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Given is the following differential equation:

$ y''=f(y)$ with $ f: \mathbb{R} \to \mathbb{R}$ and $ a: I \to \mathbb{R}$. $a$ is the solution, which can be continued and $f$ is locally Lipschitz continuous.

I want to show: $ x_1 \in I$, $a'(x_1)=0 \implies a(x_1 - c)=a(x_1 + c)$ for all $c \in (I-x_1) \cup (x_1-I)$

How can I apply $a'(x_1)=0 $ in my proof?

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  • $\begingroup$ A hint: use the uniqueness of solutions to an IVP. $\endgroup$ – user539887 Jan 11 at 10:12
  • $\begingroup$ The solution is unique because f is locally Lipschitz continuous. I don't know how this might help? $\endgroup$ – Steven33 Jan 11 at 10:22
  • $\begingroup$ Sub-hint: If $y:x\mapsto a(x_1+x)$ solves $y''=f(y)$ then $y:x\mapsto a(x_1-x)$ solves $y''=\ldots$ $\endgroup$ – Did Jan 11 at 10:29
  • $\begingroup$ I would say:...solves $ y''=f(y).$ $\endgroup$ – Steven33 Jan 11 at 10:33
  • $\begingroup$ Can somebody explain how I can proof that? $\endgroup$ – Steven33 Jan 12 at 13:45

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