# Equilibrium, Regular Polygons, and Coulomb-Type Dynamics in Different Dimensions

Authors: W. I. Skrypnik
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

The equation of motion inℝd of n generalized point charges interacting via the s-dimensional Coulomb potential, which contains for d=2 a constant magnetic field, is considered. Planar exact solutions of the equation are found if either negative n−1>2 charges and their masses are equal or n=3 and the charges are different. They describe a motion of negative charges along identical orbits around the positive immobile charge at the origin in such a way that their coordinates coincide with vertices of regular polygons centered at the origin. Bounded solutions converging to an equilibrium in the infinite time for the considered equation without a magnetic field are also obtained. A condition permitting the existence of such solutions is proposed.


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## Body

## 1. Introduction

In this paper, we find solutions of the equation of motion inℝd of n generalized point charges (or simply charges) ej∈ℝ,j=1,⋯n with masses mj>0 interacting via the pair s-dimensional Coulomb potential, which is proportional to the fundamental solutions of the s-dimensional Laplacian [1]. In the case d=2, the equation also contains a constant magnetic field hc directed along the perpendicular to the plane containing the charges. The charge coordinates xj∈ℝd obey the following equation of motion:
(1)mjd2xjdt2=−∂Uxn∂xj+δd,2ejẋj×hc,j=1,⋯,n>2,where
(2)Uxn=∑1≤j<k≤nejekφxj−xk,φx=1−δ2,sx2−s+δ2,slnx,2≤s∈ℤ+,xn=x1,⋯,xn∈ℝdn,x2=x12+⋯+xd2.ẋ=dxj/dt and the symbol × determines the skew-symmetric product. For d=2, one has ẋ×hc1=hẋ2, ẋ×hc2=−hẋ1,h∈ℝ, and δd,2 is the Kronecker symbol. The systems (1) and (2) without a magnetic field and their dynamics in the general case, case s=3, and case d=3,s=3 we call d,s-Coulomb, d-dimensional Coulomb, and Coulomb, respectively.Equation (1) is rewritten as follows:
(3)mjd2xjdt2=g∑k=1,k≠jnejekxj−xkxj−xks+δd,2ejẋj×hc,j=1,⋯,n,g=s−21−δ2,s−δ2,s.The most important among the systems are those that describe the electrodynamics, that is,d=1,2,3, s=2,3, and the sequence of ej,j=1,⋯,n contains numbers with different signs for them. We shall consider the systems with n−1 equal negative charges −e0 and a positive charge en.The profound result concerning the collision set for the dynamics determined by (3) without the magnetic field is obtained in [2] (see Remark 13).In this paper, we find the regular polygon equilibrium, exact regular polygon solutions ford=2, and bounded solutions converging to the equilibrium in the infinite time limit. The regular polygon equilibrium is created by the system of n−1 equal negative charges e0 located at all n−1 vertices of a regular polygon centered at the origin which is occupied by the central positive charge en=en0. It is given in Theorem 7 and (32) showing that en0=2−1e0n−2 for s=2 and e30=21−se0. For n=3, the regular polygon is represented by two opposite points on a circle.This equilibrium allows one to find easily the exact planar regular polygon solutions. They describe an identical motion of every negative charge around the origin where the positive charge is immobile ifgen−en0>0. This condition induces a Keplerian-type motion of the negative charges and the Keplerian motion of them for the Coulomb systems (s=3). Besides, the negative charges are tied to vertices of regular polygons centered at the origin all the time. In the simplest case, the negative charges rotate around the positive immobile charge with the same frequency. This is described by the following solution xkt∈ℂ of the equation of motion for negative charges (we consider ℝ2 as ℂ)
(4)xkt=zkeiutrt,rt>0,where zk,ut,rt satisfy the first, second, and third structure equations, respectively, the first of which does not depend on time. Here, zk are coordinates of the regular polygon vertices and the equation for u means conservation of the kinetic moment of a two-dimensional mechanical system. The third structure equation is its equation of motion for the radial variable. Earlier, we obtained the same result for the Coulomb systems with the zero magnetic field [3].For the system of the three charges, we prove the following theorem.Theorem 1.
Let (3) with hc=0,n=3 determine the dynamics of the three point charges e1=−e0,e2=−e0,e3=21−se0>0 with masses mj,j=1,2,3. Then, it possesses a bounded at positive time solution which is a real analytic function in a neighborhood of the origin in one real parameter for s>2 and d−1 real parameters for s=2,d>1 such that x−x0λ<∞, ẋλ<∞, where x0 is the equilibrium x101=−a,x201=a>0,x301=0,xj0α=0,1<α≤d,1≤j≤3, and λ<λ0,λ02=e02∣g∣s−12a−sm1−1+m2−1+4m3−1, ẋ is the velocity, and
(5)xλ=supt≥0maxj∈1,⋯,neλtxjt0,xjt0=max1≤α≤dxjαt.We were able to prove this theorem since we found all eigenvalues of the matrixM−1U0j,α;j,α, where M is the diagonal matrix in ℝdn with elements Mj,α=mj, 1≤j≤n, and 1≤α≤d and Uj,α;l,β0 is the matrix of the second partial derivatives Uj,α;j,β of the potential energy at the regular polygon equilibrium
(6)Uj,α;l,β=∂2Uxn∂xjα∂xlβ.We prove also the following theorem.Theorem 2.
LetU in (2) possess an equilibrium x0 and TrM−1U0≤0, where
(7)TrM−1U0=∑j=1nmj−1∑α=1dUj,α;j,α0.
Then, there exists an integerp, 1≤p<nd, a positive number λ, and a bounded at positive time solution xt of the equation of motion (3) with hc=0 depending on p real parameters, which is real analytic function in them in a neighborhood of the origin, such that x−x0λ<∞,ẋλ<∞.Ifs=d, then TrM−1U0=0 for an arbitrary equilibrium for our systems without a magnetic field since
(8)∑α=1dUj,α;j,α=0,Uj,α;j,β=gej∑k=1,k≠jnek−δα,βxj−xks+sxjα−xkαxjβ−xkβxj−xks+2.In our calculations of partial derivatives of the potential energy, we use the vector-valued equalities(9)∂∂x1x1−x2−k=−kx1−x2x1−x2k+2,xj∈ℝd,∂∂x1lnfx1=f−1x1∂∂x1fx1,∂∂x1ln∣x1−x2∣=x1−x2x1−x22,∂∂xjUxn=−gej∑k=1,k≠jnekxj−xkxj−xks.Theorem2 implies the following result.Theorem 3.
LetU in (2) possess an equilibrium x0 and s=d. Then, the conclusion of Theorem 2 is true.In order to estimateTrM−1U0 in the case s≠d, one has to consider a particular equilibrium. We shall choose in the next theorem the regular polygon equilibrium for n charges (see Corollary 8).Theorem 4.
For the regular polygon equilibrium forn−1 equal negative charges −e0 and central positive charge en0, the following equality is true:
(10)TrM−1U0=e02asgd−sm′e0−1en0−en,s+mn−1n−1e0−1en0,where m′=∑l=1n−1ml−1,
(11)en,s=2−s/2∑k=1n−21−cos2πkn−1−s/2=2−s∑k=1n−2sin−sπkn−1,en,s−2=2e0−1en0.An easy calculation at the end of the fourth section shows thatTrM−1U0<0 for s>d if n=3,4,5 (see Remark 14).Notethat the condition of the neutralityen=e0n−1is impossible for the regular polygon equilibrium fors=2.The conclusion of Theorem3 is true for the 3-dimensional Coulomb systems of n+2 and n+1 charges n−1 of which are equal considered in [4, 5], respectively. In their equilibria, the equal negative charges are located at all the vertices of a regular polygon centered at the origin while the rest nonplanar charges are located at the perpendicular crossing the origin. In [4], a positive charge occupies the origin and the nonplanar charges are equal and may have different signs. In [5], the nonplanar charges are equal and positive. The solutions of the equation of motion are found in these papers provided the masses of the equal negative charges are equal.Theorem2 relies on the following basic theorem proven in [6].Theorem 5.
LetU0 be the symmetric matrix of second-order partial derivatives of U at an equilibrium x0 Let also U be a real analytic function in a neighborhood of x0 and the matrix M−1U0 have p negative eigenvalues σj,j=1,⋯,p. Then, (1) with hc=0 possesses a bounded at positive time solution xjt,1≤j≤n depending on p real parameters which is a real analytic function in them in a neighborhood of the origin and x−x0λ<∞, ẋλ<∞, where ẋ is the velocity and
(12)xλ=supt≥0maxj∈1,⋯,neλtxjt,λ<λ0=minj=1,⋯,p−σj.Ford>1, the pair potentials and potential energies in (2) are real analytic functions in a neighborhood of aj∈ℝd,1≤j≤n,aj≠ak, i.e., a set where coordinates do not coincide (aj may correspond to an equilibrium). This follows from the fact that the Taylor power expansion for x−s converges absolutely and uniformly in a neighborhood of a point from ℝ+. In the one-dimensional case, the chosen order of charges is preserved by the dynamics close to the equilibrium and one can change xj−xk2−s,ln∣x∣ by the real analytic functions xj−xk2−s,lnxj−xk,xj>xk. So one can apply Theorem 5 to (1) and (2).Theorem1 follows from Theorem 12 and Theorem 5. From the equality MM−1U0M−1=M−1U0M−1, it follows that the matrix M−1U0 has the same spectrum as the symmetric matrix M−1U0M−1 and is similar to the diagonal matrix with real elements. This leads to the important conclusion: if TrM−1U0≤0, then there exists at least one negative eigenvalue of M−1U0 since the trace coincides with the sum of the eigenvalues of the matrix. This and Theorem 5 prove Theorem 2.Our paper is organized as follows. In the second section, we introduce the three structure equations and explain in Proposition6 how to construct with its help the exact solutions of (1) when there are equilibrium configurations in the form of regular polygons. Theorem 7 establishes their existence. In the third section, we consider the three charge cases and show that the exact solutions exist for which the charges are located at a common line provided the positive charge is greater than the absolute values of the rest two charges. In the fourth section, we prove Theorem 4. Eigenvalues of M−1U0 for the system of charges from Theorem 1 are calculated in the fifth section. This result is formulated in Theorem 12.


## 2. Equilibrium and Exact Planar Dynamics

Letd=2 and xk=xk1+ixk2∈ℂ. Then, the analogue of the exact Lagrange solutions of the three-body problem [7] is determined by the equalities xk=zkqt which give
(13)xj−xk=zj−zkq.From the equation of motion (3) with the contribution of magnetic field −iejhẋj, it follows that
(14)mjzjd2qdt2=gqq−s∑k=1,k≠jnejekzj−zkzj−zks−izjejhq̇.Sometimes, thes-dimensional Coulomb potential which leads to this equation is mentioned as the Riesz potential [8] if s>2 (see also [9]).Let(15)qt=eiutrt,rt>0and the following first structure equation hold
(16)−w2mjzj=g∑k=1,k≠jnejekzj−zkzj−zks,w2>0.The first structure equation reduces the equation of motion to the following single equation:(17)d2qdt2=−w2qq−s−ibq̇,b=e0hm.Letq=eiutrt, where u,r are real valued functions. Then,
(18)e−iutd2qdt2=d2rdt2+i2drdtdudt+d2udt2r−dudt2r,e−iutq̇=iu̇r+ṙ.The imaginary part of (17) multiplied by e−iu has to be zero
(19)2drdtdudt+d2udt2r+bdrdt=r−1ddtr2dudt+b2r2=0.As a result, the equation of motion yields the following second and third structure equations:(20)I=r2dudt+r2b2=η,d2rdt2−dudt2r−bdudtr+w2r1−s=0,where η∈ℝ is a constant. For b=0, they look like
(21)dudt=r−2η,d2rdt2−η2r−3+w2r1−s=0.The third structure equation is the radial Kepler’s equation fors=3. Note that if b=0,r=1, then the last equalities imply u=±tw. In the general case, if r=1, then from (17), it follows that
(22)qt=eiw′t,w′2=w2+bw′,w′=b2±b24+w2.Equation (17) for q is the equation of motion of a two-dimensional mechanical system in a constant magnetic field characterized by the kinetic moment I. Let us prove that it is defined also by
(23)I=q×q̇+br22and that it is an integral of motion. The skew-symmetric product is determined on C2 as follows:
(24)z=x+iy,q=p+iv,z×q=xv−yp,eiux×eiup=cosu+isinux×cosu+isinup=sinucosuix×p+x×ip=sinucosu−xp+xp=0,eiuiy×eiup=cosu+isinuiy×cosu+isinup=sin2uy×ip+cos2uiy×p=−sin2uyp−cos2uyp=−yp.That is(25)q×q̇=−eiuiu̇r+ṙ×eiur=u̇r2.As a result,(26)İ=q̇×q̇+q×q¨+brṙ=brṙ+q×−w2qq−s−ibq̇=brṙ−w2q−sq×q+biq̇×q=brṙ+biq̇×q=brṙ−bu̇eiur×eiur+beiuiṙ×eiur=brṙ+beiuiṙ×eiur=0.Now, we have to prove that solutions of the first structure equation exist. A crucial role in our proofs is played by the equality for the coordinates of the regular polygon vertices(27)∑k=1n−1zk=0,zk=zk,z=ei2π/n−1,which in its turn is a consequence of the equality
(28)z∑k=1n−1zk=∑k=1n−1zk.Proposition 6.
Let the equilibrium forU in (2) be determined by the positive charge en0>0 with the coordinate zn=0 and negative charges ej0=−e0<0,j=1,⋯,n−1 with the equal mass m be placed at all vertices zj of a regular polygon and ∣zn−zj∣=a if n>3. Let also z1=−z2=a if n=3. Then, the first structure equation is true if
(29)w2asme0=gen−en0>0.Proof.
Fromzn=0, it follows that for j<n the right-hand side of the first structure equation looks like
(30)∑k=1,k≠jnejekzj−zkzj−zks=∑k=1,k≠jnej0ek0zj−zkzj−zks−en−en0e0a−szj=−en−en0e0a−szj.
Fromzn=0 and (27), it follows that the first structure equation is true for j=n(31)∑k=1,k≠jnejekzj−zkzj−zks=−a−se02∑k=1n−1zk=0.
The proposition is proved.From (29), it follows that en>en0 and en<en0 for s>2 and s=2, respectively.Theorem 7.
Let the positive chargeen0>0 have the coordinate zn=0, negative charges ej0=−e0<0,j=1,⋯,n−1 be placed at all vertices zj of a regular polygon and ∣zn−zj∣=a if n>3. Let also z1=−z2=a if n=3. Then, such the configuration is an equilibrium for U in (17) for d=2 if
(32)en0=2−s2e0∑k=1n−21−cos2πkn−12−s/2=21−se0∑k=1n−2sin2−sπkn−1.Proof.
Forj=n, the equilibrium condition is satisfied
(33)∑k=1n−1en0ek0zn−zkzn−zks=e0en0a−s∑k=1n−1zk=0.
Forj<n, it yields
(34)∑k=1,k≠jnej0ek0zj−zkzj−zks=e02∑k=1,k≠jn−1zj−zkzj−zks−e0en0aszj=−zja−se0en0−e0∑k=1,k≠jn−1aszj−zks−e02∑k=1,k≠jn−1zkzj−zks=0.
If one shows that(35)∑k=1,k≠jn−1aszj−zks=an,∑k=1,k≠jn−1aszkzj−zks=an′zj,where
(36)an=∑k=1n−211−zks,an′=∑k=1n−2zk1−zks,z=ei2π/n−1,then the following equality is proved:
(37)en0=e0an−an′.
To prove these equalities, one has to prove for coordinateszj of the regular polygon such that ∣zj∣=1(38)∑k=1,k≠jn−11zj−zks=an,(39)∑k=1,k≠jn−1zkzj−zks=an′zj.
To prove (39) ((38) is proved the same way), we have to take into account that zk=zk,z=ei2π/n−1, z−l=zn−1−l(40)∑k=1,k=jn−1zkzj−zks=zj∑k=1,k=jn−1zk−j1−zk−js=zj∑k=1j−1zk−j1−zk−js+∑k=j+1n−1zk−j1−zk−js=zj∑l=1j−1z−l1−z−ls+∑l=1n−j−1zl1−zls=zj∑l=1j−1zn−l−11−zn−l−1s+∑l=1n−j−1zl1−zls=zj∑l=1n−2zl1−zls.
Hence, the multiplier ofzj does not depend on j and it is real since
(41)∑l=1n−2zl1−zls∗=∑l=1n−2z−l1−z−ls=∑l=1n−2zn−l−11−z−n−l−1s=∑l=1n−2zl1−zls.
Hence, (29) yields
(42)en0=e0∑k=1n−21−Rezk1−zks=2−s2e0∑k=1n−21−cos2πkn−1−s−2/2
sinceRezk=cos2πk/n−1,(43)1−zk2=1−2cos2πkn−1+cos22πkn−1+sin22πkn−1=21−cos2πkn−1.
But1−cos2x=2sin2x. This concludes the proof.LetU correspond to the d,s-Coulomb system and
(44)xk=xkα,α=1,⋯,d=gk,x′k,gk∈C,x′k=xk2,⋯,xkd∈Rd−2,gk1=xk1=Regk,gk2=xk2=Imgk.From(45)∂∂xjαUxn=−gej∑k=1,k≠jnekxjα−xkαxj−xks,α=1,⋯,d,it follows
(46)∂∂xjαUxn=0,α>2,xjα=0and the equilibrium is found from the equality
(47)∂∂gjUxn=∂∂gjUgn,xjα=0,α>2.Hence, we proved the corollary with the help of Theorem7.Corollary 8.
The regular polygon equilibrium determined in Theorem7 exists for the d,s-Coulomb system in ℝd in the plane xα=0, α=3,⋯.,d for d≥2.


## 3. Three Nonequal Charges on a Plane

In this section, we shall consider the first structure equation for three chargesej<0,j=1,2,e3>0. We will assume that their inertia center is immobile: m1z1+m2z2+m3z3=0. We treat zj as points of ℝ2. Note that we assume that en>en0 for s>2 and en<en0 for s=2.Proposition 9.
Solutions of the first structure equation forn=3 do not coincide with the coordinates of vertices of a triangle.Proof.
It is possible to put the first charge at a coordinate axis. Letz12=0. Then, the first structure equation for z12 gives
(48)e2z22z1−z2−s+e3z32z1−z3−s=0.
Since the center of inertia is immobile, this equality results in(49)z22∣e2∣m2z1−z2−s+e3m3z1−z3−s=0.
That is,z22=0 and z32=0. This concludes the proof.Proposition 10.
Ife1=e2=−e0,m1=m2=m0,2s−1e3>e0, then the rectilinear solution of the first structure equation for n=3 is given by zj2=0,z31=0,−z11=z21=2−1a where
(50)w2as=g2se0m0e3−21−se0>0.Proof.
Let the charges are placed at a first coordinate axis:zj2=0, q1=z11<q3=z31<q2=z21 and q2−q1=a,q3−q1=aρ,q2−q3=aσ, a>0. Then, the first structure equation leads to
(51)−e3ρa1−s−e2a1−s=−m1ge1w2q1,e1ρa1−s−e2σa1−s=−m3ge3w2q3,e1a1−s+e3σa1−s=−m2ge2w2q2.
Letρ=σ=1/2, then the second equation gives q3=0. The first and the third ones show that −q1=q2=a/2 and a,w satisfy the condition indicated in the formulation of this proposition. This concludes the proof.Note that if one multipliesa by 2 in Proposition 9, then w in it and w in Proposition 6 coincide since e30=21−se0. If the conditions of the last proposition do not hold, that is, e1≠e2, then one can apply the following proposition.Proposition 11.
Ife3≥−ej>0,j=1,2, then a rectilinear solution of the first structure equation exists.Proof.
Let us putm∗=m1+m2+m3. From the equality m1z1+m2z2+m3z3=0, it follows that
(52)m∗q1+m3ρa+m2a=0,m∗q2−m3σa−m1a=0.
Taking this into account, we see that the first and the third equalities look like (q1,q3 are excluded)
(53)e3ρ1−s+e2=−m∗−1m1ge1w2asm3ρ+m2,e3σ1−s+e1=−m∗−1m2ge2w2asm3σ+m1.
Sinceσ=1−ρ,
(54)−m∗−1w2as=ge1m1e3ρ1−s+e2m3ρ+m2=ge2m2e31−ρ1−s+e1m31−ρ+m1,0<ρ<1.
As a result in order to solve the first and third equations from the previous proposition, one has to show thatf=0 has solutions in the interval 0<ρ<1 such that e3ρ1−s>∣e2∣, e31−ρ1−s>∣e1∣ if e1,e2<0,e3>0 and
(55)fρ=e1m1e3ρ1−s−∣e2∣m3ρ+m2−e2m2e31−ρ1−s−∣e1∣m31−ρ+m1.
Ife3≥−ej>0,j=1,2, then the equation f=0 has such the solutions since f is a continuous function tending to −∞, ∞ at 0, 1, respectively. q3 is found from the second structure equation. This concludes the proof.


## 4. Equilibrium Character

In this section, we calculate the trace(56)TrM−1U0=∑j=1nmj−1∑α=1dUj,α;j,α0,where Uj,α;l,β0 is the matrix of the second derivatives of the potential energy at the regular polygon equilibrium created by the system of n−1 equal negative charges located at all n−1 vertices of a regular polygon centered at the origin where the central positive charge en0 is located.We have(57)Un,α;n,β=∂2Uxn∂xnα∂xnβ=−ge0en0∑k=1n−1−δα,βxn−xks+sxnα−xkαxnβ−xkβxn−xks+2.Further(58)Uj,α;j,β=ge02∑k=1,k≠jn−1−δα,βxj−xks+sxjα−xkαxjβ−xkβxj−xks+2−e0en0−δα,βxj−xns+sxjα−xnαxjβ−xnβxj−xns+2.As a result,(59)Uj,α;j,α=ge02∑k=1,k≠jn−1−1xj−xks+sxjα−xkα2xj−xks+2−e0en0−1xj−xns+sxjα−xnα2xj−xns+2,∑α=1dUj,α;j,α=ge02d−s−∑k=1,k≠jn−11xj−xks+e0en01xj−xns.For the equilibrium, we have(60)xj−xn=a,xj=a,xj=azj,z=ei2π/n−1∈ℂ.Then, one derives(61)∑α=1dUj,α;j,α0=gass−d∑k=1n−2e0211−zks−e0en0.Besides, we have(62)Un,α;n,α=−ge0en0∑k=1n−1−1xn−xks+sxnα−xkα2xn−xks+2,∑α=1dUn,α;n,α=ge0en0d−s∑k=1n−11xn−xks,∑α=1dUn,α;n,α0=e0en0asgd−sn−1.This gives(63)TrM−1U0=e02asgd−sm′e0−1en0−∑k=1n−211−zks+mn−1n−1e0−1en0.Now, let us apply the equalities(64)1−zk2=1−2cos2πkn−1+cos22πkn−1+sin22πkn−1=21−cos2πkn−1,∑k=1n−211−zks=2−s/2∑k=1n−21−cos2πkn−1−s/2,cos2j=121+cos2j,1−cos2j=21−cos2j=2sin2j,∑k=1n−21−cos2πkn−1−s2=2−s2∑k=1n−2sin−sπkn−1.These equalities and (63) prove (10).Forn=3, s=3,d=1, we have en0=4−1e0,e3,s=2−s=2−2e3,s−2(65)TrM−1U0=−e02a34−1m1−1+m2−1+m3−1.Forn=3, s=3,d=2, we have
(66)TrM−1U0=−e022a34−1m1−1+m2−1+m3−1.The last two equalities are derived in [6] with the help of the found eigenvalues of M−1U0 (see also Theorem 12).Fromsinπ/3=sin2π/3=√3/2, it follows
(67)e4,s=2−s+123s,e4,s−2=3e4,s.sinπ/4=sin3π/4=1/√2 implies
(68)e5,s=2−s1+21+s/2=2−s+21−s/2,e5,s−2=22−s+22−s/2>2e5,s.We also have(69)e30=2−s+1e0,e3,s=2−s,n=3.Equation (10) gives
(70)TrM−1U0=e022asgd−sm1−1+m2−1+4m3−1.This equality follows also from Theorem12.


## 5. Eigenvalues ofM−1U0 for Three Charges

The simplest example of a Coulomb system with equilibrium is thed-dimensional system of the three-point charges e1=−e0,e2=−e0,e3=21−se0>0 with masses m1,m2,m3 and the potential energy in (2) for n=3. Its equilibrium x0 is determined by x101=−a,x201=a,x301=0,xj0α=0,j=1,2,3,1<α≤d.The main result of this section is the proof of the following theorem.Theorem 12.
In the one-dimensional system,M−1U0 has the doubly degenerate zero eigenvalue and the eigenvalue −m1−1+m2−1+4m3−1gu′,u′=s−1e02/2as. In the d-dimensional systems, d>1, M−1U0 has the zero eigenvalue, which is 2d times degenerate and the eigenvalues −m1−1+m2−1+4m3−1gu′, m1−1+m2−1+4m3−1s−1−1gu′, the latter of which is d−1 times degenerate.Proof.
We find eigenvalues ofU0 at first for the one-dimensional case. We have
(71)∂∂xjUx3=−gej∑k=1,k≠j3ekxj−xkxj−xks,that is
(72)g−1∂∂x1Ux3=−e02x1−x2x1−x2s+e0e3x1−x3x1−x3s,g−1∂∂x2Ux3=−e02x2−x1x1−x2s+e0e3x2−x3x2−x3s,g−1∂∂x3Ux3=e0e3x3−x1x1−x3s+x3−x2x2−x3s.
The equality∂/∂x3Ux3=0 holds for x1=x10=−a,x2=x20=a,x30=0. This configuration is an equilibrium. This follows also from the equalities ∂/∂xjUx3=0,j=1,2.
The second derivatives of the potential energy are calculated as follows:(73)∂Ux3∂xj∂xk=∂Ux3∂xk∂xj=−gs−1ejekxj−xk−s,k≠j,∂2∂xj2Ux3=gs−1ej∑k=1,k≠j3ekxj−xk−s.
Hence, the second derivatives of the potential energy at the equilibriumUj,k0 are given by
(74)U1,20=U2,10=−gs−1e022sas=−gu′,U3,10=U1,30=U2,30=U3,20=2gu′,U1,1=U2,2=−gu′,U3,3=−4gu′.
That is(75)U0=−gu′11q11qqqq2=u′U′,q=−2.
Let us put(76)M′0λ,q=k1−λk1qk1k2k2−λqk2qk3qk3q2k3−λ,kj=mj−1.
Then, taking into consideration(77)M−1U0−λI=−gu′M0′−λgu′,q,−DetM−1U0−λI=g3u′3DetM′0−λgu′,q,q=−2and making the expansion of the determinant in the elements of the first row of M′0, we obtain
(78)DetM′0λ,q=k1−λk2−λq2k3−λ−q2k2k3−k1k2q2k3−λ−q2k2k3+qk1qk2k3−qk3k2−λ=k1−λλ2−λk2+q2k3+λk1k2+λq2k1k3=λk1−λλ−q2k3−k2+k1k2+q2k1k3.
Hence,(79)DetM0′λ,q=λ2−λ+k1+k2+q2k3,(80)DetM−1U0−λI=−λ2λ+m1−1+m2−1+4m3−1gu′.
The last formula proves the theorem for the one-dimensional case.
Let us consider the two-dimensional case. For the first partial derivatives of the planar potential energy, we obtain(81)∂∂x1αUx3=g−e02x1α−x2αx1−x2s+e0e3x1α−x3αx1−x3s,∂∂x2αUx3=g−e02x2α−x1αx1−x2s+e0e3x2α−x3αx2−x3s,∂∂x3αUx3=ge0e3x3α−x1αx1−x3s+x3α−x2αx2−x3s.
The last equality is zero at the equilibrium−x11=x21=a,x32=x31=x12=x22=0. The first two give the equilibrium relation e3=21−se0.
The second derivatives of the potential energy are given by(82)∂Ux3∂x1α∂x2β=∂Ux3∂x2β∂x1α=ge02δα,βx1−x2s−sx1α−x2αx1β−x2βx1−x2s+2,α,β=1,2,∂Ux3∂xkα∂x3β=∂Ux3∂x3β∂xkα=−ge0e3δα,βxk−x3s−sxkα−x3αxkβ−x3βxk−x3s+2,k,α,β=1,2,∂2Ux3∂xjβ∂xjα=ge02−δα,βx1−x2s+sx1α−x2αx1β−x2βx1−x2s+2+e0e3δα,βxj−x3s−sxjα−x3αxjβ−x3βxj−x3s+2,j,α,β=1,2,∂2Ux3∂x3β∂x3α=ge0e3δα,βx1−x3s−sx1α−x3αx1β−x3βx1−x3s+2+δα,βx2−x3s−sx2α−x3αx2β−x3βx2−x3s+2.
For the matrix of the second derivatives at the equilibrium, we derive(83)U1,α;1,β0=U2,α;2,β0=ge02δα,β−12as+21−sas+sδα,1δβ,112as−21−sas=ge022asδα,β1−sδα,1δβ,1=4−1U3,α;3,β0,U1,α;2,β0=U2,β;1,α0=ge022asδα,β1−sδα,1δβ,1,Uk,α;3,β0=U3,β;k,α0=−g2e022asδα,β1−sδα,1δβ,1,k,α,β=1,2.
Let us introduce the renumeration(84)1,1=1,2,1=2,3,1=3,1,2=4,2,2=5,3,2=6,m4=m1,m5=m2,m6=m3,where the first and second numbers in the round brackets correspond to the lower and upper indices of variables. Then, Uj,k0=Uk,j0=0,j≤3,k≥4 and
(85)U1,10=U2,20=4−1U3,30=−gu′,U1,20=−gu′,U1,30=U2,30=2gu′,u′=s−1e022as,U4,40=U5,50=4−1U6,60=gc′,U4,50=gc′,U4,60=U5,60=−2gc′,c′=s−1−1u′.
This means(86)U0=u′U′⊕−c′U′,where U′ is given by (75).
LetM′′=M′⊕M′ and M′ be the 3×3 diagonal matrix with the elements m1,m2,m3. Then,
(87)M′′−1U0−λI=−gu′M′0−λgu′,−2⊕gc′M′0λgc′,−2,DetM′′−1U0−λI=−u′3c′3DetM′0−λgu′,−2DetM′0λgc′,−2.
From this equality and (79), we derive
(88)−DetM−1U0−λI=λ4−λ+m1−1+m2−1+4m3−1gc′λ+m1−1+m2−1+4m3−1gu′.
This concludes the proof for the two-dimensional case.All the formulas concerning partial derivatives of the potential energy of this sections will be true if one extends the previous conditionα,β=1,2 to the condition α,β=1,2,3,⋯,d.Letd=3 and let us continue the renumeration of the variables:
(89)1,3=7,2,3=8,3,3=9,m7=m1,m8=m2,m9=m3.It is not difficult to see thatUj,k0=Uk,j0=0,j≤6,k≥7 and U7,70=U8,80=4−1U9,90=gc′,U7,80=gc′,U7,90=U8,90=−2gc′. Hence,
(90)U0=U′′⊕−c′U′,where U′′ coincides with the planar U0. Moreover,
(91)=M′′⊕M′=M′⊕M′⊕M′=⊕M′3,M−1U0=M′′−1U′′⊕−c′M′−1U′.As a result,(92)−DetM−1U0−λI=λ6−λ+m1−1+m2−1+4m3−1gc′2λ+m1−1+m2−1+4m3−1gu′.For the general case, we generalize the above renumeration as follows:(93)1,d=3d−1+1,2,d=3d−1+2,3,d=3d−1+3,m3d−1+1=m1,m3d−1+2=m2,m3d−1+3=m3.It is not difficult to see thatUj,k0=Uk,j0=0,j≤3d−1,k≥3d−1+1 and Ul+1,l+10=Ul+2,l+20=4−1Ul+3,l+30=gc′,Ul+1,l+20=gc′,Ul+1,l+20=Ul+2,l+30=−2gc′,l=3d−1.LetM=⊕M′d, then these equalities imply
(94)U0=u′U′⊕−c′U′d−1,M−1U0=u′M′−1U′⊕−c′M′−1U′d−1,(95)−DetM−1U0−λI=λ2d−λ+m1−1+m2−1+4m3−1gc′d−1λ+m1−1+m2−1+4m3−1gu′.Theorem12 is proved.Remark 13.
Existence of global bounded solutions of nonsingular ordinary differential equation is well-known [10].Remark 14.
For the systems (1) and (2) with d>1,n=3,s>1, the speed of the converges to the regular polygon equilibrium does not depend on d since M−1U0 has only three eigenvalues for all d>1. The explicit dependence of M−1U0 on d,sis given by (70) for s>1,n=3. One can also calculate it without difficulty for s>1,n=4,5.


## 6. Conclusion

We showed that for the regular polygon equilibrium for (2) determined by Theorem 7 and s=d, the conclusion of Theorem 2 is true.Our results that follow from Theorem2 and Theorem 4 for Coulomb systems and s=d are formulated in the following way.Lets=3,d=2; s=2,d=3, and n=3,4,5. Let also s=3,d=1,n=3. Then, there exists an integer p, 1≤p<nd, a positive number λ, and a bounded at positive time solution xt of the equation of motion (3) with hc=0 depending on p real parameters, which is a real analytic function in them in a neighborhood of the origin, such that x−x0λ<∞, ẋλ<∞, where x0 is a regular polygon equilibrium. This result for s=3,d=1,n=3 follows also from Theorem (1).Theorem12 shows that
(1)
for the systems (1) and (2) with d=1,n=3,s=2, the conclusion of Theorem 2 is not true since M−1U0 does not have negative eigenvalues in ℝ3(2)
for the systems (1) and (2) with d>1,n=3,s>1, the speed of the converges to the regular polygon equilibrium does not depend on d since M−1U0 has only three eigenvalues for all d>1



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