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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 397194, 10 pages
The Periodic Solutions for Planar -Body Problems
1College of Mathematics, Sichuan University, Chengdu 610064, China
2College of Mathematics and Physics, Chongqing University of Posts and Telecommunications, Chongqing 400065, China
3Department of Mathematics, Chongqing University, Chongqing 400044, China
Received 7 February 2013; Accepted 1 April 2013
Academic Editor: Baodong Zheng
Copyright © 2013 Xiaohong Hu and Changrong 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 -body problems.
1. Introduction and Main Results
Let be the mass and position of the th body. The Newtonian -body problem concerns the motion of point particles. The motion is governed by where is the Newtonian potential:
In the famous paper , Perko and Walter proved the following result.
Theorem 1 (see ). For , the bodies move with uniformly angular velocity and locate on the vertices of regular -gon if and only if .
In 1995, Moeckel and Simó  studied planar nested -body problems; they assume one regular -gon is inscribed on a unit circle, the other on a circle with radius , and ; precisely, let , where ; and the points and locate at and :
where , , , and . For short, throughout this paper, all indices and summations will range from to unless we give other restrictions.
For the nested -body problems, Moeckel and Simó proved the following theorem.
Theorem 2 (see ). If and , then for every mass ratio , there are exactly two planar central configurations consisting of two nested regular -gons. For one of them, the ratio of the sizes of the two nested -gons is less than , and for the other it is greater than .
Based on all the above works, we try to give strict proofs about the following two theorems. By the work of Moeckel and Simó , if we can get a periodic solution of the form given by (5) with , the other periodic solution with radius can be obtained by symmetry. Hence, in the following, we only discuss the periodic solution with radius . The other one with radius can be obtained by symmetry.
Theorem 6. If , and the constant is given by where and is a unique solution of the following equation: then with is a periodic solution of (1) with angular velocity .
When and , we have 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 . By symmetry, (1) has another periodic solution which rotates about the origin with radius .
2. The Proof of Theorem 5
Let , and , where Then (11) can be rewritten in the following compact form: where , and .
An matrix is called circulant (see ) if where and are equal to and , respectively. Let . If is a circulant matrix, its general formulas for the eigenvalues and eigenvectors are
Remark 7. From the formula (15), we have . The last equation implies that the eigenvalue is equal to the summation of the first row of matrix ; thus, we can get that the summation of any row, and hence, the summation of any column is equal to .
According to the definition of circulant matrix, it is easy to check that the matrixes , and are circulant. For convenience, we introduce some notations. Let , and be the th eigenvalue of matrixes , , and , respectively. Then we have the following.
Proposition 8. All of the eigenvalues of matrixes , and are real.
Proof. We only give the proof for the matrix . The proofs for the rest are similar. Since is circulant and is real number, we get from (12) that where . Thus, the matrix is Hermitian. We know that all the eigenvalues of are real since the eigenvalues of Hermitian matrix are real. The proof is completed.
From the proof of Proposition 8, we have known that , and are Hermitian. Thus, the vectors defined by (16) are basis of . It is clear that and . Let where . Substituting (18) into (13), we can get Note that . We can get from (19) that Since are basis, we can get from (20) that
Lemma 9. If , then and .
Case 1 (if ). It is clear that By Gramer's rule, we get from (22) and (23) that Note that and are real. From Proposition 8, it is clear that , and are real. Thus, we know from (21) and (23) for that and are real. Substituting (24) into (18), we get Thus, we have
Case 2. (if ). By the similar proof as to (24), we have We can get from (18) and (27) that Since and are real, it follows from (28) that if and only if or . If , from the general formulas of eigenvectors defined in (16), we know that Since , so . Hence, if and only if or . If , then from (29), which implies that . But it is impossible for . Thus, . Similarly, we can get . We get from (28) that Thus, we have
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 is the root of unity, it is easy to check that Then from the general formulas of eigenvalue of circulant matrix, we have where and (32) is used. From Proposition 8, we know that the eigenvalue is real. Thus, we only take the real part and get that
Note that . Using similar method as proving (34), we have where .
Lemma 10. .
With the similar proof as in [2, Lemma 2], we can get the following proposition.
Proposition 12. Let . Then for and , and all of its derivatives are positive.
Lemma 13. For , .
Proof. From (34) and (35), we can get Let . Then Clearly, . By Proposition 12, we get that for . Note that . We have from (40) where and . Note that and hence We know from (42) that is decreasing function in for . It is easy to check that . Thus So, since and . Note that for . From (41), we can get . The result follows.
For , let
Lemma 14. If is odd, then ; If is even, then .
Proof. If is odd, by the definition of , we have
If is even, we get that
Case 1. If is even, then
Since the signs of the terms in the sum alternate and is decreasing in for , we get that and hence .
Case 2. If is odd, then
Since the signs of the terms in summation alternate and ) is decreasing in for , we get that and hence .
By Cases 1 and 2, we see that for even .
Proposition 15. For , then .
Proof. Clearly, . We need to prove . Note that
Hence, by the definition of , we have
Then, we get that
Notice that for . Hence, for . Then we have
By (52), we get that which implies that or . If there is , we use two cases to proceed our proof.
Case 1. (If ). By (52), we have and hence . By inductions, we have which contradicts with .
Case 2. (If ). By (52), we have . Then, . By inductions, we get that for even , which contradicts with ; for odd , which contradicts with .
From Cases 1 and 2, we see that there is always a contradiction if there exists . Hence, .
For , let
Similar to Lemma 14, we have
Lemma 16. If is odd, then ; If is even, then .
Proposition 17. For , then .
Proof. Clearly, . We need to prove . Note that By the definition of , we have Then, we get that By similar proofs as in Proposition 15, we can get . The proof is completed.
Lemma 18. For and , .
Proof. Let . From (34) and (35), we can get
By the definitions of and , we see that if is even and if is odd. For odd , we get
Let for . Then
Hence, is decreasing in for . Note that the signs of the terms alternate, then the summation is negative for . By symmetry, the summation for the rest terms is also negative. Hence, we have
For any , let With direct computation, we have where Proposition 17 is used. It follows from (64) that is decreasing for . Note that . So, for . It is that where and . From (65), we can get where and .
By (59), we can get From Proposition 12, we have
where (66) is used and .
By (62) and (68), the proof is completed.
The following example illustrate that the in Lemma 18 can be obtained for some special cases.
Example 19. If , then .
Proof. For , we have
By the definition of , we have .
Similarly, we get that
By the definition of , we have . Hence, for .
3. The Proof of Theorem 6
Lemma 20. For and , the constant can be given by where is the unique solution of (75).
Proof. By the relationship between (74) and (75), it suffices to prove that (75) has solution for and . Let From (75) and (77), the roots of (75) are zeros of (77). From the definitions of defined in Proposition 12, we have . Through direct calculations, we find that From (78), (77) can be written as Notice that . Hence, we have From (79), (80), and Proposition 12, we can get that On the other hand, by (79), we get that From (82) and Proposition 12, we know that for . This, together with (81), implies that there is a unique solution