# Periodic Coulomb Dynamics of Three Equal Negative Charges in the Field of Equal Positive Charges Fixed in Octagon Vertices

Authors: W. I. Skrypnik
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

Periodic solutions of the Coulomb equation motion for three equal negative point charges in the field of six equal positive point charges fixed in vertices of a convex octagon are found. The system possesses an equilibrium configuration. The Lyapunov center theorem is applied.


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## Body

## 1. Introduction

In this paper, we find an equilibrium in the Coulomb system of three equal negative point charges in the field of six equal positive point charges fixed in vertices of a convex symmetric octagon. This gives us a possibility to find periodic solutions for one-dimensional (line), two-dimensional (planar), and three-dimensional (spacial) systems. Earlier, we found periodic and quasiperiodic solutions for the system of two and three negative charges in the field of two equal positive charges [1–4].To obtain these results, we found explicit expressions for eigenvalues of the matrixU0 of second partial derivatives of the potential energy at the equilibrium of the systems. The existence of the periodic solutions followed from the Lyapunov center theorem [5–9] whenever there is no zero eigenvalue among the eigenvalues. The existence of the quasiperiodic solutions was proved with the help of the procedure of elimination of a node [8] and an application of the Lyapunov center theorem taking into account that zero eigenvalue of U0 results from a rotation invariance of the potential energy. U0 turns out to be the direct sum of two and three three-dimensional matrices for the considered here plane and spacial systems, respectively. Their simple structure permits to find their eigenvalues. U0 of the considered here space system does not possess zero eigenvalue, and the system potential energy does not have a rotation invariance.To find periodic solutions in Coulomb systems, it is not necessary to exploit the existence of their equilibria. In [10], we found the solutions for the system of n equal negative charges in the field of n equal positive charges fixed on a line (a coordinate axis). Our technique was inspired by the Siegel advanced majorant technique [11] which permits finding solutions of the Newton equation for three gravitating bodies.The Lyapunov center theorem concerns periodic solutions of the Hamiltonian systems with an equilibrium in the origin, which belong to its neighborhood, and is formulated precisely as follows.Theorem 1.
Let ann-dimensional Hamiltonian system have real analytic Hamiltonian whose Taylor power expansion converges absolutely and uniformly at a neighborhood of the origin and begins from quadratic terms. Let also λ1,⋯,λ2n be nonzero eigenvalues of the matrix determining the linear term of the Hamiltonian vector field such that the following nonresonance relation hold for purely imaginary λs,s=1,⋯,k: λj≠n′λs, s≠j=1,⋯,2n for an arbitrary integer n′. Then, the Hamiltonian equation possesses k periodic solutions in a neighborhood of the origin such that each of them depends on a different real-valued parameter cj for some j=1,⋯,k. These solutions and their periods τ1c1,⋯,τkck are real analytic functions in the parameters in a neighborhood of the origin and τj0=2π/∣λj∣.The periodic solutions from this theorem take values in a neighborhood of the origin due to the fact that their expansion in the parameterscj does not contain a constant term.The equation of motion of mechanical systems ofN d-dimensional particles (bodies, charges) with masses mj and the potential energy U looks like
(1)mjd2xjdt2=−∂UxN∂xj,j=1,⋯,N,xN=x1,⋯,xN∈RdN,xj=xj1,⋯,xjd.If a potential energy possess an equilibriumxj0,j=1,⋯,N, then the potential energy with the new variables xj−xj0,j=1,⋯,N will have the equilibrium at the origin, and it is possible to apply the Lyapunov center theorem to (1).The Coulomb potential energyU is expressed through the pair Coulomb potential. It is well known [12] that for (1) with mj=m, the eigenvalues from Theorem 1 coincide with λj=±−m−1σj, where σj,j=1,⋯,Nd are eigenvalues of U0.We believe that the obtained results may be useful for the semiclassical and Born-Oppenheimer approximations for quantum models of ionized molecules. The fixed positive charges and the equal positive and equal negative charges are associated with heavy nuclei and light electrons, respectively.Our paper is organized as follows. In Sections2, 3, and 4, we formulate our results concerning periodic solutions of the considered Coulomb equations of motion for the line, planar, and spacial systems, respectively. They are formulated in the theorems in the end of the sections.


## 2. Line Coulomb Dynamics

We consider the line dynamics of three point equal negative charges−e0 in the field of six equal positive point charges with the same value e′>0 fixed in octagon vertices with the first coordinates −a,0,a and second coordinates ±b,±3a2+b2,±b, respectively (see the next section). The three negative point charges move along the first axis which is an invariant manifold of the planar and spacial dynamics.The potential energy for the system is given by(2)Ux3=12∑j≠k=13ejekxj−xk−2e0e′∑j=13xj−a2+b2−1+xj+a2+b2−1+xj2+3a2+b2−1,xj∈R,a,b>0.The equilibrium equations are given by∂/∂xjUx3=0,j=1,2,3.. Let us insert the equalities into it for k=1:
(3)∂∂x1x1−x2−k=−kx1−x2x1−x2k+2,∂∂x1x1−a2+b2−k=−kx1−ax12+b2k+2.That is,(4)∂∂xjUx3=−e02∑j=k,k=13xj−xkxj−xk3+2e0e′xj−axj−a2+b23+xj+axj+a2+b23+xjxj2+3a2+b23.As a result, we obtain the equilibrium relation forj=1,3 putting x1=x10=−a,x2=x20=0,x3=x30=a:
(5)e022a2+e02a2=3e′2ae02a2+b23,5e02a3=3e′2a2+b23.Equality∂/∂x2Ux3=0 is true since x20−x10=x30−x20.The most important information is a spectrum of the matrix of second derivatives ofU. Its nondiagonal elements are easily calculated as
(6)∂Ux3∂xl∂xj=∂Ux3∂xj∂xl=−2e02xj−xl−3.LetUj,l0 be this function at the equilibrium. Then,
(7)U1,30=U3,10=−e024a3=−u′,U1,20=U2,10=U2,30=U3,20=−8u′.Further,(8)∂2∂xj2Ux2=∑j=k,k=132e02xj−xk3+2e0e′1xj−a2+b23−3xj−a2xj−a2+b25+1xj+a2+b23−3xj+a2xj+a2+b25+1xj2+3a2+b23−3xj2xj2+3a2+b25.LetUj,j0 be this function at the equilibrium. Then,
(9)U1,10=U3,30=9e024a3+2e0e′b−3+22a2+b23−15a22a2+b25,U2,20=16e024a3+2e0e′2a2+b23−6a2a2+b25+13a2+b23.From the equilibrium relation, it follows that(10)5e03e′1/312a=12a2+b2,As a result,(11)a2+b2=a24η−1−1+1=a2η−14−3η,3a2+b2=a24η−1−1+3=a2η−14−η,2e0e′b−3=5232a−3e021−η−3/2=5u′31−η−3/2,4e0e′2a2+b23=4e0e′5e03e′12a3=10u′3,30e0e′a22a2+b25=4−130e0e′5e03e′5/32a−3=25u′4η,4e0e′a2+b23=4e0e′a−3η3/24−3η−3/2=803u′4−3η−3/2,12e0e′a2a2+b25=12e0e′a−3η5/24−3η−5/2=80u′η4−3η−5/2,2e0e′3a2+b23=2e0e′a−3η3/24−η−3/2=403u′4−η−3/2.These equalities allow to representu′−1Uj,k0 as simple functions of η.
(12)U1,10=U3,30=u′v,v=373+531−η−3/2−254η,U2,2=u′g=u′16+8034−3η−3/2−80η4−3η−5/2+403η4−η−3/2=u′16+8034−3η−5/24−3η−3η+403η4−η−3/2,g=3−186+202−3η4−3η−5/2+5η4−η−3/2.LetU0 be the matrix with the elements Uj,l0,j,l=1,2,3.
(13)U0=u′U′1,U1′=v−8−1−8g−8−1−8v=−2U∗1+v+1I,U∗g1=U∗1=2−118182g18181,2g1=−g+v+1,where I is unity matrix. U∗1 has identical first and third rows and this means that DetU∗1=0. This allows one to find roots of the characteristic polynomials p∗1 of U∗1 and p1′ of U1′:
(14)p∗λ,q=DetλI−U∗q=λ2−q+1λ+q−32λHere, we subtracted the third row of−U∗q+λI from the first row. The determinant does not change after the subtraction.
(15)λ−2−1−42−1−4λ−q−4−2−1−4λ−2−1⟶λ0−λ−4λ−q−4−2−1−4λ−2−1.After that, we decomposed the determinant in the elements of the first row:(16)p∗λ,q=λλ−qλ−2−1−16−16−2−1λ−q=λλ−qλ−1−32.The roots ofp∗q are given by
(17)2λ=q+1±q−12+128,λ=0.The rootsp1′ of U1′ are given by
(18)p1′λ=−23p∗−λ2+v+12,g1,λ=v−g1±g1−12+128,λ=v+1=ζ1′.Letζ2′,ζ3′ be the roots corresponding to the plus and minus before the sign of the square root, respectively:
(19)ζ2′=g+v−12+g−v+122+128,ζ3′=g+v−12−g−v+122+128.We shall always useη<1. Let 0<η≤3−1, then
(20)g≥3−186+2032=16+3−15=1723,v<373+532781/2<1213+3−11027321/2<1523.Here, we applied5/3<3<7/4. This leads to ζ2′>g>v+2>ζ1′+1 and there is no resonance in ζ2. Here, we used also
(21)g−v+122+128≥g−v+12.Ifη≥3−12, then
(22)v+5>g,since
(23)v>373+5333−254>373+253−254>373+2512>14,g<3−186+3−3/25=16+83<19.Here, we applied5/3<3<7/4. From
(24)v>g⟶g−v+122=4−1v−g+52−6v−g+5+4−162<4−1v−g+52+9⟶g−v+122+128<2−1v−g+5+12⟶ζ2′<ζ1′+13,it follows
(25)v>103⟶ζ1′>133⟶ζ2′ζ1′<1+13ζ1′<4.We have to exclude the equalityζ2′=ζ1′ with the help of
(26)ζ2′−ζ1′=g−v−32+g−v+122+128.Equalityζ2′−ζ1′=0 leads to
(27)g−v−322=g−v+122+128⟶v−g=63.We proved the following proposition.Proposition 2.
There is no resonance inζ2′>0 if 0<η≤1/3. Moreover if v−g≠63 and η≥3−12, then there is no quadratic resonance in ζ1′, i.e., ζs′ζ1′−1≠k2,s=2,3, where k is an integer.
Let’s prove the next proposition.Proposition 3.
ζ2′>0 and there is no quadratic resonance in ζ1 if 0<η≤2/3.Proof.
We shall proveζ2′ζ1′−1<4,ζ1′≠ζ2′ and take into account ζ3′<ζ2′. Let 0<η≤1/3.Then,(28)v>373+53−2512=373−512>11,g<3−186+3−5/240+3−5/25=16+3−7/2320+3−7/240=16+32045+89=16+649+89=24.Here, we used5/3<3. From v−g+13>0 and
(29)g−v+122=4−1v−g+132−7v−g+13+49<4−1v−g+132+49⟶g−v+122+128<2−1v−g+13+14→ζ2′<ζ1′+19,it follows
(30)v>163→ζ1′>193→ζ2′ζ1′<1+19ζ1′<4.Let1/3≤η≤2/3. Then,
(31)v>373+53323/2−256=496+53278>496+533>496+259>496+83=1056,g<3−186+2042+1033103/2=3−186+52+310<836+52+13<3−186+52+35<16+2032+85=27+35,v<373+5333/2−2512<373+53−2=1213+153<15+9+24,(32)g>3−186+534−3−1−3/2=16+59>16.Here, we used5/3<3,2<3/2. That is, ζ2′>0. From v−g+17>0 and g−v+1/22=4−1v−g+172−18/2v−g+17+182/4<4−1v−g+172+81⟶g−v+1/22+128<2−1v−g+17+15⟶ζ2′<ζ1′+22,it follows
(33)v>193⟶ζ1′>223⟶ζ2′ζ1′<1+22ζ1′<4.We have alsoζ1≠ζ2 since v−g<8 for 1/3≤η≤2/3 and ζ2>ζ1, if 0<η≤1/3.Now, we have to estimateζ2′ on the interval 2/3≤η′<1. Obviously g>16 if η=2/3. Let η=6−15, then
(34)g=3−186+204−52−5/22−52+2564−6−15−3/2>3−186−104293+256662020>3−186−523+56420>3−186−32+58>3−186−92+58>16−12+1=5,v=373+5366−12524>373+2009−12524>373+22−51+124>29.Letη=9−18, then
(35)g=3−186+204−83−5/24−83−5/22−832−83+409992828<3−186−4039332+409133<3−186−5334+89<3−186−254+89=−23+3−1889<2,g>3−186−5334+409144>3−186−52116+59>3−186−7+59=−3227>−2,v=373+5399−509>373+45−549>51.Here, we used4/3<2<3/2,7/4>3>5/3. From v>8,
(36)g>3−186−20>3−18−15>−40.and these inequalities follows that v+g−1>0 and ζ′2>0 since v and g are monotonically increasing and decreasing functions, respectively, on the interval 2/3≤η<1. Let us prove this, i.e., that the derivatives ∂v,−∂g are positive on the interval.
(37)∂v=521−η−5/2−254>5293−254>3252−254>0,∂g=3−181502−3η4−3η−7/2+152η4−η−5/2−604−3η−5/2<3−1815293−1−6032<3−1812−6032<0.Taking into account Proposition2, we see that ζ′2>0. Thus, we proved the proposition.To apply the Lyapunov center theorem, we have to guarantee thatζ′3≠0. Taking to the second power both terms in the expression for ζ′3, we see that this condition is satisfied if gv−1≠128. This condition is true if 0<η≤3−12 since v>9 and g>16.The order of charges is preserved due to the infinite repulsion and we can substitute the holomorphic functionsxj−xk−1 instead of xj−xk−1 in the expression for their potential energies.Since the eigenvalues ofU0 coincide with ζj=u′ζ′j, the following theorem follows from the Lyapunov center theorem [5–7].Theorem 4.
If0<η≤1/3, then the line Coulomb equation of motion (1) with mj=m,d=1,andN=3 and the potential energy (2) possesses the equilibrium x10=−a,x20=0,x30=a, j=1,2,3, and two periodic solutions in its neighborhood such that each of them depends on its own real parameter cj for j=1,2. These solutions and their periods τjc,j=1,2 are real analytic functions in a neighborhood of the origin in these parameters and τj0=2πm/ζj. If 3−1<η≤3−12 or 3−12<η<1 and gv−1≠128, v−g≠63; then, the equation possesses this equilibrium and a periodic solution in its neighborhood that depends on a real parameter c. This solution and its period τc are real analytic functions in a neighborhood of the origin in this parameter and τ0=2πm/ζ1.


## 3. Planar Coulomb Dynamics

In this section, we consider the planar system of three equal charges:−e0 in the field of six equal positive charges e′ fixed at the octagon vertices with the coordinates bj,1≤j≤6, bj=bj1,bj2∈R2.
(38)b1=a,b,b2=a,−b,b3=−a,b,b4=−a,−b,b5=0,3a2+b2,b6=0,−3a2+b2,a,b>0,with the potential energy
(39)Ux3=12∑j≠k=13ejek∣xj−xk∣−e0e′∑j=13∑k=16xj−bk−1,where
(40)xj=xj1,xj2∈R2,xj2=xj12+xj22,ej=−e0<0.The equilibrium is given byx11=x101=−a,x21=x201=0,x31=x301=a, xjα=xj0α=0,α=2. The first partial derivatives of U look like
(41)∂∂xjαUx3=−e02∑k=1,k≠j3xjα−xkαxj−xk3+e0e′∑k=16x1α−bkαxj−bk3.This gives the equilibrium relation betweene0,e′,a,andb the same as in the previous section equating to zero the right-hand sides of these equalities for j=1,3 (the result is the same), taking into account x101−x301=−2a,
(42)x30−b12=x30−b22=b2,x30−b32=x30−b42=x30−b52=x30−b62=2a2+b2,x10−b12=x10−b22=x10−b52=x10−b62=2a2+b2,x10−b32=x10−b42=b2,x20−bj2=a2+b2,j=1,2,3,4;x20−bk2=3a2+b2,k=5,6∑k=16x101−bk1x10−bk3=−6a2a2+b23,∑k=16x102−bk2x10−bk3=−∑k=16bk2x10−bk3=0.The right-hand of (41) is zero for α=2 and for j=2 since ∣x20−x10∣=∣x30−x20∣, x101=−x301,x20=0.The equilibrium relation is given by(43)e022a2+e02a2=3e′2ae02a2+b23,5e02a3=3e′2a2+b23.The second derivatives of the potential energy (39) are given by
(44)∂Ux3∂xjα∂xkβ=∂Ux3∂xkβ∂xjα=e02δα,βxj−xk3−3xjα−xkαxjβ−xkβxj−xk5,α,β=1,2,(45)∂2Ux3∂xjβ∂xjα=e02∑k=1,k≠j3−δα,βxj−xk3+3xjα−xkαxjβ−xkβxj−xk5+e0e′∑k=16δα,βxj−bk3−3xjα−bkαxjβ−bkβxj−bk5.Now, we shall find the equilibrium value for all the terms in these equalities. Letη,u′ be as in the previous section. Then, we derive the following equalities:
(46)e02∑k=1,k≠j31xj0−xk03=e022a−3+a−31−δj,2+2a−3δj,2=9u′21−δj,2+8u′δj,2,j=1,2,3,(47)e0e′∑k=16δα,βxj0−bk3=δα,β2e0e′b−3+22a2+b2−3/2=53δα,βu′1−η−3/2+2,j=1,3,(48)e0e′∑k=16δα,βx20−bk3=δα,βe0e′4a2+b2−3/2+23a2+b2−3/2=δα,β403u′24−3η−3/2+4−η−3/2relying on equalities below (10) from the previous section. Let
(49)Tjα,β=∑k=16xjα−bkαxjβ−bkβxj−bk5.Let alsoTj0α,β be the equilibrium value of Tjα,β. Now, we will prove the equalities
(50)Tj0α,β=δα,β10a22a2+b2−5/2δα,1+2b−3+3a2+2b22a2+b2−5/2δα,2,j=1,3,(51)T20α,β=δα,β4a2a2+b2−5/2δα,1+22b2a2+b2−5/2+3a2+b2−3/2δα,2.with the help of the following equalities:
(52)T301,2=−b−5a−b11b12+a−b21b22+2a2+b2−5/2a−b31b32+a−b41b42+a−b51b52+a−b61b62=0,T101,2=−2a2+b2−5/2−a−b11b12+−a−b21b22+−a−b51b52+−a−b61b62+b−5−a−b31b32+−a−b41b42=0,T301,1=b−5a−b112+a−b212+2a2+b2−5/2a−b312+a−b412+a−b512+a−b612=10a22a2+b2−5/2,T101,1=2a2+b2−5/2−a−b112+−a−b212+−a−b512+−a−b612+b−5−a−b312+−a−b412=10a22a2+b2−5/2,Tj02,2=∑k=16bk22xj0−bk5,j=1,3,T102,2=T302,2=2b2b−5+2a2+b2−5/2+23a2+b22a2+b2−5/2=2b−3+3a2+2b22a2+b2−5/2,T20α,β=∑k=16bkαbkβx20−bk5=a2+b2−5/2∑k=14bkαbkβ+3a2+b2−5/2∑k=56bkαbkβ=δα,β4a2a2+b2−5/2δα,1+4b2a2+b2−5/2+23a2+b2−3/2δα,2.Here, we took into account(53)∑k=12bk1bk2=0,∑k=34bk1bk2=0.(50) and (51) are proved. Taking into account
(54)e02∑k=1,k≠j3xj0α−xk0αxj0β−xk0βxj0−xk05=e02a−3+2a−3δα,βδα,1=9u′2δα,βδα,1,j=1,3we derive from (47), (46), and (50)
(55)U1,α;1,β0=U3,α;3,β0=u′δα,βv′−δα,1u′∗−δα,2u′′∗,where
(56)v′=1351−η−3/2−72,u′u′∗+27u′2=6e0e′5a22a2+b2−5/2,u′u′′∗=6e0e′b−3+3a2+b22a2+b2−5/2,u′∗=−272+254η,u′′∗=5u′44−η+41−η−3/2.Here, we took into account(57)6e0e5a22a2+b2−5/2=6e0e′5a25e03e′5/312a5=25u′4η,6e0e′b−3=5u′1−η−3/2,6e0e′b22a2+b2−5/2=6e0e′2a21−ηη−15e03e′5/312a5=5u′1−η,18e0e′a22a2+b25=2−19e0e′5e03e′5/32a−3=154u′η.From the equality(58)e02∑k=1,k≠23x20α−xk0αx20β−xk0βx20−xk05=e022a−3δα,βδα,1=8u′δα,βδα,1,(48), (51), and (46), it follows
(59)U2,α;2,β0=u′δα,βv1−u1δα,1−δα,2u1′,where
(60)v1=−8u′+403u′24−3η−3/2+4−η−3/2,u1=3e0e′4a2a2+b2−5/2−24u′=80u′η4−3η−5/2−24u′,u1′=3e0e′4b2a2+b2−5/2+23a2+b2−3/2=40u′81−η4−3η−5/2+4−η−3/2.Here, we used(61)2a=1−η−1/2ηb,a2+b2=a2η−14−3η,e0e′a2+b2−5/2a2=e0e′a−3η5/24−3η−5/2=203u′η4−3η−5/2,e0e′a2+b2−5/2b2=e0e′a−5η3/21−η4−3η−5/22a2=803u′1−η4−3η−5/2,e0e′3a2+b2−3/2=e0e′a−3η3/24−η−3/2=203u′4−η−3/2.We have also(62)e02x20α−xk0αx20β−xk0βx20−xk05=e02xk0αxk0βa5=e02a−3δα,βδα,1=4u′δα,βδα,1,k=1,3,e02x10α−x30αx10β−x30βx10−x305=e022a−3δα,βδα,1=u′2δα,βδα,1.From these two equalities, one derives(63)U1,α;3,β0=U3,α;1,β0=u′2δα,β1−3δα,1,U2,α;3,β0=U3,α;2,β0=U2,α;1,β0=U1,α;2,β0=4u′δα,β1−3δα,1.Now, we determine two matricesUα0,α=1,2 by the rule
(64)Uj,α;k,β0=δα,βUα;j,k0and renumerate indexes of coordinates in the following way:
(65)1,1=1;2,1=2;3,1=3;1,2=4;2,2=5;3.2=6.where the first and second indexes under the round brackets corresponds to the lower and upper indexes of coordinates. This gives
(66)U0=U10⊕U20.The elements of the symmetric matricesUj0 are defined as follows:
(67)U1;1,10=U1,1;1,10=U1;3,30=U3,1;3,10=u′v′−u∗′,U1;2,20=U2,1;2,10=u′v1−u1,U1;2,10=U1;1,20=U1,1;2,10=−8u′,U1;3,10=U1;1,30=U1,1;3,10=−u′,U1;3,20=U1;2,30=U1,1;3,10=−8u′,U2;1,10=U1,2;1,20=U2;3,30=U3,2;3,20=u′v′−u′′∗=−u′u′′+u′2,U2;2,20=U2,2;2,20=u′v1−u′1=u′g′,U2;2,10=U2;1,20=U1,2;2,20=4u′,U2;3,10=U2;1,30=U1,2;3,20=u′2,U2;3,20=U2;2,30=U2,2;3,20=4u′.The parameters of the matrix elements are defined as follows:(68)u′v′−u′∗=5u′31−η−3/2−7u′6−25u′4η+27u′2=u′v,u′v1−u1=−8u′+24u′+403u′24−3η−3/2+4−η−3/2−80u′η4−3η−5/2=u′g,−u′′u′=v′u′−u′′∗−u′2=u′1351−η−3/2−72−5u′44−η+41−η−3/2−u′2,u′′=9−5η4+1031−η−3/2,u′′>11,u′g′=u′v1−u′1,g′=−8+40324−3η−3/2+4−η−3/2−4081−η4−3η−5/2+4−η−3/2=−8−8034−η−3/2+8034−3η−5/24−3η−121−η,g′=−8−8034−η−3/2+4−3η−5/28−9η,where v,g are the same as in the previous section. As a result,
(69)U10=u′v−8−1−8g−8−1−8v=u′U′1,U1′=−2U∗g1+v+1I,2g1=−g+v+1,U20=u′2−1−2u′′+18182g′818−2u′′+1=u′U′2,U′2=U∗g2−u′′I,g2=g′+u′′,where matrix U∗ is defined in the previous section where we found its eigenvalues as the roots of p∗q. Now, it is not difficult to find eigenvalues of U′2 as roots of polynomial p′2. They are given by
(70)p′2λ=p∗λ+u′′,g2.2λ=g2−2u′′+1±g2−12+128,λ=−u′′=ζ′4<0.Letζ′5,ζ′6 be the roots corresponding to the plus and minus before the sign of the square root, respectively. Then,
(71)2ζ′5=g′−u′′+1+g′+u′′−12+128,2ζ′6=g′−u′′+1−g′+u′′−12+128.If0<η≤1/3, then
(72)−g′>8+8024+58096>8+103+4>15,∣g′∣>15.ζ′5<0 if g′<0 and
(73)∣g′∣−u′′+12+128<∣g′∣+u′′−12,u′′−1∣g′∣>32.This is true if0<η≤1/3. In this case, ζ′6<ζ′5<0. This means that there is no resonance in ζ2 and quadratic resonance in ζ′1 for the eigenvalues ζj,1≤j≤6 of U0. This and the Lyapunov center theorem imply the following theorem.Theorem 5.
If0<η≤1/3, then the planar Coulomb equation of motion (1) with mj=m,d=2,andN=3 and the potential energy (39) possesses the equilibrium x101=−a,x201=0,x301=a, xj02=0,j=1,2,3, and two periodic solutions in its neighborhood such that each of them depends on its own real parameter cj for j=1,2. These solutions and their periods τjc,j=1,2 are real analytic functions in a neighborhood of the origin in these parameters and τj0=2πm/ζj.


## 4. Spacial Coulomb Dynamics

In this section, we consider the spacial Coulomb dynamics of three equal negative charges−e0 in the field of six equal positive charges e′ fixed at vertices of the octagon with vertices bj,1≤j≤6, bj=bj1,bj2,bj3=0∈R3.
(74)b1=a,b,0,b2=a,−b,0,b3=−a,b,0,b4=−a,−b,0,b5=0,3a2+b2,0,b6=0,−3a2+b2,0,a,b>0,with the potential energy
(75)Ux3=12∑j≠k=13ejekxj−xk−e0e′∑j=13∑k=16xj−bk−1,where
(76)xj2=xj12+xj22+xj32.It is not difficult to repeat calculations of the previous section to see that the elements of the matrix of second partial derivatives at the equilibriumx11=x101=−a,x21=x201=0,x31=x301=a, xjα=xj0α=0,α=2,3 look like for α,β=1,2,3 as follows:
(77)U1,α;3,β0=U3,α;1,β0=u′2δα,β1−3δα,1,U2,α;3,β0=U3,α;2,β0=U2,α;1,β0=U1,α;2,β0=4u′δα,β1−3δα,1,U1,α;1,β0=U3,α;3,β0=δα,βu′v′−δα,1u∗′−δα,2u′′∗,U2,α;2,β0=u′δα,βv1−u1δα,1−δα,2u′1.Now, we determine three matricesUα0,α=1,2,3 by the rule
(78)Uj,α;k,β0=δα,βUα;j,k0and renumerate indexes of coordinates in the following way:
(79)1,1=1;2,1=2;3,1=3;1,2=4;2,2=5;3.2=6;1,3=7;2,3=8;3.3=9.where the first and second indexes under the round brackets correspond to the lower and upper indexes of coordinates. This gives
(80)U0=U10⊕U20⊕U30.The elements of the symmetric matricesU10,U20 are defined in the previous section and
(81)U3;1,30=U3;3,10=U1,α;3,β0=U3,α;1,β0=u′2δα,β,U3;1,20=U3;2,10=U3;2,30=U3;3,20=U2,3;3,30=U3,3;2,30=U2,3;1,30=U1,3;2,30=4u′,U3;1,10=U1,3;1,30=U3;3,30=U3,3;3,30=u′v′,U3;2,20=U2,3;2,30=u′v1,U30=u′2−12v′8182v18182v′=u′U′3,U′3=U∗g3+v′−1I,g3=v1−v′+1.The rootsp′3 of U′3 are given by
(82)p′3λ=p∗λ+1−v′,g3,2λ=g3−21−v′+1±g3−12+128,λ=v′−1=ζ′7.Letζ′8,ζ′9 be the roots corresponding to the plus and minus before the sign of the square root, respectively. Then,
(83)2ζ′8=v1+v′−1+v1−v′2+128,2ζ′9=v1+v′−1−v1−v′2+128.We remind that(84)v1=−8+40324−3η−3/2+4−η−3/2,v′=1351−η−3/2−72.Let us assume0<η≤1/3. Then,
(85)12<v′≤135278−72<1310−72=216,−3=−8+408<v1<−8+4027<0,−516<v1−v′<−12,−212<v1+v′<216,∣v1+v′−1∣<312,v1−v′2+128>112.As a result,(86)2ζ′8≥v1−v′2+128−∣v1+v′−1∣>11−312=712,2ζ′8=v1+v′−1+v1−v′2+128<116+13=1416,2ζ′9<−11+16<−10.The inequality forv′ and last three inequalities imply
(87)−12<ζ′7<116,334<ζ′8<7112,ζ′9<−5.Fromζ′1<ζ′2, ζ′1>12, it follows
(88)ζ′7<ζ′1<ζ′2,0<ζ′8<ζ′1<ζ′2.This means that there is no resonance inζ2=u′ζ′2 and quadratic resonance in ζ1=u′ζ1 for eigenvalues of U0 in the spacial Coulomb system if 0<η≤1/3 and ζ′7≠0, i.e., η≠1−10/132/3<1/3.The last inequality follows from(89)233/2=2233<35<1013,53<3,32>2.We proved the following theorem with the help of the Lyapunov center theorem.Theorem 6.
If0<η≤1/3 and η≠1−10/132/3, then the spacial Coulomb equation of motion (1) with mj=m,d=3,andN=3 and the potential energy (75) possesses the equilibrium x101=−a,x201=0,x301=a, xj0α=0,j=1,2,3, α=2,3, and two periodic solutions in its neighborhood such that each of them depends on its own real parameter cj for j=1,2. These solutions and their periods τjcj,j=1,2 are real analytic functions in a neighborhood of the origin in these parameters and τj0=2πm/ζj.


## 5. Conclusion

We have shown that the matrixU0 of second partial derivatives of the potential energy of our spacial system at the equilibrium is the direct sum of the three dimensional matrices Uj0,j=1,2,3 such that U10 and the direct sum of U10 with U20 coincide with matrices of the second partial derivatives of the potential energy at the equilibrium of one-dimensional and planar systems, respectively. We have shown also that U10 possesses two positive eigenvalues ζ1,ζ2 and that ζj is not in resonance with square roots of other eigenvalues for j=1,2 if 0<η≤1/3, η=5e0/3e′2/3. The Lyapunov center theorem implies that these eigenvalues generate periodic solutions mentioned in Theorem 6 if 0<η≤1/3 and η≠1−10/132/3. The last condition guarantees that neither of the eigenvalues are zero.



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