multiscale model. simul c vol. 14, no. 2, pp....

$: MULTISCALE MODEL. SIMUL c Vol. 14, No. 2, pp. 737{771yipn/papers/StepBunchElasticEffects_published.pdfMULTISCALE MODEL.SIMUL. c 2016 Society for Industrial and Applied Mathematics$
MULTISCALE MODEL. SIMUL. c© 2016 Society for Industrial and Applied MathematicsVol. 14, No. 2, pp. 737–771

ENERGY SCALING AND ASYMPTOTIC PROPERTIES OF STEPBUNCHING IN EPITAXIAL GROWTH

WITH ELASTICITY EFFECTS∗

TAO LUO† , YANG XIANG† , AND NUNG KWAN YIP‡

Abstract. In epitaxial growth on vicinal surfaces, elasticity effects give rise to step bunchinginstability and some self-organization phenomena, which are widely believed to be important in thefabrication of nanostructures. It is challenging to model and analyze these phenomena due to thenonlocal effects and interactions between different length scales. In this paper, we study a discretemodel for epitaxial growth with elasticity. We rigorously identify the minimum energy scaling lawand prove the formation and appearance of one bunch structure. We also provide sharp boundsfor the bunch size and the slope of the optimal step bunch profile. Both periodic and Neumannboundary conditions are considered.

Key words. epitaxial growth, elasticity, step bunching, asymptotic analysis, energy scaling law

AMS subject classifications. 74G65, 74G45, 74A50, 49K99

DOI. 10.1137/15M1041821

1. Introduction. In epitaxial film growth, elastic effects often lead to surfacemorphological instabilities. These elastic effects can be caused by surface defects suchas steps as well as the mismatch between the lattice constants of the substrate andthe film. In this paper, we concentrate on the step bunching phenomena on vicinalsurfaces [15, 18, 7]. Below the roughening transition temperature, a vicinal surfaceconsists of a succession of terraces and atomic height steps, while the angle betweenthis surface and the crystallographic plane is small, say, a few tenths of a degree. Dueto elastic interactions, the steps are attracted to each other and coalesce into stepbunches. The sizes of these step bunches increase gradually as further coalescence ofsmall step bunches in the evolution. As a self-organization phenomenon, step bunchinginstability attracts considerable research interest in the materials science communitybecause of its potential to facilitate the fabrication of nanostructures on the epitaxialsurfaces. However, the understanding of this phenomenon is still incomplete. Forinstance, the length scale and coarsening rate of the step bunching dynamics have notbeen well-studied yet.

The elastic interactions between steps that lead to such step bunching instabil-ity include the force dipole interaction arising from the surface stress and the forcemonopole interaction coming from misfit stress in the bulk due to the mismatch be-tween the lattice constants of the substrate and the film in heterogeneous epitaxy.The force dipole interaction between steps is repulsive, stabilizing the uniform steptrain, while the force monopole interaction between steps is attractive, destabilizingthe uniform step train [15, 18, 6, 7, 17]. Figure 1 is an illustration of the mechanismsof these interactions.

∗Received by the editors September 29, 2015; accepted for publication (in revised form) March 9,2016; published electronically May 3, 2016.

http://www.siam.org/journals/mms/14-2/M104182.html†Department of Mathematics, Hong Kong University of Science and Technology, Clear Water

Bay, Hong Kong ([email protected], [email protected]). Part of the first author’s work was conductedat the Department of Mathematics, Purdue University, during spring 2015. The work of the secondauthor was partially supported by HKUST School-Based Initiatives SBI14SC10.‡Department of Mathematics, Purdue University, West Lafayette, IN 47907 (yip@math.

purdue.edu).

737

http://www.siam.org/journals/mms/14-2/M104182.html

mailto:[email protected]




738 TAO LUO, YANG XIANG, AND NUNG KWAN YIP

Monopole at step Dipole at step

Fig. 1. Monopole and dipole effects of a step on the epitaxial surface, [17]. The double arrowsrepresent the compressive stress, which is in the bulk of the film in the left image and on the filmsurface in the right image.

Based on the Burton, Cabrera, and Frank (BCF) theory [3], which assumes thediffusion of adatoms on the terraces until their attachment to the steps, Tersoff andothers [18, 13] proposed a discrete model for the dynamics of steps incorporating thetwo elasticity effects. In their model, the motion of the steps is given by

(1)1

a2dxidt

= Fadli + li+1

2+ρ0D

kBT

(fi+1 − fi

li− fi − fi−1

li−1

), i ∈ Z.

In the above, xi is the position of the ith step, li = xi+1 − xi is the length of theith terrace, a is the lattice constant, Fad is adatom flux, and ρ0, D, kB , and T arethe equilibrium adatom density on a step in the absence of elastic interactions, thediffusion constant on the terrace, Boltzmann constant and temperature, respectively.The elastic force per unit length on the ith step is given by

(2) fi = −∑j 6=i

(α1

xj − xi− α2

(xj − xi)3

),

where the attractive interaction −∑j 6=i

α1

xj−xi is due to the force monopole effect of

steps, and the repulsive interaction∑j 6=i

α2

(xj−xi)3 is due to the force dipole effects of

steps. The coefficients α1, α2 are assumed to be positive constants. In their work,Tersoff and others analyzed linear instability for equispaced steps under small pertur-bation and then studied the size and spacing of the bunches as well as the coarseningrate by numerical simulation. Duport and others [6, 7] also studied the step bunchingphenomenon. Besides force monopole and dipole, they considered two more effects:the elastic interaction between the adatoms and steps and the Schweobel barrier. De-pending on the sign of the misfit, the former can stabilize or destabilize the uniformstep train, while the latter is always stabilizing.

The previously introduced models are discrete in the sense that they capture allstep positions. Continuum models of larger length scales have also been developed.However, the traditional surface morphological model [2, 10, 16] which demonstratesthe long wave length instability of a planar surface of a stressed solid is not applicablehere due to the fact that the temperature is lower than the roughening transition pointso that discrete effects remain important. To incorporate the atomic structure of thecrystal, Xiang [19] derived a continuum model by taking the continuum limit fromthe discrete models of [18, 13, 6, 7]. Xu and Xiang [21] extended this to 2+1 dimen-sions. Margetis and Kohn [14] also rigorously derived a 2+1 dimensional continuummodel from the BCF model. Their model includes the force dipole effect but does not

STEP BUNCHING WITH ELASTICITY EFFECTS 739

include the force monopole effect in the heterogeneous epitaxial films. Consideringstep line tension and force dipole interaction between adjacent steps together withthe Schweobel barriers, Fok, Rosales, and Margetis [8] derived a continuum modelfor concentric circular steps. They unified the step bunching phenomena under theseeffects by pointing out that the line tension and the other physical effects, such as des-orption, deposition, or drift, all contribute to a destabilizing, backward diffusion termin the resulting PDEs. Kukta and Bhattacharya [11] proposed a three-dimensionalcontinuum model, coupling stress with diffusion.

Xiang and E [20] performed linear stability analysis and numerical simulations tovalidate their continuum model for step bunching under elastic interactions. The con-tinuum model is able to predict linear instability of a uniform step train and evolutionof a stepped surface in the nonlinear regime, which are in agreement with the resultsof the discrete model. Zhu, Xu, and Xiang [22] performed linear instability analysisand numerical simulations using the 2+1 dimensional continuum model. Dal Maso,Fonseca, and Leoni [5] and Fonseca, Leoni, and Lu [9] proved the existence and regu-larity of weak solutions of Xiang and E’s 1+1 dimensional continuum model [19, 20].There are also several analytical results for the dynamics of epitaxial growth with-out elasticity. Al Hajj Shehadeh, Kohn, and Weare [1] considered both discrete andcontinuum models in the attachment-detachment-limited regime. Their model onlyincluded the force dipole interaction between nearest steps, and hence there is no stepbunching phenomenon. Li and Liu [12] studied well-posedness, perturbation analysis,as well as numerical simulation of epitaxial growth with or without slope selection.

As linear stability analysis is valid only for step profiles close to a uniform steptrain, it cannot provide complete information about step bunching which is a highlynonlinear phenomena. This is similar to the Cahn–Hilliard theory [4] of phase transi-tion which also involves multiple stage phenomena: the initial spinodal decompositionand the later coarsening process. To the best of our knowledge, there is not yet acomplete rigorous analysis on the step bunching phenomenon in epitaxial growth withelasticity.

In this paper, we study the Tersoff’s discrete model (equations (1) and (2)) inthe periodic setting, concentrating on the structures of the surface profiles when thenumber of steps is large. This is to mimic the behavior of such a surface underlong-range elastic interactions. Specifically, from numerical simulation, with a largenumber of initial steps, it appears that many step bunches will form as the first stage.They are roughly equidistant from each other. Hence the periodic boundary conditioncan capture the essential picture, even quantitatively. In the second stage, these stepbunches will coarsen. We believe our results can lead to better understanding of thepattern formation and self-organization phenomenon. After reformulating our modelas a gradient flow of an underlying elastic energy functional, we will investigate theproperties of the energy minimizer. Roughly speaking, we have the following questionsand answers:

1. Q: Is there a solution for the minimization and dynamical problems?A: Existence of minimizer and gradient structure of the dynamics.

2. Q: What is the energy of the minimizer?A: Energy scaling law.

3. Q: Does the minimizer have only one step bunch?A: All steps concentrate in a narrow band.

4. Q: What is the structure of this bunch?A: Size of the bunch and slope of the bunch profile.

The above questions are very subtle and yet we are able to provide quite satisfactory


answers. The main difficulty is due to the long-range interaction leading to nonlocaleffects. Estimating the energy of individual pairs of steps is far from enough incontrolling the total energy. To obtain sharp energy bounds for our model, we have totake the interrelation of step positions into account. Furthermore, due to the periodicsetting, the distance between the steps is measured on a circle. This destroys themonotonicity property of the pairwise potential and hence leads to some difficultiesin the analysis.

2. Main results and outline of proofs. We will study the step bunchinginstability based on energy consideration. The no-flux condition Fad = 0 will beassumed. In [18], Tersoff et al. reported the existence of step bunching without theadatom flux by simulations. Their model with one attraction term and one repulsionterm in (1) and (2) seems to be the simplest formulation to capture the essentialeffects. In practice, the vicinal surface is very large, and we shall use the periodicboundary condition. But our results can be extended to the Neumann boundarycondition and even more general pairwise interactions (see section 12).

We first introduce the notation for the periodic setting. Let the physical length ofone period be L. Then all the step locations {xi}∞i=−∞ can be completely describedby some N steps:

(3) X = (x1, x2, . . . , xN )T with 0 ≤ x1 < x2 < · · · < xN < L.

The periodicity is enforced by the condition xi+N − xi = L for all i ∈ Z. Hence X

lives inside the N -dimenstional torus TN = [0, L)N

. We also introduce l to be theaverage length of the terrace in one period. Then we have L = Nl. Our main analysisconcerns asymptotic properties of solutions for N � 1 or N →∞ (with l fixed).

With the above periodic boundary condition, the step motion (1) is turned into

(4)dxidt

=a2ρ0D

kBT

(fi+1 − fi

li− fi − fi−1

li−1

), i = 1, 2, . . . , N.

We denote the initial data as

xi(0) = x0i , i = 1, 2, . . . , N.(5)

The elastic force fi now can be explicitly written as(6)

fi = −∑

1≤j≤N,j 6=i

[α1π

Nlcot

π(xj − xi)Nl

− α2π3

N3l3cot

π(xj − xi)Nl

(sin

π(xj − xi)Nl

)−2]for i = 1, 2, . . . , N . This form has the advantage that we only need to calculate thelong-range summation over the finite N steps within one period while the originalform in (2) requires the summation over all steps including the steps in each periodand all their periodic images.

We relate the fi’s to some underlying elastic energy E(X) = E(x1, x2, . . . , xN ) asfollows:

(7) fi =∂E

∂xi,

where(8)

E(X) =∑

1≤i<j≤N

[α1

2log sin2 π(xj − xi)

Nl+

α2π2

2N2l2

(sin

π(xj − xi)Nl

)−2]− E0

N (l),


4

x

xx

xx

1x2

3

N−1

NxN+1

ll

l

ll

1

2

3

N−1

NL=Nl

x0

Fig. 2. An example of step configuration. Recall the definition li = xi+1 −xi for i = 1, 2, . . . Nand l(1) ≤ l(2) ≤ · · · ≤ l(N) = lN . By the periodicity assumption, we have xi+N = xi + L.

and E0N (l) (often simply denoted by E0

N ) is some renormalization constant chosen toensure that the uniform step train (with average terrace length l) has zero energy.The derivations of (6)–(8) are given in the appendix. Using F = (f1, f2, . . . fN )T , (7)can be written as

F = DXE.(9)

Note that there is no negative sign because we are not applying Newton’s second law.Indeed, E is the total elastic energy and fi is the corresponding chemical potentialfor the dynamics in (4).

For the readers’ convenience, we summarize our physical setting and notation.Physical constants. We assume that the interaction coefficients α1, α2 in (2),

the lattice constants a, and the average slope A of the surface are all constants.Consequently, the average terrace length l = a

A is also fixed. We further use r todenote the constant

(10) r :=α2π

2

α1l2,

which indicates the relative importance between the force monopole and dipole inter-actions.

Geometric setting. Recall that in the periodic setting, we have xi+N −xi = Nlfor all i ∈ Z. The terrace length is defined as li = xi+1 − xi for all i ∈ Z. A stepprofile X is defined as X = (x1, x2, . . . , xN )T with 0 ≤ x1 < x2 < · · · < xN < L, i.e.,

x ∈ TN = [0, L)N . We rearrange the terrace lengths {li}Ni=1 as l(1) ≤ l(2) ≤ · · · ≤ l(N).Without loss of generality, we set lN = l(N) = max1≤j≤N lj . See Figure 2.

The distance between any two steps xi and xj is defined as

dist(xi, xj) := min {|xi − xj |, Nl − |xi − xj |}(11)

for 0 ≤ xi, xj < Nl.Let S = {x1, x2, . . . , xN} be the set of all steps and S1, S2 be subsets of S. The

distance between two step sets S1 and S2 is defined as

dist(S1, S2) := minxi ∈ S1, xj ∈ S2

dist(xi, xj).(12)

A step chain is any subset T ⊆ S consisting of consecutive steps xi, xi+1, . . . , xj .Note that xN and x1 are regarded as two consecutive steps. We say that a step chain


l

L = Nl

H = Na

uniform step train

l0

(N − 1)l0

L = Nl

H = Na

local uniform step train

Fig. 3. Uniform step train and local uniform step train. The latter is used to construct suitableansatz to compute the mimimum energy.

T is well isolated from all the other steps if

dist(T, S\T ) > l∗∗ =Nl

πarcsin

√3r

N2 + 2r.(13)

See (20) for the definition of l∗∗.A uniform step train is any profile X with l1 = l2 = · · · = lN = l, while a local

uniform step train is any profile X with l1 = l2 = · · · = lN−1. Note that l1 may beless than l in the latter one. See Figure 3.

Interaction potential. The total elastic energy can be written as

(14) E(X) = E(x1, . . . , xN ) =∑

1≤i<j≤N

e(xj − xi)− E0N ,

where the pairwise energy density e(·) and the reference energy E0N = E0

N (l) aredefined as

(15) e(x) :=α1

2

[log sin2 πx

Nl+

r

N2

(sin

πx

Nl

)−2],

E0N :=

∑1≤i<j≤N

e((j − i)l).(16)

The first and second derivatives of e(·) are given by

e′(x) =α1π

Nl

(1− r

N2

(sin

πx

Nl

)−2)cot

πx

Nl,(17)

e′′(x) =α1π

2

N2l2

[3r

N2−(

1 +2r

N2

)sin2 πx

Nl

](sin

πx

Nl

)−4.(18)

By the periodicity assumption, we have

(19) e(x) = e(Nl − x), e′(x) = −e′(Nl − x), e′′(x) = e′′(Nl − x).

The graphs of e(·), e′(·) and e′′(·) are given in Figure 4.The following two properties of e(·) and e′(·) will be used often in our proofs.


0 L

2L

l∗x

e(x)

0 L

2L

l∗l∗∗

x

e′(x)

0 L

2Ll∗∗

x

e′′(x)

Fig. 4. Plots of energy density and its derivatives e(·), e′(·), and e′′(·). See (15), (17), and (18).

1. For N2 > r, there exist a unique l∗ and a unique l∗∗ satisfying

(20) l∗ := arg min0≤x<Nl

2

e(x), l∗∗ := arg max0≤x<Nl

2

e′(x).

Note that e′(l∗) = 0, e′′(l∗∗) = 0 and 0 < l∗ < l∗∗ <Nl2 . See Figure 4. In

fact, for N � 1, we have

l∗ =Nl

πarcsin

√r

N=

√rl

π+O(N−2),(21)

l∗∗ =Nl

πarcsin

√3r

N2 + 2r=

√3rl

π+O(N−2).(22)

2. The function e(·) is decreasing on (0, l∗) and (Nl/2, Nl− l∗) while increasingon (l∗, Nl/2) and (Nl − l∗, Nl). The function e′(·) is increasing on (0, l∗∗)and (Nl − l∗∗, Nl) while decreasing on (l∗∗, Nl − l∗∗).

Notation used in the proofs. We use C to denote positive constants whichmay depend on the physical constants α1, α2, a, and A but not on N . We also useCβ to indicate the possible dependence on some parameter β other than the physicalconstants. The value of such C may change from line to line. Next, we use N � 1 toindicate that N is sufficiently large. Finally, we use #S to denote the cardinality ofa finite set S.


As mentioned above, we will be mainly interested in the asymptotic behaviors ofminimizers as N → +∞. These include the well-posedness issue as well as the struc-ture and stability of the step profile in our model. Our main results are summarizedbelow.

Theorem 1 (existence).(a) (Initial value problem) For all N , the initial value problem (4)–(5) has a

global in time solution. In other words, there is no finite time blow-up.(b) (Minimization problem) For all N , there exists a global minimizer for the

energy E in (14).(c) (Equivalence) For all N , the stable stationary states of (4) are the same as

the local minimizers of E.

Theorem 2 (energy scaling law). For any δ > 0, there exist constants cδ andC such that

−(α1

4+ δ)N2 logN − cδN2 ≤ inf

X∈TNE(X) ≤ −α1

4N2 logN + CN2(23)

for all N . Consequently, limN→+∞inf E(X)N2 logN = −α1

4 .

Theorem 3 (one bunch structure).(a) (Closeness of steps) For all N , any step profile with a well-isolated step chain

is not a stable stationary state. In other words, any local minimizer of E hasthe property that

(24) dist(T, S\T ) ≤ l∗∗ for all T ⊂ S.

In particular, l(i) ≤ l∗∗ for all i = 1, 2, . . . , N − 1.(b) (Bunch size less than half period) For any global minimizer of E with N � 1,

we have

(25) l1 + · · ·+ lN−1 ≤Nl

2and lN ≥

Nl

2.

(c) (Asymptotically narrow bunch size) For any global minimizer of E, we have

(26) limN→+∞

l1 + · · ·+ lN−1lN

= 0.

Theorem 4 (size of the bunch). For any global minimizer of E and for all N ,we have the following:

(a) (Lower bound) There exists C such that

CN1/2 (logN)−1/2 ≤ l1 + · · ·+ lN−1.(27)

(b) (Upper bound) For any δ > 0 and any 0 < s < 1, there exists Cδ,s such that

mini

{li + · · ·+ li+[sN ]

}≤ Cδ,sN1/2+δ.(28)

Theorem 5 (slope of the bunch profile). For any global minimizer of E, wehave the following:

(a) (Estimates on the minimal terrace length l(1)) For any δ > 0, there existconstants cδ and Cδ such that

cδN−1/2−δ ≤ l(1) ≤ CδN−1/2+δ(29)

for all N .


(b) (Estimates on the next-maximal terrace length l(N−1)) For any δ > 0, thereexist constants cδ and C such that

cδN−1/6−δ ≤ l(N−1) ≤ C(30)

for all N . Furthermore, l(N−1) ≤ l∗ for N � 1.

The main strategy of proofs is to investigate the energy E. There are three keyconsiderations:

1. Construction of good approximate ansatz to give good energy upper and lowerbounds.

2. Analysis of the vanishing condition of the first variation of E, DXE = 0,which is equivalent to the force balance F = 0. The results obtained in facthold for any critical point of the energy functional. In terms of xi’s, thecondition reads(31)∂E

∂xi= 0, i.e.,

∑1≤j<i

e′(xi − xj)−∑

i<k≤N

e′(xk − xi) = 0, for i = 1, . . . , N.

This is equivalent to, in terms of li’s,(32)∂E

∂li= 0, i.e.,

∑1≤j≤i≤k≤N−1

e′(lj+· · ·+li+· · ·+lk) = 0, for i = 1, . . . , N−1.

3. Analysis of the second variation of E, D2XE. For minimizer (or stable local

minimizer), it must hold that

(33) D2XE ≥ 0.

The remainder of this paper is organized as follows. In section 3, we reformulatethe system into a gradient flow dynamics and then prove the existence of minimizers(Theorem 1). In section 4, using a suitable ansatz, we obtain a weak version ofTheorem 2 for the minimum energy scaling law. The whole theorem (the strongversion) is proved in section 9. By analyzing the Hessian matrix of a critical point,section 5 proves that an isolated bunch is unstable (Theorem 3(a)). In section 6,we first find a lower bound for the minimal terrace length l(1) by controlling thefirst variation of the energy. This gives the lower bounds in Theorem 5(a) and (b).Then an iteration scheme is applied to improve this bound. The main idea in theiteration scheme is to group the O(N2) interacting pairs into N step chains and thenestimate the contribution of each chain. Section 7 shows that all the steps of anenergy minimizer concentrate in half period (Theorem 3(b)). This is crucial to theproofs for the remaining results. Indeed, the distance dist(xi, xj) equals |xi − xj |,provided they are in the same half period. In section 8, we study the distribution ofthe terrace lengths, and then the lower bound for the bunch size is obtained (Theorem4(a)). In section 9, we complete the strong version of the minimum energy scalinglaw (Theorem 2). In section 10, we use the energy estimates on each step chain toprove the upper bound for the bunch size (Theorem 4(b)). In section 11, we combineall the previous results to prove the upper bound for terrace lengths (upper bounds inTheorem 5(a) and (b)) and the one bunch structure property (Theorem 3(c)). Finally,section 12 provides an extension to Neumann boundary condition.


3. Gradient flow structure and existence of minimizers. Mathematicalfoundation and preliminary results are established in this section, including the gra-dient flow structure and global in time solution of the initial value problem (4)–(5)and the existence of a minimizer for the energy functional E in (14). These are es-sentially the statements of Theorem 1. We start with an observation of the energydissipation law.

Proposition 1 (energy dissipation law). For the initial value problem (4)–(5)with N ≥ 2, we have the energy dissipation law

dE

dt= −a

2ρ0D

kBT

N∑i=1

1

li(fi+1 − fi)2.(34)

Therefore, the energy E decreases in time.

Proof. By using (4) and (7), this follows from a direct computation:

dE

dt=

N∑i=1

∂E

∂xixi =

a2ρ0D

kBT

N∑i=1

fi

(fi+1 − fi

li− fi − fi−1

li−1

)

= −a2ρ0D

kBT

N∑i=1

1

li(fi+1 − fi)2.

Because of the above statement, one may expect (4) to be some kind of gradientflow. In fact, we can construct an abstract gradient flow structure for this system.Notice that for our step dynamics, the center of mass of the step positions defined asx1+···+xN

N remains a constant, which without loss of generality is assumed to be L2 .

Therefore, our dynamics lives on MN−1 :={X ∈ TN : x1 + · · ·+ xN = L

2

}. In the

next proposition, we will equip the manifold MN−1 with a Riemannian metric g sothat the system is a gradient flow with respect to g.

Proposition 2 (gradient flow structure). For any N ≥ 2, there exists a Rie-mannian metric g, which can be regarded as an (N − 1) × (N − 1) positive definitematrix, such that (4) has the gradient flow structure

X|N−1 = −gradgE,(35)

where X|N−1 = (x1, x2, . . . , xN−1)T with x1+x2+· · ·+xN−1+xN = constant. Underthis metric, the dissipation law reads as E = −gX(t)(X|N−1, X|N−1).

Remark 1. The precise meaning of X|N−1 = −gradgE is that for all t > 0 and

Y ∈ TX(t)MN−1, we have

gX(t)(X|N−1(t), Y ) + 〈diffEX(t), Y 〉 = 0,(36)

where TX(t)MN−1 is the tangent space ofMN−1 at X(t) and 〈·, ·〉 is the inner producton MN−1 endowed from RN .

Proof. Let A = (A1, A2, . . . , AN−1)T with Ai = fi+1−fixi+1−xi for i = 1, 2, . . . , N − 1.

Denote AN = f1−fNNl+x1−xN . Note that

∑Ni=1 liAi = 0 and hence AN = −

∑N−1i=1 liAilN

.Using this notation, we have

x1 =a2ρ0D

kBT(A1 −AN ) , xi =

a2ρ0D

kBT(Ai −Ai−1) , i = 2, 3, . . . , N − 1, N.


We put the above into a matrix form X|N−1 = a2ρ0DkBT

JA, where J is an (N−1)×(N−1)matrix defined as follows:

J =

1 + l1

lNl2lN

. . . lN−1

lN−1 1

. . .. . .

−1 1

.

Let Λ be the (N − 1) × (N − 1) diagonal matrix Λ = diag{l1, . . . , lN−1} and Λ with

Λij = Λij +liljlN

for i, j = 1, 2, . . . , N − 1. Then (34) can be written as

E = −a2ρ0D

kBT

N∑i=1

liA2i = −a

2ρ0D

kBT

N−1∑i,j=1

AiΛijAj +

(∑N−1i=1 liAi

)2lN

= −a

2ρ0D

kBT

N−1∑i,j=1

AiΛijAj

.It is easy to check that det(J) = Nl

lN6= 0 and Λ is positive definite. Hence J and Λ

are both invertible. Therefore, we can define the positive definite metric

g =

(a2ρ0D

kBT

)−1(J−1)T ΛJ−1.(37)

Now, the energy dissipation law (34) can be written as

E = −a2ρ0D

kBTAT ΛA = −

(a2ρ0D

kBTATJT

)g

(a2ρ0D

kBTJA

)= −

(X|N−1

)TgX|N−1 = −gX(t)(X|N−1, X|N−1),

which is the desired statement.

It is important to remark that our manifold MN−1 is open in the sense thatxi < xi+1 for i = 1, 2, . . . , N − 1. The Cauchy–Lipschitz theory, in general, onlyguarantees the local in time existence of the initial value problem (4)–(5). Next weprovide a positive lower bound for xi+1− xi which is used to prove that there will beno finite time blow-up for the solution.

Proposition 3 (minimal terrace length). For any N ≥ 2 and any solutionof (4) with initial data (5) and initial energy E(X(0)), at any time t at which thesolution exists, we have the following lower bound for the terrace lengths:

lmin(t) ≥ min

{α2A

2α1aN−1,

[4

α2E(X(0)) +

4C

α2N2 logN

]− 12

},(38)

where lmin(t) = min1≤i≤N li(t) and C is independent of N and initial value. Notethat lN (t) = Nl + x1(t)− xN (t).

Proof. Notice that mins∈R+

{log s+ r

N2s

}= log r

N2 + 1. By definition (15), wehave

e(x) ≥ α1

2

(log

r

N2+ 1)

for all x > 0.(39)


By energy dissipation law (34), we deduce that

E(X(0)) ≥ E(X(t)) ≥

[α1

2log sin2 πlmin(t)

Nl+

α2π2

2N2l2

(sin

πlmin(t)

Nl

)−2]

+

[N(N − 1)

2− 1

]α1

2

(log

r

N2+ 1)− E0

N .

Note that 2lmin

Nl ≤ 1 leading to log 2lmin

Nl ≥ −(2lmin

Nl

)−1. If lmin(t) ≤ α2

2α1Nl= α2A

2α1aN−1,

then

α1

2log sin2 πlmin(t)

Nl≥ α1 log

2lmin

Nl≥ −α1

(2lmin

Nl

)−1≥ − α2π

2

4N2l2

(πlmin

Nl

)−2≥ − α2π

2

4N2l2

(sin

πlmin

Nl

)−2.(40)

Hence

E(X(0)) ≥ α2

4(lmin(t))

−2+α1

2

[N(N − 1)

2− 1

](log

r

N2+ 1)− E0

N .

By Lemma 1 (in section 4), E0N ≤ −

α1

2 N2 log 2e

π +O(N logN), there exists a constantC such that

E(X(0)) ≥ α2

4(lmin(t))

−2 − CN2 logN.

We then obtain the lower bound for the minimal terrace length

lmin(t) ≥ min

{α2A

2α1aN−1,

[4

α2E(X(0)) +

4C

α2N2 logN

]−1/2}

for all t ∈ [0, T ], where the initial value problem of ODE system has a solution.

Proof of Theorem 1(a). We have the local in time existence by means of the localLipschitz property of the right-hand side of (4). This, combined with Proposition 3,guarantees that the solution can be extended to all time.

The next result shows that the velocities of the steps decay to zero as time goesto infinity.

Proposition 4 (long time behavior of initial value problem). For any N ≥ 2and for any initial data (5), the velocity of all steps go to zero, i.e., limt→+∞ X(t) = 0.

Proof. We prove the proposition by contradiction. Suppose there exist someδ > 0 and a sequence 0 < t1 < t2 < · · · < tk < · · · with limk→+∞ tk = +∞ such that|X(tk)| > δ for all k.

We rewrite (4) into X(t) = G(X(t)), where

Gi(X(t)) =a2ρ0D

kBT

(fi+1 − fi

li− fi − fi−1

li−1

)for i = 1, 2, . . . , N.(41)

Notice that G(·) is a smooth map from MN−1 to RN . Given initial data X(0),by Proposition 3, X(t) always lives on a compact submanifold N of MN−1. Let


M = maxX∈N {|G(X)|, |DXG(X)|} < +∞ and m = minX∈N ‖g(X)‖l2 > 0, wherethe metric g is given by (37).

Since |G(X(tn))| = |X(tn)| > δ for t ∈ [tn, tn + δ2M2 ], the Taylor theorem gives

|X(t)| = |G(X(t))| ≥ |G(X(tn))| − |DXG(ξ)| |X(t)−X(tn)|

≥ |G(X(tn))| −maxξ∈N|DXG(ξ)|

∣∣∣∣∫ t

tn

G(X(τ))dτ

∣∣∣∣≥ δ −M · δ

2M2·M ≥ δ

2.

By Proposition 2, we have E(t) = −(X|N−1)T gX|N−1 ≤ −mδ2

4 for t ∈[tn, tn + δ

2M2

]and n = 1, 2, . . . . Hence, on the one hand, we have

limT→+∞

E(X(T ))− E(X(0)) = −∞.(42)

On the other hand, E0N is a fixed finite number (see Lemma 1). Hence for any

T > 0, E(X(T )) − E(X(0)) ≥ α1N(N−1)4

(log r

N2 + 1)− E0

N − E(X(0)) is finite and

independent of T . This contradicts (42). Therefore limt→+∞ X(t) = 0.

It is reasonable to believe that for most initial data, the system will tend tocertain stable stationary state X(∞). We will show the equivalence between thestable stationary states of (4) and the local minimizers of E, that is, Theorem 1(c).Then we prove the existence of the minimum energy states Theorem 1(b) and hencethe existence of the stable stationary state of (4).

Proof of Theorem 1(c). Let G(X) be the one defined in (41). Indeed we are goingto prove the equivalence between the following two stability conditions:

(1) G(X) = 0, DXG(X) ≤ 0; (2) DXE = 0, D2XE ≥ 0.

1. Critical points are the same. On the one hand, if ∂E∂xi

= 0 for i = 1, 2, . . . , N ,then fi = 0 for i = 1, 2, . . . , N . This immediately implies Gi = 0 for i = 1, 2, . . . , N .

On the other hand, if Gi = 0 for i = 1, 2, . . . , N , then f1−fNNl−(l1+···+lN−1)

= f2−f1l1

=

· · · = fN−fN−1

lN−1. We denote this number as C. Thus, 0 = (f1 − fN ) + (f2 − f1) +

· · · + (fN − fN−1) = NlC. Hence C = 0 and f1 = f2 = · · · = fN . Note thatf1 + f2 + · · · + fN = 0. Therefore f1 = f2 = · · · = fN = 0 and ∂E

∂xi= 0 for

i = 1, 2, . . . , N .2. Stability of local minimizer: D2

XE ≥ 0. Denote Hessian matrix of E as

H = (Hij)N×N := ( ∂2E∂xi∂xj

)N×N . Direct computation shows that

Hij =

−e′′(xj − xi), i 6= j,∑1≤k≤N,k 6=i

e′′(xk − xi), i = j.

For local minimizer X = (x1, x2, . . . , xN )T , the stability condition is H ≥ 0.3. Stability at critical point of ODE system: DXG ≤ 0. Linearize the equation

at any critical point. Note that f1 = f2 = · · · = fN at this critical point. Usingfi = ∂E

∂xi, we have

−(a2ρ0D

kBT

)−1∂Gi∂xj

= − 1

li

∂fi+1

∂xj+

(1

li+

1

li−1

)∂fi∂xj− 1

li−1

∂fi∂xj


+fi+1 − fi

l2i

∂li∂xj− fi − fi−1

l2i−1

∂li−1∂xj

= − 1

li

∂fi+1

∂xj+

(1

li+

1

li−1

)∂fi∂xj− 1

li−1

∂fi∂xj

= − 1

liH(i+1)j +

(1

li−1+

1

li

)Hij −

1

li−1H(i−1)j

= AikHkj ,

where H0j := HNj , H(N+1)j := H1j for j = 1, 2, . . . , N and A is defined as the N ×Nsymmetric matrix

A :=

1l1

+ 1lN

− 1l1

0 · · · 0 − 1lN

− 1l1

1l1

+ 1l2

− 1l2

0 · · · 0

0 − 1l2

1l2

+ 1l3

− 1l3

· · · 0...

......

. . ....

...0 · · · 0 − 1

lN−2

1lN−2

+ 1lN−1

− 1lN−1

− 1lN

0 · · · 0 − 1lN−1

1lN−1

+ 1lN

.

For critical point X = (x1, x2, . . . , xN )T , the stability condition is AH ≥ 0.4. Equivalence between the two stability conditions. Note that H = HT , A = AT ,

and AH = HA. Thus A and H can be simultaneously diagonalized. Moreover, Ais a rank (N − 1) positive semidefinite matrix. Therefore, there exists an invertiblematrix P such that P−1AP = Λ1 = diag(0, λ2, λ3, . . . , λN ) and P−1HP = Λ2 =diag(0, µ2, µ3, . . . , µN ) with 0 < λ2 ≤ λ3 · · · ≤ λN . Note that P−1AHP = Λ1Λ2 =diag(0, λ2µ2, λ3µ3, . . . , λNµN ). Obviously, µi ≥ 0 for i = 2, 3, . . . , N if and only ifλiµi ≥ 0 for i = 2, 3, . . . , N . This shows the equivalence between the two stabilityconditions.

The proof of the existence of minimum energy solution mainly relies on the ob-servation that whenever two adjacent steps approach each other, the energy blows upto positive infinity. This is reminiscent of the similar lower bound (38) for the initialvalue problem (4)–(5).

Proof of Theorem 1(b). Note that the energy is a continuous function of themultivariables X = (x1, x2, . . . , xN )T . The conclusion follows from the following twostatements.

1. Energy bounded from below. By (39), we obtain a uniform lower bound for the

energy E(X) ≥ α1N(N−1)4

(log r

N2 + 1)− E0

N .

2. Coercivity of the energy. Recall that l(1) is the minimal terrace length l(1) =

min1≤i≤N

li. Similar to (40), if l(1) ≤ α2

2α1Nl= α2A

2α1aN−1, then α1

2 log sin2 πl(1)

Nl ≥ −α2π

2

4N2l2

(sin πl(1)

Nl )−2. Hence we have

E(X) ≥ α2π2

4N2l2

(sin

πl(1)

Nl

)−2+α1

2

[N(N − 1)

2− 1

](log

r

N2+ 1)− E0

N

≥ α2

4(l(1))−2 +

α1

2

[N(N − 1)

2− 1

](log

r

N2+ 1)− E0

N .

Therefore liml(1)→0E(X) = +∞.

Next we turn to the main focus of this paper: energy scaling law and spatialstructure of minimizers in the asymptotic regime N � 1 or N → +∞.


4. Energy scaling law: Weak version. In this section, we analyze the energyscaling law of global minimizer as N goes to infinity. Recall the physical setting in

section 2. We have fixed α1, α2, a, A, and hence l = aA , r = α2π

2

α1l2. We further have

L = Nl → +∞. Although the exact minimum energy value is too complicated tobe obtained explicitly, we can find both its lower and upper bounds within the orderO(N2 logN).

Proposition 5 (energy scaling law, weak version). There exist constants c, C >0 such that, for all N , we have

−α1

2N2 logN − cN2 ≤ inf

X∈TNE(X) ≤ −α1

4N2 logN + CN2.(43)

Remark 2. This proposition is a weak version of Theorem 2 in the sense that theleading coefficient of the lower bound is not sharp. In section 9, we will provide asharp lower bound which is indispensible in the analysis of the optimal profile.

Inspired by numerical simulations, we construct a simple approximate solutionwhich has almost the minimum energy. This ansatz is a local uniform step train(see Figure 3) with the optimal terrace width to be determined later. The energy ofthis approximate solution provides the upper bound for minimum energy. Then weprove that the lower bound is of the same order. Before going into the details, weintroduce the following lemma giving that the energy of the uniform step train is oforder O(N2).

Lemma 1 (energy of a uniform step train). There exist c and C such that−cN2 ≤ E0

N ≤ CN2 for all N . More precisely, there exist c and C such that

−α1

2N2 − cN logN ≤ E0

N ≤ −α1

2

(1− log

π

2

)N2 + CN logN(44)

for all N .

Remark 3. The dependence of E0N = E0

N (l) on the average terrace length l is inlower order terms.

Proof. For the uniform step train, xi = (i− 1)l for i = 1, 2, . . . , N , we have

E0N =

α1

2

∑1≤i<j≤N

[log sin2 (j − i)π

N+

r

N2

(sin

(j − i)πN

)−2]

=α1

2

N−1∑k=1

(N − k)

[2 log

∣∣∣∣sin kπN∣∣∣∣+

r

N2

(sin

kπ

N

)−2]

=

α1

2

(N−1)/2∑k=1

N[2 log | sin kπ

N |+rN2

(sin kπ

N

)−2]for N odd,

α1

2

{N/2∑k=1

N[2 log | sin kπ

N |+rN2

(sin kπ

N

)−2]− r2N

}for N even.

Since α1

2 ·r

2N = O(N−1), it is sufficient to prove the result for even N . Note that2πx ≤ sinx ≤ x for 0 ≤ x = kπ

N ≤π2 is due to k ≤ N

2 . On the one hand,

E0N ≤

α1N

2

N/2∑k=1

[2 log

kπ

N+

r

N2

(2k

N

)−2]+O(N−1)


=α1N

2

N logπ

N+ 2 log (N/2)! +

N/2∑k=1

r

4k2

+O(N−1)

≤ α1N

2

[N log

π

N+N log

N

2−N +O(logN) +

rπ2

24

]+O(N−1)

= −α1

2

(1− log

π

2

)N2 + CN logN,

where we have used the Stirling formula log n! = n log n − n + O(log n) with n = N2

and the fact that∑∞k=1 k

−2 = π2

6 . On the other hand,

E0N ≥

α1N

2

N/2∑k=1

[2 log

2k

N+

r

N2

(kπ

N

)−2]+O(N−1)

≥ α1N

2

[N log

2

N+ 2 log (N/2)!

]+O(N−1)

≥ α1N

2

[N log

2

N+N log

N

2−N +O(logN)

]+O(N−1)

= −α1

2N2 − cN logN.

Proof of Proposition 5. 1. Construction of approximate solution. We first con-struct a good enough approximate solution X0 = (x01, x

02, . . . , x

0N )T to a global min-

imizer X. Take x0i = (i − 1)l0 for i = 1, 2, . . . , N for some l0 > 0 to be determined

later. Here, we assume l0 = o(1) as N → ∞ and hencexj−xiNl = o(1) as N → +∞.

This will be checked after we choose such an l0. Utilizing Lemma 1, we have

E(X0) =α1

2

∑1≤i<j≤N

log sin2π(x0j − x0i )

Nl+

r

N2

(sin

π(x0j − x0i )Nl

)−2− E0N

=α1

2

∑1≤i<j≤N

(2 log

|j − i|l0πNl

+rl2

(j − i)2l20π2

)+O(N2)

=α1

2

N−1∑k=1

[2(N − k) log

kl0π

Nl+

rl2

π2l20· N − k

k2

]+O(N2)

=α1

2

[N2 log

l0π

Nl+ 2

N−1∑k=1

(N − k) log k +rl2

π2l20· π

2

6N

]+O(N2 + l−20 logN).

The last step is due to

N−1∑k=1

N − kk2

=π2

6N +O(logN).(45)

Next, we have used the following estimate:

N−1∑k=1

(N − k) log k = N2N−1∑k=1

1

N

(1− k

N

)log

k

N+

N−1∑k=1

(N − k) logN


= N2

∫ 1

1/N

(1− x) log xdx+N(N − 1)

2logN +O(N2)

=1

2N2 logN +O(N2).(46)

Therefore,

E(X0) =α1

2

[N2 log l0 +

rl2

6l20N

]+O(N2 + l−20 logN).

2. Determination of the bunch size. To choose the optimal l0, we take ∂E∂l0

= 0

and keep the leading order. This gives N2

l0− 2rl2

6l30N = 0. Thus

l0 =

√r

3lN−1/2.

This is the optimal l0 to minimize E under the ansatz x0i = (i− 1)l0 for all i’s. Notethat our assumption l0 = o(1), as N → +∞, is satisfied.

3. Upper bound for minimum energy. Let l0 =√

r3 lN

−1/2 and x0i = (i− 1)l0 fori = 1, 2, . . . , N . We have the energy upper bound of this approximate solution

E(X0) ≤ α1

2

∑1≤i<j≤N

[2 log

|j − i|l0πNl

+rl2

4(j − i)2l20

]+O(N2)

=α1

2

N−1∑k=1

(N − k)

[2 log

k√r/3π

N3/2+

3N

4k2

]+O(N2).

By the same estimates used above (equations (45) and (46)), we obtain

infX∈TN

E(X) ≤ E(X0) ≤ −α1

4N2 logN + CN2.

4. Lower bound for minimum energy. Utilizing (39) and (44), we have

infX∈TN

E(X) ≥ α1N(N − 1)

4

(log

r

N2+ 1)− α1N

2

2log

π

2e+O(N logN)

≥ −α1

2N2 logN − cN2.

5. Instability of an isolated step bunch. We now proceed to investigate thestructure of any (local) minimizer. From the ansatz in the proof of Proposition 5, itseems that the minimizer consists of only one step bunch. This is consistent with theresults from numerical simulations. As a first result toward the one bunch structure,we show the instability of any step configuration which has a well-isolated step bunch(Theorem 3(a)). Consequently, there will not be a step bunch too far away from anyother bunches or steps. In other words, any local minimizer essentially contains onebunch. The key idea is to exploit the fact that the second variation of E at a localminimizer must be positive semidefinite. We will make use of the following lemma,which gives a sufficient condition for instability.

Lemma 2. Let H be the (n + m) × (n + m) matrix ( A BBT C

), where A = AT ,

C = CT , B are n×n, m×m, n×m matrices, respectively. Assume that each row ofH sums up to 0. If

∑i,j Bij > 0, then H is not positive semidefinite.


Proof. Choose x = (1, 1, . . . , 1, 0, 0, . . . , 0)T (first n components are 1’s). Then wehave

xTHx =

n∑i,j=1

Aij =

n∑i=1

(0−

m∑k=1

Bik

)< 0

giving the desired result.

Proof of Theorem 3(a). We prove the statement by contradiction. Without lossof generality, we assume T = {x1, . . . , xn} and dist(T, S\T ) > l∗∗. Recall that

Hij =∂2E

∂xi∂xj=

{−e′′(xj − xi), i 6= j,∑

1≤k≤N,k 6=ie′′(xk − xi), i = j,

where e′′(x) = α1π2

N2l2

[3rN2 −

(1 + 2r

N2

)sin2 πx

Nl

] (sin πx

Nl

)−4(see (18)). Note that each

row of Hij sums up to 0.By assumption, dist(xi, xj) > l∗∗ for 1 ≤ i ≤ n and n+ 1 ≤ j ≤ N . Hence,

Hij = −e′′(xj − xi) > −e′′(l∗∗) = 0.

Therefore∑ni=1

∑Nj=n+1Hij > 0. By Lemma 2, H is not positive semidefinite, and

hence the step profile X is not stable.

As another application of Lemma 2, we apply this method to show the instabilityof the uniform step train for N � 1. This instability is also studied by Tersoff et al.[18] using linear stability analysis.

Proposition 6 (instability of the uniform step train). Let X be a uniform steptrain with xi = (i − 1)l for all i = 1, 2, . . . , N . Then the Hessian matrix H is notpositive semidefinite for N � 1. In other words, the uniform step train is unstable.

We remark that from Lemma 2, the proposition is true if l > l∗∗. The point hereis that we can in fact prove the instability for any value of l.

Proof. Without loss of generality, we assume N is even. We apply Lemma 2 tothe case of m = n = N

2 . Now we estimate

∑1≤i≤N2 ,

N2 +1≤j≤N

e′′(xj − xi) =N

2e′′(Nl

2

)+ 2

N2 −1∑m=1

N2 −1∑k=m

e′′(kl)

≤ N

2· α1π

2

N2l2

(3r

N2−(

1 +2r

N2

))+

2α1π2

N2l2

N2 −1∑m=1

N2 −1∑k=m

[3r

N2· N

4

16k4−(

1 +2r

N2

)N2

4k2

]

≤ −α1π2

2Nl2

(1 +

r

N2

)+α1π

2

2l2

N2 −1∑m=1

N2 −1∑k=m

[3r

4· 1

k4− 1

k2

].

For k ≥ 4, note that 1k4 ≤

13 [ 1

(k−3)(k−2)(k−1) −1

(k−2)(k−1)k ]. Thus we have

3r

4

N2 −1∑m=4

N2 −1∑k=m

1

k4≤ r

4

N2 −1∑m=4

1

(m− 3)(m− 2)(m− 1)≤ r

16,


3r

4

3∑m=1

N2 −1∑k=m

1

k4≤ 3r

4

∞∑k=1

1

k2≤ 3r

2,

and

N2 −1∑m=1

N2 −1∑k=m

− 1

k2≤

N2 −1∑m=1

(− 1

m+

2

N

)≤ 1−

N2 −1∑m=1

1

m≤ 2 + γ − log

N

2,

where γ is Euler constant.Collecting the above estimates, for N � 1, we have∑

1≤i≤N2 ,N2 +1≤j≤N

e′′(xj − xi)

≤ −α1π2

2Nl2

(1 +

r

N2

)+α1π

2

2l2

[3r

2+

r

16+ 2 + γ − log

N

2

]< 0.

Therefore∑

1≤i≤N2 ,N2 +1≤j≤N Hij = −

∑1≤i≤N2 ,

N2 +1≤j≤N e

′′(xj−xi) > 0. Then apply

Lemma 2 to complete the proof.

6. Lower bounds for terrace lengths. In contrast to section 5, which con-siders the second variation of E, here we analyze the vanishing condition of the firstvariation (32), i.e., force balance DXE = 0. Therefore, all the results in this sectionhold for any critical point of E. Our main results are lower bounds for the terracelengths (Theorem 5(a) and (b)).

Proposition 7. Let X be a critical point of E. For any 23 < α < 1, there exists

Nα such that for all N > Nα, we have l(1) ≥ N−α.

Proof. We prove the statement by contradiction. Suppose, for any sufficientlylarge Nα, there is some N > Nα with l(1) < N−α < l∗∗. Recall that e′(·) is increasingon (0, l∗∗). We estimate e′(l(1)) as follows:

e′(l(1)) ≤ e′(N−α) =α1π

Nl

[1− r

N2

(sin

πN−α

Nl

)−2] 1− 2 sin2 πN−α

2Nl

sin πN−α

Nl

≤ α1π

Nl

[1− rl2N2α

π2

]1− π2N−2α

2N2l2

πN−α

Nl

=α1

N

[1− rl2N2α

π2

]N1+α

(1− π2N−2−2α

2l2

).

Note that 1 − rl2

π2 N2α ≤ − 1

2rl2

π2 N2α and 1 − π2

2l2N−2−2α ≥ 1

2 for N > Nα � 1.Therefore

e′(l(1)) ≤ −α1rl2

4π2N3α.(47)

Next we estimate max{e′(lj + · · ·+ lk)}. Note that l ≤ lN and hence

(48) lj + · · ·+ lk ≤ l1 + · · ·+ lN−1 = Nl − lN ≤ Nl − l = (N − 1)l.

Thus by the monotonicity property of e′(·) (see Figure 4), we have

e′(lj + · · ·+ lk) ≤ max {e′(l∗∗), e′((N − 1)l)} .


Now,

e′(l∗∗) =α1π

Nl

[1− r

N2

N2 + 2r

3r

] √1− 3rN2+2r√3r

N2+2r

≤ 2α1π

3√

3rl,(49)

e′((N − 1)l) =α1π

Nl

[r

N2

1

sin2 πN

− 1

]1− 2 sin2 π

2N

sin πN

≤ α1π

Nl

[r

N2

N2

4+ 1

]3N

2=

α13π(r + 4)

8l.(50)

Hence, for 1 ≤ j ≤ i ≤ k ≤ N − 1 with j < k, we have

e′(lj + · · ·+ lk) ≤ max {e′(l∗∗), e′((N − 1)l)}

≤ max

{2α1π

3√

3rl,α13π(r + 4)

8l

}≤ α1π

l

[2

3√

3r+

3(r + 4)

8

].(51)

Let li = l(1). Combining (47) and (51), for N > Nα � 1, we have∑1≤j≤i≤k≤N−1

e′(lj + · · ·+ lk) = e′(li) +∑

1≤j≤i≤k≤N−1,j<k

e′(lj + · · ·+ lk)

≤ −α1rl2

4π2N3α +

N2α1π

4l

[2

3√

3r+

3(r + 4)

8

]< 0,

where we have used the fact that the number of (j, k) pairs is no more than N2

4 . Thiscontradicts force balance (32).

Next, we introduce an iteration scheme to improve the exponents in the aboveestimates.

Proposition 8 (iteration scheme). Let X be a critical point of E. Supposethere exist α and Nα such that for any N > Nα, we have l(1) ≥ N−α. Then, forany β satisfying α+1

3 < β < 1, there exists Nα,β such that for N > Nα,β, we have

l(1) ≥ N−β.

We begin the proof with a lemma which gives an upper bound for the force exertedby a step chain.

Lemma 3. Suppose for some α satisfying 0 < α < 1 and for all N > Nα, we haveN−α ≤ l(1). Then, for all 1 ≤ k ≤ N − 1, we have

max0<ξ1<ξ2<···<ξk≤(N−1)l

{e′(ξ1) + e′(ξ2) + · · ·+ e′(ξk)} ≤ CαNα logN,(52)

where ξi are constrained by ξi+1 − ξi ≥ l(1) for all i = 1, 2, . . . , k− 1 and the constantCα > 0.

Later this lemma will be applied to ξ1 = lj + · · ·+ li, ξ2 = lj + · · ·+ li+1, . . . , ξk =lj + · · ·+ li+k−1. The appearance of (N − 1)l is due to the estimate (48): lj + · · ·+li+k−1 ≤ l1 + · · ·+ lN−1 = Nl − lN ≤ (N − 1)l.


Proof. Let

k1 = #{i : 0 < ξi ≤ l∗∗},k2 = #{i : l∗∗ < ξi ≤ Nl − l∗∗},k3 = #{i : Nl − l∗∗ < li ≤ (N − 1)l}.

Thus k1 + k2 + k3 = k. Without loss of generality, assume k1, k2, k3 ≥ 1. (If anyof them is 0, we still have the similar estimates and the result remains the same.)Recall that e′(·) is monotone increasing on (0, l∗∗), (Nl− l∗∗, (N − 1)l) and monotonedecreasing on (l∗∗, Nl − l∗∗). Then

e′(ξ1) + · · ·+ e′(ξk)

≤ e′(l∗∗) + e′(l∗∗ − l(1)) + · · ·+ e′(l∗∗ − (k1 − 1)l(1))

+ e′(l∗∗) + e′(l∗∗ + l(1)) + · · ·+ e′(l∗∗ + (k2 − 1)l(1))

+ e′((N − 1)l) + e′((N − 1)l − l(1)) + · · ·+ e′((N − 1)l − (k3 − 1)l(1))

≤ 2e′(l∗∗) + e′((N − 1)l) +1

l(1)

∫ l∗∗

l∗∗−(k1−1)l(1)e′(x)dx

+1

l(1)

∫ l∗∗+(k2−1)l(1)

l∗∗

e′(x)dx+1

l(1)

∫ (N−1)l

(N−1)l−(k3−1)l(1)e′(x)dx

= 2e′(l∗∗) + e′((N − 1)l) +1

l(1)

{e(l∗∗ + (k2 − 1)l(1))− e(l∗∗ − (k1 − 1)l(1))

+ e((N − 1)l)− e((N − 1)l − (k3 − 1)l(1))}.

Recall (49), (50) and estimate the following terms:

e(l∗∗ + (k2 − 1)l(1)) < e(L2 ) = O(1),

−e(l∗∗ − (k1 − 1)l(1)) ≤ −e(l∗) = α1 logN +O(1),

e((N − 1)l) = −α1 logN +O(1),

−e((N − 1)l − (k3 − 1)l(1)) ≤ −e(l∗) = α1 logN +O(1).

Collecting these estimates, we have

max0<ξ1<ξ2<···<ξk≤(N−1)l

{e′(ξ1) + e′(ξ2) + · · ·+ e′(ξk)} ≤ CαNα logN.

Proof of Proposition 8. Again we prove this proposition by contradiction. Sup-pose for any sufficiently large Nα,β , there is some N > Nα,β such that l(1) < N−β .Let li = l(1). Applying Lemma 3 with α and k = N − i to ξ1 = lj + · · · + li, ξ2 =lj + · · ·+ li+1, . . . , ξk = lj + · · ·+ lN−1, we obtain

e′(lj + · · ·+ li) + e′(lj + · · ·+ li+1) + · · ·+ e′(lj + · · ·+ lN−1) ≤ CαNα logN.

By (47), e′(li) ≤ −α1rl2

4π2 N3β . Thus for N > Nα,β � 1, by force balance (32), we have

0 =∑

1≤j≤i≤k≤N−1

e′(lj + · · ·+ li + · · ·+ lk)

= e′(li) + [e′(li + li+1) + · · ·+ e′(li + · · ·+ lN−1)]


+

i−1∑j=1

[e′(lj + · · ·+ li) + e′(lj + · · ·+ li+1) + · · ·+ e′(lj + · · ·+ lN−1)]

≤ −α1rl2

4π2N3β + iCαN

α logN

≤ −α1rl2

4π2N3β + CαN

1+α logN

< 0.

This contradiction completes the proof.

Proposition 9. Let X be a critical point of E. For any δ > 0, there exists Nδsuch that for all N > Nδ, we have N−

12−δ ≤ l(1).

Proof. Choose any δ0 such that 0 < δ0 <13 . Let

βn =1

2+

(2

3

)n−1(1

6+ δ0

).

Note that 23 < β1 = 2

3 + δ0 < 1, βn+13 < βn+1 and βn is decreasing in n.

By Proposition 7, l(1) ≥ N−β1 for sufficiently large N . Applying Proposition 8, wehave l(1) ≥ N−β2 and by induction, l(1) ≥ N−βn for sufficiently large N > Nβn . Aftera finite number of iterations, we have βn∗ <

12 +δ for n∗ ≥ min{1, [log δ

1/6+δ0/ log 2

3 ]+

2}. Hence l(1) ≥ N−βn∗ ≥ N− 12−δ for all N > Nβn∗ completing the proof.

Based on this proposition, we are able to show the lower bounds of l1 as well asl(N−1).

Proposition 10. Let X be a critical point of E. For any δ > 0, there exists Nδsuch that for all N > Nδ have N−

16−δ ≤ l1 and hence N−

16−δ ≤ l(N−1).

Proof. We prove the proposition by contradiction. Suppose there is some δ ∈(0, 12 ), and for N > Nδ, we have l1 < N−

16−δ < l∗∗. Proposition 9 shows that

N−12−δ ≤ l(1) for N > Nδ � 1. Applying Lemma 3, we have

∑N−1k=2 e

′(l1 + · · · +lk) ≤ CδN

12+δ logN . Note that e′(·) is increasing on (0, l∗∗). Thus we have e′(l1) ≤

e′(N−16−δ) ≤ −α1rl

2

4π2 N12+3δ for N > Nδ � 1 (see also the proof of (47)). Combining

these results, we have

N−1∑k=1

e′(l1 + · · ·+ lk) = e′(l1) +

N−1∑k=2

e′(l1 + · · ·+ lk)

≤ −α1rl2

4π2N

12+3δ + CδN

12+δ logN

< 0,

which contradicts force balance∑N−1k=1 e

′(l1 + · · ·+ lk) = 0. This completes the proof

for N−16−δ ≤ l1. Hence N−

16−δ ≤ l1 ≤ l(N−1).

Essentially, we have already proved the lower bounds in Theorem 5(a) and (b).The remaining parts of Theorem 5 will be completed in section 11.

7. Energy minimizing bunch concentrated in half period. In this section,we are going to show the second result (Theorem 3(b)) for the one bunch structurewhich is also the foundation of the proofs in the remaining sections. Note that we


use distance on a circle in the periodic setting. If we can show that all steps indeedconcentrate in the half period, then the distance equals the one-dimensional Euclideandistance. Hence the monotonicity of e(·) and e′(·) can be applied. Physically, theresult is also important in the sense that all steps concentrate in a band whose widthis less than half of the period. Later we will have the stronger result that the relativewidth of this band vanishes as N →∞.

We start with several useful lemmata.

Lemma 4. If N2 > r, then e(l∗ + x) < e(l∗ − x) for all 0 < x < l∗.

Proof. Let g(x) = e(l∗+x)−e(l∗−x). Then g(0) = 0, g′(x) = e′(l∗+x)+e′(l∗−x),g′(0) = 0, and

g′′(x) = e′′(l∗ + x)− e′′(l∗ − x)

=α1π

2

N2l2

[(sin

π(l∗ + x)

Nl

)−2−(

sinπ(l∗ − x)

Nl

)−2]

·

{3r

N2

[(sin

π(l∗ + x)

Nl

)−2+

(sin

π(l∗ − x)

Nl

)−2]−(

1 +2r

N2

)}.

Since 0 < π(l∗−x)Nl < π(l∗+x)

Nl and π(l∗−x)Nl < π − π(l∗+x)

Nl , we have(sin

π(l∗ + x)

Nl

)−2−(

sinπ(l∗ − x)

Nl

)−2< 0

and

3r

N2

[(sin

π(l∗ + x)

Nl

)−2+

(sin

π(l∗ − x)

Nl

)−2]−(

1 +2r

N2

)

>3r

N2

(sin

π(l∗ − x)

Nl

)−2−(

1 +2r

N2

)>

3r

N2

(sin

πl∗Nl

)−2−(

1 +2r

N2

)= 3−

(1 +

2r

N2

)=

2(N2 − r)N2

> 0.

Therefore g′′(x) < 0 for all 0 < x < l∗. For 0 < x < l∗, we have g(x) = g(0)+g′(0)x+g′′(ξ)

2 x2 < 0, where ξ ∈ (0, x). Hence e(l∗ + x) < e(l∗ − x).

Lemma 5. Suppose N−β ≤ l(1) for some 0 < β < 1 and for all N > Nβ. Then,for all 1 ≤ k ≤ N − 1, we have

min0<ξ1<ξ2<···<ξk≤Nl2

{e(ξ1) + e(ξ2) + · · ·+ e(ξk)} ≥ α1k log kN−1−β +O(k),(53)

where ξi’s are constrained by ξi+1 − ξi ≥ l(1) for all i = 1, 2, . . . , k − 1.

Proof. For k = 1, min0<ξ1≤Nl2e(ξ1) ≥ −α1 logN +O(1). For any 2 ≤ k ≤ N − 1,

suppose (ξ1, . . . , ξk) minimize∑ki=1 e(ξi) with constraints: 0 < ξ1, ξk ≤ Nl

2 and

ξi+1 − ξi ≥ l(1) for all i = 1, 2, . . . , k− 1. It is easy to see that, by mononocity of e(·),we must have

ξ2 − ξ1 = ξ3 − ξ2 = · · · = ξk − ξk−1 = l(1), ξ1 ≤ l∗ and l∗ < ξk.


Suppose ξj ≤ l∗ < ξj+1. Then 1 ≤ j ≤ k − 1. Then for all 1 ≤ i < i+ k − 1 ≤ N − 1and 2 ≤ k ≤ N − 1, using Lemma 4, we have

k∑i=1

e(ξi) ≥ e(l∗) + e(l∗ − l(1)) + · · ·+ e(l∗ − (j − 1)l(1))

+ e(l∗) + e(l∗ + l(1)) + · · ·+ e(l∗ + (k − j − 1)l(1))

≥ e(l∗) + e(l∗ + l(1)) + · · ·+ e(l∗ + (j − 1)l(1))

+ e(l∗) + e(l∗ + l(1)) + · · ·+ e(l∗ + (k − j − 1)l(1))

≥ 2e(l∗) +1

l(1)

∫ l∗+(j−1)l(1)

l∗

e(x)dx+1

l(1)

∫ l∗+(k−j−1)l(1)

l∗

e(x)dx.

Notice that for n = j − 1 and n = k − j − 1, we have∫ l∗+nl(1)

l∗

e(x)dx ≥∫ l∗+nl

(1)

l∗

(α1 log

2x

Nl+α1rl

2

2π2x2

)dx

=

[α1

(log

2

Nl

)x+ α1x log x− α1x−

α1rl2

2π2x

]∣∣∣∣l∗+nl(1)

l∗

=

(α1 log

2

Nle

)nl(1) + α1(l∗ + nl(1)) log(l∗ + nl(1))

−α1l∗ log l∗ −α1rl

2

2π2

(1

l∗ + nl(1)− 1

l∗

)≥ α1nl

(1) log2

Nle+ α1(l∗ + nl(1)) log(l∗ + nl(1))− α1l∗ log l∗.

Collecting these estimates,

k∑i=1

e(ξi) ≥ 2e(l∗) + α1(j − 1) log2

Nle

+α1(l∗ + (j − 1)l(1))

l(1)log(l∗ + (j − 1)l(1))− 2α1l∗

l(1)log l∗

+ α1(k − j − 1) log2

Nle+α1(l∗ + (k − j − 1)l(1))

l(1)log(l∗ + (k − j − 1)l(1)).

Note that 12 (j − 1 + k − j − 1) = k

2 − 1 and by convexity of function x log x, we have

α1(l∗ + (j − 1)l(1))

l(1)log(l∗ + (j − 1)l(1))

+α1(l∗ + (k − j − 1)l(1))

l(1)log(l∗ + (k − j − 1)l(1))

≥2α1(l∗ + k−2

2 l(1))

l(1)log

(l∗ +

k − 2

2l(1)).

Hence,

k∑i=1

e(ξi) ≥ 2e(l∗) + α1(k − 2) log2

Nle


+2α1(l∗ + k−2

2 l(1))

l(1)log

(l∗ +

k − 2

2l(1))− 2α1l∗

l(1)log l∗

≥ α1

(log

r

N2+ 1)− α1(k − 2) logN + α1(k − 2) log

2

le

+ α1(k − 2) log

(l∗ +

k − 2

2l(1))

+2α1l∗l(1)

logl∗ + k−2

2 l(1)

l∗

≥ α1

(log

r

N2+ 1)− α1(k − 2) logN + α1(k − 2) log

2

le

+ α1(k − 2) log

(k − 2

2l(1))

≥ −α1k logN + α1(k − 2) log(k − 2)N−β +O(k)

= α1k log kN−1−β +O(k).

Lemma 6. For any a, b satisfying 1 ≤ a ≤ b ≤ N and for all β ∈ R, we have

b∑k=a

k log(kN−1−β

)≥ −β(a+ b)(b− a+ 1)

2logN − 1

eN2(54)

for all N .

Proof. The lemma can be proved by direct computation,

b∑k=a

k log(kN−1−β

)= N

b∑k=a

k

Nlog

k

N+

b∑k=a

k logN−β

≥ Nb∑

k=a

−1

e− β

b∑k=a

k logN

≥ −β(a+ b)(b− a+ 1)

2logN − 1

eN2.

Now we are ready to prove Theorem 3(b), which basically states l1 + l2 + · · · +lN−1 ≤ 1

2Nl.

Proof of Theorem 3(b). We prove the statement by contradiction. Suppose xN −x1 >

Nl2 . Without loss of generality, we assume that x1 = 0. Then xN > Nl

2 . Let Ibe the step index set and P be the ordered step pair index set:

I = {1, 2, . . . , N} ,(55)

P = {(i, j) : 1 ≤ i < j ≤ N} .(56)

We partition I and P into six and three nonintersecting subsets:

Ij =

{i :j − 1

6Nl ≤ xi <

j

6Nl

}kj = #Ij , for j = 1, 2, . . . , 6,(57)

P1 = {(i, j) ∈ P : i, j ∈ Ik for some k} ,(58)

P2 = {(i, j) ∈ P : i ∈ Ik, j ∈ Ik+1 for some k = 1, 2, . . . , 5 or i ∈ I6, j ∈ I1} ,(59)

P3 = {(i, j) ∈ P : (i, j) ∈ P\(P1 ∪ P2)} .(60)

Recalling Proposition 9, for any 0 < δ < 12 , we have l(1) ≥ N− 1

2−δ for N � 1.


First, we estimate the energy contributions from (i, j) ∈ P1 ∪ P2. For example, ifi ∈ I1, j ∈ I1∪I2 and i < j, then we apply Lemma 5 with β = 1

2 +δ, k = k1 +k2− i tothe step chain xi, xi+1, . . . , xi+k with {ξm = xi+m − xi, m = 1, 2, . . . , k}. This gives

k1+k2∑j=i+1

e(xj − xi) ≥ α1k log kN−32−δ +O(k).

Now applying Lemma 6 with a = k2, b = k1 + k2 − 1, and β = 12 + δ, we have

∑i∈I1,j∈I1∪I2

e(xj − xi) =

k1∑i=1

k1+k2∑j=i+1

e(xj − xi)

≥k1+k2−1∑k=k2

α1k log kN−32−δ +O(k)

= −(α1

2+ α1δ

) k1(k1 + 2k2 − 1)

2logN +O(N2)

= −(α1

4+α1δ

2

)(k21 + 2k1k2

)logN +O(N2).

Similarly, we can estimate the energy contribution from i ∈ I2, j ∈ I2 ∪ I3, i < j, etc.Collecting all these contribution, we have∑

(i,j)∈P1∪P2

e(xj − xi)

=∑

i∈I1,j∈I1∪I2

e(xj − xi) +∑

i∈I2,j∈I2∪I3

e(xj − xi) +∑

i∈I3,j∈I3∪I4

e(xj − xi)

+∑

i∈I4,j∈I4∪I5

e(xj − xi) +∑

i∈I5,j∈I5∪I6

e(xj − xi) +∑

i∈I6,j∈I6∪I1

e(xj − xi)

≥ −(α1

4+α1δ

2

)(k21 + 2k1k2 + k22 + 2k2k3 + k23 + 2k3k4

+k24 + 2k4k5 + k25 + 2k5k6 + k26 + 2k6k1)

logN +O(N2).(61)

Second, we estimate the energy contributions from (i, j) ∈ P3. For example,i ∈ I1, j ∈ I3 ∪ I4 ∪ I5. Then dist(xi, xj) ≥ Nl

6 , and hence∑i∈I1,j∈I3∪I4∪I5

e(xj − xi) ≥∑

i∈I1,j∈I3∪I4∪I5

e

(Nl

6

)

=α1

2(k1k3 + k1k4 + k1k5)

(log

1

4+

4r

N2

).

Similarly, we estimate the energy contribution from i ∈ I2, j ∈ I4 ∪ I5 ∪ I6, i < j, etc.Collecting all these contribution, we have∑

(i,j)∈P3

e(xj − xi)

=1

2

∑i∈I1,j∈I3∪I4∪I5

+∑


+∑



+∑


+∑


+∑


e(xj − xi)

≥ α1

2(k1k3 + k1k4 + k1k5 + k2k4 + k2k5 + k2k6 + k3k5 + k3k6 + k3k1 + k4k6

+ k4k1 + k4k2 + k5k1 + k5k2 + k5k3 + k6k2 + k6k3 + k6k4)1

2

(log

1

4+

4r

N2

),(62)

where we have divided the sum by 2 to avoid double counting.Finally, we combine (61) and (62) to obtain∑1≤i<j≤N

e(xj − xi)

≥∑

(i,j)∈P1∪P2

e(xj − xi) +∑

(i,j)∈P3

e(xj − xi)

≥ −(α1

4+α1δ

2

)(k21 + 2k1k2 + k22 + 2k2k3 + · · ·+ k26 + 2k6k1

)logN

+α1

2(k1k3 + k1k4 + k1k5 + · · ·+ k6k2 + k6k3 + k6k4)

1

2

(log

1

4+

4r

N2

)+O(N2)

= −(α1

4+α1δ

2

)[k1 + · · ·+ k6]

2logN

+α1

2[(k1k3 + k1k4 + k1k5) + · · · ]

[(1 + 2δ) logN + log

1

2+

2r

N2

]+O(N2)

≥ −(α1

4+α1δ

2

)N2 logN +

α1

2k1k3(1 + 2δ) logN +O(N2),

where we used k1 + · · ·+ k6 = N in the last equality.

For N � 1, we have k1l∗ ≥ xk1+1 − x1 ≥ Nl6 . By (21), we have l∗ =

√rlπ +

O(N−2) ≤√rl3 for N � 1. Hence k1 ≥

(Nl6

)/l∗ ≥ N

2√r. Similarly, k3 ≥ N

2√r. Thus

k1k3 ≥ N2

4r . Therefore

−α1

4N2 logN +O(N2) ≥ E(X) =

∑1≤i<j≤N

e(xj − xi)− E0N

≥[−α1

4+α1

2

(−δ +

1 + 2δ

4r

)]N2 logN +O(N2)

≥(−α1

4+ δ)N2 logN +O(N2).

The last inequality leads to a contradiction upon choosing δ small enough such thatδ ≤ α1

2

(−δ + 1+2δ

4r

). This shows that we must have lN > Nl

2 for N � 1.

8. Lower bound for bunch size. In this section, we prove Theorem 4(a),which provides a lower bound for bunch size. The proof makes use of the notion ofthe distribution for terrace lengths

{l(k), k = 1, 2, . . . , N − 1

}.

Proposition 11 (terrace length distribution). Let X be a global minimizer ofE. There exists a constant C, independent of N and k, such that

#{j : lj > Ck1/2N−1 (logN)

−1/2, 1 ≤ j ≤ N − 1

}≥ N − k(63)

for N ≥ 2 and 1 ≤ k ≤ N − 1.


Proof. By (21), there exists a constant C, independent of N and k, such that

l∗ > CN−1/2(logN)−1/2

for N ≥ 2. If l(k) > l∗, then

l(j) ≥ l(k) > l∗ > Ck1/2N−1(logN)−1/2, j = k, k + 1, . . . , N − 1.(64)

This implies (63). In the following, we assume that k satisfies l(k) ≤ l∗. (Indeed, wewill see that by Theorem 5(b), this is the case forN � 1 and for all k = 1, 2, . . . , N−1.)Then we have

−cN2 logN ≥ E(X) ≥k∑i=1

e(l(i)) +α1

2

[N(N − 1)

2− k](

logr

N2+ 1)− E0

N

≥ ke(l(k))− CN2 logN.

The last inequality is due to the monotonicity of e(·): e(l(i)) ≥ e(l(k)) for 0 < l(i) ≤l(k) ≤ l∗. Therefore, with some C > 0,

CN2 logN ≥ ke(l(k)) =α1

2k

[log sin2 πl

(k)

Nl+

r

N2

(sin

πl(k)

Nl

)−2].

Choosing δ = 14 in Proposition 9, we obtain that l(1) ≥ CN−

34 for all N . In

addition, note that 0 < l(k) ≤ Nl2 . Thus

α1

2k log sin2 πl

(k)

Nl≥ α1

2N log sin2 πl

(1)

Nl≥ α1N log

2l(1)

Nl≥ −CN logN.

Therefore,

CN2 logN ≥ α1

2kr

N2

(sin

πl(k)

Nl

)−2≥ α1

2krl2

π2

(l(k))−2

leading to

l(k) > Ck1/2N−1 (logN)−1/2

.

Hence for k = 1, 2, . . . , N − 1, we have

#{j : lj > Ck1/2N−1 (logN)

−1/2, 1 ≤ j ≤ N − 1

}≥ N − k.(65)

Now, we are going to use this distribution result to show the lower bound of thebunch size.

Proof of Theorem 4(a). Choose k =[12N]

in Proposition 11. Then we have

#{j : lj > CN−1/2 (logN)

−1/2, 1 ≤ j ≤ N − 1

}≥ N

2.

Therefore, l1 + l2 + · · · + lN−1 ≥ N2 CN

−1/2 (logN)−1/2

= CN1/2 (logN)−1/2

forN ≥ 2.

Remark 4. By Proposition 9, we have l(1) ≥ CδN− 1

2−δ for all N . This impliesthat l1 + · · ·+ lN−1 ≥ CδN

12−δ for all N . Hence Theorem 4(a) is a somewhat stronger

version of this estimate.


9. Energy scaling law: Strong version. Based on Proposition 11, we canimprove the minimum energy lower bound in Proposition 5.

Proof of Theorem 2. Given any δ > 0, for any 0 < s < 1 and any α > 0 letk = [sN ]. Then (63) implies

#{j : lj > s1/2CαN

−1/2−α/2, 1 ≤ j ≤ N − 1}≥ (1− s)N.(66)

Let P = {(i, j) : j − i ≥ 2sN}, then #P = (1−2s)2N2

2 + O(N). For (i, j) ∈ P , wehave

xj − xi ≥ sN · s1/2CαN−1/2−α/2 + sN · 0 = s3/2CαN1/2−α/2 > l∗

for N > Nα,s � 1. Theorem 3(b) implies xj − xi ≤ Nl2 for N � 1. Thus, by the

monotonicity of e(·) on (l∗,Nl2 ), we have

e(xj − xi) ≥ e(s3/2CαN1/2−α/2)

≥ 2 log

(2s3/2Cα

lN−1/2−α/2

)+

rl2

π2s3C2α

N−1+α

= −(1 + α) logN +O(1).

Therefore,

E(X) =∑

1≤i<j≤N

e(xj − xi)− E0N

≥ α1

2

{[(1− 2s)2N2

2+O(N)

][−(1 + α) logN +O(1)]

+

[N(N − 1)

2− (1− 2s)2N2

2−O(N)

](log

r

N2+ 1)}− E0

N

=α1

2

[1− α

2(1− 2s)2 − 1

]N2 logN +O(N2)

≥ −(α1

4+ δ)N2 logN − cδN2.

The last inequality holds for sufficiently small s and α.So far, this improved lower bound holds for N > Nα,s � 1. It is easy to revise

the coefficients and obtain the same statement for all N .

10. Upper bound for bunch size. In this section, we complete the proof ofthe upper bound for bunch size. We prove the following proposition, from whichTheorem 4(b) is an immediate consequence.

Proposition 12. Let X be any global minimizer of E. Suppose that l(1) ≥ N−βfor some 1

2 < β < 1 and for sufficiently large N , i.e., N > Nβ. Then for any s, α

satisfying 0 < s < 1 −√

β−1/2β and 0 < α < β − β−1/2

(1−s)2 , there exists Nα,β, for all

N > Nα,β,

min1≤i≤N−[sN ]

xi+[sN ] − xi < N1−α.

Proof of Theorem 4(b). Let β and hence α be sufficiently close to 12 ; then Propo-

sition 12 immediately implies Theorem 4(b).


Proof of Proposition 12. Given β, from the definition of α, we have 0 < α < 12 <

β < 1. Suppose the statement is false. Then for any Nα,β , there exists an N > Nα,βsatisfying

min1≤i≤N−[sN ]

xi+[sN ] − xi ≥ N1−α.(67)

Let P = {(i, j) : 1 ≤ i < j ≤ N}. We partition P into three nonintersecting sub-sets P = P1 ∪ P2 ∪ P3, where

P1 = {(i, j) ∈ P : j − i ≥ [sN ]} ,P2 = {(i, j) ∈ P : j − i < [sN ], i ≤ N − [sN ]} ,P3 = {(i, j) ∈ P : j − i < [sN ], i > N − [sN ]} .

Then

#P1 =(1− s)2N2

2+O(N), #P2 = s(1− s)N2 +O(N), and #P3 =

s2N2

2+O(N).

For (i, j) ∈ P1, j − i ≥ [sN ], xj − xi ≥ N1−α > l∗. Recall that xj − xi ≤l1 + · · ·+ lN−1 ≤ Nl

2 for N � 1. The monotonicity of e(·) on (l∗,Nl2 ) leads to

e(xj − xi) ≥ e(N1−α) ≥ α1 log2N1−α

Nl+α1r

2N2

(πN1−α

Nl

)−2= −α1α logN +O(1).

Summing up all these pairs (i, j) ∈ P1 leads to∑(i,j)∈P1

e(xj − xi) ≥ −α1α(1− s)2

2N2 logN +O(N2).

Now we consider (i, j) ∈ P2 ∪P3. For each i, applying Lemma 5 to the step chainxi, xi+1, . . . , xi+k with {ξm = xi+m − xi, m = 1, 2, . . . , k} gives

e(xi+1 − xi) + e(xi+2 − xi) + · · ·+ e(xi+k − xi) ≥ α1k log kN−1−β +O(k).

We then estimate the contribution from P2,∑(i,j)∈P2

e(xj − xi) ≥ (N − [sN ]) ·[α1k log kN−1−β +O(k)

]∣∣k=[sN ]−1

= −α1βs(1− s)N2 logN +O(N2).

For contribution from P1, we apply Lemma 6 with a = 1, b = [sN ]− 1,

∑(i,j)∈P3

e(xj − xi) ≥[sN ]−1∑k=1

[α1k log kN−1−β +O(k)

]≥ −α1β

s2

2N2 logN +O(N2).

Now let δ = (β − α) (1−s)22 − 1

2

(β − 1

2

)> 0. Then the above estimates give

E(X) ≥ α1

{−α (1− s)2

2− βs(1− s)− β s

2

2

}N2 logN +O(N2)

= α1

(−1

4+ δ

)N2 logN +O(N2).

This contradicts the energy upper bound −α1

4 N2 logN +O(N2) ≥ E(X) for N � 1.

Hence the original statement is true.


11. Upper bound for terrace length. Here we combine the previous resultsto prove Theorem 5.

Proof of Theorem 5(a). Proposition 9 provides a lower bound for l(1). Here weonly need to show the upper bound for l(1). Let s = 1

2 (1− 1√2) ∈ (0, 1). We prove the

statement by contradiction. Suppose that for any Nδ, there exists N > Nδ such thatN−

12+δ < l(1). Now suppose Nδ � 1. Let

P1 = {(i, j) : 1 ≤ i < j ≤ N, j − i ≥ [sN ]} .

Then #P1 = (1−s)2N2

2 +O(N). For (i, j) ∈ P1, it holds that

Nl

2≥ xj − xi ≥ sNl(1) ≥ sN

12+δ > l∗.

The monotonicity of e(·) on (l∗,Nl2 ) leads to

e(xj − xi) ≥ e(sN12+δ) ≥ α1 log

2sN12+δ

Nl+α1r

2N2

(πsN1/2+δ

Nl

)−2≥ −α1

2(1− 2δ) logN + α1 log

2s

l.

Therefore

E(X) ≥ α1

2

{[(1− s)2N2

2+O(N)

] [−(1− 2δ) logN + 2 log

2s

l

]+

[N(N − 1)

2− (1− s)2N2

2+O(N)

](log

r

N2+ 1)}− E0

N

=α1

2

[−(1− s)2

(1

2− δ)− 2s+ s2

]N2 logN +O(N2)

≥(−α1

4+δ

2

)N2 logN +O(N2),

leading to a contradiction with −α1

4 N2 logN + CN2 ≥ E(X) for N � 1.

Proof of Theorem 5(b). Proposition 10 provides the lower bound for l(N−1). Herewe only need to show the upper bound for l(N−1). By Theorem 3(b), we have lN > Nl

2

for N � 1. We will prove l(N−1) ≤ l∗ for N � 1 by contradiction. Suppose li > l∗for some i, 1 ≤ i ≤ N − 1.

Note that 0 < li ≤ lj + · · ·+ li + · · ·+ lk ≤ Nl2 . Therefore∑

1≤j≤i≤k≤N−1

e′(lj + · · ·+ lk)

=∑

1≤j≤i≤k≤N−1

α1π

Nl

(1− r

N2

(sin

π(lj + · · ·+ lk)

Nl

)−2)cot

π(lj + · · ·+ lk)

Nl

≥∑

1≤j≤i≤k≤N−1

α1π

Nl

(1− r

N2

(sin

πliNl

)−2)cot

π(lj + · · ·+ lk)

Nl

> 0,

contradicting the force balance (32). Therefore, l(N−1) ≤ l∗ for N � 1 and l(N−1) ≤O(1) for all N .


Remark 5. If N � 1, the lower bound in Theorem 5(b) is stronger than the onein Theorem 3(a) because l∗ < l∗∗. However, the latter holds for all N .

Finally, we complete the proof for Theorem 3(c).

Proof of Theorem 3(c). For any 0 < s < 1, Theorems 4(b) and 5(b) togetherimply

0 ≤ lim supN→∞

l1 + · · ·+ lN−1lN

≤ lim supN→∞

Cδ,sN1/2+δ + (1− s)Nl∗

12Nl

≤ 2(1− s).

Therefore limN→∞l1+···+lN−1

lN= 0.

12. Extensions. There are several extensions of our results, including the samemodel with the Neumann boundary condition, a particle system with generic pairwiseinteractions, and higher dimensional problems. Here we state only the results corre-sponding to the same pairwise interaction with the Neumann boundary condition.

In the setting with the Neumann boundary condition, the total energy can bewritten as

E(X) =∑

1≤i<j≤N

e(xi − xj), where e(x) = α1 log x+α2

2x2.(68)

We have l∗ =√

α2

α1and l∗∗ =

√3α2

α1. Our results are stated as follows.

Theorem 6 (energy scaling law with Neumann boundary condition). For anyδ > 0, there exist constants cδ and C such that(α1

4− δ)N2 logN − cδN2 ≤ inf

X∈RNE(X) ≤ α1

4N2 logN + CN2(69)

for all N . Consequently, limN→+∞inf E(X)N2 logN = α1

4 .

Theorem 7 (size of the bunch with Neumann boundary condition). For the globalminimizer of E, we have the following:

(a) (Lower bound) There exists C such that for all N , we have

CN1/2 (logN)−1/2 ≤ l1 + · · ·+ lN−1.(70)

(b) (Upper bound) For any δ > 0 and any 0 < s < 1, there exists Cδ,s such thatfor all N , we have

mini

{li + · · ·+ li+[sN ]

}≤ Cδ,sN1/2+δ.(71)

Theorem 8 (slope of the bunch profile, with Neumann boundary condition). Forthe global minimizer of E, we have the following:

(a) (Estimates on minimal terrace length l(1)) For any δ > 0, there exist constantscδ and Cδ such that for all N , we have

cδN−1/2−δ ≤ l(1) ≤ CδN−1/2+δ.(72)

(b) (Estimates on next-maximal terrace length l(N−1)) For any δ > 0, there existsa constant cδ such that for all N , we have

cδN−1/6−δ ≤ l(N−1) ≤ l∗ = O(1).(73)


These results are almost the same as the ones in the periodic setting. It is straight-forward if we notice that we are using the Euclidean distance |xi − xj | instead of thedistance on a circle, i.e., dist(xi, xj). Indeed, the energy in the Neumann boundarysetting can be obtained by replacing sinx with x, and the model is somewhat “lin-earized.” The length scales of the bunch size and the minimal terrace remain thesame. However, we should emphasize the following two modifications for the casewith the Neumann boundary condition:

1. There is no reference state and hence no reference energy E0N .

2. The leading order of the minimum energy scaling law is α1

4 N2 logN instead of

−α1

4 N2 logN . The missing −α1

2 N2 logN can be understood in the following

way. For each pair xi, xj , the difference between α1

2 log sin2 π(xi−xj)Nl and

α1 log |xi − xj | is − logN +O(1). Totally, we have N(N−1)2 such pairs.

Our method can also be applied to more general (m,n)-type pairwise interactions,where the pairwise force is given by

f(x) =(α1|x|−m−1 − α2|x|−n−1

) x

|x|.(74)

The corresponding total energy is

E(X) =∑

1≤i<j≤N

e(xi − xj),(75)

where

(76) e(x) =

−α1

m |x|−m + α2

n |x|−n, −1 < m < n,m 6= 0, n 6= 0,

α1 log |x|+ α2

n |x|−n, 0 = m < n,

−α1

m |x|−m − α2 log |x|, −1 < m < n = 0.

Here m,n can be nonintegers. For −1 < m < 1 and 1 < n < +∞, we can also obtainthe minimum energy scaling law, the size of the system, and fine-structure lengthscales of the system. This will be reported in a future work.

Other possible generalizations of the present work include the proof of the anal-ogous results for the continuum model as derived in [19, 20] and the analysis of thedynamical equations and the temporal and spatial coarsening behaviors of the solu-tions. These will be explored in the future works.

Appendix A. Derivations of fi and E. We derive (6)–(8) in this appendix.Note that cotx =

∑k∈Z

1x+kπ . Then we have∑k∈Z

1

x+ kL=π

Lcot(πxL

),

∑k∈Z

1

(x+ kL)2=π2

L2

(sin

πx

L

)−2,

∑k∈Z

1

(x+ kL)3=π3

L3

(cot

πx

L

)(sin

πx

L

)−2.

Recall that, without loss of generality, we assume 0 ≤ x1 < x2 < · · · < xN < lN = L.Then we have the explicit form of fi and E. For i = 1, 2, . . . , N ,

fi = −∑

j∈Z,j 6=i

[α1

xj − xi− α2

(xj − xi)3

]


= −∑

1≤j≤N,j 6=i

∑k∈Z

[α1

xj − xi + kL− α2

(xj − xi + kL)3

]

= −∑

1≤j≤N,j 6=i

[α1π

Lcot

π(xj − xi)L

− α2π3

L3cot

π(xj − xi)L

(sin

π(xj − xi)L

)−2].

Let

E =∑

1≤i<j≤N

[α1

2log sin2 π(xj − xi)

L+α2

2

π2

L2

(sin

π(xj − xi)L

)−2]− E0

N

Here E0N is some constant satisfying E(uniform step train) = 0. It is easy to check

that

∂E

∂xi= −

∑1≤j≤N,j 6=i

[α1π

Lcot

π(xj − xi)L

− α2π3

L3cot

π(xj − xi)L

(sin

π(xj − xi)L

)−2]= fi.

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