Bifurcation of periodic orbits for the N-body problem, from a nongeometrical family of solutions

Abstract

Given two positive real numbers M and m and an integer $n>1$, it is well known that we can find a family of solutions of the $(n+1)$-body problem where one body with mass M stays at the origin and the other n bodies, all with the same mass m, move on the x–y plane following ellipses with eccentricity e. These periodic solutions were discovered by Lagrange and can be described analytically. In this paper, we prove the existence of periodic solutions of the $(n+1)$-body problem; they are not-trivial in the sense that none of the bodies follows conics. Besides showing the existence of these periodic solutions, we point out a trivial family of non-periodic solutions for the $(n+1)$-body problem that are easy to describe. In this way, we are considering three families of solutions of the $(n+1)$-body problem: The Lagrange family, the family of non-periodic solution and the non-trivial solutions. The authors surprisingly discovered that a numerical solution of the 4-body problem—the one displayed on the video http://youtu.be/2Wpv6vpOxXk—is part of a family of periodic solutions (those that we are calling the non-trivial) that does not approach a solution in the Lagrange family, but it approaches a solution in the family that we are calling non-periodic solutions. After pointing this out, the authors find an exact formula for the bifurcation point in the non-periodic family and use it to show the mathematical existence of nonplanar periodic solutions of the $(n+1)$-body problem for any pair of masses M, m and any integer $n>1$ (the family that we are calling non-trivial). As a particular example, we find a non-trivial solution of the 4-body problem where three bodies with mass 3 moving around a body with mass 7 that moves up and down.

Appendix

Proof of Theorem 1.1

A direct computation shows that the matrix ${\mathcal {R}}={\mathcal {R}}[2\pi /n]$ satisfies $\displaystyle {{\mathcal {R}}^{-k}={\mathcal {R}}^{n-k}}$ for all integer k. Therefore,

$$\begin{aligned} \sum _{k=1,k\ne j}^{n}\dfrac{({\mathcal {R}}^{k-j}-I_{3})\mathbf{r }}{\Vert ({\mathcal {R}}^{k-j} -I_{3})\mathbf{r }\Vert ^{3}}=\sum _{k=1}^{n-1}\dfrac{({\mathcal {R}}^{k}-I_{3})\mathbf{r }}{\Vert ({\mathcal {R}}^{k}-I_{3})\mathbf{r }\Vert ^{3}}, \end{aligned}$$

(4.3)

with $\mathbf{r }\in {\mathbb {R}}^3$ and each $j\in \left\{ 1,\ldots ,n\right\}$. In consequence, if $\mathbf{r }_{j}$, $j=1,\ldots , n+1$ is the vector position of each body, and the fist n bodies move on the vertex of a regular polygon satisfying $\mathbf{r }_{j}={\mathcal {R}}^{j-1}\mathbf{r }_{1}$ with $j=2,\ldots , n$ of then the bodies satisfies the n-body problem if and only if

$$\begin{aligned} \begin{aligned} \ddot{\mathbf{r }}_{j}=&\sum _{k=1,k\ne j}^{N-1}\dfrac{m(\mathbf{r }_{k}-\mathbf{r }_{j})}{\Vert (\mathbf{r }_{k}-\mathbf{r }_{j})\Vert ^{3}}+\dfrac{M(\mathbf{r }_{N}-\mathbf{r }_{j})}{\Vert \mathbf{r }_{N}-\mathbf{r }_{j}\Vert ^{3}}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} {\mathcal {R}}^{j-1}\ddot{\mathbf{r }}_{1}= & {} \sum _{k=1,k\ne j}^{n}\dfrac{m({\mathcal {R}}^{k-1}-{\mathcal {R}}^{j-1})\mathbf{r }_{1}}{\Vert ({\mathcal {R}}^{k-1}-{\mathcal {R}}^{j-1})\mathbf{r }_{1}\Vert ^{3}}+\dfrac{M(\mathbf{r }_{n+1}-{\mathcal {R}}^{j-1}\mathbf{r }_{1})}{\Vert \mathbf{r }_{n+1}-{\mathcal {R}}^{j-1}\mathbf{r }_{1}\Vert ^{3}},\\ {\mathcal {R}}^{j-1}\ddot{\mathbf{r }}_{1}= & {} \sum _{k=1,k\ne j}^{n}\dfrac{m{\mathcal {R}}^{j-1}\Big [({\mathcal {R}}^{j-1})^{-1}{\mathcal {R}}^{k-1}-I_{3}\Big ]\mathbf{r }_{1}}{\Vert {\mathcal {R}}^{j-1}\Big [({\mathcal {R}}^{j-1})^{-1}{\mathcal {R}}^{k-1}-I_{3}\Big ]\mathbf{r }_{1}\Vert ^{3}}+\dfrac{M(\mathbf{r }_{n+1}-{\mathcal {R}}^{j-1}\mathbf{r }_{1})}{\Vert \mathbf{r }_{n+1}-{\mathcal {R}}^{j-1}\mathbf{r }_{1}\Vert ^{3}}, \\ {\mathcal {R}}^{j-1}\ddot{\mathbf{r }}_{1}= & {} \sum _{k=1,k\ne j}^{n}\dfrac{m{\mathcal {R}}^{j-1}\Big [({\mathcal {R}}^{j-1})^{-1}{\mathcal {R}}^{k-1}-I_{3}\Big ]\mathbf{r }_{1}}{\Vert {\mathcal {R}}^{j-1}({\mathcal {R}}^{k-j}-I_{3})\mathbf{r }_{1}\Vert ^{3}}+\dfrac{M(\mathbf{r }_{n+1}-{\mathcal {R}}^{j-1}\mathbf{r }_{1})}{\Vert {\mathcal {R}}^{j-1}({\mathcal {R}}^{n+1-j}\mathbf{r }_{n+1}-\mathbf{r }_{1})\Vert ^{3}},\\ \ddot{\mathbf{r }}_{1}= & {} \sum _{k=1,k\ne j}^{n}\dfrac{m({\mathcal {R}}^{k-j}-I_{3})\mathbf{r }_{1}}{\Vert ({\mathcal {R}}^{k-j}-I_{3})\mathbf{r }_{1}\Vert ^{3}}+\dfrac{M({\mathcal {R}}^{j-1})^{-1}(\mathbf{r }_{N}-{\mathcal {R}}^{j-1}\mathbf{r }_{1})}{\Vert {\mathcal {R}}^{n+1-j}\mathbf{r }_{n+1}-\mathbf{r }_{1}\Vert ^{3}}. \end{aligned}$$

Finally, by (4.3) we obtain

$$\begin{aligned} \ddot{\mathbf{r }}_{1}=\sum _{k=1}^{n-1}\dfrac{m({\mathcal {R}}^{k}-I_{3}) \mathbf{r }_{1}}{\Vert ({\mathcal {R}}^{k}-I_{3})\mathbf{r }_{1}\Vert ^{3}}+\dfrac{M( {\mathcal {R}}^{n+1-j}\mathbf{r }_{n+1}-\mathbf{r }_{1})}{\Vert {\mathcal {R}}^{n+1-j} \mathbf{r }_{n+1}-\mathbf{r }_{1}\Vert ^{3}}. \end{aligned}$$

(4.4)

Now we study each term in (4.4). Therefore, we rewrite the vector position $\mathbf{r }_{1}$ in cylindric coordinates

$$\begin{aligned} \mathbf{r }_{1}(t)=(r(t)\cos \theta (t),r(t)\sin \theta (t),z_{1}(t)), \end{aligned}$$

and then

$$\begin{aligned} ({\mathcal {R}}^{k}-I_{3})\mathbf{r }_{1} = \begin{pmatrix} r(c_{k}-1)c_{\theta }-rs_{k}s_{\theta } \\ rs_{k}c_{\theta }+r(c_{k}-1)s_{\theta } \\ 0 \end{pmatrix} = \begin{pmatrix} r(c_{k+\theta }-c_{\theta }) \\ r(s_{k+\theta }-s_{\theta }) \\ 0 \end{pmatrix}, \end{aligned}$$

where

$$\begin{aligned} c_{k}:=\cos \big (2\pi k/n\big ), \quad c_{\theta }:=\cos \theta \quad \text {and} \quad s_{k}:=\sin \big (2\pi k/n\big ), \quad s_{\theta }:=\sin \theta . \end{aligned}$$

From here it follows directly

$$\begin{aligned} \Vert ({\mathcal {R}}^{k}-I_{3})\mathbf{r }_{1}\Vert ^{3}=\Big (2(1-c_{k})r^{2}\Big )^{3/2}=\big (4 s^{2}_{k/2}r^{2}\big )^{3/2}=8\sin ^{3}\Big (\frac{\pi k}{n}\Big ) r^{3}. \end{aligned}$$

for all $1\le k \le n-1.$ Moreover,

$$\begin{aligned} \Vert {\mathcal {R}}^{n+1-j}\mathbf{r }_{n+1}-\mathbf{r }_{1}\Vert =\Vert ({\mathcal {R}}^{j-1})^{-1}(\mathbf{r }_{n+1}-\mathbf{r }_{j}) \Vert =\Vert \mathbf{r }_{n+1}-\mathbf{r }_{j}\Vert . \end{aligned}$$

(4.5)

for all $j=1,\ldots ,n,$. Now, under the hypothesis that the center of masses stay put at the origin, we have the condition

$$\begin{aligned} m\sum _{j=1}^{n}\mathbf{r }_{j}(t)+M\mathbf{r }_{n+1}(t)=\mathbf{0 }_{{\mathbb {R}}^{3}}, \quad \forall t\in {\mathbb {R}}, \end{aligned}$$

in consequence, if $\displaystyle {\mathbf{r }_{n+1}(t)=(x_{n+1}(t),y_{n+1}(t),f(t))}$ the previous equation implies

$$\begin{aligned} x_{n+1}(t)=y_{n+1}(t)=0 \quad \text {and} \quad z_{1}(t)=-\frac{M}{nm}f(t), \quad \forall t\in {\mathbb {R}}. \end{aligned}$$

since $\displaystyle {\sum ^{n}_{j=1}c_{j-1}=\sum ^{n}_{j=1}s_{j-1}=0}$ it follows that:

In consequence,

$$\begin{aligned} \Vert \mathbf{r }_{n+1}-\mathbf{r }_{j}\Vert ^{3}=\Vert \mathbf{r }_{n+1}-\mathbf{r }_{1}\Vert ^{3}=\left[ r^{2}+\gamma ^2f^2\right] ^{3/2}, \end{aligned}$$

with $\displaystyle {\gamma =\frac{M+nm}{nm}}$ and for all $j=1,\ldots n.$ To sum up, the previous computations show that

$$\begin{aligned} \dfrac{({\mathcal {R}}^{k}-I_{3})\mathbf{r }_{1}}{\Vert ({\mathcal {R}}^{k}-I_{3})\mathbf{r }_{1}\Vert ^{3}}=\begin{pmatrix} (c_{k+\theta }-c_{\theta })/8 s^{3}_{k/2}r^{2} \\ (s_{k+\theta }-s_{\theta })/8 s^{3}_{k/2}r^{2} \\ 0 \end{pmatrix}, \quad \dfrac{{\mathcal {R}}^{N-j}\mathbf{r }_{n+1}-\mathbf{r }_{1}}{\Vert {\mathcal {R}}^{j-1}\mathbf{r }_{n+1}-\mathbf{r }_{1}\Vert ^{3}}=\begin{pmatrix} -rc_{\theta }/\left[ r^{2}+\gamma ^2f^2\right] ^{3/2}\\ -rs_{\theta }/\left[ r^{2}+\gamma ^2f^2\right] ^{3/2}\\ \gamma f/\left[ r^{2}+\gamma ^2f^2\right] ^{3/2} \end{pmatrix}. \end{aligned}$$

Finally, (4.4) in cylindric coordinates is given by

$$\begin{aligned} \begin{aligned} (\ddot{r}-r({\dot{\theta }})^{2})c_{\theta }-2(2{\dot{r}}{\dot{\theta }}+r\ddot{\theta })s_{\theta }&=m \sum _{k=1}^{n-1}\frac{c_{k+\theta }-c_{\theta }}{8 r^{2}s^{3}_{k/2}}-\frac{M r c_{\theta }}{\left[ r^{2}+\gamma ^2 f^2\right] ^{3/2}},\\ (\ddot{r}-r({\dot{\theta }})^{2})s_{\theta }-2(2{\dot{r}}{\dot{\theta }}-r\ddot{\theta })c_{\theta }&=m\sum _{k=1}^{n-1}\frac{s_{k+\theta }-s_{\theta }}{8 r^{2}s^{3}_{k/2}}-\frac{M rs_{\theta }}{\left[ r^{2}+\gamma ^2f^2\right] ^{3/2}},\\ -M \ddot{f}/n m&=M \gamma f/\left[ r^{2}+\gamma ^2f^2\right] ^{3/2}. \end{aligned} \end{aligned}$$

(4.6)

From the fact

$$\begin{aligned} \frac{c_{k+\theta }-c_{\theta }}{s^{3}_{k/2}}=-2\Big (\frac{1}{s_{k/2}}c_{\theta }+\frac{c_{k/2}}{s^{2}_{k/2}}s_{\theta }\Big ), \end{aligned}$$

it follows directly

$$\begin{aligned} m\sum _{k=1}^{n-1}\frac{c_{k+\theta }-c_{\theta }}{8 r^{2}s^{3}_{k/2}}=-\frac{m}{4r^{2}}\sum _{k=1}^{n-1}\frac{1}{s_{k/2}}c_{\theta }. \end{aligned}$$

Analogously,

$$\begin{aligned} m\sum _{k=1}^{n-1}\frac{s_{k+\theta }-s_{\theta }}{8 r^{2}s^{3}_{k/2}}=-\frac{m}{4r^{2}}\sum _{k=1}^{n-1}\frac{1}{s_{k/2}}s_{\theta }. \end{aligned}$$

Therefore, (4.6) is equivalent to the system of equations

$$\begin{aligned} \begin{aligned} \ddot{f}&=-\frac{(M+nm)f}{h^{3}}, \\ \ddot{r}&=r{\dot{\theta }}^{2}-\frac{m \lambda _n}{r^{2}}-\frac{M r }{h^{3}},\\ {\dot{\theta }}r^2&=C \end{aligned} \end{aligned}$$

(4.7)

with

$$\begin{aligned} \lambda _n=\frac{1}{4}\sum _{k=1}^{n-1}\frac{1}{s_{k/2}}, \qquad h=\left[ r^{2}+\Big (\frac{M+nm}{nm}\Big )^2 f^2\right] ^{1/2}, \end{aligned}$$

and C the constant angular momentum, implying

$$\begin{aligned} \theta (t)=\theta _{0}+\int _{0}^{t}\frac{C}{r^{2}(s)}\mathrm{d}s. \end{aligned}$$

The initial conditions

$$\begin{aligned} f(0)=0, \quad {\dot{f}}(0)=b, \quad r(0)=r_{0}, \quad {\dot{r}}(0)=0, \quad \theta (0)=0, \quad {\dot{\theta }}(0)=\frac{a}{r_{0}}, \end{aligned}$$

we obtain $C=r_{0}a$, and this completes the proof. $\square$