Abstract
In this paper, we study the existence of figure "∞"type periodic solution for 3body problems with strongforce potentials and two fixed centers, and we also give some remarks in the case with Newtonian weakforce potentials.
Mathematical Subject Classification 2000: 34C15; 34C25; 70F10.
Keywords:
3body problems with two fixed centers; "∞"type solutions; Lagrangian actions1 Introduction and Main Result
We assume two masses
Equivalently,
For the case α = 1, Euler [35] 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 m_{3 }which winds around q_{1 }and q_{2}, let
Theorem 1.1 For α ≥ 2, the minimizer of f(q) on
Figure 1. Figureeight with 2centres.
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 q_{1 }counterclockwise and q_{2 }clockwise one time during one period.
Lemma 2.2 [7] If x ∈ W^{1,2 }(ℝ/ℤ, ℝ^{2}) and ∃ t_{0 }∈ [0,1], s.t. x(t_{0}) = 0, if α ≥ 2 and a > 0, then
It's easy to see
Lemma 2.3
Lemma 2.4 f(q) is coercive and weakly lowersemicontinuous on the closure
Proof. By q(t) = q(t) and q(t) ∈ W^{1,2}(ℝ/ℤ, ℝ^{2}), we have
Lemma 2.5 [8] Let X be a reflexive Banach space,M ⊂ X be weakly closed subset,f : M → R be weakly lower semicontinous and coercive (f(x) → +∞ as ∥x∥ → +∞), then f attains its infimum on M.
According to Lemmas 2.12.5, we know that f(q) attains its infimum on
So if the minimizer of f(q) on
The most interesting case α = 1 is the case for Newtonian potential, we try to prove the minimizer is collisionfree, 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
where
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
If q(t) collides with q_{1 }at some moment t_{0 }∈ [0,1], without loss of generality, we assume t_{0 }= 0, then q(0)  q_{1 }= 0, we let x(t) = q(t)  q_{1}, y(t) = x(t), then x(0) = 0,y(0) = 0. By Jensen's inequality and HardyLittlewoodpólya inequality [9], we have
Let
Let φ'(s) = 0, we solve it to get
If we can find the test orbit
then the minimizer of f(q) on
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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