Movement of repelled particles in a ball

Question

EDIT:

Given a system of $N\geq 3$ charged point particles in $\mathbb{R}^3$ of the same charge which interact according to Coulomb law (thus they repell one from each other). Is it possible that the system remains in a fixed ball all the time? (For $N=2$ this is impossible, and this is what I expect in general.)

More precisely, denote $m_1,\dots, m_N>0$ the masses of the particles. Assume that the $i$th particle acts on $j$th one with the force

$$\vec F_{ij}=\frac{k_ee_ie_j}{|\vec x_j-\vec x_i|^3}\cdot (\vec x_j-\vec x_i), $$ where $k_e>0$ is a constant, $e_i$ is a charge of $i$th particle such that $e_ie_j>0$, $\vec x_i$ is the location of the $i$th particle. The equations of motions are $$m_j\frac{d^2 x_j}{dt^2}=\sum_{i\ne j}\vec F_{ij}, \mbox{ where } j=1,\dots,N.\,\,\,(1)$$

The question is whether there is a solution such that for some $R$ one has $$||\vec x_i(t)||<R \mbox{ for all } t>0, \, i=1,\dots, N.$$

ADDED: I expect that this is impossible. In fact I expect that not only for Coulomb law, but still in greater generality. Assume that the equations (1) are satisfied when the force $\vec F_{ij}=\vec F_{ij}(x_i,x_j)$ has the same direction as the vector $\vec x_j-\vec x_i$. Assume moreover that if all points are in a fixed ball of the radius $R$ then for some constant $\varepsilon >0$ such that $$||\vec F_{ij}||>\varepsilon.$$ Is there a solution of (1) such that all the point are in the ball of radius $R$ for all $t>0$?

Answer 1

If all the particles remained in a bounded domain, the virial theorem would apply. In the case of a radial inverse square power law, it states that twice the asymptotic time average of kinetic energy of the system equals minus the asymptotic time average of its potential energy. However, while the kinetic energy is always nonnegative, the potential energy for repulsive coulomb forces is positive, contradiction.

Answer 2

Let $B$ be the smallest ball such that all $N$ particles remain inside $B$ for all $t\geq0$. Either the trajectory of one of the particles intersects $\partial B$ at some finite time $t_0$, or there is one particle and a sequence $(t_n)_{n\in\mathbb{N}}$ with $\lim \limits_{n \to \infty} t_n ~=~\infty$ such that the particle position at $t_n$ has distance $<1/n$ from $\partial B$, and no other particle at $t_n$ is closer to $\partial B$.

In the first case the radial velocity of the particle at $t_0$ is zero, and therefore the radial component of its acceleration must be $\leq 0$, in contradiction to the fact that the radial component of all forces is positive.

In the second case for each $\epsilon>0$ one can find a time $t_n$ such that the radial acceleration of the particle is less than $\epsilon$. But the radial component of the force from the other particles has a global lower bound because they cannot get arbitrarily close to the particle due to global energy conservation, but have to stay inside the sphere. Choosing a sufficiently small $\epsilon>0$ therefore leads to a contradiction.