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I am trying to solve a question that my professor gave.

When a particle moves in one dimension $x$ in a potential $U(x)$ , the resulting motion over a very short time interval is specified by Newton’s equation, provided $x(0)$ and $\dot x(0)$ are given. Often the resulting motion can be determined over an arbitrarily long time. The resulting motion in phase space is called the phase flow. But for some potentials the motion can’t be determined arbitrarily long times: there is no phase flow. If $U(x)$ has the form ($-Cx^n$, where $C$ is an arbitrary constant), there are some values of the positive integer $n$ for which the motion doesn’t define a phase flow. What are these values and what prevents the system from having a phase flow?

Hint: From the conservation of energy trying to get $x(t)$ for all $t$ and see for what values of $n$ you get into trouble.

Could anyone please give some other hints to do this problem

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  • $\begingroup$ The problem is in the possible blow up of solutions in finite time. Check out the case $U = -x^4$ for example. $\endgroup$
    – LPZ
    Commented Jun 14, 2022 at 13:05
  • $\begingroup$ What do you mean by blowup of solutions @lpz $\endgroup$
    – Dinken
    Commented Jun 14, 2022 at 19:25
  • $\begingroup$ Solutions go to $\pm\infty$ in finite time $\endgroup$
    – LPZ
    Commented Jun 14, 2022 at 22:16
  • $\begingroup$ Is it in general n>2 doesn't have a phase flow@lpz $\endgroup$
    – Dinken
    Commented Jun 15, 2022 at 2:46
  • $\begingroup$ Depends on the sign of $C$ $\endgroup$
    – LPZ
    Commented Jun 15, 2022 at 7:02

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