Identifying a system of ODEs

Question

Studying the dynamics of the endpoints of an equilibrium measure (a minimizer of its logarithmic energy in an external field) I ran into the following system of differential equations (which I state for the case of 4 points, for simplicity): let $x_j=x_j(t)$, $j=1, \dots, 4$, be real values dependent on time $t$, all distinct at $t=0$, and satisfying the system $$ \frac{d x_j}{dt} = \frac{m_j}{q'(x_j)}=m_j \prod_{k\neq j} (x_j-x_k)^{-1}, \quad j=1, \dots, 4, $$ where $q(x)=\prod_{j=1}^4 (x-x_j)$ and $q'(x)$ is its derivative with respect to $x$. Here $m_j$ are positive numbers.

My questions (sorry if too elementary or naive) are:

1) is this kind of a system known, does it have any name attached to it?

2) I needed to prove the fact that the interior $x_j$'s collide in a finite time. Does this follow from any general fact in dynamical systems or systems of ODEs?

3) what about the more general situation, when the number of points is $n$ and the right hand sides in the system are rational functions?

Thanks in advance.

Answer 1

this is not an answer to your question. hassan aref studied the motion of point vortices.

Answer 2

Forgive the brevity - what you are interested in in something called the Osgood condition. Systems of this kind have been studied for a long time, particularly when the $m_i$'s are all equal. If you look at the $n=2$ case and look at $r= (x_1-x_2)$ the problem reduces $r_t =c/r$ and the fact that $c/r$ is not integrable at $r=0$ tells you that you have a finite time collision.

A good starting point is Ruelle's Thermodynamics text.

Answer 3

Assume $x_1(0) < x_2(0) < x_3(0) < x_4(0)$. Note that $\dfrac{dx_2}{dt}$ and $\dfrac{dx_4}{dt}$ are positive and the other two are negative. So for $t > 0$ (and before the collision), $x_1 < x_1(0) < x_2(0) < x_2 < x_3 < x_3(0) < x_4(0) < x_4$. Now $\dfrac{dx_4}{dt} \le \dfrac{m_4}{(x_4-x_3(0))^3}$. Solving the differential equation obtained by making this an equality, we find that $x_4(t) \le x_3(0) + ((x_4(0) - x_3(0))^4 + 4 m_4 t)^{1/4}$. Call the right side $B_4(t)$. Similarly $x_1(t) \ge B_1(t) = x_2(0) - ((x_2(0) - x_1(0))^4 + 4 m_1 t)^{1/4}$.

So $$ \dfrac{dx_2}{dt} > \frac{m_2}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))(x_3 - x_2)}$$ $$ \dfrac{dx_3}{dt} < \frac{m_3}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))(x_2 - x_3)}$$ $$ (x_3 - x_2) \dfrac{d}{dt} (x_3 - x_2) < - \dfrac{m_3+m_2}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))} $$ Thus a collision will occur by time $T$ if $$ \int_0^T \dfrac{ dt}{(x_3(0) -B_1(t))(B_4(t) - x_2(0))} > \frac{(x_3(0)-x_2(0))^2}{2(m_2 + m_3)} $$ The integral of the left side from $0$ to $\infty$ is infinite, since $x_3(0) - B_1(t)$ and $B_4(t) - x_2(0)$ only grow like $t^{1/4}$ as $t \to \infty$.

So there will always be a collision in finite time.

Answer 4

Oops! There was an error in my argument.