# Equilibrium points doubt (ODE system solution)

Zebx
Hi. I'm not sure about something related to the equilibrium points (or fixed points) of a non linear ode system solution. As far as I know, to check if an equilibrium point exists, I need to put the function of my ode system equal to zero. Then once the point is found, I can use it to evaluate its stability. If I have 2 equilibrium points for example, one of which is linearly stable, then it means that if I use this point as my initial condition I will get a constant solution. Moreover, if I evaluate the Jacobian of my system in the stable equilibrium point, I can use the eigenvalues to check what timesteps I have not to use in order to have stable solutions.

I was trying to apply what said before (supposing it is all correct) to my ode system. I already checked the methods used works and plot some solution, but I wanted to try to see if I could apply the above analysis properly. For instance I started by considering a 2-body problem with a star at the center of the system and a planet orbiting around it. Looking at the equations I already have a problem: the function is zero just if I use ##(x=0, y=0, z=0)##, but since I have ##\sqrt{x^2 + y^2 + z^2}## at the denominator I would get infinites. So I thought that maybe it's normal cause, if I actually had an equilibrium point for my system and I used it as starting condition, I should get a constant solution, but it's impossible to have a constant solution for my problem since I have a body moving in a star potential. So basing on this I thought that maybe this is one of these cases in which you can't have a clue about stability from jacobian study.

Is my reasoning correct?

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Homework Helper
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Describe the motion of a planet with its center of mass at the center of mass of its sun?

Zebx
Yes, I consider the star not moving at all.

Homework Helper
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Yes; there are no fixed points in the two-body problem. If you regard one mass as fixed at the origin then the domain of the problem is $\mathbb{R}^3 \setminus \{0\}$.

You can talk about the stability of a periodic orbit by considering fixed points of a return map.

Zebx
Zebx
Ok, thank you. So I suppose there are no fixed points even if I increase the number of bodies orbiting the star. I mean if I have a system of, say, 6 planets around the star still fixed in the origin. In that case I would have for every body a series of terms like ##C(\vec{x} - \vec{x}_i)/\lvert\vec{x} - \vec{x}_i\rvert^3##, where ##i = 1, \dots, 6##, so the only way to have my ##F(\vec{x}) = 0## is to have ##\vec{x} = \vec{x}_i##, which is clearly not possible, even for just two bodies (the system doesn't include close encounters).

Homework Helper
2022 Award
Consider: There are a finite number of bodies, so there is one with maximal $x_i$ coordinate. The corresponding component of the gravitational force is then negative, and thus so is the acceleration unless there is some other force to balance it. Thus there can be no fixed points.

Zebx and etotheipi
Zebx
Perfect! thank you very much.