The Periodic Solutions for Planar 2𝑁-Body Problems

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 the 2𝑁-body problems.

1. Introduction and Main Results

Let 𝑚_𝑗>0,𝑞_𝑗∈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 𝑚_𝑗𝑞̈_𝑗=𝜕𝑈𝜕𝑞_𝑗,𝑗=1,2,…,𝑁, (1) where 𝑈 is the Newtonian potential: 𝑈=_{1≤𝑗<𝑘≤𝑁}𝑚_𝑗𝑚_𝑘𝑞_𝑗−𝑞_𝑘. (2)

In the famous paper [1], Perko and Walter proved the following result.

Theorem 1 (see [1]). For 𝑁≥4, the 𝑁 bodies move with uniformly angular velocity and locate on the vertices of regular 𝑁-gon if and only if 𝑚₁=𝑚₂=⋯=𝑚_𝑁.

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 𝜌_𝑗=𝑒^{(2𝜋𝑖/𝑁)𝑗}, where 𝑖²=−1; and the points 𝑚_𝑗 and 𝑚^∼_𝑗 locate at 𝑞_𝑗 and 𝑞^∼_𝑗: 𝑞_𝑗=𝜌_𝑗−𝑧₀𝑒^𝑖𝜔𝑡,𝑞^∼_𝑗=𝑎𝜌_𝑗−𝑧₀𝑒^𝑖𝜔𝑡,𝑗=1,2,…,𝑁, (3)

where 𝑧₀=(1/(𝑀₁+𝑀₂))∑(𝑚_𝑗𝜌_𝑗+a𝑚^∼_𝑗𝜌_𝑗), 𝑀₁=∑𝑚_𝑗, 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 nested 2𝑁-body problems, Moeckel and Simó proved the following theorem.

Theorem 2 (see [2]). If 𝑚₁=⋯=𝑚_𝑁∶=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 14_𝑗≠𝑁csc𝜋𝑗𝑁−𝑁𝜔²𝑀⋅14𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁−𝑎𝑁𝜔²𝑀=1−𝑎𝜌_𝑗1−𝑎𝜌_𝑗³−𝑁𝜔²𝑀⋅𝑎−𝜌_𝑗𝑎−𝜌_𝑗³−𝑎𝑁𝜔²𝑀 (4) or equivalently 𝜔²=116𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁²−1−𝑎𝜌_𝑗1−𝑎𝜌_𝑗³𝑎−𝜌_𝑗𝑎−𝜌_𝑗³×𝑁𝑀141𝑎+𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁−𝑎1−𝑎𝜌_𝑗1−𝑎𝜌_𝑗³−𝑎−𝜌_𝑗𝑎−𝜌_𝑗³⁻¹. (5)

And in [3], they proved the following theorem [3, Theorem 2]:

Theorem 4. If 𝜔 satisfies (4) or (5) and (𝑞₁,…,𝑞_𝑁;𝑞^∼₁,…,𝑞^∼_𝑁) given by (3) is a periodic solution of (1), then 𝑚₁=⋯=𝑚_𝑁 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. For 𝑁≥3, if 𝑎∈(𝑎^∗,1), where 𝑎^∗∈(0,1), and (𝑞₁,…,𝑞_𝑁,𝑞^∼₁,…,𝑞^∼_𝑁) given by (3) is a periodic solution for (1), then 𝑚₁=⋯=𝑚_𝑁 and 𝑚^∼₁=⋯=𝑚^∼_𝑁.

Theorem 6. If 𝑚₁=⋯=𝑚_𝑁∶=𝑚,𝑚^∼₁=⋯=𝑚^∼_𝑁∶=𝑚^∼, and the constant 𝜔>0 is given by 𝜔²=𝑚4_𝑗≠𝑁csc𝜋𝑗𝑁+𝑚^∼1−𝑎₀cos2𝜋𝑗/𝑁1−𝑎₀𝜌_𝑗³, (6) where 𝜇=𝑚^∼/𝑚 and 𝑎₀ is a unique solution of the following equation: 𝑎³−𝜇4𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁−𝑎1−𝜇1−𝑎𝜌_𝑗³+1−𝑎²𝜇cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³=0, (7) then (𝑞₁,…,𝑞_𝑁,𝑞^∼₁,…,𝑞^∼_𝑁) with (𝑚₁,…,𝑚_𝑁,𝑚^∼₁,…,𝑚^∼_𝑁) is a periodic solution of (1) with angular velocity 𝜔.

When 𝑚₁=⋯=𝑚_𝑁 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 Theorem 5

Substituting (3) into (1), we have −𝜔²𝜌_𝑘−𝑧₀𝑒^𝑖𝜔𝑡=_𝑗≠𝑘𝑚_𝑗𝑞_𝑗−𝑞_𝑘𝑞_𝑗−𝑞_𝑘³+^∼𝑚𝑞_𝑗^∼_𝑗−𝑞_𝑘∼𝑞_𝑗−𝑞_𝑘³,−𝜔²𝑎𝜌_𝑘−𝑧₀𝑒^𝑖𝜔𝑡𝑚=^𝑗𝑞_𝑗−𝑞^∼_𝑘𝑞_𝑗−𝑞^∼_𝑘³+^∼_𝑗≠𝑘𝑚𝑞_𝑗^∼_𝑗−𝑞^∼_𝑘∼𝑞_𝑗−𝑞^∼_𝑘³, (8)

where 𝑘=1,2,…,𝑁.

By (3), (8) can be written as−𝜔²𝜌_𝑘−𝑧₀=_𝑗≠𝑘𝑚_𝑗𝜌_𝑗−𝜌_𝑘𝜌_𝑗−𝜌_𝑘³+^∼𝑚_𝑗𝑎𝜌_𝑗−𝜌_𝑘𝑎𝜌_𝑗−𝜌_𝑘³,−𝜔²𝑎𝜌_𝑘−𝑧₀𝑚=^𝑗𝜌_𝑗−𝑎𝜌_𝑘𝜌_𝑗−𝑎𝜌_𝑘³+1𝑎^2∼_𝑗≠𝑘𝑚_𝑗𝜌_𝑗−𝜌_𝑘𝜌_𝑗−𝜌_𝑘³, (9) where 𝑘=1,2,…,𝑁. Equation (9) is equivalent to 𝜔²𝜌_𝑘=_𝑗≠𝑘𝜌_𝑘−𝜌_𝑗𝜌_𝑘−𝜌_𝑗³𝑚_𝑗+𝜔²𝑀𝜌_𝑗𝑚_𝑗+𝜌_𝑘−𝑎𝜌_𝑗𝜌_𝑘−𝑎𝜌_𝑗^3∼𝑚_𝑗+𝑎𝜔²𝑀𝜌𝑚_𝑗^∼_𝑗,𝑎𝜔²𝜌_𝑘=𝑎𝜌_𝑘−𝜌_𝑗𝑎𝜌_𝑘−𝜌_𝑗³𝑚_𝑗+𝜔²𝑀𝜌_𝑗𝑚_𝑗+1𝑎²_𝑗≠𝑘𝜌_𝑘−𝜌_𝑗𝜌_𝑘−𝜌_𝑗^3∼𝑚_𝑗+𝑎𝜔²𝑀𝜌𝑚_𝑗^∼_𝑗, (10) where 𝑘=1,2,…,𝑁. Multiplying both sides of (10) by 𝜌_−𝑘, we get 𝜔²=_𝑗≠𝑘1−𝜌_𝑗−𝑘1−𝜌_𝑗−𝑘³𝑚_𝑗+𝜔²𝑀𝜌_𝑗−𝑘𝑚_𝑗+1−𝑎𝜌_𝑗−𝑘1−𝑎𝜌_𝑗−𝑘^3∼𝑚_𝑗+𝑎𝜔²𝑀𝜌^∼_𝑗−𝑘𝑚_𝑗,𝑎𝜔²=𝑎−𝜌_𝑗−𝑘𝑎−𝜌_𝑗−𝑘³𝑚_𝑗+𝜔²𝑀𝜌_𝑗−𝑘𝑚_𝑗+1𝑎²_𝑗≠𝑘1−𝜌_𝑗−𝑘1−𝜌_𝑗−𝑘^3∼𝑚_𝑗+𝑎𝜔²𝑀𝜌^∼_𝑗−𝑘𝑚_𝑗, (11)

where 𝑘=1,2,…,𝑁.

Let 𝐴=(𝑎_𝑘,𝑗)_𝑁×𝑁,𝐵=(𝑏_𝑘,𝑗)_𝑁×𝑁,𝐶=(𝑐_𝑘,𝑗)_𝑁×𝑁, and 𝐷=(𝑑_𝑘,𝑗)_𝑁×𝑁, where 𝑎_𝑘,𝑗∶=1−𝜌_𝑗−𝑘1−𝜌_𝑗−𝑘³,if𝑗≠𝑘,0,otherwise,𝑏_𝑘,𝑗∶=1−𝑎𝜌_𝑗−𝑘1−𝑎𝜌_𝑗−𝑘³,𝑐_𝑘,𝑗∶=𝑎−𝜌_𝑗−𝑘𝑎−𝜌_𝑗−𝑘³,𝑑_𝑘,𝑗∶=𝜌_𝑗−𝑘. (12) Then (11) can be rewritten in the following compact form: 𝜔𝐴+²𝑀𝐷𝑚^→+𝐵+𝑎𝜔²𝑀𝐷𝑚^→∼=𝜔^2→1,𝜔𝐶+²𝑀𝐷𝑚^→+1𝑎²𝐴+𝑎𝜔²𝑀𝐷𝑚^→∼=𝑎𝜔^2→1, (13) where 𝑚^→=(𝑚₁,…,𝑚_𝑁),𝑚^→∼=(𝑚^∼₁,…,𝑚^∼_𝑁), and ^→1=(1,…,1).

An 𝑁×𝑁 matrix ℛ=(𝑟_𝑗,𝑘)_𝑁×𝑁 is called circulant (see [4]) if 𝑟_𝑗,𝑘=𝑟_{𝑗−1,𝑘−1}, (14) where 𝑟_0,𝑘 and 𝑟_𝑗,0 are equal to 𝑟_𝑁,𝑘 and 𝑟_𝑗,𝑁, respectively. Let 𝜌₀=1. If ℛ is a circulant matrix, its general formulas for the eigenvalues 𝜆_𝑘(ℛ) and eigenvectors 𝑣_𝑘(ℛ) are 𝜆_𝑘(ℛ)=_𝑗𝑟_1,𝑗𝜌^𝑗−1_𝑘−1,𝑘=1,2,…,𝑁,𝑣_𝑘(ℛ)=𝜌_𝑘−1,𝜌²_𝑘−1,…,𝜌^𝑁_𝑘−1,𝑘=1,2,…,𝑁. (15)(16)

Remark 7. From the formula (15), we have 𝜆₁(ℛ)=∑^𝑁_𝑗=1𝑟_1,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 matrixes 𝐴+(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 𝜆_𝑘⁽¹⁾,𝜆_𝑘⁽²⁾,𝜆_𝑘⁽³⁾, and 𝜆_𝑘⁽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 matrixes 𝐴+(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 matrix 𝐴+(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 𝑎_1,𝑗+𝜔²𝑀𝑑_1,𝑗=1−𝜌_𝑗−11−𝜌_𝑗−1³+𝜔²𝑀𝜌_𝑗−1=𝑎_{1,𝑁−𝑗+2}+𝜔²𝑀𝑑_{1,𝑁−𝑗+2}, (17) 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 Proposition 8, 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,…,𝑣_𝑁} defined by (16) are basis of 𝐂^𝑁. It is clear that 𝑚^→∈𝐂^𝑁 and 𝑚^→∼∈𝐂^𝑁. Let 𝑚^→=𝑎₁𝑣₁+⋯+𝑎_𝑁𝑣_𝑁,𝑚^→∼=𝑏₁𝑣₁+⋯+𝑏_𝑁𝑣_𝑁, (18) where 𝑎_𝑗,𝑏_j∈𝐂. Substituting (18) into (13), we can get 𝜔^2→1=𝑎_𝑗𝜆_𝑗⁽¹⁾𝑣_𝑗+𝑏_𝑗𝜆_𝑗⁽²⁾𝑣_𝑗,𝑎𝜔^2→1=𝑎_𝑗𝜆_𝑗⁽³⁾𝑣_𝑗+𝑏_𝑗𝜆_𝑗⁽⁴⁾𝑣_𝑗. (19) Note that 𝑣₁=^→1. We can get from (19) that 𝑎₁𝜆₁⁽¹⁾+𝑏₁𝜆₁⁽²⁾−𝜔²𝑣₁+_𝑗=2𝑎_𝑗𝜆_𝑗⁽¹⁾+𝑏_𝑗𝜆_𝑗⁽²⁾𝑣_𝑗𝑎=0,₁𝜆₁⁽³⁾+𝑏₁𝜆₁⁽⁴⁾−𝑎𝜔²𝑣₁+_𝑗=2𝑎_𝑗𝜆_𝑗⁽³⁾+𝑏_𝑗𝜆_𝑗⁽⁴⁾𝑣_𝑗=0. (20) Since 𝑣₁,𝑣₂,…,𝑣_𝑁 are basis, we can get from (20) that 𝑎₁𝜆₁⁽¹⁾+𝑏₁𝜆₁⁽²⁾=𝜔²,𝑎₁𝜆₁⁽³⁾+𝑏₁𝜆₁⁽⁴⁾=𝑎𝜔²,𝑎_𝑗𝜆_𝑗⁽¹⁾+𝑏_𝑗𝜆_𝑗⁽²⁾𝑎=0,_𝑗𝜆_𝑗⁽³⁾+𝑏_𝑗𝜆_𝑗⁽⁴⁾=0,𝑗=2,3,…,𝑁. (21)(22)

Lemma 9. If 𝜆_𝑗⁽¹⁾𝜆_𝑗⁽⁴⁾−𝜆_𝑗⁽²⁾𝜆_𝑗⁽³⁾≠0,𝑗=1,2,…,𝑁−1, then 𝑚₁=𝑚₂=⋯=𝑚_𝑁 and 𝑚^∼₁= 𝑚^∼₂= ⋯=𝑚^∼_𝑁.

Proof
Case 1 (if 𝜆_𝑁⁽¹⁾𝜆_𝑁⁽⁴⁾−𝜆_𝑁⁽²⁾𝜆_𝑁⁽³⁾≠0). It is clear that 𝜆_𝑗⁽¹⁾𝜆_𝑗⁽⁴⁾−𝜆_𝑗⁽²⁾𝜆_𝑗⁽³⁾=𝜆_𝑗⁽¹⁾𝜆_𝑗⁽²⁾𝜆_𝑗⁽³⁾𝜆_𝑗⁽⁴⁾≠0,𝑗=1,2,…,𝑁. (23) By Gramer's rule, we get from (22) and (23) that 𝑎₂=⋯=𝑎_𝑁=𝑏₂=⋯=𝑏_𝑁=0. (24) 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 𝑚^→=𝑎₁𝑣₁,𝑚^→∼=𝑏₁𝑣₁. (25) Thus, we have 𝑚₁=𝑚₂=⋯=𝑚_𝑁,𝑚^∼₁=⋯=𝑚^∼_𝑁. (26)
Case 2. (if 𝜆_𝑁⁽¹⁾𝜆_𝑁⁽⁴⁾−𝜆_𝑁⁽²⁾𝜆_𝑁⁽³⁾=0). By the similar proof as to (24), we have 𝑎₂=⋯=𝑎_𝑁−1=𝑏₂=⋯=𝑏_𝑁−1=0. (27) We can get from (18) and (27) that 𝑚^→=𝑎₁𝑣₁+𝑎_𝑁𝑣_𝑁,𝑚^→∼=𝑏₁𝑣₁+𝑏_𝑁𝑣_𝑁. (28) Since 𝑎₁ and 𝑏₁ are real, it follows from (28) that 𝑚^→∈𝐑^𝑁 if and only if 𝑎_𝑁=0 or 𝑎_𝑁𝑣_𝑁∈𝐑^𝑁. If 𝑎_𝑁𝑣_𝑁∈𝐑^𝑁, from the general formulas of eigenvectors defined in (16), we know that 𝑣_𝑁=𝜌_𝑁−1,𝜌²_𝑁−1,…,𝜌^𝑁−1_𝑁−1,1. (29) Since 𝑎_𝑁𝑣_𝑁=(𝑎_𝑁𝜌_𝑁−1,…,𝑎_𝑁𝜌^𝑁−1_𝑁−1,𝑎_𝑁)∈𝐑^𝑁, so 𝑎_𝑁∈𝐑. Hence, 𝑚^→∈𝐑^𝑁 if and only if 𝑎_𝑁=0 or 𝑣_𝑁∈𝐑^𝑁. If 𝑣_𝑁∈𝐑^𝑁, then 𝜌_𝑁−1∈𝐑 from (29), which implies that sin(2𝜋(𝑁−1)/𝑁)= −sin(2𝜋/𝑁)= 0. But it is impossible for 𝑁≥3. Thus, 𝑎_𝑁=0. Similarly, we can get 𝑏_𝑁=0. We get from (28) that 𝑚^→=𝑎₁𝑣₁,𝑚^→∼=𝑏₁𝑣₁. (30) Thus, we have 𝑚₁=⋯=𝑚_𝑁,𝑚^∼₁=⋯=𝑚^∼_𝑁. (31)
From Cases 1 and 2, the proof is completed.

The rest of the proof is to verify the assumptions of Lemma 9 by the special structure of our matrixes (12). In order to proceed the proof, we must study the eigenvalues in more details. Since 𝜌_𝑗 is the root of unity, it is easy to check that _𝑗𝜌^𝑘_𝑗=0,if𝑘≠𝑁𝑁,if𝑘=𝑁. (32) Then from the general formulas of eigenvalue of circulant matrix, we have 𝜆_𝑘⁽¹⁾=_𝑗𝑎_1,𝑗+𝜔²𝑀𝑑_1,j𝜌^𝑘−1_𝑗−1=_𝑗≠1𝜌^𝑘−1_𝑗−1−𝜌^𝑘_𝑗−11−𝜌_𝑗−1³=_𝑗≠𝑁𝜌_𝑗^𝑘−1−𝜌^𝑘_𝑗1−𝜌_𝑗³, (33) where 𝑘=1,2,…,𝑁−1 and (32) is used. From Proposition 8, we know that the eigenvalue 𝜆_𝑘⁽¹⁾ is real. Thus, we only take the real part and get that 𝜆_𝑘⁽¹⁾=_𝑗≠𝑁cos2𝜋𝑗/𝑁(𝑘−1)−cos2𝜋𝑗/𝑁𝑘1−𝜌_𝑗³,for𝑘=1,2,…,𝑁−1. (34)

Note that |𝑎−𝜌_𝑗|=|1−𝑎𝜌_𝑗|. Using similar method as proving (34), we have 𝜆_𝑘⁽²⁾=𝑏_1𝑗+𝑎𝜔²𝑀𝑑_1𝑗𝜌^𝑘−1_𝑗−1()𝑘=cos2𝜋𝑗/𝑁𝑘−1−𝑎cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³,𝜆_𝑘⁽³⁾=𝑐_1𝑗+𝜔²𝑀𝑑_1𝑗𝜌^𝑘−1_𝑗−1=acos2𝜋𝑗/𝑁(𝑘−1)−cos2𝜋𝑗/𝑁𝑘1−𝑎𝜌_𝑗³,𝜆_𝑘⁽⁴⁾1=𝑎²𝑎_1𝑗+𝑎𝜔²𝑀𝑑_1𝑗𝜌^𝑘−1_𝑗−1=1𝑎²_𝑗≠𝑁cos2𝜋𝑗/𝑁(𝑘−1)−cos2𝜋𝑗/𝑁𝑘1−𝜌_𝑗³, (35) where 𝑘=1,2,…,𝑁−1.

Lemma 10. 𝜆_𝑘⁽¹⁾𝜆_𝑘⁽⁴⁾−𝜆_𝑘⁽²⁾𝜆_𝑘⁽³⁾=𝜆⁽¹⁾_{𝑁−𝑘+1}𝜆⁽⁴⁾_{𝑁−𝑘+1}−𝜆⁽²⁾_{𝑁−𝑘+1}𝜆⁽³⁾_{𝑁−𝑘+1},𝑘=2,…,𝑁−1.

Proof. We get from (34) that 𝜆⁽¹⁾_{𝑁−𝑘+1}=_𝑗≠𝑁cos2𝜋𝑗/𝑁𝑘−cos2𝜋𝑗/𝑁(𝑘−1)1−𝜌_𝑗³=−𝜆_𝑘⁽¹⁾,for𝑘=2,3,…,𝑁−1. (36) Similarly, we get from (35) that 𝜆⁽²⁾_{𝑁−𝑘+1}=−𝜆_𝑘⁽³⁾,𝜆⁽³⁾_{𝑁−𝑘+1}=−𝜆_𝑘⁽²⁾,𝜆⁽⁴⁾_{𝑁−𝑘+1}=−𝜆_𝑘⁽⁴⁾, (37) where 𝑘=2,3,…,𝑁−1. From (36) and (37), the result follows.

Remark 11. Let 𝑁ℓ=2,if𝑁iseven,(𝑁+1)2,if𝑁isodd. (38)
From Lemma 10, it is clear that to prove all the 𝑁−1 inequalities of Lemma 9 suffices to prove the ℓ inequalities 𝜆_𝑘⁽¹⁾𝜆_𝑘⁽⁴⁾−𝜆_𝑘⁽²⁾𝜆_𝑘⁽³⁾≠0,𝑘=1,2,…,ℓ.

With the similar proof as in [2, Lemma 2], we can get the following proposition.

Proposition 12. Let 𝜓_𝛽,𝑘(𝑎)=∑_𝑗(cos(2𝜋𝑗/𝑁)𝑘/|1−𝑎𝜌_𝑗|^𝛽). Then for 𝑘=1,2,…,𝑁,𝛽>0 and 𝑎∈(0,1),  𝜓_𝛽,𝑘(𝑎) and all of its derivatives are positive.

Lemma 13. For 𝑁≥3, 𝜆₁⁽¹⁾𝜆₁⁽⁴⁾−𝜆₁⁽²⁾𝜆₁⁽³⁾>0.

Proof. From (34) and (35), we can get 𝜆₁⁽¹⁾=14_𝑗≠𝑁csc𝜋𝑗𝑁,𝜆₁⁽²⁾=1−𝑎cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³,𝜆₁⁽³⁾=𝑎−cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³,𝜆₁⁽⁴⁾=14𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁. (39) Let 𝐴₁(𝑎)=𝜆₁⁽¹⁾𝜆₁⁽⁴⁾−𝜆₁⁽²⁾𝜆₁⁽³⁾. Then 𝐴₁(𝑎)=116𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁²−1−𝑎cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³×𝑎−cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³=116𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁²1−𝑎1−𝑎𝜌_𝑗³²−𝑎cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³²+1+𝑎²11−𝑎𝜌_𝑗³cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³. (40) Clearly, ∑(1/|1−𝑎𝜌_𝑗|³)>0. By Proposition 12, we get that 𝜓_3,1= ∑(cos(2𝜋𝑗/𝑁)/|1−𝑎𝜌_𝑗|³)>0 for 𝑎∈(0,1). Note that (1/4)∑_𝑗≠𝑁csc(𝜋𝑗/𝑁)=∑_𝑗≠𝑁((1−cos(2𝜋𝑗/𝑁))/|1−𝜌_𝑗|³)>0. We have from (40) 𝐴₁(𝑎)>116𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁²1−𝑎1−𝑎𝜌_𝑗³²−𝑎cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³²1+2𝑎1−𝑎𝜌_𝑗³cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³=116𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁²1−𝑎1−𝑎𝜌_𝑗³−cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³²=1𝑎²_𝑗≠𝑁1−cos2𝜋𝑗/𝑁1−𝜌_𝑗³−_𝑗≠𝑁𝑎^3/21−cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³(𝑎)=1⋅𝑔𝑎²_𝑗≠𝑁1−cos2𝜋𝑗𝑁⋅ℎ𝑗,𝑎⋅𝑔(𝑎), (41) where 𝑔(𝑎)=(1/4)∑_𝑗≠𝑁csc(𝜋𝑗/𝑁)+∑_𝑗≠𝑁(𝑎^3/2(1−cos(2𝜋𝑗/𝑁))/|1−𝑎𝜌_𝑗|³) and ℎ(𝑗,𝑎)= (1/|1−𝜌_𝑗|³)−(𝑎^3/2/|1−𝑎𝜌_𝑗|³). Note that ℎ(𝑗,𝑎)=(1/|1−𝜌_𝑗|³)−(𝑎^3/2/(1+𝑎²−2𝑎cos(2𝜋𝑗/𝑁))^3/2) and hence 𝑑ℎ𝑗,𝑎=()𝑎𝑑𝑎3/2^1/2(𝑎−1)(𝑎+1)1−𝑎𝜌_𝑗⁵<0,for𝑎∈(0,1). (42) 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 ℎ𝑗,𝑎>ℎ𝑗,1=0,for𝑎∈(0,1),𝑗=1,…,𝑁−1. (43) 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.

For 𝑘=1,2,…,ℓ, let 𝛼_𝑘(𝑎)=_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗³. (44)

Lemma 14. If 𝑁 is odd, then 𝛼_(𝑁+1)/2(1)=0; If 𝑁 is even, then 𝛼_𝑁/2(1)>0.

Proof. If 𝑁 is odd, by the definition of 𝛼_𝑘, we have 𝛼_(𝑁+1)/2(1)=_𝑗≠𝑁)))cos2𝜋𝑗(((𝑁+1/2−1𝑁))−cos2𝜋𝑗((𝑁+1/2𝑁×1−𝜌_𝑗³⁻¹=_𝑗≠𝑁cos𝜋𝑗−𝜋𝑗/𝑁−cos𝜋𝑗+𝜋𝑗/𝑁1−𝜌_𝑗³=0. (45) If 𝑁 is even, we get that 𝛼_𝑁/2(1)=_𝑗≠𝑁cos2𝜋𝑗((𝑁/2)−1)/𝑁−cos2𝜋𝑗(𝑁/2)/𝑁1−𝜌_𝑗³=14_𝑗≠𝑁(−1)^𝑗+11=1sin𝜋𝑗/𝑁4^𝑁/2_𝑗=1(−1)^𝑗+11+sin𝜋𝑗/𝑁_{𝑁/2−1𝑗=1}(−1)^𝑗+11.sin𝜋𝑗/𝑁 (46)
Case 1. If 𝑁/2 is even, then 𝛼_𝑁/2(1)=142^𝑁/2_𝑗=1(−1)^𝑗+11.sin𝜋𝑗/𝑁+1 (47)
Since the signs of the terms in the sum ∑^𝑁/2_𝑗=1(−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 ∑^𝑁/2_𝑗=1(−1)^𝑗+1(1/sin(𝜋𝑗/𝑁))>0 and hence 𝛼_𝑁/2(1)>0.
Case 2. If 𝑁/2 is odd, then 𝛼_𝑁/2(1)=142^𝑁/2−1_𝑗=1(−1)^𝑗+11.sin𝜋𝑗/𝑁+1 (48)
Since the signs of the terms in summation ∑^𝑁/2−1_𝑗=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 ∑^(𝑁/2)−1_𝑗=1(−1)^𝑗+1(1/sin(𝜋𝑗/𝑁))>0 and hence 𝛼_𝑁/2(1)>0.
By Cases 1 and 2, we see that 𝛼_𝑁/2(1)>0 for even 𝑁.

Proposition 15. For 𝑘=1,2,…,ℓ−1, then 𝛼_𝑘(1)>0.

Proof. Clearly, 𝛼₁(1)>0. We need to prove 𝛼_𝑘(1)>0,𝑘≥2. Note that 2sin𝜋𝑗/𝑁^𝑘−1_𝑚=0cos2𝜋𝑗𝑁𝑚=2sin𝜋𝑗𝑘/𝑁cos𝜋𝑗(𝑘−1)/𝑁sin²=()𝜋𝑗/𝑁sin𝜋𝑗2𝑘−1/𝑁sin²+1𝜋𝑗/𝑁.sin𝜋𝑗/𝑁 (49) Hence, by the definition of 𝛼_𝑘(𝑎), we have 𝛼_𝑘(1)=_𝑗≠𝑁2sin𝜋𝑗(2𝑘−1)/𝑁sin𝜋𝑗/𝑁2−2cos2𝜋𝑗/𝑁^3/2=14_𝑗≠𝑁−1+2sin𝜋𝑗/𝑁sin𝜋𝑗/𝑁^𝑘−1_𝑚=0cos2𝜋𝑗𝑁𝑚. (50) Then, we get that 𝛼_𝑘(1)−𝛼_𝑘−1(1)−𝛼_𝑘+1(1)−𝛼_𝑘(1)=12_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁=sin𝜋𝑗/𝑁_𝑗≠𝑁()sin𝜋𝑗2𝑘−1𝑁=cot𝜋(2𝑘−1).2𝑁 (51) 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 𝛼_𝑘(1)−𝛼_𝑘−1(1)>𝛼_𝑘+1(1)−𝛼_𝑘(1). (52)By (52), we get that 𝛼_𝑘−1(1)+𝛼_k+1(1)<2𝛼_𝑘(1) which implies that 𝛼_𝑘+1(1)<𝛼_𝑘(1) or 𝛼_𝑘−1(1)<𝛼_𝑘(1). If there is 𝛼_𝑘(1)≤0, we use two cases to proceed our proof.
Case 1. (If 𝛼_𝑘−1(1)<𝛼_𝑘(1)). By (52), we have 𝛼_𝑘−2(1)−𝛼_𝑘−1(1)<𝛼_𝑘−1(1)−𝛼_𝑘(1)<0 and hence 𝛼_𝑘−2(1)<𝛼_𝑘−1(1)<𝛼_𝑘(1)≤0. By inductions, we have 𝛼₁(1)<⋯<𝛼_𝑘(1)<0 which contradicts with 𝛼₁(1)>0.
Case 2. (If 𝛼_𝑘+1(1)<𝛼_𝑘(1)). By (52), we have 𝛼_𝑘+2(1)−𝛼_𝑘+1(1)<𝛼_𝑘+1(1)−𝛼_𝑘(1)<0. Then, 𝛼_𝑘+2(1)<𝛼_𝑘+1(1)<𝛼_𝑘(1)≤0. By inductions, we get that for even 𝑁, 𝛼_𝑁/2(1)< 𝛼_(𝑁/2)−1(1)< ⋯<𝛼_𝑘(1)<0 which contradicts with 𝛼_𝑁/2(1)>0; for odd 𝑁, 𝛼_(𝑁+1)/2(1)<𝛼_(𝑁−1)/2(1)<⋯< 𝛼_𝑘(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 exists 𝛼_𝑘(1)≤0. Hence, 𝛼_𝑘(1)>0,𝑘=1,…,ℓ.

Let _∼𝑁ℓ=2,if𝑁iseven,(𝑁−1)2,if𝑁isodd,𝑎^∗=min_{()𝑎∈0,1,𝛼𝑘(𝑎)>0,𝑘=1,2,…,∼ℓ}𝑎. (53)

By the continuity, Lemma 14 and Proposition 15, we see that 𝑎^∗∈(0,1).

For 𝑘=1,2,…,ℓ, let 𝛽_𝑘(𝑎)=_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗⁵. (54)

Similar to Lemma 14, we have

Lemma 16. If 𝑁 is odd, then 𝛽_(𝑁+1)/2(1)=0; If 𝑁 is even, then 𝛽_𝑁/2(1)>0.

Proposition 17. For 𝑘=1,2,…,ℓ−1, then 𝛽_𝑘(1)>0.

Proof. Clearly, 𝛽₁(1)>0. We need to prove 𝛽_𝑘(1)>0,𝑘≥2. Note that 2sin³𝜋𝑗/𝑁^𝑘−1_𝑚=0cos2𝜋𝑗𝑁𝑚=1sin³⋅()𝜋𝑗/𝑁2sin𝜋𝑗𝑘/𝑁cos𝜋𝑗𝑘−1/𝑁=()sin𝜋𝑗/𝑁sin𝜋𝑗2𝑘−1/𝑁sin⁴+1𝜋𝑗/𝑁sin³.𝜋𝑗/𝑁 (55) By the definition of 𝛽_𝑘(1), we have 𝛽_𝑘(1)=116_𝑗≠𝑁−1sin³+2𝜋𝑗/𝑁sin³𝜋𝑗/𝑁^𝑘−1_𝑚=0cos2𝜋𝑗𝑁𝑚. (56) Then, we get that 𝛽_𝑘(1)−𝛽_𝑘−1(1)−𝛽_𝑘+1(1)−𝛽_𝑘(1)=18_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁sin³=1𝜋𝑗/𝑁8𝛼_𝑘(1)>0. (57) By similar proofs as in Proposition 15, we can get 𝛽_𝑘>0. The proof is completed.

Let 𝑎^∼∗=min_{()𝑎∈0,1,𝛽𝑘(𝑎)>0,𝑘=1,2,…,∼ℓ}𝑎. (58) By the continuity, Lemma 16 and Proposition 17, we see that 𝑎^∼∗∈(0,1).

Lemma 18. For 𝑘=2,3,…,ℓ and 𝑎^∗=max{𝑎^∗,𝑎^∼∗}, 𝜆_𝑘⁽¹⁾𝜆_𝑘⁽⁴⁾−𝜆_𝑘⁽²⁾𝜆_𝑘⁽³⁾>0.

Proof. Let 𝐴_𝑘(𝑎)=𝜆_𝑘⁽¹⁾𝜆_𝑘⁽⁴⁾−𝜆_𝑘⁽²⁾𝜆_𝑘⁽³⁾. From (34) and (35), we can get 𝐴_𝑘(𝑎)=1𝑎²_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝜌_𝑗³²−cos2𝜋𝑗(𝑘−1)/𝑁−𝑎cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗³⋅𝑎cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁𝑎−𝜌_𝑗³. (59) 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 𝐴_(𝑁+1)/2(𝑎)=1𝑎²_𝑗≠𝑁)))cos2𝜋𝑗(((𝑁+1/2−1𝑁))−cos2𝜋𝑗((𝑁+1/2𝑁×1−𝜌_𝑗⁻³²−)))cos2𝜋𝑗(((𝑁+1/2−1𝑁))−𝑎cos2𝜋𝑗((𝑁+1/2𝑁×1−𝑎𝜌_𝑗⁻³⋅)))𝑎cos2𝜋𝑗(((𝑁+1/2−1𝑁))−cos2𝜋𝑗((𝑁+1/2𝑁×𝑎−𝜌_𝑗⁻³=(1−𝑎)²(−1)^𝑗cos𝜋𝑗/𝑁1−𝑎𝜌_𝑗³². (60) Let 𝑓(𝑥)=cos𝑥/(1+𝑎²−2𝑎cos2𝑥)^3/2 for 0<𝑥<𝜋/2. Then 𝑓^′(𝑥)=−sin𝑥1+𝑎²−2𝑎cos2𝑥−6𝑎cos𝑥sin2𝑥1+𝑎²−2𝑎cos2𝑥^5/2<0. (61) Hence,cos(𝜋𝑗/𝑁)(𝑎−1)/|𝑎−𝜌_𝑗|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 𝐴_(𝑁+1)/2(𝑎)>0. (62)
For any 𝑘,𝑘=2,3,…,ℓ^∼, let 𝐺(𝑎)=_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝜌_𝑗³−𝑎^3/2_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗³. (63) With direct computation, we have 𝐺^′(𝑎)=32𝑎^1/2(𝑎+1)(𝑎−1)×_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗⁵<0,for𝑎∈𝑎^∗,1, (64) 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 cos2𝜋𝑗(_𝑗≠𝑁𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝜌_𝑗³>𝑎^3/2_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋j𝑘/𝑁1−𝑎𝜌_𝑗³, (65) where 𝑎∈(𝑎^∗,1) and 𝑘=2,3,…,ℓ^∼. From (65), we can get 1𝑎²_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝜌_𝑗³²>𝑎_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗³², (66) where 𝑎∈(𝑎^∗,1) and 𝑘=2,3,…,ℓ^∼.
By (59), we can get 𝐴_𝑘(𝑎)=1𝑎²_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝜌_𝑗³²()−𝑎cos2𝜋𝑗𝑘−1/𝑁1−𝑎𝜌_𝑗³²+1+𝑎²cos2𝜋𝑗(𝑘−1)/𝑁1−𝑎𝜌_𝑗³×cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗³−𝑎cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗³². (67) From Proposition 12, we have 𝐴_𝑘(𝑎)>1𝑎²_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝜌_𝑗³²()−𝑎cos2𝜋𝑗𝑘−1/𝑁1−𝑎𝜌_𝑗³²()−2cos2𝜋𝑗𝑘−1/𝑁1−𝑎𝜌_𝑗³×cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗³+cos2𝜋𝑗𝑘/𝑁1−a𝜌_𝑗³²=1𝑎²_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝜌_𝑗³²−𝑎_𝑗≠𝑁cos2𝜋𝑗(𝑘−1)/𝑁−cos2𝜋𝑗𝑘/𝑁1−𝑎𝜌_𝑗³²>0, (68)
where (66) is used and 𝑘=2,3,…,ℓ^∼.
By (62) and (68), the proof is completed.

The following example illustrate that the 𝑎^∗ in Lemma 18 can be obtained for some special cases.

Example 19. If 𝑁=3, then 𝑎^∗=0.

Proof. For 𝑁=3, we have 𝛼₁(𝑎)=2(1+(𝑎/2))²+(3/4)^3/2𝛼>0,₂(𝑎)=1(1+(𝑎/2))²+(3/4)^3/2>0. (69)
By the definition of 𝑎^∗, we have 𝑎^∗=0.
Similarly, we get that 𝛽₁(𝑎)=2(1+(𝑎/2))²+(3/4)^5/2𝛽>0,₂(𝑎)=1(1+(𝑎/2))²+(3/4)^5/2>0. (70)
By the definition of 𝑎^∼∗, we have 𝑎^∼∗=0. Hence, 𝑎^∗=0 for 𝑁=3.

Lemmas 13 and 18 show that the hypotheses of Lemma 9 are satisfied. Thus the proof of the Theorem 5 was completed.

3. The Proof of Theorem 6

Since 𝑚₁=𝑚₂=⋯=𝑚_𝑁=𝑚 and 𝑚^∼₁=𝑚^∼₂=⋯=𝑚^∼_𝑁=𝑚^∼, we get that 𝑧₀=0. Hence, by (3), we have 𝑞_𝑘=𝜌_𝑘𝑒^𝑖𝜔𝑡,𝑞^∼_𝑘=𝑎𝜌_𝑘𝑒^𝑖𝜔𝑡. (71) Substituting 𝑞_𝑘 and 𝑞^∼_𝑘 into (1), then by (11) we get that 𝜔²=𝑚4_𝑗≠𝑁csc𝜋𝑗𝑁+𝑚^∼1−𝑎cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³,𝑎𝜔²=𝑚𝑎−cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³+^∼𝑚4𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁. (72)

From (72), we have 𝜔²=𝑚4_𝑗≠𝑁csc𝜋𝑗𝑁+𝑚^∼1−𝑎cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³,𝜔²=1𝑎𝑚𝑎−cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³+^∼𝑚4𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁. (73) From (73), the constant 𝜔² exists only if the following equation holds: 𝑎𝑚4_𝑗≠𝑁csc𝜋𝑗𝑁+𝑚^∼1−𝑎cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³=𝑚𝑎−cos2𝜋𝑗/𝑁1−𝑎𝜌_𝑗³+^∼𝑚4𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁. (74) Note that 𝜇=𝑚^∼/𝑚. Then we get from (74) that 𝑎³−𝜇4𝑎²_𝑗≠𝑁csc𝜋𝑗𝑁−1−𝜇𝑎𝑓(𝑎)+1−𝑎²𝜇𝑔(𝑎)=0, (75) where 𝑓(𝑎)=∑(1/|1−𝑎𝜌_𝑗|³) and 𝑔(𝑎)=∑(cos(2𝜋𝑗/𝑁)/|1−𝑎𝜌_𝑗|³).

Lemma 20. For 𝜇>0 and 𝑁≥3, the constant 𝜔² can be given by 𝜔²=𝑚4_𝑗≠𝑁csc𝜋𝑗𝑁+𝑚^∼1−𝑎₀cos2𝜋𝑗/𝑁1−𝑎₀𝜌_𝑗³, (76) 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 𝐹(𝑎)=𝜇𝑎−𝑎²14_𝑗≠𝑁csc𝜋𝑗𝑁−1−𝜇𝑎𝑓(𝑎)+1−𝑎²𝜇𝑔(𝑎). (77) From (75) and (77), the roots of (75) are zeros of (77). From the definitions of 𝜓_𝛽,𝑘(𝑎) defined in Proposition 12, we have 𝜓_1,𝑁(𝑎)=∑(1/|1−𝑎𝜌_𝑗|). Through direct calculations, we find that 𝜓_1,𝑁(𝑎)=1+𝑎²𝑓(𝑎)−2𝑎𝑔(𝑎),𝑑𝜓_1,𝑁(𝑎)(𝑎)(𝑎).𝑑𝑎=−𝑎𝑓+𝑔 (78) From (78), (77) can be written as 𝐹(𝑎)=𝜇𝑎−𝑎²14_𝑗≠𝑁csc𝜋𝑗𝑁+1+𝑎²𝜇𝑑𝜓_1,𝑁(𝑎)𝑑𝑎+𝜇𝑎𝜓_1,𝑁(𝑎). (79) Notice that 𝜓_1,𝑁(𝑎)=∑_𝑗≠𝑁(1/|1−𝑎𝜌_𝑗|)+(1/|1−𝑎𝜌_𝑁|)=∑_𝑗≠𝑁(1/|1−𝑎𝜌_𝑗|)+(1/|1−a|). Hence, we have lim_𝑎→1𝜓_1,𝑁(𝑎)=+∞. (80) From (79), (80), and Proposition 12, we can get that lim_𝑎→0𝐹(𝑎)=−∞,lim_𝑎→1𝐹(𝑎)=+∞. (81) On the other hand, by (79), we get that 𝐹^′(𝑎)=1+2𝜇𝑎³14_𝑗≠𝑁csc𝜋𝑗𝑁+𝜇𝜓_1,𝑁(𝑎)+3𝑎𝜇𝑑𝜓_1,𝑁(𝑎)+𝑑𝑎1+𝑎²𝜇𝑑²𝜓_1,𝑁(𝑎)𝑑𝑎². (82) 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.

Acknowledgments

The authors thank Professor Zhang S.Q. for his guidances, Editor Zheng B.D. for his hard work, and the anonymous referees for their comments, suggestions and, pointing out some additional references. Supported partially by Fundamental Research Funds for the Central Universities, CQDXWL-2012-006.