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Lagrangian actions on 3-body problems with two fixed centers

Xiong-rui Wang1 and Sheng He2*

Author Affiliations

1 Department of Mathematics, Yibin University, Yibin, Sichuan 644007, China

2 Department of Mathematics, Sichuan University, Sichuan 610064, P. R. China

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Boundary Value Problems 2012, 2012:28  doi:10.1186/1687-2770-2012-28


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/28


Received:1 October 2011
Accepted:28 February 2012
Published:28 February 2012

© 2012 Wang and He; licensee Springer.

This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, we study the existence of figure "∞"-type periodic solution for 3-body problems with strong-force potentials and two fixed centers, and we also give some remarks in the case with Newtonian weak-force potentials.

Mathematical Subject Classification 2000: 34C15; 34C25; 70F10.

Keywords:
3-body problems with two fixed centers; "∞"-type solutions; Lagrangian actions

1 Introduction and Main Result

We assume two masses m 1 = m 2 = 1 2 are fixed at q 1 = - 1 2 , 0 and q 2 = - q 1 = 1 2 , 0 , the third mass m3 is affected by m1 and m2 and moving according to the Newton's second law and the general gravitational law [1,2], then the position q(t) for m3 satisfies

m 3 q ¨ ( t ) = m 1 m 3 α ( q 1 - q ) q 1 - q α + 2 + m 2 m 3 α ( q 2 - q ) q 2 - q α + 2 (1.1)

Equivalently,

q ¨ ( t ) = α 2 q 1 - q q 1 - q α + 2 + q 2 - q q 2 - q α + 2 (1.2)

q ¨ ( t ) = U ( q ) q (1.3)

Where α > 0 , U ( q ) = 1 / 2 q - q 1 α + 1 / 2 q - q 2 α . (1.4)

For the case α = 1, Euler [3-5] studied (1.1)-(1.3), but didn't use variational methods to study periodic solutions.

Here we want to use variational minimizing method to look for periodic solution for m3 which winds around q1 and q2, let

f ( q ) = 0 1 1 2 q · 2 + 1 / 2 q - q 1 α + 1 / 2 q - q 2 α d t , (1.5)

q Λ = q W 1 , 2 ( / , 2 ) , q ( t ) q 1 , q 2 , q t + 1 2 = - 1 0 0 1 q ( t ) , q ( - t ) = - q ( t ) , deg ( q - q 1 ) = 1 , deg ( q - q 2 ) = - 1 (1.6)

Theorem 1.1 For α ≥ 2, the minimizer of f(q) on Λ ¯ does exist and is non-collision "∞"-type periodic solution of (1.1)-(1.3).(See Figure 1)

thumbnailFigure 1. Figure-eight with 2-centres.

2 The Proof of Theorem 1.1

Using Palais'S symmetrical Principle [6], it's easy to prove the following variational Lemma:

Lemma 2.1 The critical point of f(q) in Λ is the noncollision periodic solution winding around q1 counter-clockwise and q2 clockwise one time during one period.

Lemma 2.2 [7] If x W1,2 (ℝ/ℤ, ℝ2) and ∃ t0 ∈ [0,1], s.t. x(t0) = 0, if α ≥ 2 and a > 0, then

0 1 1 2 2 + a x α d t = + (2.1)

It's easy to see

Lemma 2.3 Λ ¯ is a weakly closed subset of the Hilbert space W1,2(ℝ/ℤ, ℝ2).

Lemma 2.4 f(q) is coercive and weakly lower-semicontinuous on the closure Λ ¯ of Λ.

Proof. By q(-t) = -q(t) and q(t) ∈ W1,2(ℝ/ℤ, ℝ2), we have 0 1 q ( t ) d t = 0 . By Wirtinger's inequality, we know f(q) is coercive. By Sobolev's embedding Theorem and Fatou's Lemma, f is weakly lower-semi-continuous on the weakly closed set Λ ¯ of W1,2.

Lemma 2.5 [8] Let X be a reflexive Banach space,M X be weakly closed subset,f : M R be weakly lower semi-continous and coercive (f(x) → +∞ as ∥x∥ → +∞), then f attains its infimum on M.

According to Lemmas 2.1-2.5, we know that f(q) attains its infimum on Λ ¯ and the minimizer of f(q) on Λ ¯ is collision-free since if let x1 = q - q1, x2 = q - q2, then

f ( q ) = 0 1 [ 1 2 | q ˙ q ˙ 1 | 2 + 1 | q q 1 | α ] d t = 0 1 [ 1 2 | q ˙ q ˙ 2 | 2 + 1 | q q 2 | α ] d t = 0 1 [ 1 2 | x ˙ 1 | 2 + 1 | x 1 | α ] d t = 0 1 [ 1 2 | x ˙ 2 | 2 + 1 | x 2 | α ] d t (2.2)

So if the minimizer of f(q) on Λ ¯ has collision at some moment, then Gordon's Lemma tell us the minimum value is +∞ which is a contradiction.

The most interesting case α = 1 is the case for Newtonian potential, we try to prove the minimizer is collision-free, but it seems very difficult, here we give some remarks.

Lemma 2.6 [9] If y(0) = 0 and 2k is an even positive integer, then

0 1 y 2 k d x c 0 1 2 k d x , (2.3)

where

c = 1 2 k - 1 2 k π sin π 2 k 2 k . (2.4)

There is equality only for a certain hyperelliptic curve.

Now we estimate the lower bound of the Lagrangian action f(q) on "∞"-type collisionorbits. Since q 2 = - q 1 = 1 2 , 0 and q t + 1 2 = - 1 0 0 1 q ( t ) , so

0 1 d t q - q 1 = 0 1 d t q - q 2 , (2.5)

f ( q ) = 0 1 1 2 q · 2 + 1 / 2 q - q 1 + 1 / 2 q - q 2 d t = 0 1 1 2 q · 2 + 1 q - q 1 d t = 0 1 1 2 q · - q · 1 2 + 1 q - q 1 d t (2.6)

If q(t) collides with q1 at some moment t0 ∈ [0,1], without loss of generality, we assume t0 = 0, then q(0) - q1 = 0, we let x(t) = q(t) - q1, y(t) = |x(t)|, then x(0) = 0,y(0) = 0. By Jensen's inequality and Hardy-Littlewood-pólya inequality [9], we have

f ( q ) = 1 2 0 1 2 d t + 0 1 d t x 1 2 0 1 d d t x 2 d t + 1 3 / 2 0 1 x 2 d t - 1 / 2 = 1 2 0 1 2 d t + 0 1 y 2 d t - 1 / 2 1 2 π 2 2 0 1 y 2 d t + 0 1 y 2 d t - 1 / 2 (2.7)

Let 0 1 y 2 d t = s 0 , φ ( s ) = π 2 8 s 2 + s - 1 , then φ ( s ) = π 2 4 + 2 s - 3 > 0 , that is φ is strictly convex.

Let φ'(s) = 0, we solve it to get s 0 = π 2 4 - 1 / 3 is the critical point for φ(s), and φ ( s 0 ) = 3 2 π 2 2 3 , which is the maximum value for φ(s) on s > 0 since φ is convex and φ(s) → +∞ as s → 0+.

If we can find the test orbit q ̃ ( t ) Λ such that

f q ̃ ( t ) < 3 2 π 2 2 / 3 (2.8)

then the minimizer of f(q) on Λ ¯ is collision-free.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

All authors read and approved the final manuscript.

Acknowledgements

The authors would like to thank the anonymous referees for their valuable suggestions which improve this work. This work was supported by Scientific Research Fund of Sichuan Provincial Education Department (11ZA172).

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