# New Classes of Spatial Central Configurations forN + N + 2-Body Problem

Authors: Liu Xuefei; Zhang Chuntao; Luo Jianmei; Zhang Gan
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

Under arbitrary masses, in this paper, we discuss the existence of new families of spatial central configurations for theN + N + 2-body problem, N≥2. We study some necessary conditions and sufficient conditions for a families of spatial double pyramidical central configurations (d.p.c.c.), where 2N bodies are at the vertices of a nested regular N-gons Γ1∪Γ2, and the other two bodies are symmetrically located on the straight line that is perpendicular to the plane that contains Γ1∪Γ2 and passes through the geometric center of Γ1∪Γ2. We prove that if the bodies are in a d.p.c.c., then the masses on each N-gon are equal, and the other two are also equal. And also we prove the existence and uniqueness of the central configurations for any given ratios of masses.


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## Body

## 1. Main Results

The Newtoniann-body problem (see [1–7]) concerns the motion of n point particles with masses mj∈R+ and positions q̅j∈R3(j=1,…,n). The motion is governed by Newton’s law:(1.1)mjq̅̈j=∂U(q̅)∂q̅j,
where q̅=(q̅1,…,q̅n) and U(q̅) is the Newtonian potential:(1.2)U(q̅)=∑1≤k<j≤nmkmj|q̅k-q̅j|.
Consider the space(1.3)X={q̅=(q̅1,…,q̅n)∈R3n:∑k=1nmkq̅k=0},
that is, suppose that the center of mass is fixed at the origin of the space. Because the potential is singular when two particles have the same position, it is natural to assume that the configuration avoids the set Δ={q̅:q̅k=q̅j for some k≠j}. The set X∖Δ is called the configuration space.Definition 1.1 (see [4]).
A configurationq̅=(q̅1,…,q̅n)∈X∖Δ   is called a central configuration (c.c.) if there exists a constant λ  such that q̅̈k=-λq̅k,k=1,2,…,n, or
(1.4)∑j=1,j≠knmjmk|q̅j-q̅k|3(q̅j-q̅k)=-λmkq̅k,1≤k≤n,
and the value of the constant λ in (1.4) is uniquely determined by
(1.5)λ=UI,
where
(1.6)I=∑k=1nmk|q̅k|2.For any coordinate system, we have that, if the center of massesm1,m2,…,mn with position vectors r1,r2,…,rn is not at the origin, central configuration equations (1.4) are equivalent to the following:(1.7)∑j=1,j≠knmjmk|rj-rk|3(rj-rk)=-λmk(rk-r0),1≤k≤n,
where r0=∑k=1nmkrk/∑k=1nmk is the center of masses m1,m2,…,mn.The knowledge of central configurations allows us to compute homographic solutions (see [8]); there is a relation between central configurations and the bifurcations of the hypersurfaces of constant energy and angular momentum (see [9]); if the N bodies are moving towards a simultaneous collision, then the bodies tend to a central configuration (see [10]). See also [11, 12].Some examples of spatial central configurations are a regular tetrahedron with arbitrary positive masses at the vertices [13] and a regular octahedron with six equal masses at the vertices [12]. Double nested spatial central configurations for 2N bodies were studied for two nested regular polyhedra in [14]. More recently, the same authors studied central configurations of three regular polyhedra for the spatial 3N-body problem in [15]. See also [16], where nested regular tetrahedrons are studied.Recently, Hampton and Santoprete [17] provided new examples of stacked spatial central configurations—central configurations for the N-body problem where a proper subset of the N bodies are already on a central configuration—for the 7-body problem where the bodies are arranged as concentric three- and two-dimensional simplex. New classes of stacked spatial central configurations for the 6-body problem which have four bodies at the vertices of a regular tetrahedron and the other two bodies on a straight line connecting one vertex of the tetrahedron with the center of the opposite face were studied in [18].In this paper, we study new classes of spatial double pyramidical central configurations (d.p.c.c) for theN+N+2-body that satisfy the following.(1)
The position vectorsr1,r2,…,rN of masses m1,m2,…,mN are at the vertices of a regular N-gon Γ1, whose sides have length ι1. The position vectors rN+1,rN+2,…,r2N of masses mN+1,mN+2,…,m2N are at the vertices of another regular N-gon Γ2, whose sides have length ι2. Two N-gons have a same geometric center and form a affine nested N-gons (base plane Π).(2)
LetL be the straight line perpendicular to the base plane Π that contains Γ1 and Γ2 and passes through the geometric center of Γ1∪Γ2. The position vectors r2N+1 and r2N+2 of masses m2N+1 and m2N+2 are on L and on opposite sides with respect to the plane Π.The central configurations studied in this paper are in some measure related to the double pyramidal central configurations (d.p.c.c) studied in [19] and the paper in [20]. The configuration in [19] consists of n masses on a plane that are located at the vertices of a regular n-gon and two equal masses located on the line perpendicular to passing through the geometric center of the N-gon. And the authors in [20] assumed that the center of the N-gon is at the origin, that is, Σj=1nmjrj=0, and more that the origin of the inertial system is the center of mass of the system, that is, Σj=1n+3mjrj=0 and rn+3=0. In fact the origin is the geometric center of the N-gon. Hence the configuration in [20] is only to append a mass at the geometric center of the N-gon in [19].As far as we know, the spatial central configurations studied here are very new. The number of bodies (masses) is increased to2N+2, it is not to suppose the origin of the inertial system, and the proofs are more difficult than those in [19, 20].The main results of this paper are the following.Theorem 1.2.
ConsiderN+N+2 bodies with masses m1,m2,…,m2N,m2N+1,m2N+2 located according to the following.(i)
r1,r2,…,rN are at the vertices of a regular n-gon Γ1 inscribed on a circle of radius α.(ii)
rN+1,rN+2,…,r2N are at the vertices of another regular n-gon Γ2 inscribed on a circle of radius a.(iii)
r2N+1 and r2N+2 are on the straight line L, on opposite sides with respect to the plane Π, where L is the straight line that is perpendicular to Π and passes through the geometric center of Γ1∪Γ2. Let h1=distance(r2N+1,Π) and h2=distance(r2N+2,Π).
In order that theN+N+2 bodies can be in a central configuration (c.c.), the following statements hold.
(1)
Ifh1=h2=:h, then there is m2N+1=m2N+2.(2)
Ifh1=h2=:h, then not only m2N+1=m2N+2, but also(1.8)m1=m2=⋯=mN=:m,mN+1=mN+2=⋯=m2N=:m̃.(3)
The origin is the mass center ofm1,m2,…,m2N and also the mass center of m1,m2,…,m2N+2, that is,
(1.9)∑j=12Nmjrj=0,∑j=12N+2mjrj=0.(4)
Get rid of massesm2N+1 and m2N+2, when ratio of masses b=mN+1/m1 in some interval and the masses m1,m2,…,m2N may form a central configuration.Remark 1.3.
Letm2N+1=m2N+2, and the origin is the mass center of m1,m2,…,m2N, also it is the mass center of m1,m2,…,m2N+2, and then we have that h1=h2 and (1.8) hold. The conclusions are the oppose problem of some items in Theorem 1.2, which is similar to that in [20]. We may similarly prove h1=h2. The proof of (1.8) still see Theorem 1.2 in this paper.Theorem 1.4.
Under the suppositions of the positions for masses, andm1=m2=⋯=mN=:m, mN+1=mN+2=⋯=m2N=:m̃, m2N+1=m2N+2 and h1=h2=:h, then m1,…,mN,mN+1,…,m2N,m2N+1,m2N+2 are in a c.c., if and only if the parameters b,a,c and h satisfy the following relationships:
(1.10)λM=1(N+Nb+2c)(∑j≠N1-ρj|1-ρj|3+∑jb(1-aρj)|1-aρj|3+2c(1+h2)3/2),λM=1h(N+Nb+2c)[Nh(1+h2)3/2+Nbh(a2+h2)3/2+c4h2],b=∑j((a-ρj)/|a-ρj|3)-a∑j≠N((1-ρj)/|1-ρj|3)+(2c/(a2+h2)3/2)-(2ca/(1+h2)3/2)a∑j((1-aρj)/|1-aρj|3)-a-2∑j≠N((1-ρj)/|1-ρj|3),
where b=m̃/m,c=m2N+1/m  and  ρj=exp(i(2πj/N)).Theorem 1.5.
Under the suppositions of the positions for masses, andm1=m2=⋯=mN=:m, mN+1=mN+2=⋯=m2N=m̃, m2N+1=m2N+2 and h1=h2=:h, then for any ratios of masses b=m̃/m and c=m2N+1/m,m1,…,mN,mN+1,…,m2N,m2N+1,m2N+2 may form a unique c.c. such that a∈(0,1) and h∈(0,+∞).


## 2. Some Lemmas

Definition 2.1 (see [21]).
IfN×N matrix A=(ai,j) satisfies
(2.1)ai,j=ai-1,j-1,1≤i,j≤N,
where we assume ai,0=ai,N and a0,j=aN,j, then one calls that A is a circular matrix.Lemma 2.2 (see [21]).
(i) IfA and B are N×N circular matrices, for any numbers γ and β, then A+B,A-B,AB and γA+βB are also circular matrices, and AB=BA.
(ii)
LetA=(ai,j) be an  N×N circular matrix; the eigenvalues λk and the eigenvectors v⃗k of A are
(2.2)λk(A)=∑ja1,jρk-1j-1,v⃗k=(1,ρk-1,ρk-12,…,ρk-1N-1)T.(iii)
LetA and B be circular matrices, and let  λk(A) and λk(B) be eigenvalues of A and B. Then the eigenvalues of A+B,A-B and A·B are λk(A)+λk(B),λk(A)-λk(B) and λk(A)·λk(B).It is clear that.Lemma 2.3.
IfA=(ai,j) is an N×N circular matrices, and AX=0, where X=(x1,…,xn)T, xi>0(i=1,…,N), then
(2.3)a1,j+⋯+aN,j=0,1≤j≤N,ai,1+⋯+ai,N=0,1≤i≤N.Lemma 2.4.
LetA and B be N×N Hermite circular matrices; then A+B, A-B, AB and γA+βB(γ,β∈R) are also Hermite circular matrices.Lemma 2.5.
LetA be a Hermite circular matrix; then the eigenvalues of A are real number and(i)
whenn=2m+1(m≥1),A can be denoted by A=A2m+1=cir(a,b1,b2,…,bm,b̅m,…,b̅2,b̅1), where a∈R and b̅l is a conjugate complex number of bl. It has eigenvalues
(2.4)λ0=a+2∑l=1mRebl,λk=a+2∑l=1m[Reblcos⁡2kπl2m+1-Im⁡blsin⁡2kπl2m+1],1≤k≤2m,(ii)
whenn=2m(m≥1), A can be denoted by A=A2m=cir(a,b1,b2,…,bm-1,bm,b̅m-1,…,b̅2,b̅1,). It has eigenvalues
(2.5)λ0=a+2∑l=1m-1Rebl+bm,λm=a+2∑l=1m-1(-1)lRebl+(-1)mbm,λk=a+2∑l=1m-1[Reblcos⁡2kπl2m-Im⁡blsin⁡2kπl2m]+(-1)kbm,1≤k≤2m-1,k≠m.Lemma 2.6.
The complex subspaceL of CN generated by X1=(1,1,…,1) and X2=(1,ρ,…,ρN-1), where N=2k>2(ρ=exp2πI/N), and the complex subspace L̃ generated by X1,X2, and X3=(1,ρk+1,…,ρ(N-1)(k+1)), where N=2k+1>3, all contain no real vectors other than the multiples of (1,1,…,1).Remark 2.7.
Lemmas2.3–2.5 can be simply proved by properties of circular matrix and Hermite matrix, and after some algebraic computations, Lemma 2.6 can be also simply proved.Lemma 2.8 (see [5]).
LetA=(1/4)∑j≠Ncsc(πj/N); then A(N) has the following asymptotic expansion for N large:
(2.6)A(N)~N2π(γ+log⁡2Nπ)+∑k≥0(-1)k(22k-1-1)B2k2π2k-1(2k)(2k)!1N2k-1,
where γ stands for the Euler-Mascheroni constant and B2k stands for the Bernoulli numbers.Lemma 2.9 (see [5]).
LetΦν(x)=∑j1/djν, where ν>0 anddj=1+x2-2xcos(2πj/N); then, for 0<x<1, Φν(x) and all of its any order derivatives are positive. Moreover, the same is thus forΨν(x)=∑jcos(2πj/N)/djν.


## 3. The Proof of Theorem

Proposition 3.1.
The central configuration equations (1.4) or (1.5) is equivalent to the following:(3.1)∑j=1,j≠knmj(rj-rk)(Rj,k-λM)=0,1≤k≤n,
where M=∑k=1nmk,Rj,k=|rj-rk|-3.Proof.
From central configuration equation (1.4) or (1.5), we easily prove.Denoterk=(xk,yk,zk)∈R3, and z⃗=(0,0,1). Observing (1.5), one could have a free choice of the origin for a configuration. Without loss of generality, consider that the origin is at the geometric center of Γ1∪Γ2, and let zk=0,1≤k≤2N.Proposition 3.2.
Under the hypotheses of Theorem1.2, if h1=h2=:h, then the following equations are verified:
(3.2)m2N+1=m2N+2,(3.3)R2N+1,j=R2N+1,k=R2N+2,j=R2N+2,k,1≤j≠k≤N,(3.4)R2N+1,N+j=R2N+1,N+k=R2N+2,N+j=R2N+2,N+k,1≤j≠k≤N,
where Rj,k=|rj-rk|-3.Proof.
In (3.1), considering the equations along the direction z⃗, and k=1,N+1, we have
(3.5)m2N+1(z2N+1-z1)[R1,2N+1-λM]+m2N+2(z2N+2-z1)[R1,2N+2-λM]=0,m2N+1(z2N+1-zN+1)[RN+1,2N+1-λM]+m2N+2(z2N+2-zN+1)[RN+1,2N+2-λM]=0,
where
(3.6)R1,2N+1=(|r1|2+|z2N+1|2)-3/2,R1,2N+2=(|r1|2+|z2N+2|2)-3/2,RN+1,2N+1=(|rN+1|2+|z2N+1|2)-3/2,RN+1,2N+2=(|rN+1|2+|z2N+2|2)-3/2,z1=zN+1=0.
By theh1=h2, we have
(3.7)z2N+1=-z2N+2,R1,2N+1=R1,2N+2,RN+1,2N+1=RN+1,2N+2.
Hence
(3.8)(m2N+1-m2N+2)[R1,2N+1-λM]=0,(m2N+1-m2N+2)[RN+1,2N+1-λM]=0.
Obviously R1,2N+1≠RN+1,2N+1, so
(3.9)m2N+1=m2N+2.
By our assumptions for the origin, (3.3), (3.4) clearly holds.Equation (3.9) of Proposition 3.2 proves item (1) of Theorem 1.2.Proposition 3.3.
Under the hypotheses of Theorem1.2, if  m2N+1=m2N+2, the origin is the mass center of m1,m2,…,m2N, and it also is the mass center of m1,m2,…,m2N+2, then
(3.10)h1=h2=:h
and (3.3), (3.4) hold.Proof.
The proof is very similar to that in [20].Proposition 3.4.
Under the hypotheses of  Theorem1.2, if h1=h2=:h, then the masses at the vertices of circle Γ1 are equal, and also the masses at the vertices of circle Γ2 are equal, that is,
(3.11)m1=m2=⋯=mN,mN+1=mN+2=⋯=m2N.Proof.
Because ifr→εr is a transformation in a central configuration, then λ→(1/ε2)λ can be a new parameter of a central configuration. We say that the old and the new are equivalent. Hence without loss of generality, we may let α=1,0<a<1. Then the vectors of positions based on the previous assumptions can be interpreted by the following:
(3.12)rk=(ρk,0)(1≤k≤N),rN+k=(ρ̃k,0)(1≤k≤N),r2N+1=(0⋅ρ0,h)(h>0),r2N+2=(0⋅ρ0,-h)(h>0),
where ρk=exp((2πk/N)i),ρ̃k=aρk,a>0, and i=-1,ρk denote the N complex kth roots of unity, that is, that mk(1≤k≤N) each locates at the vertices rk of the one regular N-gon Γ1,mN+k(1≤k≤N) each locates at the vertices rN+k of the other regular N-gon Γ2, and m2N+1 and m2N+2 lie on the vertices of r2N+1,r2N+2. Then the center of masses is
(3.13)r0=∑j(mjrj+mN+jrN+j)+m2N+1r2N+1+m2N+2r2N+2M,
where
(3.14)M=∑j(mj+mN+j)+m2N+1+m2N+2.
In (3.13)-(3.14) and throughout this paper, unless other restricted, all indices and summations will range from 1 to N.
Leth1=h2=:h; then m2N+1=m2N+2. Now we discuss all equations for the 2N masses on the base plane Π. According to (3.1), then
(3.15)∑j≠kmj(1|ρk-ρj|3-λM)(ρk-ρj)+∑jmN+j(1|ρk-ρ̃j|3-λM)(ρk-ρ̃j)+2m2N+1(1(1+h2)3/2-λM)ρk=0,∑jmj(1|ρ̃k-ρj|3-λM)(ρ̃k-ρj)+∑j≠kmN+j(1|ρ̃k-ρ̃j|3-λM)(ρ̃k-ρ̃j)+2m2N+1(1(a2+h2)3/2-λM)aρk=0.
Multiplying both sides byρN-k, noting that |ρk-ρj|=|ρk||1-ρj-k|=|1-ρj-k| and ρ̃k=aρk, we see that (3.15) may be written as
(3.16)∑j≠kmj(1|1-ρj-k|3-λM)(1-ρj-k)+∑jmN+j(1|1-aρj-k|3-λM)(1-aρj-k)+2m2N+1(1(1+h2)3/2-λM)=0,∑jmj(1|a-ρj-k|3-λM)(a-ρj-k)+∑j≠kmN+j(1|a-aρj-k|3-λM)(a-aρj-k)+2m2N+1(1(a2+h2)3/2-λM)a=0,
where k=1,2,…,N.
Now we define theN×N circular matrices C=[ck,j], A=[ak,j], B=[bk,j], and D=[dk,j] as follows:
(3.17)ck,j=0,fork=j,ck,j=(1|1-ρj-k|3-λM)(1-ρj-k),fork≠j,ak,j=(1|1-aρj-k|3-λM)(1-aρj-k),bk,j=(1|a-ρj-k|3-λM)(a-ρj-k),dk,j=0,fork=j,dk,j=(1|a-aρj-k|3-λM)(a-aρj-k)fork≠j.
Also define
(3.18)1⃗=(1,…,1)T,E=2(1(1+h2)3/2-λM)⋅1⃗,F=2a(1(a2+h2)3/2-λM)⋅1⃗.
We see that (3.16) holds if and only if the matrix equation
(3.19)(CAEBDF)(m1⋮mNmN+1⋮m2Nm2N+1)=0
has a positive solution. Let
(3.20)m⃗=(m1,…,mN)T,m̃⃗=(mN+1,…,m2N)T.
Then (3.19) is equivalent to
(3.21)Cm⃗+Am̃⃗+m2N+1E=0⃗,(3.22)Bm⃗+Dm̃⃗+m2N+1F=0.
Noticing that A,B,C, and D are N×N circular matrix, using the properties of circular matrix, we know that they must have positive real eigenvector 1⃗. Each of (3.21) and (3.22) left multiplies 1T⃗=(1,1,…,1); there are
(3.23)(∑kmk)∑j≠N(1|1-ρj|3-λM)(1-ρj)+(∑kmN+k)∑j(1|1-aρj|3-λM)(1-aρj)+2m2N+1⋅N(1(1+h2)3/2-λM)=0,(∑kmk)∑j(1|a-ρj|3-λM)(a-ρj)+(∑kmN+k)∑j≠N(1|a-aρj|3-λM)(a-aρj)+2m2N+1⋅N(1(a2+h2)3/2-λM)a=0.
By (3.21) and (3.22) we have
(3.24)(CD-AB)m⃗+2m2N+1[(1(1+h2)3/2-λM)D-a(1(a2+h2)3/2-λM)A]1⃗=0⃗,(AB-CD)m̃⃗+2m2N+1[(1(1+h2)3/2-λM)B-a(1(a2+h2)3/2-λM)C]1⃗=0⃗.
From Lemma 2.2 we see that ((1/(1+h2)3/2)-(λ/M))D-a((1/(a2+h2)3/2)-(λ/M))A, ((1/(1+h2)3/2)-(λ/M))B-a((1/(a2+h2)3/2)-(λ/M))C, and CD-AB are circular matrix, and we know that they must have positive real eigenvector 1⃗. By the properties of circular matrix, (3.24) can be written as
(3.25)(CD-AB)⋅m⃗+γ1⋅1⃗=0⃗,(AB-CD)⋅m̃⃗+γ2⋅1⃗=0⃗,
where
(3.26)γ1⋅1⃗=2m2N+1[(1(1+h2)3/2-λM)D-a(1(a2+h2)3/2-λM)A]1⃗,γ2⋅1⃗=2m2N+1[(1(1+h2)3/2-λM)B-a(1(a2+h2)3/2-λM)C]1⃗,γ1=2m2N+1[(1(1+h2)3/2-λM)∑j≠N(1|a-aρj|3-λM)(a-aρj)-a(1(a2+h2)3/2-λM)∑j(1|1-aρj|3-λM)(1-aρj)],γ2=2m2N+1[(1(1+h2)3/2-λM)∑j(1|a-ρj|3-λM)(a-ρj)-a(1(a2+h2)3/2-λM)∑j≠N(1|1-ρj|3-λM)(1-ρj)].
We easily prove γ1,γ2∈R, and from (3.23), we have
(3.27)(∑kmN+k){(1(1+h2)3/2-λM)∑j≠N(1|a-aρj|3-λM)(a-aρj)-a(1(a2+h2)3/2-λM)∑j(1|1-aρj|3-λM)(1-aρj)}+(∑kmk){(1(1+h2)3/2-λM)∑j(1|a-ρj|3-λM)(a-ρj)-a(1(a2+h2)3/2-λM)∑j≠N(1|1-ρj|3-λM)(1-ρj)}=0,
that is,
(3.28)γ1∑kmN+k+γ2∑kmk=0,mj>0,m̃j>0.
Hence one has the following.
(1) Ifγ1=0, then γ2=0. By (3.25), there are
(3.29)(CD-AB)⋅m⃗=0⃗,(3.30)(CD-AB)⋅m̃⃗=0⃗.
We notice that (3.29) or (3.30) must have positive real solutions, which is equivalent to that CD-AB has positive real eigenvectors corresponding to eigenvalue 0.
But we notice that (3.29) and (3.30) must hold, and for k=1, we have an eigenvalue λ1=0, and an matching eigenvector v⃗1=(1,1,…,1)T of CD-AB. Noticing that A,B,C,and D are Hermite circular matrices, from the properties of circular and Hermite matrix in Lemmas 2.2 and 2.4, then CD-AB=G is also a Hermite circular matrix. We may denote G by cir(a0,g1,g2,…,gm-1,gm,g̅m-1,…,g̅2,g̅1) when N=2m, where a0,gm∈R. We also denote G by cir(a0,g1,g2,…,gm-1,gm,g̅m,…,g̅2,g̅1) when N=2m+1, where a0∈R. Using Lemmas 2.4 and 2.5, after complex computation, we may prove that the kernel of G is at most a subspace in L when N=2m>2, and a subspace in L̃ when N=2m+1>3 (see [22]), where the meanings of L and L̃ are in Lemma 2.6. Hence the kernel of G does not contain any positive real vectors other than multiplication of 1⃗=(1,1,…,1).
WhenN=2,3, we may easily prove the conclusion.
(2) Ifγ1≠0, then γ2≠0; from (3.25) and (3.28),we get
(3.31)(AB-CD)[(∑kmk)m̃⃗-(∑kmN+k)m⃗]=0⃗.
If
(3.32)(∑kmk)m̃⃗-(∑kmN+k)m⃗≠0,
let G=AB-CD=(gij), which is not zero circular matrix; by Lemmas 2.2 and 2.3 we see that ∑jgij=0 and G has an eigenvalue λ1=0. Using the properties of circular matrix, we have (AB-CD)1⃗=0⃗ or 1⃗T(AB-CD)=0⃗T. Let 1⃗T left multiplies (3.25); we get γ1=γ2=0, which is a contradiction to the supposition. So
(3.33)(∑kmk)m̃⃗-(∑km̃k)m⃗=0,
and then mN+j=bmj, where b=∑km̃k/∑kmk, that is, m̃⃗=bm⃗. Substituting it into (3.21)-(3.22), we have
(3.34)(C+bA)m⃗+2m2N+1(1(1+h2)3/2-λM)1⃗=0,(B+bD)m⃗+2a⋅m2N+1(1(a2+h2)3/2-λM)1⃗=0.
Then
(3.35)[a⋅(1(a2+h2)3/2-λM)(C+bA)-(1(1+h2)3/2-λM)(B+bD)]m⃗=0.
Noticing that a·((1/(a2+h2)3/2)-(λ/M))(C+bA)-((1/(1+h2)3/2)-(λ/M))(B+bD) is a Hermite circular matrix, similarly we also have m⃗=(m1,m1,…,m1)T and m̃⃗=(mN+1,…,mN+1), where m1,mN+1>0.Remark 3.5.
Proposition3.4 proves item (2) of Theorem 1.2.Proposition 3.6.
Under the hypothesis of Theorem1.2, if h1=h2, then the origin is the mass center of m1,m2,…,m2N  and also is the mass center of m1,m2,…,m2N+2.Proof.
We have already(3.36)m2N+1=m2N+2,(3.37)m1=m2=⋯=mN,(3.38)mN+1=mN+2=⋯=m2N.
So there are
(3.39)∑j=12Nmjrj=0,∑j=12N+2mjrj=0,
by the positions of rj,j=1,2,…,2N+2.Remark 3.7.
Proposition3.6 proves item (3) of Theorem 1.2. By (3.37) and (3.38), item (4) of Theorem 1.2 may be proved.Proposition 3.8.
Under the hypothesis of Theorem1.4, the conclusion of Theorem 1.4 holds.Proof (The Proof of the Necessary).
By the hypothesis of Theorem1.4, let m̃=bm,m2N+1=cm; from (3.23), there are
(3.40)[∑j≠N(1|1-ρj|3-λM)(1-ρj)+b∑j(1|1-aρj|3-λM)(1-aρj)]+2c(1(1+h2)3/2-λM)=0,[∑j(1|a-ρj|3-λM)(a-ρj)+b∑j≠N(1|a-aρj|3-λM)(a-aρj)]+2c(1(a2+h2)3/2-λM)a=0.
We know that
(3.41)∑j(1-ρj)=N,∑jb(1-aρj)=bN,∑j(a-ρj)=aN,∑jb(a-aρj)=abN.
From (3.40) we have
(3.42)λM=1N+Nb+2c[∑j≠N1-ρj|1-ρj|3+b∑j1-aρj|1-aρj|3+2c(1+h2)3/2],λM=1a(N+Nb+2c)[∑ja-ρj|a-ρj|3+b∑j≠Na-aρj|a-aρj|3+2c(a2+h2)3/2],
and similarly we also have
(3.43)λM=1h(N+Nb+2c)[Nh(1+h2)3/2+Nbh(a2+h2)3/2+ch4h3]
by (3.1). Then
(3.44)a(∑j≠N1-ρj|1-ρj|3+b∑j1-aρj|1-aρj|3+2c(1+h2)3/2)=∑ja-ρj|a-ρj|3+b∑j≠Na-aρj|a-aρj|3+2c(a2+h2)3/2,(3.45)b=∑j((a-ρj)/|a-ρj|3)-a∑j≠N((1-ρj)/|1-ρj|3)+(2c/(a2+h2)3/2)-(2ca/(1+h2)3/2)a∑j((1-aρj)/|1-aρj|3)-a-2∑j≠N((1-ρj)/|1-ρj|3),
where (3.42), (3.43), (3.45) imply (1.10).Proof of Sufficient.
ForN≥2, under the suppositions of Theorem 1.4, then m1,m2,…,m2N+2 in a c.c. if and only if (3.42)–(3.43) hold, which are equivalent to (1.10). Hence the proof of the sufficient was finished.Proposition 3.9.
Under the hypothesis of Theorem1.5, for any ratios of masses b=m̃/m and c=m2N+1/m,m1,m2,…,m2N+2 may form a unique c.c. such that a∈(0,1) and h∈(0,+∞).Here is Theorem1.5.Proof.
Under the suppositions of the positions for masses, andh1=h2=:h,m1=m2=⋯=mN=:m,mN+1=m2=⋯=m̃2N=:m̃, and m2N+1=m2N+2, then for any ratios of masses b=m̃/m and c=m2N+1/m,m1,m2,…,m2N+2 are in a unique c.c. such that a∈(0,1),h>0, if and only if that (1.10)–(1.14) or (3.42)–(3.43) have a unique positive solution on 0<a<1 and h>0 for any given numbers b,c(>0).
Leta=x,h=y, and
(3.46)K(x,y)=x(∑j≠N1-ρj|1-ρj|3+b∑j1-xρj|1-xρj|3+2c(1+y2)3/2)-∑jx-ρj|x-ρj|3-b∑j≠Nx-xρj|x-xρj|3-2c(x2+y2)3/2,L(x,y)=(∑j≠N1-ρj|1-ρj|3+b∑j1-xρj|1-xρj|3+2c(1+y2)3/2)-N(1+y2)3/2-Nb(x2+y2)3/2-c4y3.
It suffices to prove that K(x,y)=0,and L(x,y)=0 have a unique positive solution (x,y) for any given ratios b,c(>0) in x∈(0,1), and y∈(0,+∞).
Now let(3.47)dj2=1+x2-2xcos⁡(2πjN),P(x)=∑j1dj3,Q(x)=∑jcos⁡(2πj/N)dj3,A=∑j≠N1-ρj|1-ρj|3={14(2∑j=1(N/2)-1csc⁡(πjN)+1)whenNiseven,12∑j=1(N-1)/2csc⁡(πjN)whenNisodd,Φ(x)=∑j1dj.
It follows from the definitions that
(3.48)Φ(x)=(1+x2)P(x)-2xQ(x),
and it implies
(3.49)P(x)-xQ(x)=Q(x)+x(Q(x)-xP(x)).
Since
(3.50)dΦdx=Q(x)-xP(x),
then now K and L in (3.46) can be written as follows:
(3.51)K(x,y)=(x-bx2)A+bx(P(x)-xQ(x))+(Q(x)-xP(x))+2cx(1+y2)3/2-2c(x2+y2)3/2=(x-bx2)A+bxΦ(x)+(1+bx2)dΦdx+2cx(1+y2)3/2-2c(x2+y2)3/2,L(x,y)=(∑j≠N1-ρj|1-ρj|3+b∑j1-xρj|1-xρj|3+2c(1+y2)3/2)-N(1+y2)3/2-Nb(x2+y2)3/2-c4y3=A+b[Q(x)+xdΦdx]+2c(1+y2)3/2-N(1+y2)3/2-Nb(x2+y2)3/2-c4y3,
where x=a∈(0,1). From Lemmas 2.8 and 2.9 and their proofs, and with implicit function theory, after some complex calculation (some ideas partially see [23]), we can prove that K=0, and L=0 have only one solution for any given ratios of masses b(>0) and c(>0) such that 0<x<1, and 0<h<+∞.



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*Source: 948356-2012-05-08.xml*