# The Periodic Solutions for Planar2N-Body Problems

Authors: Xiaohong Hu; Changrong Zhu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Based on the works of Perko and Walter, Moeckel and Simo, and Zhang and Zhou, we
study the necessary conditions and suffcient conditions for the uniformly rotating planar
nested regular polygonal periodic solutions for the2N-body problems.


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## Body

## 1. Introduction and Main Results

Letmj>0,qj∈R2 be the mass and position of the jth body. The Newtonian N-body problem concerns the motion of N point particles. The motion is governed by
(1)mjq¨j=∂U∂qj,j=1,2,…,N,
where U is the Newtonian potential:
(2)U=∑1≤j<k≤N‍mjmk|qj-qk|.In the famous paper [1], Perko and Walter proved the following result.Theorem 1 (see [1]).
ForN≥4, the N bodies move with uniformly angular velocity and locate on the vertices of regular N-gon if and only if m1=m2=⋯=mN.In 1995, Moeckel and Simó [2] studied planar nested 2N-body problems; they assume one regular N-gon is inscribed on a unit circle, the other on a circle with radius a, a>0 and a≠1; precisely, let ρj=e(2πi/N)j, where i2=-1; and the points mj and m~j locate at qj and q~j:
(3)qj=(ρj-z0)eiωt,q~j=(aρj-z0)eiωt,j=1,2,…,N,wherez0=(1/(M1+M2))∑‍(mjρj+am~jρj), M1=∑‍mj, M2=∑‍m~j, and M=M1+M2. For short, throughout this paper, all indices and summations will range from 1 to N unless we give other restrictions.For the nested2N-body problems, Moeckel and Simó proved the following theorem.Theorem 2 (see [2]).
Ifm1=⋯=mN∶=m and m~1=⋯=m~N∶=m~, then for every mass ratio r=m~/m, there are exactly two planar central configurations consisting of two nested regular N-gons. For one of them, the ratio of the sizes of the two nested N-gons is less than 1, and for the other it is greater than 1.In 2003, Zhang and Zhou [3] studied the inverse problem of Moeckel and Simó’s theorem [2], and they got the following results.Theorem 3 ([3], the case of θ=0).
If (3) is a solution of (1), then ω2 satisfies
(4)[14∑j≠N‍cscπjN-Nω2M]·[14a2∑j≠N‍cscπjN-aNω2M]=[∑‍1-aρj|1-aρj|3-Nω2M]·[∑‍a-ρj|a-ρj|3-aNω2M]
or equivalently
(5)ω2=(116a2(∑j≠N‍cscπjN)2-(∑‍1-aρj|1-aρj|3)(∑‍a-ρj|a-ρj|3))×(NM[14(a+1a2)(∑j≠N‍cscπjN)-a∑‍1-aρj|1-aρj|3-∑‍a-ρj|a-ρj|3])-1.And in [3], they proved the following theorem [3, Theorem 2]:Theorem 4.
Ifω satisfies (4) or (5) and (q1,…,qN;q~1,…,q~N) given by (3) is a periodic solution of (1), then m1=⋯=mN and m~1=⋯=m~N.In the proof of [3, Theorem 2], the authors claimed that the first eigenvalue λ1(AB-CD) of the matrix AB-CD is simple [3, page 2168]; this is not obvious. In fact, it seems very difficult to prove.Based on all the above works, we try to give strict proofs about the following two theorems. By the work of Moeckel and Simó [2], if we can get a periodic solution of the form given by (5) with a∈(0,1), the other periodic solution with radius 1/a can be obtained by symmetry. Hence, in the following, we only discuss the periodic solution with radius a∈(0,1). The other one with radius 1/a can be obtained by symmetry.Theorem 5.
ForN≥3, if a∈(a*,1), where a*∈(0,1), and (q1,…,qN,q~1,…,q~N) given by (3) is a periodic solution for (1), then m1=⋯=mN and m~1=⋯=m~N.Theorem 6.
Ifm1=⋯=mN∶=m,m~1=⋯=m~N∶=m~, and the constant ω>0 is given by
(6)ω2=m4∑j≠N‍cscπjN+m~∑‍1-a0cos(2πj/N)|1-a0ρj|3,
where μ=m~/m and a0 is a unique solution of the following equation:
(7)a3-μ4a2∑j≠N‍cscπjN-(1-μ)∑‍a|1-aρj|3+(1-a2μ)∑‍cos(2πj/N)|1-aρj|3=0,
then (q1,…,qN,q~1,…,q~N) with (m1,…,mN,m~1,…,m~N) is a periodic solution of (1) with angular velocity ω.Whenm1=⋯=mN and m~1=⋯=m~N, we have z0=0 which implies that the center of the masses locates at the origin. By Theorem 6, we get a periodic solution of (1) which rotates about the origin with radius a0∈(0,1). By symmetry, (1) has another periodic solution which rotates about the origin with radius 1/a0∈(1,∞).


## 2. The Proof of Theorem5

Substituting (3) into (1), we have
(8)-ω2(ρk-z0)eiωt=∑j≠k‍mj(qj-qk)|qj-qk|3+∑‍m~j(q~j-qk)|q~j-qk|3,-ω2(aρk-z0)eiωt=∑‍mj(qj-q~k)|qj-q~k|3+∑j≠k‍m~j(q~j-q~k)|q~j-q~k|3,wherek=1,2,…,N.By (3), (8) can be written as(9)-ω2(ρk-z0)=∑j≠k‍mj(ρj-ρk)|ρj-ρk|3+∑‍m~j(aρj-ρk)|aρj-ρk|3,-ω2(aρk-z0)=∑‍mj(ρj-aρk)|ρj-aρk|3+1a2∑j≠k‍m~j(ρj-ρk)|ρj-ρk|3,
where k=1,2,…,N. Equation (9) is equivalent to
(10)ω2ρk=(∑j≠k‍ρk-ρj|ρk-ρj|3mj+ω2M∑‍ρjmj)+(∑‍ρk-aρj|ρk-aρj|3m~j+aω2M∑‍ρjm~j),aω2ρk=(∑‍aρk-ρj|aρk-ρj|3mj+ω2M∑‍ρjmj)+(1a2∑j≠k‍ρk-ρj|ρk-ρj|3m~j+aω2M∑‍ρjm~j),
where k=1,2,…,N. Multiplying both sides of (10) by ρ-k, we get
(11)ω2=(∑j≠k‍1-ρj-k|1-ρj-k|3mj+ω2M∑‍ρj-kmj)+(∑‍1-aρj-k|1-aρj-k|3m~j+aω2M∑‍ρj-km~j),aω2=(∑‍a-ρj-k|a-ρj-k|3mj+ω2M∑‍ρj-kmj)+(1a2∑j≠k‍1-ρj-k|1-ρj-k|3m~j+aω2M∑‍ρj-km~j),wherek=1,2,…,N.LetA=(ak,j)N×N,B=(bk,j)N×N,C=(ck,j)N×N, and D=(dk,j)N×N, where
(12)ak,j∶={1-ρj-k|1-ρj-k|3,ifj≠k,0,otherwise,bk,j∶=1-aρj-k|1-aρj-k|3,ck,j∶=a-ρj-k|a-ρj-k|3,dk,j∶=ρj-k.
Then (11) can be rewritten in the following compact form:
(13)(A+ω2MD)m→+(B+aω2MD)m~→=ω21→,(C+ω2MD)m→+(1a2A+aω2MD)m~→=aω21→,
where m→=(m1,…,mN),m~→=(m~1,…,m~N), and 1→=(1,…,1).AnN×N matrix ℛ=(rj,k)N×N is called circulant (see [4]) if
(14)rj,k=rj-1,k-1,
where r0,k and rj,0 are equal to rN,k and rj,N, respectively. Let ρ0=1. If ℛ is a circulant matrix, its general formulas for the eigenvalues λk(ℛ) and eigenvectors vk(ℛ) are
(15)λk(ℛ)=∑j‍r1,jρk-1j-1,k=1,2,…,N,(16)vk(ℛ)=(ρk-1,ρk-12,…,ρk-1N),k=1,2,…,N.Remark 7.
From the formula (15), we have λ1(ℛ)=∑j=1N‍r1,j. The last equation implies that the eigenvalue λ1(ℛ) is equal to the summation of the first row of matrix ℛ; thus, we can get that the summation of any row, and hence, the summation of any column is equal to λ1(ℛ).According to the definition of circulant matrix, it is easy to check that the matrixesA+(ω2/M)D, B+(aω2/M)D,C+(ω2/M)D, and (1/a2)A+(aω2/M)D are circulant. For convenience, we introduce some notations. Let λk(1),λk(2),λk(3), and λk(4) be the kth eigenvalue of matrixes A+(ω2/M)D,  B+(aω2/M)D,C+(ω2/M)D, and (1/a2)A+(aω2/M)D, respectively. Then we have the following.Proposition 8.
All of the eigenvalues of matrixesA+(ω2/M)D,B+(aω2/M)D,C+(ω2/M)D, and (1/a2)A+(aω2/M)D are real.Proof.
We only give the proof for the matrixA+(ω2/M)D. The proofs for the rest are similar. Since A+(ω2/M)D is circulant and (ω2/M) is real number, we get from (12) that
(17)a-1,j+ω2Md-1,j=1-ρ-j-1|1-ρ-j-1|3+ω2Mρ-j-1=a1,N-j+2+ω2Md1,N-j+2,
where j=1,2,…,N. Thus, the matrix A+(ω2/M)D is Hermitian. We know that all the eigenvalues of A+(ω2/M)D are real since the eigenvalues of Hermitian matrix are real. The proof is completed.From the proof of Proposition8, we have known that A+(ω2/M)D,B+(aω2/M)D,C+(ω2/M)D, and (1/a2)A+(aω2/M)D are Hermitian. Thus, the vectors {v1,…,vN} defined by (16) are basis of CN. It is clear that m→∈CN and m~→∈CN. Let
(18)m→=a1v1+⋯+aNvN,m~→=b1v1+⋯+bNvN,
where aj,bj∈C. Substituting (18) into (13), we can get
(19)ω21→=∑‍ajλj(1)vj+∑‍bjλj(2)vj,aω21→=∑‍ajλj(3)vj+∑‍bjλj(4)vj.
Note that v1=1→. We can get from (19) that
(20)(a1λ1(1)+b1λ1(2)-ω2)v1+∑j=2‍(ajλj(1)+bjλj(2))vj=0,(a1λ1(3)+b1λ1(4)-aω2)v1+∑j=2‍(ajλj(3)+bjλj(4))vj=0.
Since v1,v2,…,vN are basis, we can get from (20) that
(21)a1λ1(1)+b1λ1(2)=ω2,a1λ1(3)+b1λ1(4)=aω2,(22)ajλj(1)+bjλj(2)=0,ajλj(3)+bjλj(4)=0,j=2,3,…,N.Lemma 9.
Ifλj(1)λj(4)-λj(2)λj(3)≠0,j=1,2,…,N-1, then m1=m2=⋯=mN and m~1= m~2= ⋯=m~N.Proof
Case1 (if λN(1)λN(4)-λN(2)λN(3)≠0). It is clear that
(23)λj(1)λj(4)-λj(2)λj(3)=|λj(1)λj(2)λj(3)λj(4)|≠0,j=1,2,…,N.
By Gramer's rule, we get from (22) and (23) that
(24)a2=⋯=aN=b2=⋯=bN=0.
Note that ω2 and aω2 are real. From Proposition 8, it is clear that λ1(1),λ1(2),λ1(3), and λ1(4) are real. Thus, we know from (21) and (23) for j=1 that a1 and b1 are real. Substituting (24) into (18), we get
(25)m→=a1v1,m~→=b1v1.
Thus, we have
(26)m1=m2=⋯=mN,m~1=⋯=m~N.
Case2. (if λN(1)λN(4)-λN(2)λN(3)=0). By the similar proof as to (24), we have
(27)a2=⋯=aN-1=b2=⋯=bN-1=0.
We can get from (18) and (27) that
(28)m→=a1v1+aNvN,m~→=b1v1+bNvN.
Since a1 and b1 are real, it follows from (28) that m→∈RN if and only if aN=0 or aNvN∈RN. If aNvN∈RN, from the general formulas of eigenvectors defined in (16), we know that
(29)vN=(ρN-1,ρN-12,…,ρN-1N-1,1).
Since aNvN=(aNρN-1,…,aNρN-1N-1,aN)∈RN, so aN∈R. Hence, m→∈RN if and only if aN=0 or vN∈RN. If vN∈RN, then ρN-1∈R from (29), which implies that sin(2π(N-1)/N)= -sin(2π/N)= 0. But it is impossible for N≥3. Thus, aN=0. Similarly, we can get bN=0. We get from (28) that
(30)m→=a1v1,m~→=b1v1.
Thus, we have
(31)m1=⋯=mN,m~1=⋯=m~N.
From Cases 1 and 2, the proof is completed.The rest of the proof is to verify the assumptions of Lemma9 by the special structure of our matrixes (12). In order to proceed the proof, we must study the eigenvalues in more details. Since ρj is the root of unity, it is easy to check that
(32)∑j‍ρjk={0,ifk≠NN,ifk=N.
Then from the general formulas of eigenvalue of circulant matrix, we have
(33)λk(1)=∑j‍(a1,j+ω2Md1,j)ρj-1k-1=∑j≠1‍ρj-1k-1-ρj-1k|1-ρj-1|3=∑j≠N‍ρjk-1-ρjk|1-ρj|3,
where k=1,2,…,N-1 and (32) is used. From Proposition 8, we know that the eigenvalue λk(1) is real. Thus, we only take the real part and get that
(34)λk(1)=∑j≠N‍cos(2πj/N)(k-1)-cos(2πj/N)k|1-ρj|3,fork=1,2,…,N-1.Note that|a-ρj|=|1-aρj|. Using similar method as proving (34), we have
(35)λk(2)=∑‍(b1j+aω2Md1j)ρj-1k-1=∑‍cos(2πj/N)(k-1)-acos(2πj/N)k|1-aρj|3,λk(3)=∑‍(c1j+ω2Md1j)ρj-1k-1=∑‍acos(2πj/N)(k-1)-cos(2πj/N)k|1-aρj|3,λk(4)=∑‍(1a2a1j+aω2Md1j)ρj-1k-1=1a2∑j≠N‍cos(2πj/N)(k-1)-cos(2πj/N)k|1-ρj|3,
where k=1,2,…,N-1.Lemma 10.
λ
k
(
1
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λ
k
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4
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λ
k
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2
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λ
k
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3
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=
λ
N
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k
+
1
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1
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λ
N
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k
+
1
(
4
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λ
N
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k
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1
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2
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λ
N
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+
1
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3
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,
k
=
2
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…
,
N
-
1.Proof.
We get from (34) that
(36)λN-k+1(1)=∑j≠N‍cos(2πj/N)k-cos(2πj/N)(k-1)|1-ρj|3=-λk(1),fork=2,3,…,N-1.
Similarly, we get from (35) that
(37)λN-k+1(2)=-λk(3),λN-k+1(3)=-λk(2),λN-k+1(4)=-λk(4),
where k=2,3,…,N-1. From (36) and (37), the result follows.Remark 11.
Let(38)ℓ={N2,ifNiseven,(N+1)2,ifNisodd.
From Lemma10, it is clear that to prove all the N-1 inequalities of Lemma 9 suffices to prove the ℓ inequalities λk(1)λk(4)-λk(2)λk(3)≠0,k=1,2,…,ℓ.With the similar proof as in [2, Lemma 2], we can get the following proposition.Proposition 12.
Letψβ,k(a)=∑j‍(cos(2πj/N)k/|1-aρj|β). Then for k=1,2,…,N,β>0 and a∈(0,1),  ψβ,k(a) and all of its derivatives are positive.Lemma 13.
ForN≥3, λ1(1)λ1(4)-λ1(2)λ1(3)>0.Proof.
From (34) and (35), we can get
(39)λ1(1)=14∑j≠N‍cscπjN,λ1(2)=∑‍1-acos(2πj/N)|1-aρj|3,λ1(3)=∑‍a-cos(2πj/N)|1-aρj|3,λ1(4)=14a2∑j≠N‍cscπjN.
Let A1(a)=λ1(1)λ1(4)-λ1(2)λ1(3). Then
(40)A1(a)=116a2(∑j≠N‍cscπjN)2-(∑‍1-acos(2πj/N)|1-aρj|3)×(∑‍a-cos(2πj/N)|1-aρj|3)=116a2(∑j≠N‍cscπjN)2-a(∑‍1|1-aρj|3)2-a(∑‍cos(2πj/N)(1-aρj)3)2+(1+a2)(∑‍1|1-aρj|3)(∑‍cos(2πj/N)|1-aρj|3).
Clearly, ∑‍(1/|1-aρj|3)>0. By Proposition 12, we get that ψ3,1= ∑‍(cos(2πj/N)/|1-aρj|3)>0 for a∈(0,1). Note that (1/4)∑j≠N‍csc(πj/N)=∑j≠N‍((1-cos(2πj/N))/|1-ρj|3)>0. We have from (40)
(41)A1(a)>116a2(∑j≠N‍cscπjN)2-a(∑‍1|1-aρj|3)2-a(∑‍cos(2πj/N)|1-aρj|3)2+2a(∑‍1|1-aρj|3)(∑‍cos(2πj/N)|1-aρj|3)=116a2(∑j≠N‍cscπjN)2-a(∑‍1|1-aρj|3-∑‍cos(2πj/N)|1-aρj|3)2=1a2{∑j≠N‍1-cos(2πj/N)|1-ρj|3nnnnnn-∑j≠N‍a3/2(1-cos(2πj/N))|1-aρj|3}·g(a)=1a2{∑j≠N‍(1-cos2πjN)·h(j,a)}·g(a),
where g(a)=(1/4)∑j≠N‍csc(πj/N)+∑j≠N‍(a3/2(1-cos(2πj/N))/|1-aρj|3) and h(j,a)= (1/|1-ρj|3)-(a3/2/|1-aρj|3). Note that h(j,a)=(1/|1-ρj|3)-(a3/2/(1+a2-2acos(2πj/N))3/2) and hence
(42)dh(j,a)da=(3/2)a1/2(a-1)(a+1)|1-aρj|5<0,fora∈(0,1).
We know from (42) that h(j,a) is decreasing function in a for a∈(0,1). It is easy to check that h(j,1)=0. Thus
(43)h(j,a)>h(j,1)=0,fora∈(0,1),j=1,…,N-1.
So, ∑j≠N‍(1-cos(2πj/N))·h(j,a)>0 since h(j,a)>0 and 1-cos(2πj/N)>0. Note that g(a)>0 for a∈(0,1). From (41), we can get A1(a)>0. The result follows.Fork=1,2,…,ℓ, let
(44)αk(a)=∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-aρj|3.Lemma 14.
IfN is odd, then α(N+1)/2(1)=0; If N is even, then αN/2(1)>0.Proof.
IfN is odd, by the definition of αk, we have
(45)α(N+1)/2(1)=∑j≠N‍(cos2πj(((N+1)/2)-1)Nnnnnnnnnn-cos2πj((N+1)/2)N)×(|1-ρj|3)-1=∑j≠N‍cos(πj-(πj/N))-cos(πj+(πj/N))|1-ρj|3=0.
If N is even, we get that
(46)αN/2(1)=∑j≠N‍cos(2πj((N/2)-1)/N)-cos(2πj(N/2)/N)|1-ρj|3=14∑j≠N‍(-1)j+11sin(πj/N)=14[∑j=1N/2‍(-1)j+11sin(πj/N)nnnnnnn+∑j=1N/2-1‍(-1)j+11sin(πj/N)].
Case1. If N/2 is even, then
(47)αN/2(1)=14[2∑j=1N/2‍(-1)j+11sin(πj/N)+1].
Since the signs of the terms in the sum∑j=1N/2‍(-1)j+1(1/sin(πj/N)) alternate and 1/sin(πj/N) is decreasing in j for 1≤j≤N/2, we get that ∑j=1N/2‍(-1)j+1(1/sin(πj/N))>0 and hence αN/2(1)>0.
Case2. If N/2 is odd, then
(48)αN/2(1)=14[2∑j=1N/2-1‍(-1)j+11sin(πj/N)+1].
Since the signs of the terms in summation∑j=1N/2-1‍(-1)j+1(1/sin(πj/N)) alternate and 1/sin(πj/N) is decreasing in j for 1≤j≤N/2-1, we get that ∑j=1(N/2)-1‍(-1)j+1(1/sin(πj/N))>0 and hence αN/2(1)>0.
By Cases 1 and 2, we see thatαN/2(1)>0 for even N.Proposition 15.
Fork=1,2,…,ℓ-1, then αk(1)>0.Proof.
Clearly,α1(1)>0. We need to prove αk(1)>0,k≥2. Note that
(49)2sin(πj/N)∑m=0k-1‍cos2πjNm=2sin(πjk/N)cos(πj(k-1)/N)sin2(πj/N)=sin(πj(2k-1)/N)sin2(πj/N)+1sin(πj/N).
Hence, by the definition of αk(a), we have
(50)αk(1)=∑j≠N‍2sin(πj(2k-1)/N)sin(πj/N)(2-2cos(2πj/N))3/2=14∑j≠N‍[-1sin(πj/N)nnnnnnnn+2sin(πj/N)∑m=0k-1‍cos2πjNm].
Then, we get that
(51)(αk(1)-αk-1(1))-(αk+1(1)-αk(1))=12∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)sin(πj/N)=∑j≠N‍sinπj(2k-1)N=cotπ(2k-1)2N.
Notice that 0<π(2k-1)/(2N)<π/2 for k=2,…,ℓ-1. Hence, cotπ(2k-1)/2N>0 for k=2, …, ℓ-1. Then we have
(52)αk(1)-αk-1(1)>αk+1(1)-αk(1).By (52), we get that αk-1(1)+αk+1(1)<2αk(1) which implies that αk+1(1)<αk(1) or αk-1(1)<αk(1). If there is αk(1)≤0, we use two cases to proceed our proof.
Case1. (If αk-1(1)<αk(1)). By (52), we have αk-2(1)-αk-1(1)<αk-1(1)-αk(1)<0 and hence αk-2(1)<αk-1(1)<αk(1)≤0. By inductions, we have α1(1)<⋯<αk(1)<0 which contradicts with α1(1)>0.
Case2. (If αk+1(1)<αk(1)). By (52), we have αk+2(1)-αk+1(1)<αk+1(1)-αk(1)<0. Then, αk+2(1)<αk+1(1)<αk(1)≤0. By inductions, we get that for even N, αN/2(1)< α(N/2)-1(1)< ⋯<αk(1)<0 which contradicts with αN/2(1)>0; for odd N, α(N+1)/2(1)<α(N-1)/2(1)<⋯< αk(1)≤0 which contradicts with α(N+1)/2(1)=0.
From Cases 1 and 2, we see that there is always a contradiction if there existsαk(1)≤0. Hence, αk(1)>0,k=1,…,ℓ.Let(53)ℓ~={N2,ifNiseven,(N-1)2,ifNisodd,a^*=mina∈(0,1),αk(a)>0,k=1,2,…,ℓ~a.By the continuity, Lemma14 and Proposition 15, we see that a^*∈(0,1).Fork=1,2,…,ℓ, let
(54)βk(a)=∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-aρj|5.Similar to Lemma14, we haveLemma 16.
IfN is odd, then β(N+1)/2(1)=0; If N is even, then βN/2(1)>0.Proposition 17.
Fork=1,2,…,ℓ-1, then βk(1)>0.Proof.
Clearly,β1(1)>0. We need to prove βk(1)>0,k≥2. Note that
(55)2sin3(πj/N)∑m=0k-1‍cos2πjNm=1sin3(πj/N)·2sin(πjk/N)cos(πj(k-1)/N)sin(πj/N)=sin(πj(2k-1)/N)sin4(πj/N)+1sin3(πj/N).
By the definition of βk(1), we have
(56)βk(1)=116∑j≠N‍[-1sin3(πj/N)nnnnnnnnn+2sin3(πj/N)∑m=0k-1‍cos2πjNm].
Then, we get that
(57)(βk(1)-βk-1(1))-(βk+1(1)-βk(1))=18∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)sin3(πj/N)=18αk(1)>0.
By similar proofs as in Proposition 15, we can get βk>0. The proof is completed.Let(58)a~*=mina∈(0,1),βk(a)>0,k=1,2,…,ℓ~a.
By the continuity, Lemma 16 and Proposition 17, we see that a~*∈(0,1).Lemma 18.
Fork=2,3,…,ℓ and a*=max{a^*,a~*}, λk(1)λk(4)-λk(2)λk(3)>0.Proof.
LetAk(a)=λk(1)λk(4)-λk(2)λk(3). From (34) and (35), we can get
(59)Ak(a)=1a2(∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-ρj|3)2-(∑‍cos(2πj(k-1)/N)-acos(2πjk/N)|1-aρj|3)·(∑‍acos(2πj(k-1)/N)-cos(2πjk/N)|a-ρj|3).
By the definitions of ℓ and ℓ~, we see that ℓ~=ℓ if N is even and ℓ~=ℓ-1 if N is odd. For odd N, we get
(60)A(N+1)/2(a)=1a2(∑j≠N‍(cos2πj(((N+1)/2)-1)Nnnnnn-cos2πj((N+1)/2)N)×|1-ρj|-3)2-(∑j≠N‍∑(cos2πj(((N+1)/2)-1)N‍nnnnn-acos2πj((N+1)/2)N)×|1-aρj|-3∑j≠N‍)·(∑j≠N‍∑‍(acos2πj(((N+1)/2)-1)Nnnnnn-cos2πj((N+1)/2)N)×|a-ρj|-3∑j≠N‍)=(1-a)2[∑‍(-1)jcos(πj/N)|1-aρj|3]2.
Let f(x)=cosx/(1+a2-2acos2x)3/2 for 0<x<π/2. Then
(61)f′(x)=-sinx(1+a2-2acos2x)-6acosxsin2x(1+a2-2acos2x)5/2<0.
Hence,cos(πj/N)(a-1)/|a-ρj|3 is decreasing in j for 0<πj/N<π/2. Note that the signs of the terms alternate, then the summation is negative for 0<πj/N<π/2. By symmetry, the summation for the rest terms is also negative. Hence, we have
(62)A(N+1)/2(a)>0.
For anyk,k=2,3,…,ℓ~, let
(63)G(a)=∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-ρj|3-a3/2∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-aρj|3.
With direct computation, we have
(64)G′(a)=32a1/2(a+1)(a-1)×∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-aρj|5<0,nnnnnnnnnnnnnnnnnnnnnnnnnnfora∈(a*,1),
where Proposition 17 is used. It follows from (64) that G(a) is decreasing for a∈(a*,1). Note that G(1)=0. So, G(a)>G(1)=0 for a∈(0,1). It is that
(65)∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-ρj|3>a3/2∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-aρj|3,
where a∈(a*,1) and k=2,3,…,ℓ~. From (65), we can get
(66)1a2(∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-ρj|3)2>a(∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-aρj|3)2,
where a∈(a*,1) and k=2,3,…,ℓ~.
By (59), we can get
(67)Ak(a)=1a2(∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-ρj|3)2-a(∑‍cos(2πj(k-1)/N)|1-aρj|3)2+(1+a2)(∑‍cos(2πj(k-1)/N)|1-aρj|3)×(∑‍cos(2πjk/N)|1-aρj|3)-a(∑‍cos(2πjk/N)|1-aρj|3)2.
From Proposition 12, we have
(68)Ak(a)>1a2(∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-ρj|3)2-a{(∑‍cos(2πj(k-1)/N)|1-aρj|3)2mmmmm-2(∑‍cos(2πj(k-1)/N)|1-aρj|3)mmmmm×(∑‍cos(2πjk/N)|1-aρj|3)mmmmm+(∑‍cos(2πjk/N)|1-aρj|3)2}=1a2(∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-ρj|3)2-a(∑j≠N‍cos(2πj(k-1)/N)-cos(2πjk/N)|1-aρj|3)2>0,
where (66) is used and k=2,3,…,ℓ~.
By (62) and (68), the proof is completed.The following example illustrate that thea* in Lemma 18 can be obtained for some special cases.Example 19.
IfN=3, then a*=0.Proof.
ForN=3, we have
(69)α1(a)=2((1+(a/2))2+(3/4))3/2>0,α2(a)=1((1+(a/2))2+(3/4))3/2>0.
By the definition ofa^*, we have a^*=0.
Similarly, we get that(70)β1(a)=2((1+(a/2))2+(3/4))5/2>0,β2(a)=1((1+(a/2))2+(3/4))5/2>0.
By the definition ofa~*, we have a~*=0. Hence, a*=0 for N=3.Lemmas13 and 18 show that the hypotheses of Lemma 9 are satisfied. Thus the proof of the Theorem 5 was completed.


## 3. The Proof of Theorem6

Sincem1=m2=⋯=mN=m and m~1=m~2=⋯=m~N=m~, we get that z0=0. Hence, by (3), we have
(71)qk=ρkeiωt,q~k=aρkeiωt.
Substituting qk and q~k into (1), then by (11) we get that
(72)ω2=m4∑j≠N‍cscπjN+m~∑‍1-acos(2πj/N)|1-aρj|3,aω2=m∑‍a-cos(2πj/N)|1-aρj|3+m~4a2∑j≠N‍cscπjN.From (72), we have
(73)ω2=m4∑j≠N‍cscπjN+m~∑‍1-acos(2πj/N)|1-aρj|3,ω2=1a[m∑‍a-cos(2πj/N)|1-aρj|3+m~4a2∑j≠N‍cscπjN].
From (73), the constant ω2 exists only if the following equation holds:
(74)a{m4∑j≠N‍cscπjN+m~∑‍1-acos(2πj/N)|1-aρj|3}=m∑‍a-cos(2πj/N)|1-aρj|3+m~4a2∑j≠N‍cscπjN.
Note that μ=m~/m. Then we get from (74) that
(75)a3-μ4a2∑j≠N‍cscπjN-(1-μ)af(a)+(1-a2μ)g(a)=0,
where f(a)=∑‍(1/|1-aρj|3) and g(a)=∑‍(cos(2πj/N)/|1-aρj|3).Lemma 20.
Forμ>0 and N≥3, the constant ω2 can be given by
(76)ω2=m4∑j≠N‍cscπjN+m~∑‍1-a0cos(2πj/N)|1-a0ρj|3,
where a0 is the unique solution of (75).Proof.
By the relationship between (74) and (75), it suffices to prove that (75) has solution a=a0 for μ>0 and N≥3. Let
(77)F(a)=(a-μa2)14∑j≠N‍cscπjN-(1-μ)af(a)+(1-a2μ)g(a).
From (75) and (77), the roots of (75) are zeros of (77). From the definitions of ψβ,k(a) defined in Proposition 12, we have ψ1,N(a)=∑‍(1/|1-aρj|). Through direct calculations, we find that
(78)ψ1,N(a)=(1+a2)f(a)-2ag(a),dψ1,N(a)da=-af(a)+g(a).
From (78), (77) can be written as
(79)F(a)=(a-μa2)14∑j≠N‍cscπjN+(1+a2μ)dψ1,N(a)da+μaψ1,N(a).
Notice that ψ1,N(a)=∑j≠N‍(1/|1-aρj|)+(1/|1-aρN|)=∑j≠N‍(1/|1-aρj|)+(1/|1-a|). Hence, we have
(80)lima→1ψ1,N(a)=+∞.
From (79), (80), and Proposition 12, we can get that
(81)lima→0F(a)=-∞,lima→1F(a)=+∞.
On the other hand, by (79), we get that
(82)F′(a)=(1+2μa3)14∑j≠N‍cscπjN+μψ1,N(a)+3aμdψ1,N(a)da+(1+a2μ)d2ψ1,N(a)da2.
From (82) and Proposition 12, we know that F′(a)>0 for a∈(0,1). This, together with (81), implies that there is a unique solution a=a0 for F(a)=0. The proof is completed.



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*Source: 397194-2013-04-27.xml*