# Arnold´s Math Methods of Classical Mechs - a question on Newtonian Mechanics

There is the following question/answer on Arnold´s book Mathematical Methods of Classical page 10.

Arnold's Question A mechanical system consists of two points. At the initial moment their velocities (in some inertial coordinate system) are equal to zero. Show that the points will stay on the line which connected them in the initial moment.

Proof A solution to this problem would be: Any rotation of the system around the line connecting the initial configuration is a Galilean transformation and so sends a solution to the differential equation of motion to another solution. Since these rotations fix the initial conditions, by the uniqueness of solutions of differential equations, it is easy to see that the motion of the points must be constrained to the afore mentioned line. qed.

My qualms with this solution is that it assumes the equation of motion is nice enough and hence has unique solutions. My question is then:

Question Is there a force field with a solution to the equation of motion that contradicts the above exercise question of Arnold? This force field will have to be "pathological" enough so that the solutions to the equations of motion are not unique - is there a known physical configuration (i.e. an existing real physical configuration) with this property?

I came up with the given solution after trying, and not suceeding, to solve this problem via the creation of some conserved quantity (inner product) that would restrict the motion to the given line. Perhaps such an argument would rule out the above pathologies, and hence would be a stronger/better argument.

There are no examples for a rotationally invariant potential, no matter how horrible, because the force law is then a function of $x^2 + y^2$, and you can prove conservation of angular momentum for solutions in the weakest sense you can think of, because it follows algebraically from the equations of motion.
But if you allow a 2-d potential which is of the seperable form $V(x) + V(y)$, you can also prove the collinear theorem for starting using reflection symmetry in y. But for $V(y)=-\sqrt{y}$, there are two solutions, $y=0$ and $y= (t +C)^{4/3}$, where the linear constant multiplying t has been made 1 by a judicious choice of mass.