# Reversal of time in solutions of autonomous differential equation

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?

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