Lagrangian actions on 3-body problems with two fixed centers

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.

1 Introduction and Main Result

We assume two masses $m_1=m_2=\frac{1}{2}$ are fixed at $q_1=\left(\frac{-1}{2},0\right)$ and $q_2=-q_1=\left(\frac{1}{2},0\right)$, 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

$$m_3\ddot{q}\left(t\right)=\frac{m_1{m}_3\alpha \left(q_1-q\right)}{{\left|q_1-q\right|}^{\alpha +2}}+\frac{m_2{m}_3\alpha \left(q_2-q\right)}{{\left|q_2-q\right|}^{\alpha +2}}$$

(1.1)

Equivalently,

$$\ddot{q}\left(t\right)=\frac{\alpha }{2}\left[\frac{q_1-q}{{\left|q_1-q\right|}^{\alpha +2}}+\frac{q_2-q}{{\left|q_2-q\right|}^{\alpha +2}}\right]$$

(1.2)

$$\ddot{q}\left(t\right)=\frac{\partial U\left(q\right)}{\partial q}$$

(1.3)

$$\mathsf{\text{Where}}\alpha >0,U\left(q\right)=\frac{1/2}{{\left|q-q_1\right|}^{\alpha }}+\frac{1/2}{{\left|q-q_2\right|}^{\alpha }}.$$

(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

$$f\left(q\right)=\underset{0}{\overset{1}{\int }}\left[\frac{1}{2}{\left|\stackrel{·}{q}\right|}^2+\frac{1/2}{{\left|q-q_1\right|}^{\alpha }}+\frac{1/2}{{\left|q-q_2\right|}^{\alpha }}\right]dt,$$

(1.5)

$$q\in \Lambda =\left\{\begin{array}{c}q\in W^{1,2}\left(ℝ/\mathbb{Z},{ℝ}^2\right),q\left(t\right)\ne q_1,q_2,\hfill \\ q\left(t+\frac{1}{2}\right)=\left(\begin{array}{cc}-1\hfill & \hfill 0\hfill \\ 0\hfill & \hfill 1\hfill \end{array}\right)q\left(t\right),q\left(-t\right)=-q\left(t\right),\hfill \\ \text{deg}\left(q-q_1\right)=1,\text{deg}\left(q-q_2\right)=-1\hfill \end{array}\right\}$$

(1.6)

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

Figure 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 ∈ W^1,2 (ℝ/ℤ, ℝ²) and ∃ t₀ ∈ [0,1], s.t. x(t₀) = 0, if α ≥ 2 and a > 0, then

$$\underset{0}{\overset{1}{\int }}\left[\frac{1}{2}{\left|ẋ\right|}^2+\frac{a}{{\left|x\right|}^{\alpha }}\right]dt=+\infty$$

(2.1)

It's easy to see

Lemma 2.3$\overline{\Lambda }$ is a weakly closed subset of the Hilbert space W^1,2(ℝ/ℤ, ℝ²).

Lemma 2.4 f(q) is coercive and weakly lower-semicontinuous on the closure $\overline{\Lambda }$ of Λ.

Proof. By q(-t) = -q(t) and q(t) ∈ W^1,2(ℝ/ℤ, ℝ²), we have $\underset{0}{\overset{1}{\int }}q\left(t\right)dt=0$. 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 $\overline{\Lambda }$ of W^1,2.

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 $\overline{\Lambda }$ and the minimizer of f(q) on $\overline{\Lambda }$ is collision-free since if let x₁ = q - q₁, x₂ = q - q₂, then

$$\begin{array}{ll}f(q)\hfill & =\underset{0}{\overset{1}{{\displaystyle \int }}}\left[\frac{1}{2}|\dot{q}-{\dot{q}}_1{|}^2+\frac{1}{|q-q_1{|}^{\alpha }}\right]dt=\underset{0}{\overset{1}{{\displaystyle \int }}}\left[\frac{1}{2}|\dot{q}-{\dot{q}}_2{|}^2+\frac{1}{|q-q_2{|}^{\alpha }}\right]dt\hfill \\ =\underset{0}{\overset{1}{{\displaystyle \int }}}\left[\frac{1}{2}|{\dot{x}}_1{|}^2+\frac{1}{|x_1{|}^{\alpha }}\right]dt=\underset{0}{\overset{1}{{\displaystyle \int }}}\left[\frac{1}{2}|{\dot{x}}_2{|}^2+\frac{1}{|x_2{|}^{\alpha }}\right]dt\hfill \end{array}$$

(2.2)

So if the minimizer of f(q) on $\overline{\Lambda }$ 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

$$\underset{0}{\overset{1}{\int }}{y}^{2k}dx\le c\underset{0}{\overset{1}{\int }}{ẏ}^{2k}dx,$$

(2.3)

where

$$c=\frac{1}{2k-1}{\left(\frac{2k}{\pi }\text{sin}\left(\frac{\pi }{2k}\right)\right)}^{2k}.$$

(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 $q_2=-q_1=\left(\frac{1}{2},0\right)$ and $q\left(t+\frac{1}{2}\right)=\left(\begin{array}{cc}-1\hfill & \hfill 0\hfill \\ 0\hfill & \hfill 1\hfill \end{array}\right)q\left(t\right)$, so

$$\underset{0}{\overset{1}{\int }}\frac{dt}{\left|q-q_1\right|}=\underset{0}{\overset{1}{\int }}\frac{dt}{\left|q-q_2\right|},$$

(2.5)

$$\begin{array}{ll}\hfill f\left(q\right)& =\underset{0}{\overset{1}{\int }}\left[\frac{1}{2}{\left|\stackrel{·}{q}\right|}^2+\frac{1/2}{\left|q-q_1\right|}+\frac{1/2}{\left|q-q_2\right|}\right]dt\\ =\underset{0}{\overset{1}{\int }}\left[\frac{1}{2}{\left|\stackrel{·}{q}\right|}^2+\frac{1}{\left|q-q_1\right|}\right]dt\\ =\underset{0}{\overset{1}{\int }}\left[\frac{1}{2}{\left|\stackrel{·}{q}-{\stackrel{·}{q}}_1\right|}^2+\frac{1}{\left|q-q_1\right|}\right]dt\end{array}$$

(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

$$\begin{array}{ll}\hfill f\left(q\right)& =\frac{1}{2}\underset{0}{\overset{1}{\int }}\left({\left|ẋ\right|}^{2}dt+\underset{0}{\overset{1}{\int }}\frac{dt}{\left|x\right|}\right)\\ \ge \frac{1}{2}\underset{0}{\overset{1}{\int }}{\left[\frac{d}{dt}\left|x\right|\right]}^{2}dt+1^{3/2}{\left(\underset{0}{\overset{1}{\int }}{\left|x\right|}^{2}dt\right)}^{-1/2}\\ =\frac{1}{2}\underset{0}{\overset{1}{\int }}{ẏ}^{2}dt+{\left(\underset{0}{\overset{1}{\int }}{y}^{2}dt\right)}^{-1/2}\\ \ge \frac{1}{2}{\left(\frac{\pi }{2}\right)}^2\underset{0}{\overset{1}{\int }}{y}^{2}dt+{\left(\underset{0}{\overset{1}{\int }}{y}^{2}dt\right)}^{-1/2}\end{array}$$

(2.7)

Let $\sqrt{{\int }_0^1{y}^{2}dt}=s\ge 0,\phi \left(s\right)=\frac{{\pi }^2}{8}{s}^2+s^{-1}$, then ${\phi }^{″}\left(s\right)=\frac{{\pi }^2}{4}+2s^{-3}>0$, that is φ is strictly convex.

Let φ'(s) = 0, we solve it to get $s_0={\left(\frac{{\pi }^2}{4}\right)}^{-1/3}$ is the critical point for φ(s), and $\phi \left(s_0\right)=\frac{3}{2}{\left(\frac{\pi }{2}\right)}^{\frac{2}{3}}$, 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 $\tilde{q}\left(t\right)\in \Lambda$ such that

$$f\left(\tilde{q}\left(t\right)\right)<\frac{3}{2}{\left(\frac{\pi }{2}\right)}^{2/3}$$

(2.8)

then the minimizer of f(q) on $\overline{\Lambda }$ is collision-free.