Recently I came across the problem of Norton's dome.
I thought of two questions, for which I found no answer.

  1. Does there exist a newtonian initial value problem, where the total force on each body is non-zero everywhere, with more than one solution and/or broken symmetry?
  2. Given a newtonian system, is the property of an initial condition to have a unique solution generic?
    (i.e. is the subset of initial conditions with a unique solution open and dense?)


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    $\begingroup$ Related to non-uniqueness & Norton's dome: physics.stackexchange.com/q/39632, physics.stackexchange.com/q/141111 $\endgroup$ – Kyle Kanos Feb 6 '16 at 16:24
  • $\begingroup$ @KyleKanos, thanks for you comment. I read this thread a few days ago. It was useful, in understanding Norton's dome, but unfortunately it does not answer my questions. $\endgroup$ – Shay Ben Moshe Feb 6 '16 at 16:25
  • $\begingroup$ Where did I say it would answer your questions? $\endgroup$ – Kyle Kanos Feb 6 '16 at 16:26
  • $\begingroup$ Never said you did :), I also thanked you for it, and said that is a useful thread. Take it easy. $\endgroup$ – Shay Ben Moshe Feb 6 '16 at 16:29
  • $\begingroup$ You made the claim that it does not answer my questions, as if I had posted it to be an answer to the queries. All I said was that those two links were related to this one due to their contents. $\endgroup$ – Kyle Kanos Feb 6 '16 at 16:30

In order to have non-uniqueness, you must have a discontinuity in the derivative of the potential. As mentioned in the dome paper, a local maximum, or saddle point will not be sufficient, as any perturbation would take an infinite time to manifest. Only initial conditions that result in exactly reaching the discontinuity produce not unique outcomes. Thus in order for the set of initial conditions that result in unique outcomes to not be dense, the subset of the potential where the derivative is continuous would need to not be dense. While there are curves whose derivative is not densely continuous, I'm struggling trying to get a newtonian potential with the same property. I would guess that it's not possible.


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