Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 397194, 10 pages
http://dx.doi.org/10.1155/2013/397194
Research Article
The Periodic Solutions for Planar 2𝑁-Body Problems
Xiaohong Hu1,2 and Changrong Zhu3
1
College of Mathematics, Sichuan University, Chengdu 610064, China
College of Mathematics and Physics, Chongqing University of Posts and Telecommunications,
Chongqing 400065, China
3
Department of Mathematics, Chongqing University, Chongqing 400044, China
2
Correspondence should be addressed to Xiaohong Hu; huxh@cqupt.edu.cn
Received 7 February 2013; Accepted 1 April 2013
Academic Editor: Baodong Zheng
Copyright © 2013 X. Hu and C. Zhu. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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 𝑚𝑗 > 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
𝑎 ≠ 1; precisely, let 𝜌𝑗 = 𝑒(2𝜋𝑖/𝑁)𝑗 , where 𝑖2 = −1; and the
̃𝑗 locate at 𝑞𝑗 and 𝑞̃𝑗 :
points 𝑚𝑗 and 𝑚
𝑞𝑗 = (𝜌𝑗 − 𝑧0 ) 𝑒𝑖𝜔𝑡 ,
𝑞̃𝑗 = (𝑎𝜌𝑗 − 𝑧0 ) 𝑒𝑖𝜔𝑡 ,
(3)
𝜕𝑈
,
𝑚𝑗 𝑞𝑗̈ =
𝜕𝑞𝑗
𝑗 = 1, 2, . . . , 𝑁,
(1)
where 𝑈 is the Newtonian potential:
𝑈=
𝑚𝑗 𝑚𝑘
󵄨󵄨
󵄨󵄨 .
1≤𝑗<𝑘≤𝑁 󵄨󵄨󵄨𝑞𝑗 − 𝑞𝑘 󵄨󵄨󵄨
∑
(2)
𝑗 = 1, 2, . . . , 𝑁,
̃𝑗 𝜌𝑗 ), 𝑀1 = ∑ 𝑚𝑗 ,
where 𝑧0 = (1/(𝑀1 + 𝑀2 )) ∑(𝑚𝑗 𝜌𝑗 + a𝑚
̃𝑗 , and 𝑀 = 𝑀1 + 𝑀2 . For short, throughout this
𝑀2 = ∑ 𝑚
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 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 = ⋅ ⋅ ⋅ = 𝑚𝑁.
̃1 = ⋅ ⋅ ⋅ =
Theorem 2 (see [2]). If 𝑚1 = ⋅ ⋅ ⋅ = 𝑚𝑁 := 𝑚 and 𝑚
̃ then for every mass ratio 𝑟 = 𝑚/𝑚,
̃
̃𝑁 := 𝑚,
there are exactly
𝑚
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 1995, Moeckel and Simó [2] studied planar nested 2𝑁body problems; they assume one regular 𝑁-gon is inscribed
on a unit circle, the other on a circle with radius 𝑎, 𝑎 > 0 and
In 2003, Zhang and Zhou [3] studied the inverse problem
of Moeckel and Simó’s theorem [2], and they got the following
results.
In the famous paper [1], Perko and Walter proved the following result.
2
Abstract and Applied Analysis
Theorem 3 ([3], the case of 𝜃 = 0). If (3) is a solution of (1),
then 𝜔2 satisfies
𝜔2 =
2
2
[ 1 ∑ csc 𝜋𝑗 − 𝑁𝜔 ] ⋅ [ 1 ∑ csc 𝜋𝑗 − 𝑎𝑁𝜔 ]
4
𝑁
𝑀
4𝑎2 𝑗 ≠ 𝑁
𝑁
𝑀
] [
]
[ 𝑗 ≠ 𝑁
𝜋𝑗
𝑚
∑ csc
4 𝑗 ≠ 𝑁
𝑁
̃∑
+𝑚
1 − 𝑎𝜌𝑗
𝑎 − 𝜌𝑗
𝑁𝜔2 ] [
𝑎𝑁𝜔2 ]
−
−
= [∑ 󵄨
⋅
∑
󵄨󵄨3
󵄨󵄨
󵄨󵄨3
󵄨
𝑀
𝑀
[ 󵄨󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
] [ 󵄨󵄨󵄨𝑎 − 𝜌𝑗 󵄨󵄨󵄨
]
(4)
1 − 𝑎0 cos (2𝜋𝑗/𝑁)
,
󵄨󵄨
󵄨󵄨3
󵄨󵄨󵄨1 − 𝑎0 𝜌𝑗 󵄨󵄨󵄨
(6)
̃ and 𝑎0 is a unique solution of the following
where 𝜇 = 𝑚/𝑚
equation:
𝜋𝑗
𝑎3 − 𝜇
∑ csc
4𝑎2 𝑗 ≠ 𝑁
𝑁
or equivalently
𝑎
− (1 − 𝜇) ∑ 󵄨
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨󵄨3
𝑗 󵄨󵄨
󵄨󵄨
𝜔2
2
=(
̃1 = ⋅ ⋅ ⋅ = 𝑚
̃𝑁 := 𝑚,
̃
Theorem 6. If 𝑚1 = ⋅ ⋅ ⋅ = 𝑚𝑁 := 𝑚, 𝑚
and the constant 𝜔 > 0 is given by
𝜋𝑗
1
( ∑ csc )
2
16𝑎 𝑗 ≠ 𝑁
𝑁
(7)
cos (2𝜋𝑗/𝑁)
+ (1 − 𝑎2 𝜇) ∑ 󵄨
= 0,
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨󵄨3
󵄨󵄨
𝑗 󵄨󵄨
1 − 𝑎𝜌𝑗
𝑎 − 𝜌𝑗
− (∑ 󵄨
) (∑ 󵄨
))
3
󵄨
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨
󵄨󵄨𝑎 − 𝜌 󵄨󵄨󵄨3
󵄨󵄨
󵄨󵄨
𝑗 󵄨󵄨
𝑗 󵄨󵄨
̃1 , . . . , 𝑚
̃𝑁) is
then (𝑞1 , . . . , 𝑞𝑁, 𝑞̃1 , . . . , 𝑞̃𝑁) with (𝑚1 , . . . , 𝑚𝑁, 𝑚
a periodic solution of (1) with angular velocity 𝜔.
̃1 = ⋅ ⋅ ⋅ = 𝑚
̃𝑁, we have 𝑧0 = 0
When 𝑚1 = ⋅ ⋅ ⋅ = 𝑚𝑁 and 𝑚
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, ∞).
𝜋𝑗
𝑁 1
1
× ( [ (𝑎 + 2 ) ( ∑ csc )
𝑀 4
𝑎
𝑁
𝑗 ≠ 𝑁
[
−1
𝑎 − 𝜌𝑗
1 − 𝑎𝜌𝑗
]) .
−∑󵄨
−𝑎 ∑ 󵄨
3
󵄨
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨
󵄨󵄨𝑎 − 𝜌 󵄨󵄨󵄨3
󵄨󵄨
󵄨󵄨
𝑗 󵄨󵄨
𝑗 󵄨󵄨 ]
(5)
And in [3], they proved the following theorem [3, Theorem 2]:
Theorem 4. If 𝜔 satisfies (4) or (5) and (𝑞1 , . . . , 𝑞𝑁; 𝑞̃1 ,
. . . , 𝑞̃𝑁) given by (3) is a periodic solution of (1), then 𝑚1 =
̃1 = ⋅ ⋅ ⋅ = 𝑚
̃𝑁 .
⋅ ⋅ ⋅ = 𝑚𝑁 and 𝑚
In the proof of [3, Theorem 2], the authors claimed that
the first eigenvalue 𝜆 1 (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 , . . . , 𝑞𝑁, 𝑞̃1 , . . . , 𝑞̃𝑁) given by (3) is a periodic solution for
̃1 = ⋅ ⋅ ⋅ = 𝑚
̃𝑁 .
(1), then 𝑚1 = ⋅ ⋅ ⋅ = 𝑚𝑁 and 𝑚
2. The Proof of Theorem 5
Substituting (3) into (1), we have
̃𝑗 (̃
𝑚𝑗 (𝑞𝑗 − 𝑞𝑘 )
𝑞𝑗 − 𝑞𝑘 )
𝑚
+
−𝜔2 (𝜌𝑘 − 𝑧0 ) 𝑒𝑖𝜔𝑡 = ∑ 󵄨
∑
󵄨󵄨3
󵄨󵄨
󵄨3 ,
󵄨
󵄨󵄨𝑞̃𝑗 − 𝑞𝑘 󵄨󵄨󵄨
𝑗 ≠ 𝑘 󵄨󵄨𝑞𝑗 − 𝑞𝑘 󵄨󵄨
󵄨
󵄨
󵄨
󵄨
̃𝑗 (̃
𝑚𝑗 (𝑞𝑗 − 𝑞̃𝑘 )
𝑞𝑗 − 𝑞̃𝑘 )
𝑚
−𝜔2 (𝑎𝜌𝑘 − 𝑧0 ) 𝑒𝑖𝜔𝑡 = ∑ 󵄨
+
∑
󵄨󵄨
󵄨󵄨3 ,
󵄨󵄨𝑞 − 𝑞̃ 󵄨󵄨󵄨3
󵄨
𝑗 ≠ 𝑘 󵄨󵄨𝑞
󵄨󵄨 𝑗
𝑘 󵄨󵄨
󵄨 ̃𝑗 − 𝑞̃𝑘 󵄨󵄨
(8)
where 𝑘 = 1, 2, . . . , 𝑁.
By (3), (8) can be written as
̃𝑗 (𝑎𝜌𝑗 − 𝜌𝑘 )
𝑚𝑗 (𝜌𝑗 − 𝜌𝑘 )
𝑚
−𝜔2 (𝜌𝑘 − 𝑧0 ) = ∑ 󵄨
+∑ 󵄨
,
3
󵄨
󵄨
󵄨
󵄨󵄨𝑎𝜌 − 𝜌 󵄨󵄨󵄨3
𝑗 ≠ 𝑘 󵄨󵄨𝜌𝑗 − 𝜌𝑘 󵄨󵄨
󵄨
󵄨
𝑗
𝑘
󵄨
󵄨
󵄨
󵄨
̃𝑗 (𝜌𝑗 − 𝜌𝑘 )
𝑚𝑗 (𝜌𝑗 − 𝑎𝜌𝑘 ) 1
𝑚
−𝜔2 (𝑎𝜌𝑘 − 𝑧0 ) = ∑ 󵄨
+ 2∑ 󵄨
,
3
󵄨
󵄨󵄨𝜌 − 𝑎𝜌 󵄨󵄨
𝑎 𝑗 ≠ 𝑘 󵄨󵄨𝜌 − 𝜌 󵄨󵄨󵄨3
󵄨󵄨 𝑗
󵄨󵄨 𝑗
𝑘 󵄨󵄨
𝑘 󵄨󵄨
(9)
Abstract and Applied Analysis
3
̃⃗ = (𝑚
̃1 , . . . , 𝑚
̃𝑁), and 1⃗ = (1,
where 𝑚⃗ = (𝑚1 , . . . , 𝑚𝑁), 𝑚
. . . , 1).
An 𝑁 × 𝑁 matrix R = (𝑟𝑗,𝑘 )𝑁×𝑁 is called circulant (see
[4]) if
where 𝑘 = 1, 2, . . . , 𝑁. Equation (9) is equivalent to
𝜌𝑘 − 𝜌𝑗
𝜔2
𝑚𝑗 +
𝜔2 𝜌𝑘 = ( ∑ 󵄨
∑ 𝜌𝑗 𝑚𝑗 )
3
󵄨
󵄨
󵄨
𝑀
𝑗 ≠ 𝑘 󵄨󵄨𝜌𝑘 − 𝜌𝑗 󵄨󵄨
󵄨
󵄨
𝜌𝑘 − 𝑎𝜌𝑗
𝑎𝜔2
̃𝑗 +
̃𝑗 ) ,
+ (∑ 󵄨
𝑚
∑ 𝜌𝑗 𝑚
3
󵄨
󵄨󵄨𝜌 − 𝑎𝜌 󵄨󵄨
𝑀
󵄨󵄨 𝑘
󵄨
𝑗󵄨
𝑟𝑗,𝑘 = 𝑟𝑗−1,𝑘−1 ,
(10)
𝑎𝜌𝑘 − 𝜌𝑗
𝜔2
𝑎𝜔2 𝜌𝑘 = (∑ 󵄨
𝑚
+
∑ 𝜌𝑗 𝑚𝑗 )
󵄨󵄨𝑎𝜌 − 𝜌 󵄨󵄨󵄨3 𝑗 𝑀
󵄨󵄨 𝑘
𝑗 󵄨󵄨
𝑗−1
2
𝑁
V𝑘 (R) = (𝜌𝑘−1 , 𝜌𝑘−1
, . . . , 𝜌𝑘−1
),
1 − 𝜌𝑗−𝑘
𝜔2
𝜔2 = ( ∑ 󵄨
𝑚𝑗 +
∑ 𝜌𝑗−𝑘 𝑚𝑗 )
3
󵄨
󵄨
󵄨
𝑀
𝑗 ≠ 𝑘 󵄨󵄨1 − 𝜌𝑗−𝑘 󵄨󵄨
󵄨
󵄨
(11)
𝑎 − 𝜌𝑗−𝑘
𝜔2
𝑎𝜔2 = (∑ 󵄨
𝑚
+
󵄨3 𝑗 𝑀 ∑ 𝜌𝑗−𝑘 𝑚𝑗 )
󵄨󵄨󵄨𝑎 − 𝜌𝑗−𝑘 󵄨󵄨󵄨
󵄨
󵄨
+(
where 𝑘 = 1, 2, . . . , 𝑁.
Let 𝐴 = (𝑎𝑘,𝑗 )𝑁×𝑁, 𝐵 = (𝑏𝑘,𝑗 )𝑁×𝑁, 𝐶 = (𝑐𝑘,𝑗 )𝑁×𝑁, and 𝐷 =
(𝑑𝑘,𝑗 )𝑁×𝑁, where
𝑎𝑘,𝑗
(12)
𝜔2
𝑎𝜔2
1
̃⃗ = 𝑎𝜔2 1,⃗
(𝐶 + 𝐷) 𝑚⃗ + ( 2 𝐴 +
𝐷) 𝑚
𝑀
𝑎
𝑀
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
𝜔2
𝑑
𝑀 1,𝑗
= 𝑎1,𝑁−𝑗+2 +
(17)
𝜔2
,
𝑑
𝑀 1,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.
Then (11) can be rewritten in the following compact form:
𝜔2
𝑎𝜔2
̃⃗ = 𝜔2 1,⃗
𝐷) 𝑚⃗ + (𝐵 +
𝐷) 𝑚
𝑀
𝑀
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
(2) (3)
convenience, we introduce some notations. Let 𝜆(1)
𝑘 , 𝜆𝑘 , 𝜆𝑘 ,
2
and 𝜆(4)
𝑘 be the 𝑘th eigenvalue of matrixes 𝐴 + (𝜔 /𝑀)𝐷, 𝐵 +
2
2
2
2
(𝑎𝜔 /𝑀)𝐷, 𝐶 + (𝜔 /𝑀)𝐷, and (1/𝑎 )𝐴 + (𝑎𝜔 /𝑀)𝐷, respectively. Then we have the following.
1 − 𝜌𝑗−1
𝜔2
=󵄨
+
𝜌
󵄨󵄨1 − 𝜌 󵄨󵄨󵄨3 𝑀 𝑗−1
󵄨󵄨
𝑗−1 󵄨󵄨
otherwise,
𝑑𝑘,𝑗 := 𝜌𝑗−𝑘 .
(𝐴 +
(16)
Remark 7. From the formula (15), we have 𝜆 1 (R) = ∑𝑁
𝑗=1 𝑟1,𝑗 .
The last equation implies that the eigenvalue 𝜆 1 (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 𝜆 1 (R).
𝑎1,𝑗 +
if 𝑗 ≠ 𝑘,
1 − 𝑎𝜌𝑗−𝑘
𝑏𝑘,𝑗 := 󵄨
,
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨󵄨3
󵄨󵄨
󵄨
𝑗−𝑘 󵄨
𝑎 − 𝜌𝑗−𝑘
𝑐𝑘,𝑗 := 󵄨
󵄨3 ,
󵄨󵄨󵄨𝑎 − 𝜌𝑗−𝑘 󵄨󵄨󵄨
󵄨
󵄨
𝑘 = 1, 2, . . . , 𝑁.
(15)
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.
1 − 𝜌𝑗−𝑘
1
𝑎𝜔2
̃
̃𝑗 ) ,
+
𝑚
∑
∑ 𝜌𝑗−𝑘 𝑚
𝑎2 𝑗 ≠ 𝑘 󵄨󵄨󵄨1 − 𝜌 󵄨󵄨󵄨3 𝑗 𝑀
󵄨󵄨
𝑗−𝑘 󵄨󵄨
1 − 𝜌𝑗−𝑘
{
{󵄨
󵄨3 ,
:= { 󵄨󵄨󵄨1 − 𝜌𝑗−𝑘 󵄨󵄨󵄨
󵄨
{󵄨
{0,
𝑘 = 1, 2, . . . , 𝑁,
𝑗
where 𝑘 = 1, 2, . . . , 𝑁. Multiplying both sides of (10) by 𝜌−𝑘 ,
we get
1 − 𝑎𝜌𝑗−𝑘
𝑎𝜔2
̃𝑗 +
̃𝑗 ) ,
+ (∑ 󵄨
𝑚
∑ 𝜌𝑗−𝑘 𝑚
3
󵄨
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨
𝑀
󵄨󵄨
𝑗−𝑘 󵄨󵄨
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 𝜆 𝑘 (R) and eigenvectors V𝑘 (R) are
𝜆 𝑘 (R) = ∑𝑟1,𝑗 𝜌𝑘−1 ,
𝜌𝑘 − 𝜌𝑗
1
𝑎𝜔2
̃
̃𝑗 ) ,
+( 2 ∑󵄨
+
𝑚
∑ 𝜌𝑗 𝑚
𝑎 𝑗 ≠ 𝑘 󵄨󵄨𝜌 − 𝜌 󵄨󵄨󵄨3 𝑗 𝑀
󵄨󵄨 𝑘
𝑗 󵄨󵄨
(14)
(13)
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 , . . . , V𝑁}
defined by (16) are basis of C𝑁. It is clear that 𝑚⃗ ∈ C𝑁 and
̃⃗ ∈ C𝑁. Let
𝑚
𝑚⃗ = 𝑎1 V1 + ⋅ ⋅ ⋅ + 𝑎𝑁V𝑁,
̃⃗ = 𝑏1 V1 + ⋅ ⋅ ⋅ + 𝑏𝑁V𝑁,
𝑚
(18)
4
Abstract and Applied Analysis
where 𝑎𝑗 , 𝑏j ∈ C. Substituting (18) into (13), we can get
We can get from (18) and (27) that
(2)
𝜔2 1⃗ = ∑ 𝑎𝑗 𝜆(1)
𝑗 V𝑗 + ∑ 𝑏𝑗 𝜆 𝑗 V𝑗 ,
(4)
𝑎𝜔2 1⃗ = ∑ 𝑎𝑗 𝜆(3)
𝑗 V𝑗 + ∑ 𝑏𝑗 𝜆 𝑗 V𝑗 .
Note that V1 = 1.⃗ We can get from (19) that
2
𝑁−1
V𝑁 = (𝜌𝑁−1 , 𝜌𝑁−1
, . . . , 𝜌𝑁−1
, 1) .
𝑗=2
(4)
2
(3)
(4)
(𝑎1 𝜆(3)
1 + 𝑏1 𝜆 1 − 𝑎𝜔 ) V1 + ∑ (𝑎𝑗 𝜆 𝑗 + 𝑏𝑗 𝜆 𝑗 ) V𝑗 = 0.
𝑗=2
(20)
Since V1 , V2 , . . . , V𝑁 are basis, we can get from (20) that
𝑎1 𝜆(3)
1
+
𝑏1 𝜆(4)
1
(2)
𝑎𝑗 𝜆(1)
𝑗 + 𝑏𝑗 𝜆 𝑗 = 0,
(4)
𝑎𝑗 𝜆(3)
𝑗 + 𝑏𝑗 𝜆 𝑗 = 0,
= 𝑎𝜔 ,
𝑗 = 2, 3, . . . , 𝑁.
(4)
(2) (3)
Case 1 (if 𝜆(1)
𝑁 𝜆 𝑁 − 𝜆 𝑁 𝜆 𝑁 ≠ 0). It is clear that
−
󵄨󵄨 (1) (2) 󵄨󵄨
󵄨󵄨𝜆 𝑗 𝜆 𝑗 󵄨󵄨
󵄨
󵄨
= 󵄨󵄨󵄨󵄨 (3) (4) 󵄨󵄨󵄨󵄨 ≠ 0,
󵄨󵄨𝜆 𝑗 𝜆 𝑗 󵄨󵄨
󵄨󵄨
󵄨󵄨
𝑚1 = ⋅ ⋅ ⋅ = 𝑚𝑁,
(22)
Proof
(3)
𝜆(2)
𝑗 𝜆𝑗
𝑚⃗ = 𝑎1 V1 ,
(31)
0, if 𝑘 ≠ 𝑁
∑𝜌𝑗𝑘 = {
𝑁, if 𝑘 = 𝑁.
𝑗
(23)
𝜆(1)
𝑘 = ∑ (𝑎1,𝑗 +
(24)
𝑗
Note that 𝜔 and 𝑎𝜔 are real. From Proposition 8, it is clear
(2) (3)
(4)
that 𝜆(1)
1 , 𝜆 1 , 𝜆 1 , and 𝜆 1 are real. Thus, we know from (21)
and (23) for 𝑗 = 1 that 𝑎1 and 𝑏1 are real. Substituting (24) into
(18), we get
̃⃗ = 𝑏1 V1 .
𝑚
(25)
̃1 = ⋅ ⋅ ⋅ = 𝑚
̃𝑁 .
𝑚
(26)
Thus, we have
(4)
(2) (3)
Case 2. (if 𝜆(1)
𝑁 𝜆 𝑁 − 𝜆 𝑁 𝜆 𝑁 = 0). By the similar proof as to
(24), we have
𝑎2 = ⋅ ⋅ ⋅ = 𝑎𝑁−1 = 𝑏2 = ⋅ ⋅ ⋅ = 𝑏𝑁−1 = 0.
(27)
(32)
Then from the general formulas of eigenvalue of circulant
matrix, we have
2
𝑚1 = 𝑚2 = ⋅ ⋅ ⋅ = 𝑚𝑁,
̃1 = ⋅ ⋅ ⋅ = 𝑚
̃𝑁 .
𝑚
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 𝜌𝑗 is the root of unity, it is easy to check
that
𝑗 = 1, 2, . . . , 𝑁.
𝑎2 = ⋅ ⋅ ⋅ = 𝑎𝑁 = 𝑏2 = ⋅ ⋅ ⋅ = 𝑏𝑁 = 0.
𝑚⃗ = 𝑎1 V1 ,
(30)
From Cases 1 and 2, the proof is completed.
By Gramer’s rule, we get from (22) and (23) that
2
̃⃗ = 𝑏1 V1 .
𝑚
Thus, we have
(4)
(2) (3)
Lemma 9. If 𝜆(1)
𝑗 𝜆 𝑗 − 𝜆 𝑗 𝜆 𝑗 ≠ 0, 𝑗 = 1, 2, . . . , 𝑁 − 1, then
̃1 = 𝑚
̃2 = ⋅ ⋅ ⋅ = 𝑚
̃𝑁 .
𝑚1 = 𝑚2 = ⋅ ⋅ ⋅ = 𝑚𝑁 and 𝑚
(4)
𝜆(1)
𝑗 𝜆𝑗
(29)
𝑁−1
Since 𝑎𝑁V𝑁 = (𝑎𝑁𝜌𝑁−1 , . . . , 𝑎𝑁𝜌𝑁−1
, 𝑎𝑁) ∈ R𝑁, so 𝑎𝑁 ∈ R.
Hence, 𝑚⃗ ∈ 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
(21)
2
(28)
Since 𝑎1 and 𝑏1 are real, it follows from (28) that 𝑚⃗ ∈ 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
(2)
2
(1)
(2)
(𝑎1 𝜆(1)
1 + 𝑏1 𝜆 1 − 𝜔 ) V1 + ∑ (𝑎𝑗 𝜆 𝑗 + 𝑏𝑗 𝜆 𝑗 ) V𝑗 = 0,
(2)
2
𝑎1 𝜆(1)
1 + 𝑏1 𝜆 1 = 𝜔 ,
̃⃗ = 𝑏1 V1 + 𝑏𝑁V𝑁.
𝑚
𝑚⃗ = 𝑎1 V1 + 𝑎𝑁V𝑁,
(19)
𝜔2
𝑑 ) 𝜌𝑘−1
𝑀 1,j 𝑗−1
𝑘−1
𝑘
𝜌𝑗−1
− 𝜌𝑗−1
= ∑ 󵄨
󵄨󵄨3
󵄨
𝑗 ≠ 1 󵄨󵄨1 − 𝜌𝑗−1 󵄨󵄨
󵄨
󵄨
(33)
𝜌𝑗𝑘−1 − 𝜌𝑗𝑘
= ∑ 󵄨
󵄨󵄨3 ,
󵄨
𝑗 ≠ 𝑁 󵄨󵄨1 − 𝜌𝑗 󵄨󵄨
󵄨
󵄨
where 𝑘 = 1, 2, . . . , 𝑁−1 and (32) is used. From Proposition 8,
we know that the eigenvalue 𝜆(1)
𝑘 is real. Thus, we only take the
real part and get that
cos (2𝜋𝑗/𝑁) (𝑘 − 1) − cos (2𝜋𝑗/𝑁) 𝑘
,
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
𝜆(1)
𝑘 = ∑
for 𝑘 = 1, 2, . . . , 𝑁 − 1.
(34)
Abstract and Applied Analysis
5
Note that |𝑎 − 𝜌𝑗 | = |1 − 𝑎𝜌𝑗 |. Using similar method as
proving (34), we have
𝜆(2)
𝑘 = ∑ (𝑏1𝑗 +
=∑
𝑎𝜔2
𝑑 ) 𝜌𝑘−1
𝑀 1𝑗 𝑗−1
(4)
(2) (3)
Lemma 13. For 𝑁 ≥ 3, 𝜆(1)
1 𝜆 1 − 𝜆 1 𝜆 1 > 0.
cos (2𝜋𝑗/𝑁) (𝑘 − 1) − 𝑎 cos (2𝜋𝑗/𝑁) 𝑘
,
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
𝜆(3)
𝑘 = ∑ (𝑐1𝑗 +
=
Proof. From (34) and (35), we can get
𝜆(1)
1 =
2
𝜔
𝑑 ) 𝜌𝑘−1
𝑀 1𝑗 𝑗−1
acos (2𝜋𝑗/𝑁) (𝑘 − 1) − cos (2𝜋𝑗/𝑁) 𝑘
=∑
,
󵄨󵄨
󵄨󵄨3
󵄨󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝜆(4)
𝑘 = ∑(
Proposition 12. Let 𝜓𝛽,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.
𝜆(3)
1
1
𝑎𝜔2
𝑎
+
𝑑 ) 𝜌𝑘−1
𝑎2 1𝑗 𝑀 1𝑗 𝑗−1
𝜆(4)
1
Lemma 10.
𝑘 = 2, . . . , 𝑁 − 1.
2
(4)
(2)
(3)
𝜆(1)
𝑁−𝑘+1 𝜆 𝑁−𝑘+1 −𝜆 𝑁−𝑘+1 𝜆 𝑁−𝑘+1 ,
Proof. We get from (34) that
𝜆(1)
𝑁−𝑘+1
cos (2𝜋𝑗/𝑁) 𝑘 − cos (2𝜋𝑗/𝑁) (𝑘 − 1)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
= ∑
= −𝜆(1)
𝑘 ,
(36)
for 𝑘 = 2, 3, . . . , 𝑁 − 1.
𝜋𝑗
1
𝐴 1 (𝑎) =
( ∑ csc )
2
16𝑎 𝑗 ≠ 𝑁
𝑁
− (∑
1 − 𝑎 cos (2𝜋𝑗/𝑁)
)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
× (∑
𝑎 − cos (2𝜋𝑗/𝑁)
)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
2
𝜋𝑗
1
=
( ∑ csc )
2
16𝑎 𝑗 ≠ 𝑁
𝑁
Similarly, we get from (35) that
(3)
𝜆(2)
𝑁−𝑘+1 = −𝜆 𝑘 ,
(2)
𝜆(3)
𝑁−𝑘+1 = −𝜆 𝑘 ,
2
1
(37)
(4)
𝜆(4)
𝑁−𝑘+1 = −𝜆 𝑘 ,
where 𝑘 = 2, 3, . . . , 𝑁 − 1. From (36) and (37), the result
follows.
Remark 11. Let
𝑁
{
,
{
{
{2
ℓ={
{
{
{ (𝑁 + 1)
,
{ 2
(39)
(4)
(2) (3)
Let 𝐴 1 (𝑎) = 𝜆(1)
1 𝜆 1 − 𝜆 1 𝜆 1 . Then
where 𝑘 = 1, 2, . . . , 𝑁 − 1.
(4)
(2) (3)
𝜆(1)
𝑘 𝜆 𝑘 −𝜆 𝑘 𝜆 𝑘 =
1 − 𝑎 cos (2𝜋𝑗/𝑁)
,
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
𝑎 − cos (2𝜋𝑗/𝑁)
= ∑
,
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
𝜋𝑗
1
= 2 ∑ csc .
4𝑎 𝑗 ≠ 𝑁 𝑁
𝜆(2)
1 = ∑
(35)
cos (2𝜋𝑗/𝑁) (𝑘 − 1) − cos (2𝜋𝑗/𝑁) 𝑘
1
,
∑
󵄨󵄨
󵄨3
𝑎2 𝑗 ≠ 𝑁
󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
𝜋𝑗
1
∑ csc ,
4 𝑗 ≠ 𝑁
𝑁
if 𝑁 is even,
(38)
if 𝑁 is odd.
From Lemma 10, it is clear that to prove all the 𝑁 − 1
inequalities of Lemma 9 suffices to prove the ℓ inequalities
(4)
(2) (3)
𝜆(1)
𝑘 𝜆 𝑘 − 𝜆 𝑘 𝜆 𝑘 ≠ 0, 𝑘 = 1, 2, . . . , ℓ.
With the similar proof as in [2, Lemma 2], we can get the
following proposition.
cos (2𝜋𝑗/𝑁)
2
− 𝑎(∑ 󵄨
) − 𝑎(∑
)
3
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨󵄨3
(1 − 𝑎𝜌𝑗 )
𝑗 󵄨󵄨
󵄨󵄨
cos (2𝜋𝑗/𝑁)
1
+ (1 + 𝑎2 ) (∑ 󵄨
) (∑ 󵄨
).
3
󵄨
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨󵄨3
󵄨󵄨
󵄨󵄨
𝑗 󵄨󵄨
𝑗 󵄨󵄨
(40)
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 ≠ 𝑁 csc (𝜋𝑗/𝑁) = ∑𝑗 ≠ 𝑁((1−cos(2𝜋𝑗/𝑁))/|1−𝜌𝑗 |3 ) >
0. We have from (40)
𝐴 1 (𝑎) >
𝜋𝑗
1
( ∑ csc
)
16𝑎2 𝑗 ≠ 𝑁
𝑁
2
2
cos (2𝜋𝑗/𝑁)
− 𝑎(∑ 󵄨
) − 𝑎(∑ 󵄨
3
󵄨
󵄨3 )
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨
󵄨󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨󵄨
𝑗 󵄨󵄨
󵄨
󵄨
1
2
6
Abstract and Applied Analysis
Proof. If 𝑁 is odd, by the definition of 𝛼𝑘 , we have
cos (2𝜋𝑗/𝑁)
1
+ 2𝑎 (∑ 󵄨
) (∑ 󵄨
)
3
󵄨
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨󵄨3
󵄨󵄨
󵄨󵄨
𝑗 󵄨󵄨
𝑗 󵄨󵄨
𝛼(𝑁+1)/2 (1)
2
𝜋𝑗
1
=
( ∑ csc )
2
16𝑎 𝑗 ≠ 𝑁
𝑁
1
− 𝑎(∑ 󵄨
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨󵄨3
󵄨󵄨
𝑗 󵄨󵄨
=
= ∑ (cos
𝑗 ≠ 𝑁
cos (2𝜋𝑗/𝑁)
−∑ 󵄨
)
󵄨󵄨1 − 𝑎𝜌 󵄨󵄨󵄨3
󵄨󵄨
𝑗 󵄨󵄨
2
− cos
−1
cos (𝜋𝑗 − (𝜋𝑗/𝑁)) − cos (𝜋𝑗 + (𝜋𝑗/𝑁))
= 0.
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
= ∑
𝑎3/2 (1 − cos (2𝜋𝑗/𝑁)) }
−∑
} ⋅ 𝑔 (𝑎)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
}
󵄨
󵄨
(45)
If 𝑁 is even, we get that
𝛼𝑁/2 (1)
}
2𝜋𝑗
1 {
) ⋅ ℎ (𝑗, 𝑎)} ⋅ 𝑔 (𝑎) ,
= 2 { ∑ (1 − cos
𝑎
𝑁
{𝑗 = ̸ 𝑁
}
cos (2𝜋𝑗 ((𝑁/2) − 1) /𝑁) − cos (2𝜋𝑗 (𝑁/2) /𝑁)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
= ∑
(41)
3/2
where 𝑔(𝑎) = (1/4) ∑𝑗 ≠ 𝑁 csc (𝜋𝑗/𝑁) + ∑𝑗 ≠ 𝑁(𝑎 (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 −
2𝑎 cos(2𝜋𝑗/𝑁))3/2 ) and hence
=
1
1
∑ (−1)𝑗+1
4 𝑗 ≠ 𝑁
sin (𝜋𝑗/𝑁)
=
1
1[
∑ (−1)𝑗+1
4 𝑗=1
sin (𝜋𝑗/𝑁)
[
𝑁/2
𝑁/2−1
for 𝑎 ∈ (0, 1) .
+ ∑ (−1)𝑗+1
𝑗=1
1
].
sin (𝜋𝑗/𝑁)
]
(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
ℎ (𝑗, 𝑎) > ℎ (𝑗, 1) = 0,
2𝜋𝑗 ((𝑁 + 1) /2)
)
𝑁
󵄨3
󵄨
× (󵄨󵄨󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨󵄨 )
1 − cos (2𝜋𝑗/𝑁)
1 {
∑
{
2
󵄨󵄨
󵄨3
𝑎
󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
{𝑗 = ̸ 𝑁
󵄨
󵄨
𝑑ℎ (𝑗, 𝑎) (3/2) 𝑎1/2 (𝑎 − 1) (𝑎 + 1)
< 0,
=
󵄨󵄨
󵄨5
𝑑𝑎
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
2𝜋𝑗 (((𝑁 + 1) /2) − 1)
𝑁
for 𝑎 ∈ (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 𝐴 1 (𝑎) > 0. The result follows.
(46)
Case 1. If 𝑁/2 is even, then
𝛼𝑁/2 (1) =
𝑁/2
1
1[
+ 1] .
2 ∑ (−1)𝑗+1
4
sin
(𝜋𝑗/𝑁)
]
[ 𝑗=1
(47)
𝑗+1
Since the signs of the terms in the sum ∑𝑁/2
(1/
𝑗=1 (−1)
sin(𝜋𝑗/𝑁)) alternate and 1/ sin(𝜋𝑗/𝑁) is decreasing in 𝑗 for
𝑗+1
1 ≤ 𝑗 ≤ 𝑁/2, we get that ∑𝑁/2
(1/ sin(𝜋𝑗/𝑁)) > 0 and
𝑗=1 (−1)
hence 𝛼𝑁/2 (1) > 0.
Case 2. If 𝑁/2 is odd, then
For 𝑘 = 1, 2, . . . , ℓ, let
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
.
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
𝛼𝑘 (𝑎) = ∑
𝛼𝑁/2 (1) =
(44)
Lemma 14. If 𝑁 is odd, then 𝛼(𝑁+1)/2 (1) = 0; If 𝑁 is even, then
𝛼𝑁/2 (1) > 0.
𝑁/2−1
1
1[
+ 1] .
2 ∑ (−1)𝑗+1
4
sin
(𝜋𝑗/𝑁)
𝑗=1
]
[
(48)
𝑗+1
Since the signs of the terms in summation ∑𝑁/2−1
𝑗=1 (−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
(−1)𝑗+1 (1/
𝑗=1
sin(𝜋𝑗/𝑁)) > 0 and hence 𝛼𝑁/2 (1) > 0.
By Cases 1 and 2, we see that 𝛼𝑁/2 (1) > 0 for even 𝑁.
Abstract and Applied Analysis
7
Proposition 15. For 𝑘 = 1, 2, . . . , ℓ − 1, then 𝛼𝑘 (1) > 0.
Proof. Clearly, 𝛼1 (1) > 0. We need to prove 𝛼𝑘 (1) > 0, 𝑘 ≥ 2.
Note that
𝑘−1
2𝜋𝑗
2
𝑚
∑ cos
𝑁
sin (𝜋𝑗/𝑁) 𝑚=0
=
2 sin (𝜋𝑗𝑘/𝑁) cos (𝜋𝑗 (𝑘 − 1) /𝑁)
sin2 (𝜋𝑗/𝑁)
=
sin (𝜋𝑗 (2𝑘 − 1) /𝑁)
1
.
+
sin (𝜋𝑗/𝑁)
sin2 (𝜋𝑗/𝑁)
𝛽𝑘 (𝑎) = ∑
(50)
Proof. Clearly, 𝛽1 (1) > 0. We need to prove 𝛽𝑘 (1) > 0, 𝑘 ≥ 2.
Note that
𝑘−1
2𝜋𝑗
2
𝑚
∑ cos
3
𝑁
sin (𝜋𝑗/𝑁) 𝑚=0
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
1
∑
2 𝑗 ≠ 𝑁
sin (𝜋𝑗/𝑁)
𝑗 ≠ 𝑁
= cot
(51)
𝜋 (2𝑘 − 1)
.
2𝑁
Notice that 0 < 𝜋(2𝑘 − 1)/(2𝑁) < 𝜋/2 for 𝑘 = 2, . . . , ℓ − 1.
Hence, cot 𝜋(2𝑘 − 1)/2𝑁 > 0 for 𝑘 = 2, . . . , ℓ − 1. Then we
have
𝛼𝑘 (1) − 𝛼𝑘−1 (1) > 𝛼𝑘+1 (1) − 𝛼𝑘 (1) .
Similar to Lemma 14, we have
Proposition 17. For 𝑘 = 1, 2, . . . , ℓ − 1, then 𝛽𝑘 (1) > 0.
(𝛼𝑘 (1) − 𝛼𝑘−1 (1)) − (𝛼𝑘+1 (1) − 𝛼𝑘 (1))
𝜋𝑗 (2𝑘 − 1)
𝑁
(54)
Lemma 16. If 𝑁 is odd, then 𝛽(𝑁+1)/2 (1) = 0; If 𝑁 is even, then
𝛽𝑁/2 (1) > 0.
Then, we get that
= ∑ sin
𝑎.
𝑎∈(0,1), 𝛼𝑘 (𝑎)>0, 𝑘=1,2,...,ℓ̃
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
.
󵄨󵄨
󵄨5
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
3/2
𝑘−1
2𝜋𝑗
2
+
𝑚] .
∑ cos
𝑁
sin (𝜋𝑗/𝑁) 𝑚=0
=
min
(53)
By the continuity, Lemma 14 and Proposition 15, we see that
𝑎̂∗ ∈ (0, 1).
For 𝑘 = 1, 2, . . . , ℓ, let
(2 − 2 cos (2𝜋𝑗/𝑁))
1
1
∑ [−
4 𝑗 ≠ 𝑁 sin (𝜋𝑗/𝑁)
=
𝑁
{
,
if 𝑁 is even,
{
{
{2
̃
ℓ ={
{
{
{ (𝑁 − 1)
, if 𝑁 is odd,
{ 2
𝑎̂∗ =
2 sin (𝜋𝑗 (2𝑘 − 1) /𝑁) sin (𝜋𝑗/𝑁)
𝑗 ≠ 𝑁
Let
(49)
Hence, by the definition of 𝛼𝑘 (𝑎), we have
𝛼𝑘 (1) = ∑
From Cases 1 and 2, we see that there is always a contradiction if there exists 𝛼𝑘 (1) ≤ 0. Hence, 𝛼𝑘 (1) > 0, 𝑘 = 1,
. . . , ℓ.
(52)
=
2 sin (𝜋𝑗𝑘/𝑁) cos (𝜋𝑗 (𝑘 − 1) /𝑁)
1
⋅
sin (𝜋𝑗/𝑁)
sin (𝜋𝑗/𝑁)
=
sin (𝜋𝑗 (2𝑘 − 1) /𝑁)
1
+
.
sin4 (𝜋𝑗/𝑁)
sin3 (𝜋𝑗/𝑁)
3
(55)
By the definition of 𝛽𝑘 (1), we have
𝛽𝑘 (1) =
1
1
∑ [−
16 𝑗 ≠ 𝑁 sin3 (𝜋𝑗/𝑁)
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.
Case 1. (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.
Case 2. (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.
𝑘−1
2𝜋𝑗
2
+
𝑚] .
∑ cos
𝑁
sin3 (𝜋𝑗/𝑁) 𝑚=0
(56)
Then, we get that
(𝛽𝑘 (1) − 𝛽𝑘−1 (1)) − (𝛽𝑘+1 (1) − 𝛽𝑘 (1))
=
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
1
∑
8 𝑗 ≠ 𝑁
sin3 (𝜋𝑗/𝑁)
(57)
1
= 𝛼𝑘 (1) > 0.
8
By similar proofs as in Proposition 15, we can get 𝛽𝑘 > 0. The
proof is completed.
8
Abstract and Applied Analysis
Let 𝑓(𝑥) = cos 𝑥/(1 + 𝑎2 − 2𝑎 cos 2𝑥)3/2 for 0 < 𝑥 < 𝜋/2. Then
Let
𝑎̃∗ =
𝑎.
min
𝑎∈(0,1),𝛽𝑘 (𝑎)>0, 𝑘=1,2,...,ℓ̃
(58)
By the continuity, Lemma 16 and Proposition 17, we see that
𝑎̃∗ ∈ (0, 1).
𝑎∗ , 𝑎̃∗ },
Lemma 18. For 𝑘 = 2, 3, . . . , ℓ and 𝑎∗ = max{̂
(1) (4)
(2) (3)
𝜆 𝑘 𝜆 𝑘 − 𝜆 𝑘 𝜆 𝑘 > 0.
(4)
(2) (3)
Proof. Let 𝐴 𝑘 (𝑎) = 𝜆(1)
𝑘 𝜆 𝑘 − 𝜆 𝑘 𝜆 𝑘 . From (34) and (35), we
can get
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
1
)
𝐴 𝑘 (𝑎) = 2 ( ∑
󵄨󵄨
󵄨󵄨3
𝑎 𝑗 ≠ 𝑁
󵄨󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
− (∑
⋅ (∑
2
󸀠
𝑓 (𝑥) =
̃ we see that ℓ̃ = ℓ if 𝑁 is even
By the definitions of ℓ and ℓ,
and ℓ̃ = ℓ − 1 if 𝑁 is odd. For odd 𝑁, we get
𝐴 (𝑁+1)/2 (𝑎) > 0.
2𝜋𝑗 ((𝑁 + 1) /2)
󵄨−3 2
󵄨
) × 󵄨󵄨󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨󵄨 )
𝑁
̃ let
For any 𝑘, 𝑘 = 2, 3, . . . , ℓ,
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
𝐺 (𝑎) = ∑
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
.
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
(63)
− 𝑎3/2 ∑
With direct computation, we have
3
𝐺󸀠 (𝑎) = 𝑎1/2 (𝑎 + 1) (𝑎 − 1)
2
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
< 0,
󵄨󵄨
󵄨5
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
× ∑
⋅ ( ∑ (𝑎 cos
= (1 − 𝑎)
2𝜋𝑗 ((𝑁 + 1) /2)
󵄨−3
󵄨
) × 󵄨󵄨󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨󵄨 )
𝑁
(65)
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋j𝑘/𝑁)
> 𝑎3/2 ∑
,
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
2
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
1
(∑
)
󵄨󵄨
󵄨󵄨3
𝑎2 𝑗 ≠ 𝑁
󵄨󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
2𝜋𝑗 ((𝑁 + 1) /2)
󵄨−3
󵄨
) × 󵄨󵄨󵄨󵄨𝑎 − 𝜌𝑗 󵄨󵄨󵄨󵄨 )
𝑁
cos (𝜋𝑗/𝑁) ]
∑ (−1)𝑗 󵄨
󵄨3
󵄨󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨 ]
[
󵄨
󵄨
2[
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
̃ From (65), we can get
where 𝑎 ∈ (𝑎∗ , 1) and 𝑘 = 2, 3, . . . , ℓ.
2𝜋𝑗 (((𝑁 + 1) /2) − 1)
𝑁
− cos
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
∑
2𝜋𝑗 (((𝑁 + 1) /2) − 1)
− ( ∑ (cos
𝑁
−𝑎 cos
(62)
for 𝑎 ∈ (𝑎∗ , 1) ,
(64)
2𝜋𝑗 (((𝑁 + 1) /2) − 1)
1
( ∑ (cos
2
𝑎 𝑗 ≠ 𝑁
𝑁
− cos
< 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
𝐴 (𝑁+1)/2 (𝑎)
=
5/2
(1 + 𝑎2 − 2𝑎 cos 2𝑥)
(61)
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − 𝑎 cos (2𝜋𝑗𝑘/𝑁)
)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
𝑎 cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
).
󵄨󵄨
󵄨3
󵄨󵄨𝑎 − 𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
(59)
− sin 𝑥 (1 + 𝑎2 − 2𝑎 cos 2𝑥) − 6𝑎 cos 𝑥 sin 2𝑥
2
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
> 𝑎( ∑
),
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
2
.
(60)
̃
where 𝑎 ∈ (𝑎∗ , 1) and 𝑘 = 2, 3, . . . , ℓ.
(66)
Abstract and Applied Analysis
9
Example 19. If 𝑁 = 3, then 𝑎∗ = 0.
By (59), we can get
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
1
)
𝐴 𝑘 (𝑎) = 2 ( ∑
󵄨󵄨
󵄨󵄨3
𝑎 𝑗 ≠ 𝑁
󵄨󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
− 𝑎(∑
cos (2𝜋𝑗 (𝑘 − 1) /𝑁)
)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
+ (1 + 𝑎2 ) (∑
× (∑
2
Proof. For 𝑁 = 3, we have
𝛼1 (𝑎) =
2
𝛼2 (𝑎) =
cos (2𝜋𝑗 (𝑘 − 1) /𝑁)
)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
cos (2𝜋𝑗𝑘/𝑁)
󵄨󵄨
󵄨3 )
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
𝛽2 (𝑎) =
2
5/2
> 0,
5/2
> 0.
(69)
((1 + (𝑎/2))2 + (3/4))
2
((1 + (𝑎/2))2 + (3/4))
(70)
1
((1 + (𝑎/2))2 + (3/4))
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
̃1 = 𝑚
̃2 = ⋅ ⋅ ⋅ = 𝑚
̃𝑁 =
Since 𝑚1 = 𝑚2 = ⋅ ⋅ ⋅ = 𝑚𝑁 = 𝑚 and 𝑚
̃ we get that 𝑧0 = 0. Hence, by (3), we have
𝑚,
2
𝑞𝑘 = 𝜌𝑘 𝑒𝑖𝜔𝑡 ,
𝑞̃𝑘 = 𝑎𝜌𝑘 𝑒𝑖𝜔𝑡 .
(71)
Substituting 𝑞𝑘 and 𝑞̃𝑘 into (1), then by (11) we get that
𝜔2 =
cos (2𝜋𝑗𝑘/𝑁)
󵄨󵄨
󵄨3 )
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
cos (2𝜋𝑗𝑘/𝑁)
)
+ (∑ 󵄨
󵄨󵄨1 − a𝜌 󵄨󵄨󵄨3
󵄨󵄨
𝑗 󵄨󵄨
3/2
> 0.
1
2
cos (2𝜋𝑗 (𝑘 − 1) /𝑁)
− 2 (∑
)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
× (∑
> 0,
(67)
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
1
(∑
)
󵄨󵄨
󵄨3
𝑎2 𝑗 ≠ 𝑁
󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
{
{
cos (2𝜋𝑗 (𝑘 − 1) /𝑁)
− 𝑎 {(∑
)
󵄨󵄨
󵄨3
{
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
{
3/2
By the definition of 𝑎̃∗ , we have 𝑎̃∗ = 0. Hence, 𝑎∗ = 0 for
𝑁 = 3.
From Proposition 12, we have
𝐴 𝑘 (𝑎) >
((1 + (𝑎/2))2 + (3/4))
By the definition of 𝑎̂∗ , we have 𝑎̂∗ = 0.
Similarly, we get that
𝛽1 (𝑎) =
cos (2𝜋𝑗𝑘/𝑁)
− 𝑎(∑ 󵄨
󵄨3 ) .
󵄨󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
2
1 − 𝑎 cos (2𝜋𝑗/𝑁)
𝜋𝑗
𝑚
̃∑
,
+𝑚
∑ csc
󵄨󵄨
󵄨3
4 𝑗 ≠ 𝑁
𝑁
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
𝑎 − cos (2𝜋𝑗/𝑁)
𝜋𝑗
̃
𝑚
𝑎𝜔 = 𝑚 ∑
+ 2 ∑ csc .
󵄨󵄨
󵄨󵄨3
4𝑎
𝑁
𝑗 ≠ 𝑁
󵄨󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
(72)
2
2
}
}
}
}
}
From (72), we have
𝜔2 =
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
1
= 2( ∑
)
󵄨󵄨
󵄨󵄨3
𝑎 𝑗 ≠ 𝑁
󵄨󵄨󵄨1 − 𝜌𝑗 󵄨󵄨󵄨
2
cos (2𝜋𝑗 (𝑘 − 1) /𝑁) − cos (2𝜋𝑗𝑘/𝑁)
− 𝑎( ∑
)
󵄨󵄨
󵄨3
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝑗 ≠ 𝑁
󵄨
󵄨
𝜔2 =
2
> 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.
1 − 𝑎 cos (2𝜋𝑗/𝑁)
𝜋𝑗
𝑚
̃∑
,
+𝑚
∑ csc
󵄨󵄨
󵄨󵄨3
4 𝑗 ≠ 𝑁
𝑁
󵄨󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
𝑎−
1[
𝑚∑
𝑎
[
cos (2𝜋𝑗/𝑁)
𝜋𝑗
̃
𝑚
+ 2 ∑ csc ] .
󵄨󵄨
󵄨󵄨3
4𝑎 𝑗 ≠ 𝑁
𝑁
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨
]
󵄨
󵄨
(73)
From (73), the constant 𝜔2 exists only if the following equation holds:
{𝑚
1 − 𝑎 cos (2𝜋𝑗/𝑁) }
𝜋𝑗
̃∑
𝑎 { ∑ csc
+𝑚
}
󵄨󵄨
󵄨3
4
𝑁
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨󵄨
}
{ 𝑗 ≠ 𝑁
󵄨
󵄨
𝑎 − cos (2𝜋𝑗/𝑁)
𝜋𝑗
̃
𝑚
= 𝑚∑
+ 2 ∑ csc .
󵄨󵄨
󵄨󵄨3
4𝑎 𝑗 ≠ 𝑁 𝑁
󵄨󵄨1 − 𝑎𝜌𝑗 󵄨󵄨
󵄨
󵄨
(74)
10
Abstract and Applied Analysis
̃ Then we get from (74) that
Note that 𝜇 = 𝑚/𝑚.
𝑎3 − 𝜇
𝜋𝑗
− (1 − 𝜇) 𝑎𝑓 (𝑎) + (1 − 𝑎2 𝜇) 𝑔 (𝑎) = 0,
∑ csc
4𝑎2 𝑗 ≠ 𝑁
𝑁
(75)
where 𝑓(𝑎) = ∑(1/|1 − 𝑎𝜌𝑗 |3 ) and 𝑔(𝑎) = ∑(cos(2𝜋𝑗/𝑁)/|1 −
𝑎𝜌𝑗 |3 ).
Lemma 20. For 𝜇 > 0 and 𝑁 ≥ 3, the constant 𝜔2 can be given
by
1 − 𝑎0 cos (2𝜋𝑗/𝑁)
𝜋𝑗
𝑚
̃∑
,
+𝑚
∑ csc
󵄨󵄨
󵄨3
4 𝑗 ≠ 𝑁
𝑁
󵄨󵄨1 − 𝑎0 𝜌𝑗 󵄨󵄨󵄨
󵄨
󵄨
where 𝑎0 is the unique solution of (75).
𝜔2 =
(76)
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) = (𝑎 −
𝜋𝑗
𝜇 1
) ∑ csc
2
𝑎 4 𝑗 ≠ 𝑁
𝑁
(77)
− (1 − 𝜇) 𝑎𝑓 (𝑎) + (1 − 𝑎2 𝜇) 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
𝜓1,𝑁 (𝑎) = (1 + 𝑎2 ) 𝑓 (𝑎) − 2𝑎𝑔 (𝑎) ,
𝑑𝜓1,𝑁 (𝑎)
= − 𝑎𝑓 (𝑎) + 𝑔 (𝑎) .
𝑑𝑎
From (78), (77) can be written as
𝐹 (𝑎) = (𝑎 −
𝜇 1
𝜋𝑗
) ∑ csc
2
𝑎 4 𝑗 ≠ 𝑁
𝑁
𝑑𝜓1,𝑁 (𝑎)
+ (1 + 𝑎2 𝜇)
+ 𝜇𝑎𝜓1,𝑁 (𝑎) .
𝑑𝑎
(78)
(79)
Notice that 𝜓1,𝑁(𝑎) = ∑𝑗 ≠ 𝑁(1/|1 − 𝑎𝜌𝑗 |) + (1/|1 − 𝑎𝜌𝑁|) =
∑𝑗 ≠ 𝑁(1/|1 − 𝑎𝜌𝑗 |) + (1/|1 − a|). Hence, we have
lim 𝜓1,𝑁 (𝑎) = +∞.
𝑎→1
(80)
From (79), (80), and Proposition 12, we can get that
lim 𝐹 (𝑎) = −∞,
𝑎→0
lim 𝐹 (𝑎) = +∞.
𝑎→1
(81)
On the other hand, by (79), we get that
𝐹󸀠 (𝑎) = (1 +
2𝜇 1
𝜋𝑗
) ∑ csc
𝑎3 4 𝑗 ≠ 𝑁
𝑁
+ 𝜇𝜓1,𝑁 (𝑎) + 3𝑎𝜇
+ (1 + 𝑎2 𝜇)
𝑑 𝜓1,𝑁 (𝑎)
𝑑𝑎
𝑑2 𝜓1,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, CQDXWL2012-006.
References
[1] L. M. Perko and E. L. Walter, “Regular polygon solutions of the
𝑁-body problem,” Proceedings of the American Mathematical
Society, vol. 94, no. 2, pp. 301–309, 1985.
[2] R. Moeckel and C. Simó, “Bifurcation of spatial central configurations from planar ones,” SIAM Journal on Mathematical
Analysis, vol. 26, no. 4, pp. 978–998, 1995.
[3] S. Zhang and Q. Zhou, “Periodic solutions for planar 𝑁-body
problems,” Proceedings of the American Mathematical Society,
vol. 131, no. 7, pp. 2161–2170, 2003.
[4] M. Marcus and H. Minc, A Survey of Matrix Theory and Matrix
Inequalities, Allyn and Bacon, Boston, Mass, USA, 1964.
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