Distributions of points on non-extensible closed curves in R3 realizing maximum total energies

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Distributions of points on non-extensible closed curves in R ³ realizing maximum total energies

参赛学⽣

王中⼦ Zhongzi Wang

北京中国⼈民⼤学附中⾼⼀ 24 班

指导⽼师

郑绍远 Shiu-Yuen Cheng

北京清华⼤学数学系

带队⽼师

待定

北京中国⼈民⼤学附中

Abstract: Distributions of points under certain conditions are widely concerned by people. Our motivation: Let G_n be a non-extensible, flexible closed curve of lengthn in R³ with n particles A1,... ,A_n evenly fixed (according to arc length of G_n) on the curve. Let f_α(x) =x^α for α >0,f₀(x) = lnx,f_α(x) =−x^α for α <0, wherex≥0. Letdbe the distance in R³. Define the total energy

E_n^α(G_n) = 1 2

∑

p̸=q

f_α(d(A_p, A_q)).

Problem 1.1. What is the shape ofG_nwhen the total energy reaches the maximum?

Note E_n^α(Gn) relies only the positions of particles ofGn, but positions of those particles are constrained by the non-extensible curve.

A₁

A₂ A_n

A_i

The famous Thomson type problem, which considers the distribution ofnpoints on the unit sphere in R³ under essentially the same energy functions f_α, is an inspiration of the distribution problem we studied here.

We denote the maximum of the total energy E_n^α by maxE_n^α. We will verify the existence of maxE_n^α (Theorem 2.1) and prove each G_n realizing maxE^α_n must be aΓ_n, a convexn-gon (may be degenerated) with edge length 1 (Theorem 3.1).

Problem 1.2. What is the shape ofΓnwhen the total energy reaches the maximum?

There are two special shapes for Γ_n: the regular n-gon Γ^o_n, and the double straight arc Γ⁻_n (only defined for evenn).

Forn= 4, the Problem is completely solved (Example 4.3).

We will prove for given n, E_n^α(Γ_n) is maximum if and only ifΓ_n = Γ^o_n for large enough negative α (Theorem 5.6); and for given even n, E_n^α(Γn) is maximum if and only ifΓn= Γ⁻_n for large enough positiveα. (Theorem 6.1)

Theorem 5.6 follows from Theorem 5.1: If Γ_n satisfies a bending condition, then E_n^α(Γn) is maximum if and only if Γn = Γ^o_n for α ≤ 1. All central symmetry Γn satisfy this bending condition.

For each evenn, Γ_n realizing maxE_n^α we found so far are only Γ^o_n and Γ⁻_n. But there are infinitely manyΓ5 realizing maxE₅^α asα varies (Proposition 6.7).

We also add some information on the Thomson type problem (Theorem 7.2).

Distributions of points on non-extensible closed curves in R

realizing maximum total energies

Contents

1 Introduction 4

2 The existence of the maximum for E_n^α 7

3 E_n^α(G_n) maximum implies G_n isΓ_n, a convexn-gon of edge length 1 9 4 Basic facts, the classification for n= 4 15 5 When the regular n-gonΓ^o_n realizing maxE_n^α 17 5.1 Γ^o_n realizing maxE_n^α ifΓnnot bending fast for α≤1 . . . 17 5.2 Γ^o_n realizing maxE_n^α for large negative α, more corollaries . . . . 22 6 Which Γn realizing maxE_n^α for large positive α 25 6.1 For evenn, the double straight arc Γ⁻_n realizing maxE_n^α for large α >0 25 6.2 A sample of oddn: Infinitely manyΓ5 realizing maxE₅^α for largeα >0 31

7 Back to Thomson type problems 33

8 Some miscellaneous results 36

8.1 E₆^α reaches max at Γ⁻₆ for central symmetry Γ₆ when α≥6 . . . 36 8.2 A physics meaning ofE_n²(Γ_n) . . . 41

9 文献 42

10队员简介 43

1 Introduction

Distributions of points under certain conditions are widely concerned by people. A motivation of our study is as below:

Letf be an energy function which is increasing about the distance dinR³. Let G_nbe a non-extensible, flexible closed curve withnparticlesA₁,... ,A_n evenly fixed on the curve. PutGninto R³. Define the total energy

E_n^f(Gn) = 1 2

∑

p̸=q

f(d(Ap, Aq)), wheredis the distance inR³.

Problem 1.1. What is the shape ofG_nwhen the total energy reaches the maximum?

Note En^f(Gn) relies only the positions of particles of Gn, but positions of those particles are constrained by the non-extensible curve.

A₁

A₂

A_n

A_i

Figure 1

Mathematically, let G_n be a circle G of length n with n vertices A₁, ..., A_n attached consecutively so that the distance between A_i and A_i+1 is 1 along G, Ai+n=Ai. BothGandR³have their own standard metrics. Call a mapg:Gn→R³ is non-extensible, ifgdoes not extend the length of any portion ofG. More precisely, we assume g:G_n→ R³ is differentiable with finitely many exception points of G_n. Callg :Gn→R³ is non-extensible, if the modulus of first derivative|g^′(x)|= 1for any differentiable pointx∈G. We will considerEn^f ong(Gn)for any non-extensible mapg:Gn→R³, and call g(Gn)⊂R³ is allowable.

For simplicity, we will often use G_n ⊂R³ to denote g(G_n) ⊂ R³, A_i to denote g(A_i) and so on. In particulard(A_i, A_i+1)≤1. We will rewrite (1.1) as

E_n^f(G_n) = 1 2

∑

p̸=q

f(|A_p−A_q|) =∑

p<q

f(|A_p−A_q|) (1.2) where eachAp is considered as a vector inR³ and|Ap|is the length of Ap.

The following family of energy functionsf_α, α∈Rare often appeared in geom- etry and physics.

fα(x) =





x^α, α >0;

lnx, α= 0;

−x^α, α <0.

(1.3)

Remark 1.2. Cases α= −1 and α = 2 have physics means: Case α =−1 was first consider by Thomson for points in the unit 2-sphere [To]. In our setting−E_n⁻¹(G_n) is the total electric potential energy ofGn, where each vertex ofGnhas unit charge, and there is no charge on the edges. For α = 2, E_n²(Gn) is the moment of inertia of G_n about its mass center (Remark 8.9), where each vertex ofG_n has unit mass, and there is no mass on the edges. According to [HLP], a most direct case α = 1 was first considered by Toth for points with mutully distances≤1 [To].

Below we will simply denote E_n^fα by E_n^α. First we will verify the following two results.

Theorem 2.1. For each α and n, the maximum of E_n^α(G_n) exists among all allowableG_n⊂R³.

Theorem 3.1. Suppose En^f(Gn)reaches the maximum. Then Gn is a convexn-gon (may be degenerated) with edge length 1.

The verification of Theorem 3.1 is longer and subtler than we first thought.

Below we use∏

n={Γn}to denote the set of all convexn-gons with edge length 1 in the plane. With Theorem 3.1, Problem 1.1 is transformed to the following Problem 1.3. What is the shape ofΓnwhen the total energy reaches the maximum?

Below we always assume that the integern≥4. (Forn= 2 orn= 3the answer is obvious). We often use maxE_n^α to denote the maximum of E_n^α.

There are two special shapes for Γn: the regular n-gon Γ^o_n, and the double straight arc Γ⁻_n (only for even n, see Figure 2 right for n = 6, where two lines are coincided indeed. See Section 4 for the precise definition). We will see E₄^α(Γ4) reaches the maximum atΓ^o₄ forα <2and atΓ⁻₄ forα >2, andE₄²(Γ4)is a constant for allΓ₄ (Example 4.3).

Figure 2 For general nwe have

Theorem 5.6. For given n, there is an α_∗ < 0 (depends on n) such that E_n^α(Γn) reaches the maximum if and only if Γ_n is the regular n-gon for α < α_∗.

Theorem 6.1. For given even n >0, there is an α^∗ >0 (depends on n) such that E_n^α(Γ_n) reaches the maximum if and only if Γ_n= Γ⁻_n when α > α^∗.

The proofs of Theorem 5.6 and Theorem 6.1 are quite different: Theorem 6.1 follows from a rather complicated estimation (see Section 6), while Theorem 5.6 follows from Theorem 5.1 below whose proof needs a decomposition ofE_n^α(Γ_n) (see Section 5). Let [x]be the maximum integer not bigger thanx.

Theorem 5.1: Suppose the sum of any consecutive [n/2]−1 exterior angles of Γn

is no more thanπ. Then E_n^α(Γ_n) is the maximum if and only ifΓ_n= Γ^o_n, the regular n-gon of edge length 1 forα≤1.

A direct consequences of Theorem 5.1 is that if Γn is central symmetry, then E_n^α(Γ_n)is maximum if and only if Γ_n= Γ^o_n forα≤1. (Corollary 5.8).

By Theorem 5.6 and Theorem 6.1, for each even number n larger than 2, the equation E^αn(Γ^o_n) = E^αn(Γ⁻_n) always has solutions. So far the Γn reaches maxE_n^α we proved are only Γ^o_n and Γ⁻_n. We are interested to find some other Γ_n realizing maxE_n^α.

For odd n, the situation is different. LetΓ^∆_n denote the unique Γn which is an isoceles triangle with base length 1. We observed thatE₅^α never reaches maximum atΓ^∆₅ for anyα (Propositions 6.6), and based on this observation we have

Proposition 6.7. There are infinitely many Γ5 realizing maxE₅^α as α varies.

We believe this is true for any odd n.

Beyond the several results listed above, Problem 1.3 is open in general. Some efforts are made to get some local results. An example is below. Note 6 is the next even number after 4, and central symmetry condition allow us to deal with calculus of only three variables, then some elementary tricks can apply.

Proposition 8.1. If a central symmetry Γ₆ realizes maxE₆^α for α≥6, then Γ₆ = Γ⁻₆.

The Thomson type problem, which considers the distribution ofnpoints on the unit sphere inR³ under the energy functionsf_α given by (1.3), is an inspiration of the distribution problem we studied in this note. The problem was first raised by Thomson forα=−1[Th], and later generalized to allα∈R. Smale put Thomson’s problem in his problem list for 21st century [Sm]. There many studies on Thomson type problem, see [AP], [BH], [PB] and their references. For concreten, the precise distributions ofnpoint which realize the extremum is known only for few small n.

Mathematically, Thomson type problems can be raised for unit sphere S^m of R^m+1for any integerm >0. Now we can also add some information to the Thomson type problem. One sample result is the following (where we use E_n^α(m) to denote the corresponding totoal energy).

Theorem 7.2. Let A1, ..., An be npoints on the unit sphere S^m. Then (1) For α = 2, A1, ..., An realize the maxE_n²(m) if and only ∑_n

i=1Ai = 0, in particular there are infinitely many distributions to realize maxE_n²(m).

(2) For α >2 and n even, A1, ..., An realize the maxE_n^α(m) if and only if they stay evenly in the two ends of a diameter ofS^m.

(3) For α < 2 and n ≤ m+ 2, A₁, ..., A_n realize the maxE_n^α(m) if and only if they are the vertices regular (n−1)-simplex inscribed in Sⁿ⁻² =S^m∩Rⁿ⁻¹, where Rⁿ⁻¹ is a subspace of R^m+1 passing the origin.

(3) is known at least forα=−1, m= 2andα= 1, any integerm >0,n=m+2, see [PB], [BH] for example.

The paper is developed as the table of content. All calculus used can be found in [St], or [LZ] in Chinese; some basic topology of Euclidean spaces can be found in [Ar], or [Yo] in Chinese. Several classical inequalities are well known, can be found in [HLP].

2 The existence of the maximum for E

Theorem 2.1. For eachα andn, the maximum ofE^α_n(Gn) exists among all allow- ableGn⊂R³.

We use some basic topology of Euclidean space (see [Ar] or [Yo]) to prove The- orem 2.1.

First note every subset of Rⁿ with the metric given by Rⁿ become a metric space. For x ∈ Rⁿ and ϵ > 0, an open ϵ-neighborhood of x in Rⁿ is defined as Uϵ(x) = {y|d(x, y) < ϵ}. Call a subset X ⊂ Rⁿ is open, if each x ∈ X has an U_ϵ(x)⊂X for someϵ >0. LetX¯ denote the complement ofX inRⁿ. Call a subset X ⊂Rⁿ is closed if X¯ is open. It is easy to verify that the union (intersection) of open (closed) sets is open (closed), and intersection (union) of finitely many open (closed) set is open (closed).

Call a subset X ⊂ Rⁿ is compact, if every infinite sequence in X contains a convergent sub-sequence with limit inX. X compact implies that X is closed, and the intersection of a compact subset and a closed set is compact.

Theorem 2.2. (1) (Heine-Borel theorem) For a subset X ⊂ Rⁿ, X is compact if and only if X is closed and bounded.

(2) A continuous real-valued function defined on a compact subset is bounded and reaches its bounds.

Proof of theorem 2.1. In this proof, we use (x1, x2, ..., xn), where xi is a vector in R³, to denote of the image of the vertices under allowable maps g:Gn→R³.

Now we define

B1 :|x2−x1| ≤1, B2 :|x3−x2| ≤1,

...,

Bn−1 :|xn−xn−1| ≤1, B_n:|x₁−x_n| ≤1.

Positions of (x₁, x₂, ..., x_n) form a subset U^′ ⊂(R³)ⁿ which defined by B^′ =∩ⁿi=1B_i.

Since each B_i, defined by≤, is closed, their intersectionB^′ is closed.

Since E_n^α(G_n) is invariant under Euclidean transformations, so we may assume thatx1= 0. NoteB^′′⊂(R³)ⁿ defined byx1= 0 is also a closed subset. Let

B =B^′∩B^′′. B is also closed.

To consider the value of E_n^α, we need only restrict our attention onB. Since

|x_i−x₁| ≤i−1< n, we have |xi|< n, so

d((x1, x2, ...., xn),0)² =x²₁+x²₂+...+x²_n≤n³, henceB is bounded.

By Heine-Borel theorem, a closed bounded subset of Euclidean space is compact.

SoB is compact.

If α > 0, then E_n^α(Gn) = ∑

i<j|xi−xj|^α is continuous function defined on B.

By Theorem 2.2 (2),B is compact implies that E_n^α has a maximum on B.

If α ≤ 0, then E_n^α(Gn) is defined only on B_D = B\D (those points in B but not inD), where

D=∪i̸=jDi,j,

D_i,j ={(x₁, x₂, ..., x_n)|x_i=x_j, i̸=j}.

Clearly Di,j is closed, soD, as a finite union of closed set is also closed. Hence B\D is not closed.

Now fori̸=j and some ϵ >0, let

B^ϵ_i,j ={(x₁, x₂, ..., x_n)| |x_i−x_j| ≥ϵ}, thenB_i,j^ϵ is a closed subset. Let

C^ϵ=∩i,jB_i,j^ϵ . ThenC^ϵ is a closed subset. Let

B^ϵ=B∩C^ϵ.

A closed subset of a compact set is compact. So B^ϵ is compact. For any (x₁, x₂, ..., x_n) ∈ U^ϵ, by definition |x_i −x_j| ≥ ϵ for any i ̸= j. So E^α_n(G_n) is de- fined on B^ϵ ⊂ B \D for all α ≤ 0. By the same reason as before, E_n^α has a maximum onB^ϵ.

Once the (ordered) vertices ofG_nbelong toB^ϵ, we will simply to writeG_n∈B^ϵ. Below we assume thatϵ <1. Let Γ^o_n denote regular n-gon of edge length 1. Then Γ^o_n∈B^ϵ. Suppose E_n^α(G^o_n) =l.

Whenα <0, pickϵso that−ϵ^α< l. IfGn∈/ B^ϵ, then we have somek̸=m such that|x_k−x_m|< ϵ. Therefore

E_n^α(G_n) =−∑

i<j

|x_i−x_j|^α≤ −|x_k−x_m|^α <−ϵ^α< l=E_n^α(Γ^o_n).

When α= 0, pick ϵso that lnϵ+ (ⁿ⁽ⁿ₂⁻¹⁾ −1)lnn < l. IfG_n∈/B^ϵ, then we have somek < msuch that |x_k−x_m|< ϵ. Clearly|x_i−x_j| ≤n. Therefore

E_n⁰(Gn) =∑

i<j

ln|xi−xj|= ln|x_k−x_m|+ ∑

i<j,(i,j)̸=(k,m)

ln|x_i−x_j|

<lnϵ+ (n(n−1)

2 −1)lnn < l=E_n^α(Γ^o_n).

In either case, the value of E_n^α(G_n) on B_D\B^ϵ is bounded by E_n^α(G^o_n). Since G^o_nis in B^ϵ, the maximum value of E_n^α on B^ϵ is the maximum value onB\D.

So for each α≤0 the maximum value ofE_n^α exists.

Finally for each α∈R, the maximum value ofE^α_n exists.

3 E

(G

) maximum implies G

is Γ

, a convex n-gon of edge length 1

A subsetX ⊂Rⁿ is convex if it contains the line segments connecting each pair of its points. The convex hull ofX is the (unique) minimal convex set containing X.

Suppose S is a set of finitely many points. The convex hull of S forms a convex polygon if S⊂R² and forms a convex polytope if S⊂R³.

The concept ”convex polygon” are often used in two ways: either a 2-dimensional convex polygon, or its 1-dimensional boundary. People usually can understand the means from the context. Some time, we will indicate a convex polygon is 1 or 2 dimensional. We call polygon with n sides an-gon.

For our purpose, we allow convex polygon to be degenerated. Precisely if the points of S are in a line in R², the convex hull is the line segment joining the outermost two points P1 and P2. However we will consider it as a degenerated convex polygon, rather than a straight arc. Its boundary is still a closed curve which consists of two coincided straight arcs connectingP₁ and P₂, and the exterior angles atP1 and P2 areπ.

Some explanations will be helpful: WhenS is inR², we may imagine stretching a rubber band Gso that it surrounds the set S and then releasing it, finallyG will become a convex polygon which encloses the convex hull ofS (see the right of Figure 3). WhenS is in a line, the rubber band becomes a two stretched segments between the leftmost and rightmost points (see the left of Figure 3).

Figure 3

Theorem 3.1. Suppose En^f(G_n)reaches the maximum. Then G_n is aΓ_n, a convex n-gon with edge length 1.

Recall the vertices A₁, ..., A_n are cyclicly consecutive inG.

Theorem 3.1 follows from the following proposition whose statement gives the steps of the proof.

Proposition 3.2. Suppose E_n^f reaches the maximum at G_n (the image of some allowable mapg:Gn→R³). Then

(i) All vertices of G_n are in the same plane;

(ii) All vertices of G_n are vertices of a convex polygon C.

(iii) There is another Γn (the image of another allowable map g^′ : Gn → R³) such that

(a) the vertices of Γ_n and the vertices of G_n are coincided, counting the multi- plicity, in particularEn^f also reaches the maximum at Γ_n;

(b) the vertices ofA₁, ...., A_n of Γ_n is cyclicly consecutive in the boundary of the convex polygonC in (ii).

(iv) Γ_n in (iii) is a convex n-gon of edge length 1.

(v) The original G_n is a convex n-gon of edge length 1.

Proof. (i) Let C¯ be the convex hull of those vertices ofG_n (the edges ofG_nusually are not in C). If those vertices are not contained in any plane, then¯ C¯ is a 3- dimensional polyhedron, and we pick a face of C¯ and denote the plane containing this face by Π.

Denoted the vertices in Π by P₁, P₂, ... , P_k according to their orders in the curveG. Note all remaining vertices are in one side of Π.

If P1, P2,..., P_k are not consecutive in G, we may assume that P1, P2 are not consecutive inG. Then P₁ and P₂ divide G_n into two parts G^′ and G^′′, each part contains some vertices not inΠ, see Figure 4. Now reflectG^′ aboutΠ we get a new distrubution of Gn. To compare with the old distribution, the distances d(P^′, P^′′) increases for each vertexP^′ ofG^′ andP^′′ of G^′′ who are not in Π; and the distance of any remaining two vertices are not changed. So for the new distributionEn^f(Gn) is larger.

P₁ P₂

P₃

P_k P'

P''

Figure 4

Suppose nowP1,P2, ...,P_kare consecutive inG. LetCbe the convex hull ofP1, P₂,..., P_k inΠ. Then ∂C, the boundary ofC, is a convex polygon in Π. There are two vertices, sayP_i and P_j, consecutive in ∂C but not consecutive in G(otherwise all vertices ofGn are already inΠ). Then we can rotate Π along the line L passes P_i and P_j a very small angle so that except P_i and P_j, all vertices ofG_n are below Π (noteG_n is invariant when we rotate Π), see Figure 5. P_i and P_j divideG_n into two parts G^′ and G^′′, each part contains some point not in Π. Now we can repeat the same argument in the last graph to showEn^f(G_n) can not be the maximum.

P'

P''

'

L

Figure 5

We have proved that all vertices of Gn are in the same plane when En^f(Gn) reaches the maximum.

(ii) By (1), we assume now all vertices ofGn are in the plane Π. Suppose some vertex P^′ of G_n is in the interior of C (still refer Figure 5). Then again some line L^′ in Π (see Figure 5) contains an edge of C which divides Gn into two parts G^′ and G^′′, each part contains some point not inL^′. Since the position of edges of Gn

do not affectsEn^f(G_n), for convenience, we may assume thatG_n is inΠ. Reflect G^′ aboutL^′, we can repeat the same argument as in (i) to showEn^f(G_n)can not be the

maximum.

We have proved that all vertices ofG_n are vertices of the convex polygon.

Below we will still use A1, ..., An to replace P1, ...., Pk.

(iii) In the conclusion of (ii), the cyclic order of vertices in ∂C usually are not the same as the their cyclic order inG_n, see Figure 6. Also may beA_i=A_j on∂C, and C can be degenerated, see Figure 8. RecallE_n^α(Gn) relies only the positions of all vertices (counting the multiplicity) of Gn. To prove (iii), we first to prove the following

A₃

A₅ A₄ A₁ ^A2

A₃

A₅ A₄ A₁

A₂

Figure 6

Lemma 3.3. Suppose C is not degenerated. Then any two consecutive vertices in

∂C has distance no more than 1.

Proof. In this case ∂C is convex polygon as in Figure 7. Suppose Lis a maximum straight arc in ∂C. Then the two end points of L are vertices of ∂C. To prove the lemma, we need only to show that any two consecutive vertices inLhas distance no more than 1.

LetS be all vertices ofC inL. We first claim the points ofS are consecutive in Gn. Precisely, for any two verticesAiandAi+kare inL, then eitherAi+1, ..., Ai+k−1, orA_i+k+1, ...., Ai−1 must be inL. Otherwise we have someAj ∈ {Ai+1, ..., A_i+k−1} andA_l∈ {A_i+k+1, ...., A_i−1}, bothA_j andA_lare not inL. ThenA_iandA_i+kdivide Gn into G^′ and G^′′ with Aj ∈G^′ and Al∈ G^′′. Both of Aj and Al must be in one side ofL. Just reflectG^′ about the line containingL, we can argue as before to get En^f(G_n)is not the maximum.

A

^A

A

A

Figure 7

Now we prove that any two consecutive vertices inL has distance no more than 1. SupposeA_i andA_i+k are two consecutive vertices inL. We may supposeL is in

horizontal position and A_i is on the left of A_i+k, see Figure 7. By the claim in the last paragraph, either A_i+1, ..., A_i+k−1, or A_i+k+1, ...., A_i−1 must be in L. We may assume thatAi+1, ..., Ai+k−1 are in L. Letj be the minimal integer such that Ai+j

is not in the left side ofA_i+k,j= 1, ..., k. ThenA_i+j−1 must be in left side ofA_i+k. This implies thatA_i+j−1A_i+j coversA_iA_i+k. Thend(A_i+j−1, A_i+j)≤1implies that d(Ai, Ai+k)≤1. This finishes the proof of the lemma.

Remark3.4. In the last paragraph, we verified if the vertices ofG_non the maximum straight arc L in∂C are consecutive in Gn, then any two consecutive vertices has distance no more than 1. This fact will also be used for degenerated case.

Suppose C is non-degenerated and the vertices of Gn appear in ∂C consecu- tively as Q₁, Q₂, ..., Q_l with multiplicity q₁, q₂, ...., q_l, ∑_l

i=1q_i = n. By Lemma 3.3, d(Q_i, Q_i+1) ≤1. Then there is a non-extensible map g^′ :G_n → Π which sends the first q1 vertices A1,... ,A_q1 to Q1, the next q2 vertices Aq1+1, ..., Aq1+q2 to Q2,...., and the lastq_l vertices are sent toQ_l. ClearlyΓ_n, the image ofg^′, satisfies both (a) and (b) of (iii).

Suppose nowC is degenerated. As we discussed in the begin of this section, ∂C consists of two coincided straight arcsC₁ and C₂, and ∂C first travel first along C₁ then along C₂. So it makes sense to about the cyclic order in ∂C in degenerated case.

We may assume that A₁ at one end and A_k at another end. The right-up of Figure 8 illustrates howG₇ maps to C when we view C as a straight arc.

A

A

A

= A

_A

2

= A

A

= A

A

A

= A

A

A

A

= A

A

A

A

= A

A

A

A

A

A

A

A

A

A

Figure 8

Since C1 and C2 are coincide, we assume that the image of all vertices from A1

toA_k inG_nstay in C₁ and the image of all vertices from A_k+1 toA_n inG_n stay in C₂. The the right-middle of Figure 8 illustrates how G₇ maps to ∂C =C₁∪C₂ (in the figure we slightly bendC1 andC2 so that their interiors are disjoint).

Suppose the vertices in C₁ from A₁ toA_k appears as Q₁, Q₂, ..., Q_l with multi- plicityq₁, q₂, ...., q_l,∑_l

i=1q_i =k. Since all vertices in C₁ (resp. C₂) are consecutive in Gn, by Remark 3.4, we have d(Qi, Qi+1) ≤ 1 for i = 1, ..., l−1. So there is a non-extensible mapg^′ :G_n→Πwhich send the part of G_nfrom A₁ toA_k toC₁ so that the first q₁ vertices are sent toQ₁, the nextq₂ vertices are sent to Q₂,...., and

the lastq_l vertices are sent toQ_l. Theng^′ maps the vertices inG_n fromA_k+1 toA_n toC₂ in similar way. Clearly Γ_n, the image ofg^′, satisfies the conclusion of (iii).

The right-down of Figure 8 illustrates how G7 maps to C which preserves the cyclic orders.

(iv) By (iii) we may assume that the vertices of in∂C are in the cyclic orderA₁, A2,...,An.

A

A

A

A

A

A

A

A

A

A

Figure 9

Suppose the distance of two consecutive vertices in Γ_n, say A₁ and A₂, is less than 1, that is the unique edgee of Γn connecting A1 and A2 is not straight. Let A_i be the vertex such that d(A_i, A₁) is maximum. Then the angle ∠A_i−1A_iA_i+1 must be less thanπ (otherwise contradicts thatd(A_i, A₁) is maximum). Noweand Ai divide Γn into two parts G^′ and G^′′, G^′ contains A1 and G^′′ contains A2. Let G¹ be the union of G^′ and the segment A₁A_i and G² be the union of G^′′ and the segment A₂A_i. Now keep both G¹ and G² rigid. Then rotate slightly G² around Ai to increase the angle∠Ai−1AiAi+1 slightly but still less thenπ. We can do this since the unique edge e connecting G¹ and G² is not straight. Since each G¹ and G²are rigid, and the angle∠A_i−1A_iA_i+1 is increasing but still less thenπ, it is easy to see the distance for points in G¹ is not changed, the distance for points in G² is not changed, but for eachA_k in G¹, A_l inG², k, l ̸=i, the distance d(A_k, A_l) is increasing by using cosine theorem. SoEn^f(Γn) can not be the maximum.

We have proved (iv), that isΓ_n is a convex n-gon of edge length 1.

(v) The idea of the verification is easy:

1. Γ_n is a convex n-gonC with edge length 1, the length of ∂C isn.

2. ∂C is the unique shortest loop passing all vertices of Γ_n.

3. The vertices of Gn are coincided the vertices of Γn, and and Gn is a loop of lengthn.

SoGn must be coincided with∂C, and indeedGn must be coincided with Γnas a polygon.

We finish the verification of (v), therefore the proof of the proposition.

When Gn is a (1-dimensional) convex n-gon with edge length 1, we will denote G_nby Γ_n and Gby Γ.

4 Basic facts, the classification for n = 4

From now on, we always assume thatΓn is a convexn-gon with each edge of length 1 inR². We often use maxE_n^α to denote the maximum value ofE_n^α below.

SupposeΓ_n has verticesA₁, ...., A_n and the exterior angle atA_i isθ_i. There are two extreme shapes for Γ_n: one is the regular n-gon, denoted as Γ^o_n, which can be defined by θ1 =.... =θn; another the double straight arc, defined for onlyn = 2m , denoted asΓ⁻_n, which can be defined by eitherθ_i =θ_i+m =π for somei, or some diagonal has lengthm. Γ⁻_2m is shown in Figure 10.

A _i

A_i+5

Figure 10

Proposition 4.1. (1) (Jensen inequality) Suppose f is a convex function on [a, b], θi ∈[a, b]. Then ∑_n

i=1f(θ_i) n ≤f(

∑_n

i=1θ_i n ), and the equality holds if and only ifθ1 =θ2=...=θn.

(2) (Karamata inequality) Supposeg is a concave function on[a, b]and there are nvariablesx1, x2, ..., xn∈[a, b]with a fixed sum. Then the value ∑_n

i=1g(xi) reaches the maximum if and only if at least n−1 variables are at endpoints.

Lemma 4.2. f_α(x) is an increasing function; furthermore f_a(x) is convex when α <1 and is concave whena >1.

Proof. First calculate the first derivative of fα:

f_α^′(x) =





αx^α−1, α >0;

1/x, α= 0;

−αx^α−1, α <0.

f_α^′ is always positive, hence f^α is an increasing function.

Then calculate the second derivative of fα

f_α^′′(x) =





α(α−1)x^α−2, α >0;

−1/x², α= 0;

−α(α−1)x^α−2, α <0.

f_α^′′ is negative when α <1, hence f_α is convex whenα <1. f_α^′′ is positive when α >1, hence fα is concave when α >1.

The following decomposition ofE_n^α plays significant roles in this note.

E_n^α(Γn) =

[n/2]∑

k=1

µn,kE_n,k^α (Γn), (4.1)

where

E^α_n,k(Γn) =

∑n i=1

fα(|Ai−A_i+k|). (4.2)

whereµn,k= 1/2 ifnis even andk=n/2 and = 1 for the remaining cases.

Figure 11

In Figure 11, the interactions of E_n,k^α along the black lines for (n, k) = (6,1),(7,1), along the blue lines for (n, k) = (6,2),(7,2), and along the red lines for(n, k) = (6,3),(7,3).

Note E_n,1^α (Γn) is a constant. Precisely

E_n,1^α (Γn) =

∑n i=1

f^α(|Ai−Ai+1|) =





n, α >0;

0 α= 0;

−n, α <0.

(4.3)

Some times it is more brief just to considerE¯_n^α(Γn) =∑[n/2]

k=2 E_n,k^α (Γn).

Example 4.3. We will classify whenΓ4 realizing maxE₄^α.

As we just discussed, E₄^α(Γ4) =E_4,1^α (Γn) +E_4,2^α (Γn) and E_4,1^α (Γn) is a constant for given(n, α). So we need only to classify whenΓ₄ realizing maxE_4,2^α = ¯E₄^α.

A

A

A

A

Figure 12

Let the inner angle atA₁ beϕ. Then Γ₄ is determined by ϕ, and we have E_4,2^α (Γ₄) =|A₁A₃|^α+|A₂A₄|^α

= (2cosϕ/2)^α+ (2sinϕ/2)^α

= 2^α(cos²ϕ/2)^α/2+ (sin²ϕ/2)^α/2)

= 2^α(t^α/2+ (1−t)^α/2) wheret=cos²ϕ/2.

Notef_α is a convex function ifα <1and a concave function ifα >1by Lemma 4.2 .

If α <2, then α/2<1, we can apply by Jenson inequality to get that E_4,2^α (Γ4) reached the maximum if and only ift= 1/2, that is cos²ϕ/2 = 1/2, that isϕ=π/2 and therefore E_4,2^α (Γ₄) reaches the maximum if and only if Γ₄ = Γ^o_n. Moreover E_4,2^α (Γ4) = 2^α(¹₂^α/2+ ¹₂^α/2) = 2^α/2+1 if α > 0, E^α_4,2(Γ4) = −2^α/2+1 if α < 0 and E₄⁰(Γ^o_n) =ln2.

If α > 2, then α/2 > 1, we can apply Karamata inequality to get E_4,2^α (Γ₄) reached the maximum if and only if t = 0 or 1, that is cos²ϕ/2 = 0 or 1, that is ϕ = 0 or π, and therefore E_4,2^α (Γ4) reaches the maximum if and only if Γ4 = Γ⁻_n. Moreover E_4,2^α (Γ⁻₄) = 2^α.

When α= 2, then α/2 = 1, andE₄^α(Γ₄) is constant 8.

By the discussion above and (4.3) we have the following classification for n= 4:

maxE₄^α(Γ4) =













4 + 2^α, Γ4 = Γ⁻₄ α >2;

8 any Γ₄ α= 2

4 + 2^α2+1 Γ₄ = Γ^o₄ 0< α <2;

ln2, Γ4 = Γ^o₄ α= 0.

−4−2^α2+1 Γ₄ = Γ^o₄ α <0.

(4.4)

Example 4.4. This example provides some solutions of E_n^αn(Γ^o_n) = E^α_nⁿ(Γ⁻_n) for n= 6,8.

E₆²(Γ^o₆) = 36>33 =E₆²(Γ⁻₆) E₆³(Γ^o₆)<63<67 =E₆³(Γ⁻₆).

E₆^α6(Γ^o₆) =E₆^α6(Γ⁻₆) for someα6 ∈(2.5525,2.5529).

E₈²(Γ^o₈)>109>97 =E₈²(Γ⁻₈) E₈³(Γ^o₈)<243<248 =E₆³(Γ⁻₆).

E₈^α8(Γ^o₈) =E₈^α8(Γ⁻₈) for someα₈ ∈(2.878,2.879).

5 When the regular n-gon Γ

realizing maxE

5.1 Γ^o_n realizing maxE_n^α if Γ_n not bending fast for α≤1 Let[x]be the maximum integer not bigger thanx.

Theorem 5.1. Suppose the sum of any consecutive [n/2]−1 exterior angles of Γ_n is no more thanπ. Then E_n^α(Γn) is the maximum if and only ifΓn= Γ^o_n, the regular n-gon of edge length 1 forα≤1.