Spatial periodic orbits in the equilateral circular restricted four-body problem: computer-assisted proofs of existence

Abstract

We use validated numerical methods to prove the existence of spatial periodic orbits in the equilateral restricted four-body problem. We study each of the vertical Lyapunov families (up to symmetry) in the triple Copenhagen problem, as well as some halo and axial families bifurcating from planar Lyapunov families. We consider the system with both equal and non-equal masses. Our method is constructive and non-perturbative, being based on a posteriori analysis of a certain nonlinear operator equation in the neighborhood of a suitable approximate solution. The approximation is via piecewise Chebyshev series with coefficients in a Banach space of rapidly decaying sequences. As by-product of the proof, we obtain useful quantitative information about the location and regularity of the solution.

Appendix: Proof of Lemma 3

Since $\gamma$ is a solution of the differential equation, we consider the first, third, and fifth components of the vector field and have that $u_2 = \dot{u}_1$, $u_4 = \dot{u}_3$ and $u_6 = \dot{u}_5$. Moreover, considering the seventh component gives

$$\begin{aligned} \dot{u}_7 = \left( -u_1u_2 - u_3 u_4 - u_5 u_6 + x_1 u_2 + y_1 u_4 + \alpha _1 \right) u_7^3, \end{aligned}$$

and since $u_7 > 0$, we divide by $u_7^3$ and rewrite this as

$$\begin{aligned} \frac{1}{u_7^3} \frac{\mathrm{d}}{\mathrm{d}t}u_7&= -(u_1 - x_1)u_2 - (u_3 - y_1)u _4 - u_5 u_6 + \alpha _1 \\&= -(u_1 - x_1)u_1' - (u_3 - y_1)u _3' - u_5 u_5' + \alpha _1 \\&= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t}(u_1 - x_1)^2 - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t}(u_2 - y_1)^2 - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t}(u_3)^2 + \alpha _1 \\&= -\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \left( (u_1 - x_1)^2 + (u_2 - y_1)^2 + u_3^2 \right) + \alpha _1, \end{aligned}$$

or

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} G_1(t) = \frac{\mathrm{d}}{\mathrm{d}t} F_1(t) + \alpha _1, \end{aligned}$$

(15)

where

$$\begin{aligned} F_1(t) -\frac{1}{2}\left( (u_1(t) - x_1)^2 + (u_3(t) - y_1)^2 + (u_1(t) - x_1)^2 \right) , \end{aligned}$$

and

$$\begin{aligned} G_1(t) -\frac{1}{2u_7(t)^2}, \end{aligned}$$

are both periodic functions. Taking the average of Eq. (15) over the interval [0, T] leads to

$$\begin{aligned} 0 = \frac{1}{T}\int _{0}^T \alpha _1 \, \mathrm{d}t = \alpha _1, \end{aligned}$$

as the derivatives of $F_1$ and $G_1$ (indeed the derivatives of any periodic function) have average zero. Since $T > 0$, we conclude that $\alpha _1 = 0$ as desired. Nearly identical arguments, applied to the eighth and ninth component equations, show that $\alpha _2 = \alpha _3 = 0$.

Now define

$$\begin{aligned} {\hat{u}}(t) \frac{1}{\sqrt{(u_1(t) - x_1)^2 + (u_3(t) - y_1)^2 + u_5(t)^2}}, \end{aligned}$$

and note that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} {\hat{u}}(t)&= -\frac{\frac{\mathrm{d}}{\mathrm{d}t} (u_1(t) - x_1)^2 + (u_3(t) - y_1)^2 + u_5(t)^2}{2 \left( \sqrt{(u_1(t) - x_1)^2 + (u_3(t) - y_1)^2 + u_5(t)^2}\right) ^3 } \\&= -\,{\hat{u}}^3 \left( (u_1 - x_1) u_2 + (u_3 - y_1)u_4 + u_5 u_6 \right) \\&=-\,{\hat{u}}^3 u_1 u_2 - {\hat{u}}^3 u_3 u_4 -{\hat{u}}^3 u_5 u_6 + {\hat{u}}^3 x_1 u_2 + {\hat{u}}^3 y_1 u_4 . \end{aligned}$$

Then, we see that $u_7(t)$ and ${\hat{u}}(t)$ satisfy the same differential equation with the same initial condition. By existence and uniqueness for ODEs, we have that $u_7(t) = {\hat{u}}(t)$, i.e.,

$$\begin{aligned} u_7(t) = \frac{1}{\sqrt{(u_1(t) - x_1)^2 + (u_3(t) - y_1)^2 + u_5(t)^2}}. \end{aligned}$$

(16)

for all $t\in [0,T]$. Similarly,

$$\begin{aligned} u_8(t) = \frac{1}{\sqrt{(u_1(t) - x_2)^2 + (u_3(t) - y_2)^2 + u_5(t)^2}}. \end{aligned}$$

(17)

and

$$\begin{aligned} u_9(t) = \frac{1}{\sqrt{(u_1(t) - x_3)^2 + (u_3(t) - y_3)^2 + u_5(t)^2}}. \end{aligned}$$

(18)

for all $t\in [0,T]$.

The argument that $\beta = 0$ is similar to the above but different enough that we include it for the sake of completeness. Inspired by the energy functional for the circular restricted four-body problem, we define the function

$$\begin{aligned} H(u) = -(u_2^2 + u_4^2 + u_6^2) + u_1^2 + u_3^2 + 2\left( m_1 u_7 + m_2 u_8 + m_3 u_9 \right) \end{aligned}$$

and observe that $H(\gamma (t))$ is a periodic function. We have that

$$\begin{aligned} \nabla H = \left( \begin{array}{c} 2 u_1 \\ -2 u_2 \\ 2 u_3 \\ -2 u_4 \\ 0 \\ -2 u_6 \\ 2 m_1 \\ 2 m_2 \\ 2 m_3 \end{array} \right) . \end{aligned}$$

Since we have already established that $\alpha _1 = \alpha _2 = \alpha _3 = 0$, we have that

$$\begin{aligned} \left<\nabla H, g \right>&= {\langle }{ \begin{pmatrix} 2 u_1 \\ -2 u_2 \\ 2 u_3 \\ -2 u_4 \\ 0 \\ -2 u_6 \\ 2 m_1 \\ 2 m_2 \\ 2 m_3 \end{pmatrix} \begin{pmatrix} u_2 \\ 2u_4 + u_1 - m_1 u_1 u_7^3 - m_2 u_1 u_8^3 - m_3 u_1 u_9^3 + m_1 x_1 u_7^3 + m_2 x_2 u_8^3 + m_3 x_3 u_9^3 + \beta u_2 \\ u_4 \\ -2 u_2 + u_3 - m_1 u_3 u_7^3 - m_2 u_3 u_8^3 - m_3 u_3 u_9^3 + m_1 y_1 u_7^3 + m_2 y_2 u_8^3 + m_3 y_3 u_9^3 \\ u_6 \\ -m_1 u_5 u_7^3 - m_2 u_5 u_8^3 - m_3 u_5 u_9^3 + m_1 z_1 u_7^3 + m_2 z_2 u_8^3 + m_3 z_3 u_9^3 \\ - u_1 u_2 u_7^3 - u_3 u_4 u_7^3- u_5 u_6 u_7^3 + x_1 u_2 u_7^3 +y_1 u_4 u_7^3 + z_1 u_6 u_7^3 \\ - u_1 u_2 u_8^3 - u_3 u_4 u_8^3- u_5 u_6 u_8^3 + x_2 u_2 u_8^3 +y_2 u_4 u_8^3 + z_2 u_6 u_8^3 \\ - u_1 u_2 u_9^3 - u_3 u_4 u_9^3- u_5 u_6 u_9^3 + x_3 u_2 u_9^3 +y_3 u_4 u_9^3 + z_3 u_6 u_9^3 \end{pmatrix}} {\rangle } \\&= 2 u_1 u_2 -2 u_2 (2u_4 + u_1 - m_1 u_1 u_7^3 - m_2 u_1 u_8^3 - m_3 u_1 u_9^3 + m_1 x_1 u_7^3 + m_2 x_2 u_8^3 + m_3 x_3 u_9^3) - 2 \beta u_2^2 \\&\quad + 2 u_3 u_4 - 2 u_4 (-2 u_2 + u_3 - m_1 u_3 u_7^3 - m_2 u_3 u_8^3 - m_3 u_3 u_9^3 + m_1 y_1 u_7^3 + m_2 y_2 u_8^3 + m_3 y_3 u_9^3) \\&\quad -2u_6 (-m_1 u_5 u_7^3 - m_2 u_5 u_8^3 - m_3 u_5 u_9^3 + m_1 z_1 u_7^3 + m_2 z_2 u_8^3 + m_3 z_3 u_9^3) \\&\quad +2m_1(- u_1 u_2 u_7^3 - u_3 u_4 u_7^3- u_5 u_6 u_7^3 + x_1 u_2 u_7^3 +y_1 u_4 u_7^3 + z_1 u_6 u_7^3) \\&\quad + 2 m_2 (- u_1 u_2 u_8^3 - u_3 u_4 u_8^3- u_5 u_6 u_8^3 + x_2 u_2 u_8^3 +y_2 u_4 u_8^3 + z_2 u_6 u_8^3) \\&\quad +2 m_3 (- u_1 u_2 u_9^3 - u_3 u_4 u_9^3- u_5 u_6 u_9^3 + x_3 u_2 u_9^3 +y_3 u_4 u_9^3 + z_3 u_6 u_9^3 ) \\&= 2 u_1 u_2 -4 u_2 u_4 -2 u_2 u_1 + 2 m_1 u_2 u_1 u_7^3 + 2 m_2 u_2 u_1 u_8^3 + 2 m_3 u_2 u_1 u_9^3 \\&\quad -2 m_1 u_2 x_1 u_7^3 -2 m_2 u_2 x_2 u_8^3 -2 m_3 u_2 x_3 u_9^3 - 2 \beta u_2^2 \\&\quad + 2 u_3 u_4 + 4 u_2 u_4 - 2 u_3 u_4 + 2 m_1 u_3 u_4 u_7^3 + 2 m_2 u_3 u_4 u_8^3 + 2 m_3 u_3 u_4 u_9^3 \\&\quad - 2 m_1 y_1 u_4 u_7^3 - 2 m_2 y_2 u_4 u_8^3 - 2 m_3 y_3 u_4 u_9^3 \\&\quad + 2 m_1 u_5 u_6 u_7^3 + 2 m_2 u_5 u_6 u_8^3 + 2 m_3 u_5 u_6 u_9^3 - 2 m_1 z_1 u_6 u_7^3 - 2 m_2 z_2 u_6 u_8^3 - 2 m_3 z_3 u_6 u_9^3 \\&\quad - 2 m_1 u_1 u_2 u_7^3 - 2 m_1 u_3 u_4 u_7^3 - 2 m_1 u_5 u_6 u_7^3 + 2 m_1 x_1 u_2 u_7^3 + 2 m_1 y_1 u_4 u_7^3 + 2 m_1 z_1 u_6 u_7^3 \\&\quad - 2 m_2 u_1 u_2 u_8^3 - 2 m_2 u_3 u_4 u_8^3 - 2 m_2 u_5 u_6 u_8^3 + 2 m_2 x_2 u_2 u_8^3 + 2 m_2 y_2 u_4 u_8^3 + 2 m_2 z_2 u_6 u_8^3 \\&\quad - 2 m_3 u_1 u_2 u_9^3 - 2 m_3 u_3 u_4 u_9^3 - 2 m_3u_5 u_6 u_9^3 + 2 m_3 x_3 u_2 u_9^3 + 2 m_3 y_3 u_4 u_9^3 \\&\quad + 2 m_3 z_3 u_6 u_9^3. = - 2 \beta u_2^2. \end{aligned}$$

Then, we note that $H(\gamma (t))$ is a periodic function and that the above computation gives

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} H(t)&= \nabla H(\gamma (t)) \gamma '(t) \\&= \nabla H(\gamma (t)) g(\gamma (t)) \\&= -2 \beta u_2^2(t). \end{aligned}$$

Taking the average and exploiting that the average of the derivative of a periodic function is zero give

$$\begin{aligned} 0 = \frac{1}{T}\int _0^T \frac{\mathrm{d}}{\mathrm{d}t} H(t) \, \mathrm{d}t = -\frac{2 \beta }{T} \int _0^T u_2(t)^2 \, \mathrm{d}t. \end{aligned}$$

Since $T > 0$ and $u_2(t)^2$ do not change sign, it follows that $\beta = 0$.

Finally, we recall Eqs. (16), (17), and (18) as well as the fact that $\beta = 0$ and have that

$$\begin{aligned} \dot{u}_2&= 2u_4 + u_1 - m_1 u_1 u_7^3 - m_2 u_1 u_8^3 - m_3 u_1 u_9^3 + m_1 x_1 u_7^3 + m_2 x_2 u_8^3 + m_3 x_3 u_9^3 \\&= 2u_4 + u_1 - m_1 (u_1 - x_1) u_7^3 - m_2 (u_1 - x_2) u_8^3 - m_3 (u_1 - x_3) u_9^3 \\&= 2u_4 + u_1 - \frac{m_1 (u_1 - x_1)}{\left( (u_1(t) - x_1)^2 + (u_3(t) - y_1)^2 + u_5(t)^2 \right) ^{3/2}} \\&\quad - \frac{m_2 (u_1 - x_2)}{\left( (u_1(t) - x_2)^2 + (u_3(t) - y_2)^2 + u_5(t)^2 \right) ^{3/2}} \\&\quad - \frac{m_3 (u_1 - x_3)}{\left( (u_1(t) - x_3)^2 + (u_3(t) - y_3)^2 + u_5(t)^2 \right) ^{3/2}} \\&= 2u_4 + {\varOmega }_x. \end{aligned}$$

A similar computation shows that

$$\begin{aligned} \dot{u}_4 = -2 u_2 + {\varOmega }_y \end{aligned}$$

and that

$$\begin{aligned} \dot{u}_6 = {\varOmega }_z. \end{aligned}$$

Then, ${\hat{\gamma }}(t) (u_1(t), u_2(t), u_3(t), u_4(t), u_5(t), u_6(t))$ is a periodic solution of the circular restricted four-body problem as desired.