0) = max w(y), w(y) → 0 as |y| → ∞. y∈R2

(1.2)

In this paper, we shall prove results on the existence and stability of a spike cluster of (1.1) located around p 0 = 0 with two spikes. Our first result is on the existence: Theorem 1.1. Let p 0 be a non-degenerate local maximum point of the Gaussian curvature K(p) of M. Assume that 0 < ε <<

√

D << 1, 0 <

√

D log

1 ε 2 D log

√

D ε

<< 1,

(1.3)

and K22 = 1. K11

(1.4)

Then the Gierer-Meinhardt system (1.1) has at least two different 2-spike cluster solutions (Ai , Hi ) for i = 1, 2, which both concentrate near p 0 . In particular, each of these solutions satisfies A∼

 Dξε x x Dξε ) + w( ) , H (±qi ) ∼ 2 , + q − q w( i i 2 ε ε ε ε

where εqi → 0 as ε → 0 and ξε ∼

1√ log

D ε

for i = 1, 2.

Remark 1.2. The limit √ε → 0 means that the diffusivity of the activator u is asymptotically D smaller than the diffusivity of the inhibitor √ v. If this is not assumed, then the pattern will no longer have a spike profile. The second limit D log 2 1 √D → 0 is the condition which guarantees ε D log

that the spikes form a cluster, i.e. εqi → 0 as ε → 0.

ε

Remark 1.3. As one will see from the proof, we will construct an approximate solution which concentrates on a regular k-polygon shrinking to the point 0 for general k ≥ 2. But when solving the reduced problem we can only handle the case k = 2. The condition (1.4) is imposed to make sure that the reduced problem is solvable.

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Next we study the stability of the 2-spike cluster constructed in Theorem 1.1. Our second result on the stability is the following: Theorem 1.4. Let p 0 be a non-degenerate local maximum point of the Gaussian curvature K(p). Assume (1.3), (1.4) and let (Ai , Hi ) for i = 1, 2 be the solutions constructed in Theorem 1.1. Then one of the solutions is stable and the other one is unstable. Using the transformation x = εy, u =

ε2 ε2 A, v = H, D D

equation (1.1) becomes ⎧ ⎨ gε u − u + ⎩ where σ =

√ε . D

u2 v

=0

gε v − σ 2 v + u2 = 0

in Mε

(1.5)

In the rest of this paper, we will work with (1.5).

1.4. Related work and motivation We now comment on some related work. Generally speaking, the Gierer-Meinhardt system is difficult to solve since it does neither have a variational structure nor a priori estimates. One way to study it is to examine the so-called shadow system. Namely, we let D → +∞ first. It is known (see [21,28,35]) that the study of the shadow system amounts to the study of the following single equation for p = 2: 

ε 2 u − u + up = 0, ∂u ∂ν

=0

on ∂.

u > 0 in ,

(1.6)

Equation (1.6) has a variational structure and has been studied by numerous authors. It is known that equation (1.6) has both boundary spike solutions and interior spike solutions. For existence of boundary spike solutions, see [16,29–31,46,47] and the references therein. For existence of interior spike solutions, see [17,33] and the references therein. For stability of spike solutions see [32,44,45]. Next we review some results for bumps, spikes and related patterns in the Gierer-Meinhardt system. Ground states on the real line are studied in [9,11,12,58] and for the whole R2 in [10]. Spikes for an interval are studied in [18,19,25,37,43] and for bounded two-dimensional domains in [23,24,31,48–52]. Hopf bifurcation of spikes is investigated in [7,41,42]. For dynamics we refer to [5,6,13,20,36]. Steady states with spherical layers have been constructed in [25,34]. Stripes have been studied in [22]. Nonlocal eigenvalue problems related to the one in this paper have been studied in [44,45,57]. The existence of spikes for single semilinear elliptic PDEs on manifolds has been investigated in [4,8,27]. Existence and stability of a single spike solution for the Gierer-Meinhardt system on a Riemannian manifold has been shown in [38].

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In [52] the existence and stability of N -peaked steady states for the Gierer-Meinhardt system with precursor inhomogeneity has been explored. The spikes in the patterns can vary in amplitude. In particular, the results imply that a precursor inhomogeneity can induce instability. Single-spike solutions for the Gierer-Meinhardt system with precursor including spike dynamics have been studied in [40]. For more background, modeling, analysis and computation on the Gierer-Meinhardt system, we refer to [54] and the references therein. Previous results on stable spike clusters include a stable spike cluster for a consumer chain model [53]. For the Gierer-Meinhardt system spike clusters have been established in the following situations: stable interior spike clusters for the one-dimensional Gierer-Meinhardt system with precursor inhomogeneity [55], stable interior spike clusters for the two-dimensional GiererMeinhardt system with precursor inhomogeneity [56] and stable boundary spike clusters for the two-dimensional Gierer-Meinhardt system [2]. In the last paper the boundary curvature plays the role of the precursor in the previous papers. In the current paper we will see that the Gaussian curvature takes over that role for the Gierer-Meinhardt system on a compact two-dimensional Riemannian manifold without boundary. We would like to summarize this role as follows: the spikes in the cluster are mutually repelling and also each spike is attracted to a local maximum point of the Gaussian curvature (or to a local minimum of the precursor gradient or local maximum of the boundary curvature, respectively). This balance between attracting and repelling interactions can lead to a stable spike cluster. This paper is organized as follows. In Section 2, we give some preliminaries and describe the construction of the approximate cluster solution. In Section 3, we use the Liapunov-Schmidt method to reduce the existence problem to finite dimensions. In Section 4 we solve this reduced problem. In Sections 5-6, we study the stability of the spike cluster steady states. In Section 5 we consider large eigenvalues. In Section 6 we study small eigenvalues. In Section 7 we discuss the results of the paper. In the appendix we give some identities needed in the main part of the paper and we calculate the eigenvalues of the reduced matrix in main order for a general number of spikes. Acknowledgments W. Ao was supported by NSFC (No. 11801421 and No. 11631011). J. Wei is partially supported by NSERC of Canada (No. RGPIN-2018-03773). M. Winter thanks the Department of Mathematics at Wuhan University and the Institut für Analysis, Dynamik und Modellierung at the Universität Stuttgart for their kind hospitality. 2. Preliminaries and construction of the approximate solution 2.1. Expansion of the Laplacian Let the local normal coordinates around point p be x. For a function u in the rescaled coordinates y = xε , one has the following expansion of the Laplace-Beltrami operator (see appendix A of [38] and also [1]): gε u(y) = y u(y)  1 1 1 + K(p)ε2 + (∇K(p) · y)ε 3 + (y∇ 2 K(p)y t )ε 4 (Q[u] − 2P [u]) 3 6 20

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1 2 K (p)|y|2 ε 4 (3Q[u] − 4P [u]) 45 1 1 + ε 3 R1 [u] + ε 4 R2 [u] + O(ε 5 ) 6 10 +

(2.1)

where Q[u] = y22

2 ∂ 2u ∂ 2u 2∂ u − 2y y + y , 1 2 1 ∂y1 ∂y2 ∂y12 ∂y22

P [u] = y1

∂u ∂u + y2 , ∂y1 ∂y2

R1 [u] =

y22 − y12 ∂K ∂u ∂K ∂u (p) − (p) ] [ 2 ∂x1 ∂y1 ∂x2 ∂y2 ∂K ∂u ∂K ∂u −y1 y2 [ (p) + (p) ], ∂x2 ∂y1 ∂x1 ∂y2

R2 [u] = [

y22 − y12 ∂u ∂u ∂ 2K ∂ 2K − y1 y 2 ][y1 2 (p) + y2 (p)] 2 ∂y1 ∂y2 ∂x1 ∂x2 ∂x1

−[

y22 − y12 ∂u ∂u ∂ 2K ∂ 2K + y1 y 2 ][y2 2 (p) + y1 (p)]. 2 ∂y2 ∂y1 ∂x1 ∂x2 ∂x2 ⎛

∂K ∂K Note that ∇K(p) = ( ∂x , )(p), and ∇ 2 K(p) = ⎝ 1 ∂x2

∂2K ∂2K ∂x1 ∂x1 ∂x1 ∂x2 ∂2K ∂2K ∂x1 ∂x2 ∂x2 ∂x2

⎞ ⎠ (p) are not rescaled.

2.2. The Green’s function Now we introduce a Green’s function Gσ which is needed for our analysis. Let Gσ be the Green’s function given by gε Gσ (p, q) − σ 2 Gσ (p, q) + δq = 0 in Mε .

(2.2)

For properties of this Green’s function please see [3]. From (2.2), one has  Gσ dvgε (p) = Mε

Setting Gσ (p, q) =

1 σ 2 |Mε |



¯ σ (p, q), then G ¯ σ satisfies +G ¯ σ − σ 2G ¯ σ − 1 + δq = 0 in Mε gε G |Mε |  ¯ σ dvgε = 0. G Mε

˜ σ be defined by Let G

1 . σ2

(2.3)

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˜ σ − σ 2G ˜ σ + δ0 = 0 in R2 . G By the expansion of the Laplace Beltrami operator, one has ¯ σ (q, r) = G(y, ˜ G z) + ε2 G1 (y, z) + O(ε 3 )

(2.4)

where y = Xp (q), z = Xp (r) and G1 (y, z) is even function in |y − z|. ˜ σ (y, z) := G ˜ 1 (σy, σ z), one has For G Lemma 2.1. If |y − z| << 1, ˜ 1 (y, z) = 1 log 1 + H˜ 1 (y, z) G 2π |y − z| where H˜ 1 is the regular part of the Green’s function and ∇y H˜ 1 (y, z)|y=z = 0. If |y − z| >> 1, ˜ 1 (y, z)(1 + o(1)) ˜ 1 (y, z) = c|y − z|− 2 e−|y−z| (1 + o(1)), |∇y G ˜ 1 (y, z)| = G G 1

for some constant c > 0. 2.3. The construction of the approximate solutions In this subsection, we describe the approximate solution we will use. Given k ≥ 2, define qj0 = (R cos θj , R sin θj ) for j = 1, · · · , k where θj = α + 2π k (j − 1) in geodesic normal coordinates. Here α is the parameter for the angle representing the degeneracy due to rotations. The constant R for the radius will be determined later in the leading order of the reduced problem. Since our manifold is not rotationally symmetric α will be derived below in a higher order of the reduced problem. Next we introduce suitable coordinates in a neighborhood of q0 = (q10 , · · · , qk0 ). Let f˜i , g˜ i ∈ R, i = 1, · · · , k, we define qi = qi0 + f˜i n i + g˜ i t i

(2.5)

where t i = (− sin θi , cos θi ), n i = (cos θi , sin θi ). So f˜i , g˜ i measure the displacements in the normal and tangential directions, respectively. Denote Qε = {qi , i = 1, · · · , k, σ |f˜i | + σ |g˜ i | ≤ C}.

(2.6)

Now we introduce wj to be the unique radially symmetric solution of the equation 1 y wj − wj − K(εqj )ε 2 rwj (r) + wj2 (r) = 0 in R2 3

(2.7)

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where K(q) is the Gaussian curvature at q ∈ M. Existence and uniqueness of wj can be derived using the implicit function theorem and the non-degeneracy of the positive solution w to the equation w − w + w2 = 0. Moreover, one has wj − wH 2 (R2 ) = O(ε 2 ) if |εqj | is bounded. The readers are referred to [38] for more details. Then we set our approximate solution to be

U=

k 

ξε,qi wi (y − qi )χε (y − qi )

(2.8)

i=1 i

where χ = 1 for |x| ≤ 4g and χ = 0 for |x| ≥ mined in the following subsection.

ig 2

and χε = χ( xε ), the height ξε,qj is to be deter-

2.4. Calculating the height of the peaks In this subsection, we formally calculate the height of the peaks. It turns out that the height of the peaks does not depend on the spike location in leading order but only in higher order. For a function u ∈ H 2 (Mε ), let T [u] be the unique solution to the equation gε T [u] − σ 2 T [u] + u = 0. Then from the equation satisfied by v, one can choose the approximate solution as u = U, v = T (U 2 ) = V . Next we calculate the height of the peaks ξε,qj = V (qj )  = Gσ (qj , q)U 2 (q)dvgε (q) Mε

 Gσ (qj , y)wj (y − qj )2 χε,j dvgε

2 = ξε,q j

Mε

+

 i=j

+



 Gσ (qj , y)wi (y − qi )2 χε,i dvgε

2 ξε,q i

Mε

i 2 = ξε,q j

π

2 O(ξε,q e−2R sin k ) i

1 1 log 2π σ

 R2

one has

 2 wj2 dy + O( ξε,q ) i i

(2.9)

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⎛ 1 1 1⎜ = log ⎝ ξε,qj 2π σ



 wj2 dy + O



9

⎞

1 ⎟ ⎠. log σ

(2.10)

R2

Denote ⎛ 1 ⎜ 1 ξε = ⎝ log 2π σ



⎞−1 ⎟ w 2 dy ⎠

.

R2

Then one has ξε,qj = ξε (1 + O( log1 σ )). 3. Existence: reduction to finite dimension 3.1. Error of the approximate solution Let us start to prove Theorem 1.1. The first step is choosing a good approximate solution which was done in the last section. The second step is to use the Liapunov-Schmidt reduction to reduce the problem to a finite dimension problem which we do in this section. First we need to calculate the error of the approximate solution (U, V ) given in (2.9). S1 (U, V ) = gε U − U +

U2 V

U2  ξε,qi wi2 (y − qi )χε,i − V k

=

i=1

+

k 

ε3

i=1

+

k  i=1

+

ε4

 1 ∇K(εqi ) · (y − qi )(Q[wi ] − 2P [wi ]) + R1 [wi ] 6 6

1

1 (y − qi )∇ 2 K(εqi )(y − qi )t (Q[wi ] − 2P [wi ]) 20

 1 1 2 K (εqi )|y − qi |2 (3Q[wi ] − 4P [wi ]) + R2 [wi ] + O(ε 5 ). 45 10

Next we calculate for j = 1, · · · , k, and y = qj + z with |εz| ≤ V (qj + z) − V (qj )  = [Gσ (qj + z, p) − Gσ (qj , p)]U 2 (p)dvgε Mε



2 = ξε,q j

[Gσ (qj + z, p) − Gσ (qj , p)]wj (y − qj )2 χj2 dvgε Mε

ig 2,

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+

 l=j



 2 = ξε,q [ j

 l=j

+

 l=j

π

Mε

˜ σ (qj + z, p) − G ˜ σ (qj , p))wj (y − qj )2 χj2 dvgε ] (G

Mε

+

[Gσ (qj + z, p) − Gσ (qj , p)]wl (y − ql )2 χl2 dvgε + O(ξε2 e−2R sin k )

2 ξε,q l



˜ σ (qj , ql ) · zwl (y − ql )2 χl2 dvgε ∇qj G

2 ξε,q l

Mε 2 ξε,q l

1 2



˜ σ (qj , ql )zt wl (y − ql )2 χl2 dvgε z∇q2j G

Mε

1

+O(ξε2 R − 2 e−2R sin = ξε2

 log R2

π k

−1

+ ξε2 ε 2 Rσ 2 e−Rσ |z|2 + ξε2 σ 3

|y| w 2 (y)dy + ∇qj F (q) · z |y − z| j





˜ σ (qj , ql )|z|3 ) G

j =l

1 wj2 dy + z∇q2j F (q)zt 2

R2

 wj2 dy



R2

−1

+O(ξε2 [σ 3 |z|3 + ε 2 |z|2 ]Rσ 2 e−Rσ ) where F (q) =

k  i=1

H˜ σ (qi , qi ) +

 i=j

˜ 1 (σ qi , σ qj ), Rσ = 2Rσ sin π . G k

(3.1)

Using this estimate and the expansion (2.1), we have the following estimate for the error: S1 (U, V )(z)  2 2 log = −ξε wj (z) R2

+

k 

ξε ε 4

i=1

+

k  i=1

ξε ε 4

|y| w 2 (y)dy + ∇qj F (q) · z |y − z| j

where

1 w dy + z∇q2j F (q)zt 2

1 z∇ 2 K(εqi )zt (Q[w] − 2P [w]) 20



2

R2

 1 qj ∇ 2 K(0)zt (Q[w] − 2P [w]) + R˜ 1 [w] 6 6

1

 1 2 1 K (εqi )|z|2 (3Q[w] − 4P [w]) + R2 [w] 45 10 

1 − +O ξε2 ([σ 3 |z|3 + ε 2 |z|2 ]Rσ 2 e−Rσ ) + ξε ε 5

+



w 2 dy R2



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R˜ 1 [u] =

11

z22 − z12 ∂ ∂u ∂u ∂ ∇K(0) · qj − ∇K(0) · qj ) ( 2 ∂x1 ∂z1 ∂x2 ∂z2 ∂u ∂u ∂ ∂ +z1 z2 ( ∇K(0) · qj − ∇K(0) · qj ). ∂x2 ∂z1 ∂x1 ∂z2

It is easy to see from the above estimate that for y = qj + z, and |εz| ≤

ig 2.

Lemma 3.1. S1 (U, V )(z) = S11 + S12 where S11 is an even function in z given by S11 = ξε2 wj2 (z)R1 (z) + ξε ε 4 R2 (z)wj (z) and R1 (z) = O(log(1 + |z|)), R2 (z) = O(|z|2 ), while 

S12 = −ξε2 wj2 (z)

 ∇qj F (q) · z

1 w dy + z∇q2j F (q)zt 2



2

R2

w 2 dy



R2

 1 qj ∇ 2 K(0)zt (Q[w] − 2P [w]) + R˜ 1 [w] 6 6 i=1

 −1 +O ξε2 ([σ 3 |z|3 + ε 2 |z|2 ]Rσ 2 e−Rσ ) + ξε ε 5 . +

k 

ξε ε 4

1

δ

Furthermore, S1 (U, V ) = O(ξε e− σ ) for |z| > σδ . 3.2. Linear theory In this section, we study the linearized operator Lε,q : H 2 (Mε ) × H 2 (Mε ) → L2 (Mε ) × ε ) defined by

L2 (M

 Lε,q = DS1

U V

.

To denote the dependence on ε and q we will also use the notation S1 = Sε,q . First define Zi,j (y) =

∂wi (y − qi )χε (y − qi ) ∂yj

where the coordinates are the geodesic normal coordinates. Set Kε,q = {Zi,j , i = 1, · · · , k, j = 1, 2} ⊂ H 2 (Mε ), Cε,q = {Zi,j , i = 1, · · · , k, j = 1, 2} ⊂ L2 (Mε ).

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We define our approximate kernels and cokernels as Kε,q := Kε,q × {0} ⊂ H 2 (Mε ) × H 2 (Mε ), Cε,q := Cε,q × {0} ⊂ L2 (Mε ) × L2 (Mε ). ⊥ and C ⊥ denote the orthogonal complement with respect to the scalar product Then we let Kε,q ε,q 2 2 L (Mε ) in H (Mε ) and L2 (Mε ), respectively. Define ⊥ ⊥ Kε,q := Kε,q × {0} ⊂ H 2 (Mε ) × H 2 (Mε ), ⊥ ⊥ := Cε,q × {0} ⊂ L2 (Mε ) × L2 (Mε ). Cε,q ⊥ . We are going to show that the equation Let πε,q denote the projection in L2 (Mε ) onto Cε,q

 πε,q ◦ Sε,q  has a unique solution  =

φ ψ



U +φ V +ψ

=0

(3.2)

⊥ . ∈ Kε,q

Set ⊥ ⊥ Lε,q = πε,q ◦ Lε,q : Kε,q → Cε,q .

(3.3)

The following proposition shows the invertibility of Lε,q . The proofs are quite standard now and so we omit the details here. We refer to [2] for details. Proposition 3.2. Let Lε,q be defined in (3.3). Then there exists a positive constant δ0 such that for √ε < δ0 , there is a constant C > 0 such that D

Lε,q L2 (Mε ) ≥ CH 2 (Mε ) ,

(3.4)

⊥ . Moreover, the map L for any q ∈ Qε ,  ∈ Kε,q ε,q is surjective.

3.3. Solving the nonlinear problem module the cokernel From the above proposition, we know that Lε,q is invertible (denote the inverse by L−1 ε,q ). Then we can rewrite the equation (3.2) as  = −(L−1 ε,q

 

U ◦ πε,q ) Sε,q − (L−1 ε,q ◦ πε,q )Nε,q () := Mε,q () V

where  =

φ ψ



 , Nε,q = Sε,q

U +φ V +ψ



 − Sε,q

U V



 − Sε,q



U V

.

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13

We are going to show that Mε,q () is a contraction mapping on Bε,η = { ∈ H 2 (Mε ) × H 2 (Mε )|H 2 (Mε ) ≤ η}. We have by Lemma 3.1 and Proposition 3.2 that 

Mε,q ()H 2 (Mε ) ≤ C πε,q ◦ Nε,q ()L2 (Mε ) + πε,q ◦ Sε,q

U V

L2 (Mε )



≤ C(c(η)η + cε,D ) where C > 0 is independent of η, c(η) → 0 as η → 0 and cε,D → 0 as max{ √ε , D √ D log 2 1 √D } → 0. Moreover, we have ε D log

ε

Mε,q () − Mε,q ( )H 2 (Mε ) ≤ Cc(η) −  H 2 (Mε ) . We choose η such that Cc(η) < 13 and Ccε,D ≤ 13 η. Such a choice of η is possible if we have √ taken max{ √ε , D log 2 1 √D } small enough. Then Mε,q is a contraction mapping in Bε,η . D

ε D log

ε

By the contraction mapping principle, there exists a solution to (3.2). Thus we have √ Proposition 3.3. There exists δ0 > 0 such that for max{ √ε , D log D

⊥ satisfying q ∈ Qε , we can find a unique solution (φ, ψ) ∈ Kε,q

 Sε,q

U +φ V +ψ

1 ε 2 D log

√

D ε

} ∈ (0, δ0 ), and

∈ Cε,q

and −1

(φ, ψ)H 2 (Mε ) ≤ C(ξε2 + ξε ε 4 R + ξε2 σ Rσ 2 e−Rσ ). For our purpose, we need more refined estimates on φ. Recall that S1 can be decomposed as S11 + S12 , where S11 in leading order is an even function in z while S12 in leading order is an odd function in z. So we can decompose φ = φε,q as in the following lemma. Lemma 3.4. Let φ = φ = ε, q be defined in Proposition 3.3. Then for y = pj + z, |σ z| ≤ δ0 , we have φ = φ1 + φ2 where φ1 is radially symmetric in z and 

−1 φ2 H 2 (Mε ) ≤ Cξε ξε σ Rσ 2 e−Rσ + ε 4 R .

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14

Proof. Let S[u] = S1 (u, T (u2 )), we first solve S[U + φ1 ] − S[U ] +

k 

S11 (y − qj ) ∈ Cε,q ,

j =1 ⊥ . Then we solve for φ1 ∈ Kε,q

S[U + φ1 + φ2 ] − S[U + φ1 ] +

k 

S12 (y − qj ) ∈ Cε,q ,

j =1 ⊥ . Using the same proof as in Lemma 3.3, both the above two equations a have for φ2 ∈ Kε,q √ unique solution for max{ √ε , D log 2 1 √D } small enough. This implies the uniqueness of D

ε D log

ε

φ = φ1 + φ2 . Moreover, it is easy to see from the estimate of S12 that 

−1 S12 L2 (Mε ) = ξε ξε σ Rσ 2 e−Rσ + ε 4 R ⊥ since S and S11 ∈ Cε,q 11 is an even function. Then we conclude that φ1 , φ2 have the required properties. 2

4. The reduced problem 4.1. Deriving the reduced problem By Proposition 3.3, for each q ∈ Qε , there exists (u, v) = (U + φ, V + ψ) such that  u Sε,q ∈ Cε,q . v Now, to solve the equation exactly, we have to further choose q such that Sε,q

 u ⊥ . ∈ Cε,q v

Lemma 4.1. Under the assumption of Proposition 3.3, the following expansion holds:  S1 (U + φ, V + ψ)Zi,j dvgε Mε

= −c2 ξε2 σ

 c ε 4 ∂K(0) 1 · qi − ∇ c2 ξε σ ∂xj

 l=i+1,i−1

  ˜ 1 (σ |qi − ql |) qi − ql G |qi − ql | j

+O(Eε )    ˜ σ (qi , qj ) . where c1 , c2 are given in (4.1) and (4.2), O(Eε ) = O ξε2 ε 4 R + ξε3 σ i=j G

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Proof. We compute  S1 (U + φε,q , V + ψε,q )Zi,j dvgε Mε

=

 

gε (U + φε,q ) − (U + φε,q ) +

(U + φε,q )2  Zi,j dvgε V + ψε,q

gε (U + φε,q ) − (U + φε,q ) +

(U + φε,q )2  Zi,j dvgε V

Mε

=

 

Mε

+

  (U + φε,q )2 (U + φε,q )2  − Zi,j dvgε V + ψε,q V

Mε

:= I1 + I2 . We decompose  I1 =

[gε (ξε,qi wi + φε,q ) − (ξε,qi wi + φε,q ) +

Mε



− Mε

(ξε,qi wi + φε,q )2 ]Zi,j dvgε V (qi )

1 π (ξε,qi wi + φε,q )2 [V (qi + z) − V (qi )]Zi,j dvgε + O(ξε R − 2 e−2R sin k ) 2 V (qi )

= I11 + I12 . Note that φε,q = φ1 + φ2 which implies that  [gε φε,q − φε,q + 2wi φε,q ]Zi,j dvgε Mε



=

(φ1 + φ2 )∂yj [gε wi − wi + wi2 ]dvgε

Mε

=O



  −1 ξε Rσ 2 σ e−Rσ + ε 4 R ε 2 ξε ,

and 

2 φε,q

Mε

ξε,qi 

= Mε

Zi,j dvgε

2φ1 φ2 + φ22 Zi,j dvgε ξε,qi

15

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16

 −1 = O ξε2 ξε Rσ 2 σ e−Rσ + ε 4 R since φ1 is an even function. From the expression for R˜ 1 and using Lemma B.2 in [38], one has  ξε,qi [gε wi − wi + wi2 ]Zi,j dvgε Mε



= ξε,qi ε 4

 ∂w 1 dy + O(ξε ε 5 ) qi ∇ 2 K(0)y t (Q[w] − 2P [w]) + R˜ 1 [w] 6 ∂yj

R2

π = (− 4

∞ ∂K(0) (w  )2 r 3 dr)ξε,qi ε 4 ∇ · qi + O(ξε ε 5 ) ∂xj 0

= −c1 ξε,qi ε 4 ∇

∂K(0) · qi + O(ξε ε 5 ) ∂xj

where π c1 = 4

∞ (w  )2 r 3 dr > 0.

(4.1)

0

Combining the above estimates, one has I11 = −c1 ξε,qi ε 4 ∇

∂K(0) · qi ∂xj

 −1 +O ξε2 ξε Rσ 2 σ e−Rσ + ε 4 R . Next for I12 , one has  I12 = − Mε

=−



(ξε,qi wi + φε,q )2 [V (qi + z) − V (qi )]Zi,j dvgε V (qi )2 wi2 (V (qi + z) − V (qi ))Zi,j dvgε

Mε

 +

Mε



+ Mε

2φwi (V (qi + z) − V (qi ))Zi,j dvgε ξε,qi  φ2 (V (q + z) − V (q ))Z dv i i i,j g ε 2 ξε,q i 

2 ∂ F (q) = −ξε,q i qi,j R2

 w 2 dy R2

w2

 ∂w −1 yj dy + O ξε2 ξε Rσ 2 σ e−Rσ + ξε ε 2 σ 2 + ε 4 R ∂yj

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17

 − 12 2 2 −Rσ 4 = c2 ξε,q ∂ F (q) + O ξ R σ e + ε R , ξ q ε σ ε i,j i where

 c2 = −

 2

w dy

R2

w2

∂w yj dy > 0. ∂yj

R2

In conclusion, one has   ∂K(0) I1 = −ξε,qi c1 ε 4 ∇ · qi − c2 ξε,qi ∂qi,j F (q) + O(Eε ). ∂xj For I2 , recall that ψε,q satisfies 2 gε ψε,q − σ 2 ψε,q + 2U φε,q + φε,q = 0.

We can make the following decomposition 2 gε ψε,q,1 − σ 2 ψε,q,1 + 2U φε,q,1 + φε,q,1 =0

and 2 gε ψε,q,2 − σ 2 ψε,q,2 + 2U φε,q,2 + φε,q,2 + 2φε,q,1 φε,q,2 = 0.

Then one can see that ψε,q,1 is radially symmetric with respect to z, and

 −1 ψε,q,2 H 2 (Mε ) = O ξε ξε Rσ 2 σ e−Rσ + ε 4 R . Moreover, from the Green’s representation formula,    2 ψε,q (qi + z) − ψε,q (qi ) = )dvgε (p) Gσ (qi + z, p) − Gσ (qi , p) (2U φε,q + φε,q Mε

= O(ξε3 )∇qi F (q)|z| + Re (z) where Re (z) is even function in z. This implies I2 =

  (U + φε,q )2 (U + φε,q )2  − Zi,j dvgε V + ψε,q V

Mε



=− Mε



=−

(U + φε,q )2 ψε,q Zi,j dvgε + O(Eε ) V2 1 ∂wi3 (ψε,q − ψε,q (qi ))dy + O(Eε ) 3 ∂yj

R2

= O(Eε ).

(4.2)

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18

Thus one has  Mε

  ∂K(0) S1 (U + φ, V + ψ)Zi,j dvgε = −ξε c1 ε 4 ∇ · qi − c2 ξε,qi ∂qi,j F (q) + O(Eε ). ∂xj

Recall the definition of F (q) from (3.1):

F (q) :=

k 

H˜ 1 (σ qi , σ qi ) +



˜ 1 (σ qi , σ qj ) G

i=j

i=1

and ∇y H˜ (y, z)|y=z = 0. Using the asymptotic behavior of ˜ 1 (x, y) = c|x − y|− 2 e−|x−y| (1 + o(1)), G 1

(4.3)

one has  S1 (U + φ, V + ψ)Zi,j dvgε Mε

= −c2 ξε2 σ

 c ε 4 ∂K(0) 1 · qi − ∇ c2 ξε σ ∂xj

 l=i+1,i−1

  ˜  (σ |qi − ql |) qi − ql + O(Eε ). G |qi − ql | j

Define ∂U = (Zq1 · n 1 , · · · , Zqk · n k , Zq1 · t 1 , · · · , Zqk · t k )t ∂q and Qi = σ qi = Q0i + σ f˜i n i + σ g˜ i t i . In the following, we denote q˜ = σ (f˜1 , · · · , f˜k , g˜ 1 , · · · , g˜ k )t = (f1 , · · · , fk , g1 , · · · , gk )t and R0 = |Q0i | = σ R.

2

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19

4.2. The reduced problem for general k = 3, 4, . . .  Next we analyze Mε S1 (U + φ, V + ψ) ∂U ∂q dvgε . We have the following:  Lemma 4.2. Mε S1 (U + φ, V + ψ) ∂U ∂q dvgε = 0 is equivalent to the following system for the ˜ perturbation q:

dˆ d

M1 +

 1 C1 M2 + M3 q˜ = C2 b0 + O(E) d d

where d = 2R0 sin πk , and dˆ is defined in (4.5),  E=

E1 E2



and Ei are k-dimensional vectors of the form E1 = O

     ξε ˜ 2  1 1 |q| 2 ˜ 1 , E2 = O 1 . ξε + 2 + |q| + + R0 R02 R0 R0

 

11 11 Further, C1 = 4 sin2 πk K22K−K , C2 = −2 sin πk K22K−K are two constants and the matrices 11 11 M1 , M2 , M3 and the vector b0 are given as follows:

 M1 =  M2 =

(A1 + 4I ) sin2 −A2 sin πk cos

A1 cos2

π k

π k π k

A2 sin πk cos πk

+ 4 sin2 πk I

−A1 cos2



π k

−A2 sin πk cos πk

A2 sin πk cos πk −A1 sin2 πk   B1 B2 B1 0 M3 = , b0 = 1 B2 B3 0 B2

where ⎛

−2 ⎜ 1 ⎜ A1 = ⎜ . ⎝ .. ⎛

1

1 0 ··· −2 1 · · · .. .. . . . . . 0 0 ···

0 1 0 ··· ⎜ −1 0 1 · · · ⎜ A2 = ⎜ . .. .. . . ⎝ .. . . . 1 0 0 ···

0 0 .. .

1 0 .. .

1

−2

0 0 .. . −1

⎞ ⎟ ⎟ ⎟, ⎠

⎞ −1 0 ⎟ ⎟ .. ⎟ , . ⎠ 0

B1 = diag{sin θ1 , · · · , sin θk }, 2

,

2

B2 = diag{sin θ1 cos θ1 , · · · , sin θk cos θk },

 ,

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20

B3 = diag{cos2 θ1 , · · · , cos2 θk }. Proof. Wlog, assume that  ∇ 2 K(0) =

K11 0

0 K22



where K11 , K22 < 0. By direct calculation, one has ∇ 2 K(0) · Qi = ∇ 2 K(0) · (Q0i + fi n i + gi t i ) = K11 (Q0i + fi n i + gi t i )   +(K22 − K11 ) (R0 + fi )(sin2 θi n i + sin θi cos θi t i ) + gi (sin θi cos θi n i + cos2 θi t i ) = K11 R0 n i +R0 (K22 − K11 )(sin2 θi n i + sin θi cos θi t i )   + ni K11 fi + (K22 − K11 )(sin2 θi fi + sin θi cos θi gi )   +t i K11 gi + (K22 − K11 )(sin θi cos θi fi + cos2 θi gi .

(4.4)

Next using the facts that 2π 2π 2π 2π n i + sin t i , t i+1 = − sin n i + cos t i , k k k k 2π 2π 2π 2π n i−1 = cos n i − sin t i , t i−1 = sin n i + cos t i , k k k k n i+1 = cos

and for |a| >> |b| a+b a b a·b a |b| = + − + O( 2 ), |a + b| |a| |a| |a|2 |a| |a| one has Qi+1 − Qi |Qi+1 − Qi | =

Q0i+1 − Q0i + fi+1 n i+1 + gi+1 t i+1 − fi n i − gi t i

|Q0i+1 − Q0i + fi+1 n i+1 + gi+1 t i+1 − fi n i − gi t i | π π = − sin n i + cos t i k k   1 2π 2π 2π 2π + − gi+1 sin − fi ) + t i (fi+1 sin + gi+1 cos − gi )

i (fi+1 cos π n 2R0 sin k k k k k

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21

 1 2π 2π π − gi+1 sin − fi ) sin π − (fi+1 cos 2R0 sin k k k k  2π 2π π + (fi+1 sin + gi+1 cos − gi ) cos k k k π π |q| × (− ni sin + t i cos ) + O( 2 ), k k R0 −

and Qi−1 − Qi |Qi−1 − Qi | =

Q0i−1 − Q0i + fi−1 n i−1 + gi−1 t i−1 − fi n i − gi t i

|Q0i−1 − Q0i + fi−1 n i−1 + gi−1 t i−1 − fi n i − gi t i | π π = − sin n i − cos t i k k   1 2π 2π 2π 2π

+ (f cos sin ) + t (−f sin cos ) + g − f + g − g n

i i−1 i−1 i i i−1 i−1 i 2R0 sin πk k k k k  2π 2π π 1 + gi−1 sin − fi ) sin (fi−1 cos − 2R0 sin πk k k k  2π 2π π + gi−1 cos − gi ) cos + (−fi−1 sin k k k π π |q| × ( ni sin + t i cos ) + O( 2 ). k k R0 Moreover, we define dˆ = −

˜  (d) G 1 d = d + O(1). ˜  (d) G

(4.5)

1

We expand ˜ 1 (|Qi+1 − Qi |) = G ˜ 1 (|Q0 − Q0 |) + G ˜ 1 (|Q0 − Q0 |) G i i i+1 i+1  2π 2π π 2π 2π π − gi+1 sin − fi ) sin + (fi+1 sin + gi+1 cos − gi ) cos × − (fi+1 cos k k k k k k and ˜ 1 (|Qi−1 − Qi |) = G ˜ 1 (|Q0 − Q0 |) + G ˜ 1 (|Qi−1 − Qi |) G i−1 i  2π 2π π 2π 2π π + gi−1 sin − fi ) sin − (−fi−1 sin + gi−1 cos − gi ) cos . × − (fi−1 cos k k k k k k

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22

Combining all the above expansions, one has i+1 

˜ 1 (|Qi − Qj |) G

j =i−1

Qj − Qi |Qj − Qi |

 ˜ 1 (d) − 2 sin π n i =G k 

ˆ d π 2 π − − fi+1 + fi−1 + 2fi + (gi+1 − gi−1 ) cot sin n i d k k

 π  π + (fi+1 − fi−1 ) tan + gi+1 + gi−1 − 2gi cos2 t i k k   1 π π + fi+1 + fi−1 − 2fi − (gi+1 − gi−1 ) tan cos2 n i d k k

  π π + (fi+1 − fi−1 ) cot − (gi+1 + gi−1 + 2gi ) sin2 t i k k  ! ˜2 |q| |q| ˜ 1 (d) |q| ˜ 2 n i + t i + 2 +O G . d R0 Now let us define R0 such that −2 sin

π ˜ π  c1 ε 4 K11 R0 = 0 G1 2R0 sin + k k c2 ξε σ 2

˜  < 0 and K11 < 0. which is possible since G 1 Then  ∂U S1 dvgε = 0 ∂q Mε

is reduced to the following linear system for the perturbation q˜ = (f1 , · · · , fk , g1 , · · · , gk )t

dˆ d

M1 +

 1 C1 M2 + M3 q˜ = C2 b0 + O(E) d d

where  E=

E1 E2



and Ei are k-dimensional vectors of the form  E1 = O

 !  !  ˜2 |q| ξε |q| |q| 2 ˜ ξε + 2 + |q| + + 1 , E2 = O 1 . R0 R02 R0 R0

Further, we have C1 = 4 sin2

π K22 −K11 k K11 ,

C2 = −2 sin πk

K22 −K11 K11 ,

(4.6)

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 M1 =  M2 =

(A1 + 4I ) sin2 −A2 sin πk cos

A1 cos2

π k

π k π k



A2 sin πk cos πk

+ 4 sin2 πk I

−A1 cos2

23

,

π k

−A2 sin πk cos πk



A2 sin πk cos πk −A1 sin2 πk   B1 B2 B1 0 , b0 = 1 M3 = B2 B3 0 B2

,

where ⎛

−2 ⎜ 1 ⎜ A1 = ⎜ . ⎝ .. ⎛

1 0 ··· −2 1 · · · .. .. . . . . . 0 0 ···

1

0 1 0 ··· ⎜ −1 0 1 · · · ⎜ A2 = ⎜ . .. .. . . ⎝ .. . . . 1 0 0 ···

0 0 .. .

1 0 .. .

1

−2

0 0 .. . −1

⎞ ⎟ ⎟ ⎟, ⎠

⎞ −1 0 ⎟ ⎟ .. ⎟ , . ⎠ 0

B1 = diag{sin2 θ1 , · · · , sin2 θk }, B2 = diag{sin θ1 cos θ1 , · · · , sin θk cos θk }, B3 = diag{cos2 θ1 , · · · , cos2 θk }.

2

Remark 4.3. Since for general k ≥ 2, the linear system (4.6) is not easy to solve, we now compute q˜ for k = 2. In this case, only two spikes interact with each other, and one has | sin θ1 | = | sin θ2 |, | cos θ1 | = | cos θ2 |. This will simplify our computations a lot. 4.3. The reduced problem for k = 2 The reduced problem for k = 2 is given by the following result:  Lemma 4.4. When k = 2, Mε S1 (U + φ, V + ψ) ∂U ∂q dvgε = 0 is equivalent to the following ˜ system for the perturbation q: ⎛ ⎞  ˜ 2 (1, 1)t ξε + |q|2 + |q|  1 1 R0 ⎠ Mq˜ := M2 + M3 q˜ = b0 + O ⎝  ξε M1 + ˜ 2 |q| |q| t d R0 R0 (1, 1) + + 2 R R

dˆ

0

R0

where 



β1 I β3 A1 , β3 A1 β2 I   0 0 0 0

M3 = , b0 = −β3 1, 0 A1 0 − 12 β1 A0 M1 =

β1 A0 0

0 0



, M2 =

0

(4.7)

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24

 A0 =

1 1

1 1



 , A1 =

1 0

0 −1



and β1 = K11 +(K22 −K11 ) sin2 θ1 , β2 = K11 +(K22 −K11 ) cos2 θ1 , β3 = (K22 −K11 ) sin θ1 cos θ1 . Proof. The proof is similar to Lemma 4.2. First we get ˜ 1 (|Q1 − Q2 |) Q2 − Q1 G |Q2 − Q1 |

  ˆ ˜ 1 (d) 1 − d (f1 + f2 ) − n 1 − 1 (g1 + g2 )t 1 =G d 2R0  " # 2 ˜ |q| ˜ 1 (d) |q| ˜ 2 n i + +O G t i d where d = 2R0 . Combining with (4.4), we have  S1 (U + φ, V + ψ)∇wdvgε Mε

=

c1 ε 4 R0 c2 ξε σ 2

× (K11 + (K22 − K11 ) sin2 θ1 ) n1 + (K22 − K11 ) sin θ1 cos θ1 t 1 ) 1 (K11 f1 + (K22 − K11 )(sin2 θ1 f1 + sin θ1 cos θ1 g1 )) n1 R0  1 + (K11 g1 + (K22 − K11 )(sin θ1 cos θ1 f1 + cos2 θ1 g1 ))t 1 R0   dˆ 1  ˜ n 1 + (g1 + g2 )t 1 −G1 (d) 1 − (f1 + f2 ) d d  ! ˜2 | q| |q|  2 ˜ 1 (d) |q| ˜ n i + t i + 2 +O G + O(E). d R0 +

˜ << 1, Here by carefully checking the error estimates and using the facts that for k = 2, if |q|  Q1 − Q2 1 = (1 + o(1)) n1 + O t 1 , |Q1 − Q2 | d   ˜  (d) σG  1 ˜ 1 (d)z · n i ) ˜ · z = O(σ G ni + O z · t i t i , ∂q1,j F (q) d

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25

and 1 Q1 ∇ 2 K(0)z(Q[w] − 2P [w]) + R˜ 1 [w] = O(R0 ) n1 + O(1)t 1 , 6 one can have a more accurate estimate for the error term E, i.e.  E=

E1 E2



⎞   ˜ σ (qi , qj ) (1, 1)t ξε2 ε 4 R + ξε3 σ i=j Gs ⎟ ⎜  =O⎝  ⎠.  1 2 4 3 t ˜ ξ ε R + ξ σ (q , q ) (1, 1) G σ i j ε i=j R0 ε ⎛ 

Define R0 as

 c1 ε 4 ˜ 1 (2R0 ). R0 K11 + (K22 − K11 ) sin2 θ1 = G 2 c2 ξε σ Considering the leading order matrix M1 , the kernel in leading order is spanned by the vectors α(1, −1, 0, 0)t , β(0, 0, 1, 0)t , γ (0, 0, 0, 1)t . Since the righthand side in leading order is b0 = −β3 (0, 0, 1, −1)t we get the solvability condition β3 = 0. Therefore we have to choose θ1 = 0 or π2 . By Taylor expansion,

log(log 2 )  1 1 3 1 1 c3 ε D log 2 − log(log 2 ) − log + O 1 2 ε D 4 ε D 2 ξε log 2 1

R0 =

ε D

where c3 = −

c1 β1 >0 2cc2

since β1 < 0. So the reduced system becomes ⎛ ⎞  ˜ 2 (1, 1)t ξε + |q|2 + |q|  1 1 R0 ⎠ Mq˜ := M2 + M3 q˜ = b0 + O ⎝  ξε M1 + ˜ 2 |q| |q| d R0 R0 (1, 1)t + 2 + R R

dˆ

0

R0

given in (4.7), where 



β1 I β3 A1 , β3 A1 β2 I   0 0 0 0

= −β M3 = , b 1, 0 3 0 A1 0 − 12 β1 A0   1 1 1 0 , A1 = A0 = 1 1 0 −1 M1 =

β1 A0 0

0 0



, M2 =

0

(4.8)

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26

and β1 = K11 + (K22 − K11 ) sin2 θ1 , β2 = K11 + (K22 − K11 ) cos2 θ1 , β3 = (K22 − K11 ) sin θ1 cos θ1 . This finishes the proof.

2

Remark 4.5. From the definition of R0 , one can check that ε|q1 | ∼

εR0 √ 1 √ . ∼ D log σ ε 2 D log D ε

So under assumption (1.3), one can easily see that ε|q1 | → 0 as ε → 0. Finally, for k = 2 we solve the reduced problem and complete the proof Theorem 1.1. Proof of Theorem 1.1. First since β3 = 0, we have to choose θ1 = 0 or reduced system becomes  Mq˜ =

ˆ

( dd A0 + R10 I )β1 0

 0 1 1 R0 (β2 I − 2 β1 A0 )

π 2.

In this case, the

⎛

 |q| ˜ 2 (1, 1)t + |q| R02 q˜ = O ⎝  ξε ˜ 2 |q| |q| t R0 + R 2 + R0 (1, 1) ξε +

⎞ ⎠.

(4.9)

0

If β2 − β1 = 0, the matrix is invertible, and one can check that M−1 ≤ CR0 . Our idea is to first improve the top line of the right hand side of (4.9) to O( ξε2 ) from O(ξε ). R0

This is done in the following way. Since when θ1 = 0 or π2 , this approximate solution has some symmetry around each spike in main order, by carefully checking the calculation in Section 3 and 4, one can decompose Eε in Lemma 4.1 as [δ1 ξε + δ2 Rξε0 + O( ξε2 )]ξε ε 4 R for some δ1 , δ2 R0

which is tedious but standard. So one can decompose fi = f 0 + f 1 + fˆi , where f 0 , f 1 are chosen to match the O(ξε ) and O( Rξε0 ) term on the right hand side of the reduced problem. First ˜  (2R0 + 2f 0 ) = G ˜  (2R0 )(1 + δ1 ξε ), which implies that |f 0 | = O(ξε ). f 0 is chosen such that G 1 1  1 ˜ (2R0 + 2f 0 + 2f 1 ) − G ˜  (2R0 + 2f 0 ) = G ˜  (2R0 )δ2 ξε and Then we choose f such that G 1

1

1

R0

|f 1 | = O( Rξε0 ). In this way we can get the reduced problem for {fˆi , gi } (we still denote its solution ˜ as follows: by q)  Mq˜ =

ˆ

( dd A0 + R10 I )β1 0

 0 1 1 R0 (β2 I − 2 β1 A0 )

⎛ξ

ε

q˜ = O ⎝

R2  ξε0 R0

+ +

|q| R02 |q| R02

⎞  ˜ 2 (1, 1)t + |q| ⎠. ˜ 2 t (1, 1) + |Rq| 0

Since M−1  ≤ CR0 , one can find a solution q˜ to by contraction mapping such that ˜ ≤ Cξε . |q|

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27

In conclusion, we find a solution such that maxi (|fˆi | + |gi |) = O(ξε ). It is easy to check that when θ1 = 0, then β2 − β1 = K22 − K11 ; while when θ1 = π2 , β2 − β1 = K22 K11 − K22 . So if K = 1, one can solve the equation and get two solutions which correspond to 11 π θ1 = 0 and θ1 = 2 , respectively. 2 5. Stability study I: study of the large eigenvalues We consider the stability of the steady-state (uε , vε ) constructed in Theorem 1.1. In this section, we first study the large eigenvalues which satisfy λε → λ0 = 0 in the limit as √ max{ √ε , D log 2 1 √D } → 0. D

ε D log

ε

Linearizing the system around the equilibrium states (uε , vε ) obtained in Theorem 1.1, we obtain the following eigenvalue problem: ⎧ ⎨  φ − φ + 2uε φ − u2ε ψ = λφ, gε vε vε2 (5.1) ⎩  ψ − σ 2 ψ + 2u φ = τ λσ 2 ψ, gε

ε

for (φ, ψ) ∈ H 2 (Mε ) × H 2 (Mε ). In this section, √ since we study the large eigenvalues, we may assume that |λε | ≥ c > 0 for max{ √ε , D log 2 1 √D } small enough. If Re(λε ) ≤ −c < 0, then λε is a stable large D

ε D log

ε

eigenvalue,√we are done. Therefore, we may assume that Re(λε ) ≥ −c and for a subsequence max{ √ε , D log 2 1 √D } → 0, λε → λ0 = 0. We shall derive the limiting eigenvalue probD

ε D log

ε

lem which is given by a coupled system of NLEPs. The second equation of (5.1) is equivalent to gε ψ − σ 2 (1 + τ λε )ψ + 2uε φ = 0 on Mε .

(5.2)

We introduce the following notation: $ σλ = σ 1 + τ λε , √ where in 1 + τ λε , we take the principal part of the square root. Let us assume that φH 2 (Mε ) = 1. We cut off φ = φε as follows: φε,j = φε χε (z − qj ), j = 1, · · · , k,

(5.3)

where the cutoff function χε has been defined in (2.8). From (5.1) and the exponential decay of w, it follows that φε =

k  j =1

φε,j (1 + o(1)) in H 2 (Mε ).

(5.4)

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Then by a standard procedure (see [15], Section 7.12), we extend φε,j to a function defined on R2 such that φε,j H 2 (R2 ) ≤ Cφε,j H 2 (Mε ) , j = 1, · · · , k. Since φε H 2 (Mε ) = 1, φε,j H 2 (R2 ) ≤ C. By taking a subsequence, we may assume that √ φε,j → φj as max{ √ε , D log 2 1 √D } → 0 in H 1 (R2 ) for some φj ∈ H 1 (R2 ) for j = D

ε D log

1, · · · , k. By (5.1), we have

ε

 ψε (qj ) =

Gσλ (qj , y)2uε φε (y) dy

Mε

 =

k  Gσλ (qj , y)2( ξε,qi wj (y − qi )φε,i + O(ξε2 )) dy i=1

Mε

=



1 1 log 2π σλ

2ξε,j wj φε,j (1 + o(1)) dx. R2

Substituting the above √ equation into the first equation of (5.1) and using the expansion of ξε,j , in the limit max{ √ε , D log 2 1 √D } → 0 we arrive at the following nonlocal eigenvalue D

ε D log

problem (NLEP):

ε

 2 R2 wφj dx 2 + w = λ0 φj , j = 1, · · · , k. φj − φj + 2wφj − 1 + τ λ0 R2 w 2 dx

(5.5)

+

By Theorem 3.5 in [54], (5.5) has only stable eigenvalues if τ is small enough. In conclusion, we have shown that the large eigenvalues of the solutions given in Theorem 1.1 are all stable if τ is small enough. 6. Stability study II: study of the small eigenvalues Now we study the eigenvalue √ problem (5.1) with respect to small eigenvalues. Namely, we assume that λε → 0 as max{ √ε , D log 2 1 √D } → 0. D

ε D log

ε

Our main result in the section says that if λε → 0, then λε ∼ ε 4 Rσ0 (M) where σ0 (M) is an eigenvalue of M defined in (4.9). So the stability of the solutions depends on K22 . the eigenvalues of M. It turns out that it is related to the ratio K 11

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29

6.1. Eigenfunctions and error estimates Let (uε , vε ) be the equilibrium state constructed for equation (1.5), and define uε,j = ξε,qj uε (y), j = 1, · · · , k, where ξε,qj is defined in (2.8) and calculated in (2.10). It is easy to see that uε =

k 

uε,j (1 + o(1)) in H 2 (Mε ).

j =1

Now let us set λ0 = 0 in (5.5), we have  φj − φj + 2wφj − 2w

wφj dy = 0, 2 R2 w dy

2 R2

which is equivalent to   2 wφj dy L0 φj − 2 R w = 0, j = 1, · · · , k, 2 R2 w dy

where L0 =  − 1 + 2w. We have  % & ∂w R2 wφj dy  w ∈ span φj − 2 , i = 1, 2 , j = 1, · · · , k. 2 ∂yi R2 w dy This implies that

 R2

wφj dy = 0, and we can decompose φε as φε =

2 aε k   j,i ∂uε,j + φε⊥ ξε ∂yi j =1 i=1

where φε⊥

⊥ K˜ ε,q := span

%

& ∂uε,j , j = 1, · · · , k, i = 1, 2 . ∂yi

The decomposition of φε implies that ψε =

2 k  

ε aj,i ψε,j,i + ψε⊥

j =1 i=1

where ψε,j,i is the unique solution of gε ψε,j,i − σλ2 ψε,j,i + 2ξε−1 uε

∂uε,j = 0, ∂yi

(6.1)

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30

and gε ψε⊥ − σλ2 ψε⊥ + 2ξε−1 uε φε⊥ = 0. ε = O(1). Substituting the decomposition of φ Supposing φε H 2 (Mε ) = 1, then we have aj,i ε and ψε into (5.1), using the fact that

gε uε,j − uε,j +

u2ε,j vε

= h.o.t.,

we have 2 k  

2 ε uε ∂yi vε aj,i ( vε2 ξε j =1 i=1

+ gε φε⊥

− φε⊥

− ψε,j,i ) +

2 aε k   j,i j =1 i=1

ξε

[gε

∂uε,j ∂ − g uε,j ] ∂yi ∂yi ε

  aj,i ∂uε,j 2uε ⊥ u2ε ⊥ + φε − 2 ψε − λφε⊥ + h.o.t. = λ . vε vε ξε ∂yi k

2

ε

(6.2)

j =1 i=1

We set I1 =

2 k  

  aj,i ∂uε,j u2ε ∂yi vε ∂ ( − ψε,j,i ) + [gε − g uε,j ] 2 ξε ∂yi ∂yi ε v ε ξε k

ε aj,i

j =1 i=1

ε

2

j =1 i=1

:= I11 + I12 , I2 = gε φε⊥ − φε⊥ +

2uε ⊥ u2ε ⊥ φ − 2 ψε − λφε⊥ . vε ε vε

First we shall derive the estimate for φε⊥ . Since φε⊥ ⊥ K˜ ε,q , we have φε⊥ H 2 ≤ CI1 L2 . By the expansion of gε in (2.1), one knows that I12  ≤ Cε

2

2 k  

ε |aj,i |.

(6.3)

j =1 i=1

For I11 , using the equation satisfied by ψε,j,i , we get 

˜ σλ (y, z)[2ξε−1 uε G

ψε,j,i (y) = Mε



= ξε Mε

˜ σλ (y, z) G

∂uε,j ]dz + h.o.t. ∂zi

∂w(z − qj )2 dz + h.o.t., ∂zi

(6.4)

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31

and using the equation satisfied by vε , we have ∂vε = ∂yi



Mε

˜σ ∂G (y, z)u2ε (z)dz ∂yi 

= ξε2 Mε

(6.5)

k  ˜σ ∂G (y, z)( w(z − ql )2 )dz + h.o.t. ∂yi l=1

Combining (6.4) and (6.5), one has 1 ∂vε (y) − ψε,j,i (y) ξε ∂yi  k    ∂G  ˜σ ∂w(z − qj )2 2 ˜ σλ (y, z) = ξε (y, z)( w(z − ql ) )dz − dz + h.o.t. (6.6) G ∂yi ∂zi l=1

Mε

= ξε

 1 2π 

+ Mε

 Mε

Mε

∂ 1 1 ∂w 2 (z − qj ) ln dz w 2 (z − qj ) − ln ∂yi |y − z| |y − z| ∂zi

(6.7)

∂w 2 (z − qj ) ∂ H˜ σ (y, z)w 2 (y − qj ) − H˜ σ (y, z) dz ∂yi ∂zi

(6.8)

   ∂G ˜σ + (y, z)w 2 (z − ql )dz + h.o.t. . ∂yi l=j M

(6.9)

ε

∂ Using the fact that ( ∂y +

∂ ∂z ) log |y

− z| = 0 for y = z, we have

∂Fj (y) 1 ∂vε (y) − ψε,j,i (y) = ξε ( ξε ∂yi ∂yi

 w 2 dz + O(σ ))

R2

where Fj (y) = H˜ σ (y, qj ) +



˜ σ (y, q ). G

=j

From this estimate, using the fact that

I11 =

∂Fj (qj ) ∂yi

=

1 ∂F (q) 2 ∂qj,i ,

we have

2 aε k  2  j,i uε ∂vε [ − ξε ψε,j,i ] ξε vε2 ∂yi j =1 i=1

=

2 k   j =1 i=1

ε aj,i ξε

∂Fj (qj ) ( ∂yi



R2

w 2 dz)(1 + O(σ |y − qj |))

u2ε vε2

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32

 k 2 ∂F (q)   ε = O ξε |aj,i | ∂qj,i  =O

ε 4 Rσ σ

j =1 i=1



(6.10)

.

Combining (6.3) and (6.10), one has I1 L2 (Mε ) ≤ Cε 2 . So φε⊥ H 2 (Mε )

≤ Cε

2

2 k  

ε |aj,i |.

(6.11)

ε |aj,i |.

(6.12)

j =1 i=1

Using the equation satisfied by ψε⊥ , 2 k  

ψε⊥ H 2 (Mε ) ≤ Cε 2

j =1 i=1

6.2. Derivation of the finite-dimensional eigenvalue problem Multiplying (6.2) by

1 ∂uε,m ξε ∂y

and integrating over Mε , one has

r.h.s = λ

2 aε  k   ∂uε,j ∂uε,m j,i dy 2 ξε ∂yi ∂y j =1 i=1

=λ

k  2 

Mε

ε aj,i δj,m δi,

j =1 i=1 ε = λam,

 ∂w 2 dy + o(1) ∂y1

R2

 ∂w 2 dy + o(1). ∂y1

R2

For the l.h.s., we get  Mε

∂uε,m I2 ξε−1 dy ∂y



ξε−1 [φε⊥ − φε⊥ +

= Mε



= −λ

ξε−1 φε⊥

Mε

 +

Mε



= −λ Mε

2uε ⊥ u2 ∂uε,m φε − λφε⊥ − 2ε ψε⊥ ] dy vε vε ∂y

∂uε,m dy ∂y

u2ε,m −1 ∂vε ⊥ ∂uε,m ⊥  ε 6 Rσ ξε φε − ψε dy + O( ) 2 ∂y ∂y σ vε ξε−1 φε⊥

∂uε,m dy ∂y

(6.13)

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 + Mε



+ Mε



− Mε



− Mε

33

u2ε,m −1 ⊥ ∂vε ∂vε ξ φ ( (qm + y) − (qm ))dy ∂y vε2 ε ε ∂y u2ε,m −1 ⊥ ∂vε ξ φ (qm )dy vε2 ε ε ∂y u2ε,m −1 ∂uε,m ⊥ ξ (ψε (qm + y) − ψε⊥ (qm ))dy vε2 ε ∂yl u2ε,m −1 ∂uε,m ⊥ ξ ψ (qm )dy + o(ε 4 ) vε2 ε ∂y ε

= J1 + J2 + J3 + J4 + J5 + o(ε 4 ).

(6.14)

Using the equation for ψε⊥ , one has ψε⊥ (qm ) =



Gσλ (y, z)(2ξε−1 uε φε⊥ )(z)dz = O(φε⊥ H 2 ) = O(ε 2 ),

Mε

ψε⊥ (qm

+ y) − ψε⊥ (qm ) =

 =2



˜ σλ (y + qm , z) − G ˜ σλ (qm , z)]2ξε−1 uε φε⊥ (z)dz [G

Mε

(6.15)

˜ σλ (qm , z) · yξε−1 uε φε⊥ dz ∇qm G

Mε

= O(ε 2

ε 4 Rσ |y|) = o(ε 4 |y|). σ ξε

Similarly, using the equation satisfied by vε , one has  ∂vε ∂F (q) (qm ) = O ξε2 , ∂y ∂qm,

 2 ∂vε ∂vε 2 ∂ F (q) (qm + y) − (qm ) = O ξε |y| . ∂y ∂y ∂qm, ∂qj,i

So using the definition of φε⊥ , one has J1 = 0. Using (6.16),  ε 4 Rσ J 2 + J3 = O ε 2 = o(ε 4 ), σ while using (6.15), one has J4 + J5 = o(ε 4 ). Combining all the above estimates, one has

(6.16)

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I2 ξε−1

Mε

Next recall the estimate for 

I11 ξε−1

Mε

=

j =1 i=1

=

Mε ε aj,i ξε

j =1 i=1

= −c2



⎜ ε aj,i ξε ⎝

2 k  

1 ∂vε ξε ∂yi (y) − ψε,j,i

(6.17)

in (6.9), we have

∂uε,m dy ∂y ⎛

k  2 

∂uε,m dy = o(ε 4 ). ∂yell

⎞ ∂Fj (y) u2ε ∂wj (y − qm ) ⎟ dy + h.o.t.⎠ ∂yi vε2 ∂y ⎛

∂F (q) ⎜ δi, ⎝ ∂qj,i ∂qm,

ε aj,i ξε

j =1 i=1

w 2 dy

R2

⎞⎛

 w 2 (y)yi R2

k  2 



∂w ⎟⎜ dy + o(1)⎠ ⎝ ∂yi



⎞

(6.18)

⎟ w 2 dy + o(1)⎠

R2

∂ 2 F (q) (δi, + o(1)) ∂qj,i ∂qm,

where c2 is defined in (4.2). For I12 , we get 

I12 ξε−1

Mε

=

k  2  j =1 i=1

∂uε,m dy ∂y  [gε

ε aj,i

Mε

∂w ∂ ∂w(y − qm ) (y − qj ) − gε w(y − qj )] dyδm,j δi, + o(1). ∂yi ∂yi ∂y (6.19)

Consider the expansion of gε around each point qj , i.e. replacing 0 by εqj in (2.1), we have   ∂ ∂ ∂w 1 g ε w −  g ε = K(εqj )ε 2 (Q[w] − 2P [w]) − (Q[∂i w] − 2P [∂i w]) ∂yi ∂yi 3 ∂yi   ∂ 1 + (∇K(εqj ) · y)ε 3 (Q[w] − 2P [w]) − (Q[∂i w] − 2P [∂i w]) 6 ∂yi 1 ∂ (∇K(εqj ) · y)ε 3 (Q[w] − 2P [w]) 6 ∂yi   ∂ 1 + (y∇ 2 K(εqj )y t )ε 4 (Q[w] − 2P [w]) − (Q[∂i w] − 2P [∂i w]) 20 ∂yi +

+

1 ∂ (y∇ 2 K(εqj )y t )ε 4 (Q[w] − 2P [w]) 20 ∂yi

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  ∂ 1 2 (3Q[w] − 4P [w]) − (3Q[∂i w] − 4P [∂i w]) K (εqj )|y|2 ε 4 45 ∂yi 2 2 + K (εqj )yi ε 4 (3Q[w] − 4P [w]) 45  1 3 ∂ R1 [w] − R1 [∂i w] + ε 6 ∂yi 1 4 ∂ R2 [w] − R2 [∂i w]] + o(ε 4 ). + ε [ 10 ∂yi +

Using Lemma A.1, one has  [ R2

∂w ∂w ∂ 1 ∂ 2K g ε w −  g ε ] dy = − (εqj ) ∂yi ∂yi ∂yi 4 ∂xi2



(w  )2 yi2 dy

R2

=−

ε4

∂ 2K

4 ∂xi2

= −c1 ε 4

 (0)

∂ 2K ∂xi2

(w  )2 yi2 dy(1 + o(1))

(6.20)

R2

(0)(1 + o(1)).

Combining (6.18) and (6.20),  Mε

I1 ξε−1

 ∂uε,m ∂ 2 F (q) ∂ 2K ε dy = aj,i [−c2 ξε + c1 ε 4 2 (0)δj,m ](δi, + o(1)). (6.21) ∂y ∂qj,i ∂qm, ∂xi k

2

j =1 i=1

So one has l.h.s =

k  2 

ε aj,i [−c2 ξε

j =1 i=1

∂F (q) ∂ 2K δi, + c1 ε 4 2 (0)δi, δj,m + o(1)]. ∂qj,i ∂qm, ∂xi

(6.22)

Combining the l.h.s. and r.h.s., k  2 

ε aj,i [−c2 ξε

j =1 i=1 ε ( = λam,

∂ 2 F (q) ∂ 2K δi, + c1 ε 4 2 (0)δj,m δi, ] + o(ε 4 ) ∂qj,i ∂qm, ∂xi

 ∂w 2 dy + o(1)) ∂y1

(6.23)

R2

Finally, for k = 2 we solve the finite-dimensional eigenvalue problem and complete the proof of Theorem 1.4. Proof of Theorem 1.4. Equation (6.23) shows that the small eigenvalues λε of (5.1) are given by

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 ∂ 2 F (q) ∂ 2K λ ε ∼ σ 0 − c2 ξ ε δi, + c1 ε 4 2 (0)δj,m δi, ∼ c1 ε 4 Rσ0 (M) j,m=1,··· ,k,i,=1,2 ∂qj,i ∂qm, ∂xi (6.24) where M is given in (4.7). From the expression of M, we know that if θ1 = 0, the eigenvalues are ˆ given by λ1 ∼ KR110 , λ2 ∼ ( 2dd + R10 )K11 , λ3 ∼ KR220 , λ4 ∼ R10 (K22 − K11 ); while when θ1 = π2 , ˆ

the eigenvalues are given by λ1 ∼ KR220 , λ2 ∼ ( 2dd + R10 )K22 , λ3 ∼ KR110 , λ4 ∼ R10 (K11 − K22 ). So since K11 = K22 , it follows that one of the solutions is stable and the other one is unstable. 2 7. Discussion In this section we discuss the main results given in Theorems 1.1 and 1.2. We consider specific two-dimensional Riemannian manifolds without boundary. In particular let us choose the surface of a three-dimensional ellipsoid. First we study the surface of a tri-axial ellipsoid with semi-axes a1 < a2 < a3 . There are two maximum points of the Gaussian curvature near each of which two different two-spike cluster solutions exist. The orientation of the stable cluster is towards the smaller principal curvature and the orientation of the unstable cluster is towards the larger principal curvature. There are also two saddle points of the Gaussian curvature for which a single two-spike cluster exists whose spikes are orientated in the direction in which the saddle point is a local maximum of the Gaussian curvature. These spike clusters are unstable. Finally, there are two minimum points of the Gaussian curvature near which no two-spike cluster exists. Second we consider an American football for which the semi-axes are a1 = a2 < a3 . This surface has two maximum points of the Gaussian curvature. Near each of them multiple twospike clusters exist. Since the manifold is invariant under rotation around the maximum points any orientation is possible. All of these two-spike clusters are stable. This result is not proved in the current paper but it will follow by adapting our analysis to the case of rotationally symmetric manifolds (which is simpler than the more general non-rotationally symmetric setting considered here), then the finite-dimensional problems for existence and stability can be handled as in [56]. Further, for the American football case there is also a minimum point of the Gaussian curvature near which no spike cluster exists. The degenerate case of a point for which the two principal curvatures are the same but the manifold is not rotationally symmetric is more difficult to handle. Further expansions are required which will determine the existence and stability of two-spike cluster solutions near this point. Spike clusters of more than two spikes have not been considered in this paper since higherorder expansions of the contributions from the local geometry of the manifold are required to determine the orientation of the cluster. We are currently investigating this problem. Appendix A In this appendix, we will give some useful identities and we will compute the eigenvalues of the matrix M. A.1. Some identities By direct calculation (following Appendix B of [38]), one has the following lemma:

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Lemma A.1. If w is a radial function, then the following identities hold: 

∂w (Q[w] − 2P [w])yj dy = − ∂yj

R2



(w  )2 yi2 dy

∞ = −π (w  )2 r 3 dr, 0

R2



(3Q[w] − 4P [w])yi

∂w dy = − ∂yi

R2



(w  (r))2 yi2 dy,

R2



[

∂ ∂w R1 [w] − R1 [∂i w]] dy = 0, ∂yi ∂yj

R2



R2

∂ ∂w 3 ∂ 2K [ R2 [w] − R2 [∂i w]] dy = − (εqj ) ∂yi ∂yi 2 ∂xi2



(w  )2 yi2 dy,

R2

∂ (Q[w] − 2P [w]) − (Q[∂i w] − 2P [∂i w]) = 0. ∂yi A.2. Eigenvalues of the matrix M Next we will compute the eigenvalues of the matrix M = M1 + 1ˆ (M2 + C1 M3 ) given in d Lemma 4.2. By direct calculation, the eigenvalues of A1 are given by λ1,l = −2 + εl−1 + ε (k−1)(l−1) = −4 sin2

(l − 1)π k

and the eigenvalues of A2 by λ2,l = ε l−1 − ε (k−1)(l−1) = 2i sin

2(l − 1)π k

for l = 1, · · · , k. Denote the diagonal matrices of A1 and A2 by D1 = diag(λ1,1 , · · · , λ1,k ) and D2 = diag(λ2,1 , · · · , λ2,k ), respectively. Using the matrix Pk of eigenvectors for a k × k circulant matrix, we have  −1    1 1 Pk Pk 0 0 P −1 M1 + (M2 + C1 M3 ) P = + + C M ) M (M 1 2 1 3 0 Pk 0 Pk−1 dˆ dˆ ⎞ ⎛ D2 sin πk cos πk (1 − 1ˆ ) + 1ˆ C1 B2 (D1 + 4I ) sin2 πk + 1ˆ (D1 cos2 πk + 4 sin2 πk I + C1 B1 ) d d d ⎠. =⎝ −D2 sin πk cos πk (1 − 1ˆ ) + 1ˆ C1 B2 −D1 (cos2 πk + 1ˆ sin2 πk ) + 1ˆ C1 B3 d

d

d

d

Since the matrix M1 + 1ˆ (M2 + C1 M3 ) is symmetric and its entries are all real numbers, its d eigenvalues are also real and satisfy the equations 2l + bl l + cl = 0,

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where bl = λ1,l cos

2π k

 1−

1 dˆ

− 4 sin2

π k

 1+

1 dˆ

−

1 C1 dˆ

  (l − 1)π π (l − 1)π π 2π 4 2π C1 = −4 sin2 cos + sin2 + cos − sin2 − sin2 k k k k k k 4 dˆ and  π 1 2 2π sin cos2 1− ˆ k k d   1 2π 1 2 2 π 2 π 2 π −λ1,l cos + sin sin + cos k k k k dˆ dˆ   π π π 1 1 −4λ1,l 1 + cos2 + sin2 sin2 k k k dˆ dˆ  # " 4 π (l − 1)π (l − 1)π π (l − 1)π 4 + C1 sin2 cos2 + C1 − sin2 + sin2 1 + sin2 k k k k k dˆ dˆ 2 " #

 16 2 (l − 1)π π π (l − 1)π π (l − 1)π 4 sin = sin2 1 + cos2 − sin2 + C1 sin2 cos2 k k k k k k dˆ dˆ

cl = λ22,l

16 2 (l − 1)π π 2π sin sin2 cos 2 ˆ k k k d  # " 4 (l − 1)π π (l − 1)π + C1 − sin2 + sin2 1 + sin2 . k k k dˆ 2

−

For k ≥ 3, we get bl ≤ − 8ˆ < 0. Denote the solutions by d

bl 1,l = − 2



' 1−

4cl 1− 2 bl



bl and 2,l = − 2

 1+

'

4cl 1− 2 bl

For k = 3, 5, 6, 7, . . ., we have 1,1 = 0,

 π 1 sin2 > 0 2,1 = 4 1 + k dˆ

and for l = 2, . . . , k, i = 1, 2 it follows that |l,i | > for some c4 > 0.

c4 dˆ 2

 , respectively.

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