Abstract and Applied Analysis

Volume 2013 (2013), Article ID 397194, 10 pages

http://dx.doi.org/10.1155/2013/397194

Research Article

The Periodic Solutions for Planar -Body Problems

Xiaohong Hu1,2 and Changrong Zhu³

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.

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 -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 [1], Perko and Walter proved the following result.

Theorem 1 (see [1]). 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 [2] 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 [2]). 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 .

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 ). If (3) is a solution of (1), then satisfies

or equivalently

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

Theorem 4. If satisfies (4) or (5) and given by (3) is a periodic solution of (1), then and .

In the proof of [3, Theorem 2], the authors claimed that the first eigenvalue of the matrix 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 , 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 5. For , if , where , and given by (3) is a periodic solution for (1), then and .

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

Substituting (3) into (1), we have

where .

By (3), (8) can be written as

where . Equation (9) is equivalent to

where . Multiplying both sides of (10) by , we get

where .

Let , and , where

Then (11) can be rewritten in the following compact form:

where , and .

An matrix is called circulant (see [4]) 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   .

Proof

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. .

Proof. We get from (34) that

Similarly, we get from (35) that

where . From (36) and (37), the result follows.

Remark 11. Let

From Lemma 10, it is clear that to prove all the inequalities of Lemma 9 suffices to prove the inequalities .

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, .

Let

By the continuity, Lemma 14 and Proposition 15, we see that .

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.

Let

By the continuity, Lemma 16 and Proposition 17, we see that .

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 .

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 and , we get that . Hence, by (3), we have

Substituting and into (1), then by (11) we get that

From (72), we have

From (73), the constant exists only if the following equation holds:

Note that . Then we get from (74) that

where and .

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 for . 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.

References

L. M. Perko and E. L. Walter, “Regular polygon solutions of the -body problem,” Proceedings of the American Mathematical Society, vol. 94, no. 2, pp. 301–309, 1985.

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.

S. Zhang and Q. Zhou, “Periodic solutions for planar -body problems,” Proceedings of the American Mathematical Society, vol. 131, no. 7, pp. 2161–2170, 2003.

M. Marcus and H. Minc, A Survey of Matrix Theory and Matrix Inequalities, Allyn and Bacon, Boston, Mass, USA, 1964.