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Jul 12, 2023 at 17:28 comment added jd27 @Chordx Yes that is correct, if $F$ depends on time we can see from what you wrote that $\tilde{\phi}$ will not be a solution in general by comparing it to the equation i wrote below equation 1.
Jul 12, 2023 at 16:23 comment added Chordx Just to clarify, why is it true that $\tilde{\phi}$ is not always a solution to Newton's equation? Because $\ddot{\phi}(t)=F(\phi(t),\dot{\phi}(t),t) \implies \ddot{\phi}(t+s)=F(\phi(t+s),\dot{\phi}(t+s),t+s),$ correct? The example I gave seems to demonstrate that as well.
Jul 12, 2023 at 16:20 vote accept Chordx
Jul 12, 2023 at 15:38 comment added Valter Moretti To conclude you should also assume that $F$ is continuous, in order to have your last assumption: there is at least a solution for generic initial conditions. Uniqueness is not necessary here.
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Jul 12, 2023 at 14:24 history answered jd27 CC BY-SA 4.0