# If $f$ is even and $y'=f(y)$ then $y$ is odd

Let $f\in C^1(\mathbb{R}, \mathbb{R})$ be an even function.

Consider the maximal solution $y\colon\left]\alpha ,\beta\right[\to \mathbb{R}$ of the IVP $$y'=f(y),\ y(0)=0$$

Prove that $y$ is an odd function and $\beta =-\alpha$.

To be able to prove that $y$ is odd, I first need its domain to be symmetric ($x\in \left]\alpha ,\beta\right[\implies -x\in \left]\alpha ,\beta\right[$), from here I can conclude that $\alpha=-\beta$. But how to prove the domain is symmetric?

And how to prove that $y(-x)=-y(x)$ for all $x\in \left]\alpha,\beta\right[$? I suspect it has something to do with the fact that the derivative of an even function is odd and $y'=f(y)$, but I can't see how to get the desired result.

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Clearly, $\alpha<0<\beta$. Let $r=\min\{-\alpha,\beta\}$ so that we at least have a solution on $\left]-r,r\right[$.

Then show: If $y$ is a solution on $]\alpha,\beta[$, then $x\mapsto-y(-x)$ is a solution on $\left]-\beta,-\alpha\right[$.

By uniqueness, these coincide on $\left]-r,r\right[$.

Then you can combine $y$ with $x\mapsto- y(-x)$ to find a solution on $\left]-\max\{-\alpha,\beta\},\max\{-\alpha,\beta\}\right[$. By maximality, we conclude $\alpha=-\max\{\alpha,\beta\}=-\beta$.

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Why does uniqueness hold? –  Git Gud May 21 '13 at 21:19
@GitGud First of all, if it doesn't, the claim is false ;) –  Hagen von Eitzen May 21 '13 at 21:20
^_^, thanks. Can you provide a counterexample? –  Git Gud May 21 '13 at 21:21
@GitGud Of course not. As $f$ is $C^1$, it is Lipschitz and we optain uniqueness from Picard-Lindelöff. –  Hagen von Eitzen May 21 '13 at 21:33
Thanks.${}{}{}{}$ –  Git Gud May 21 '13 at 21:39

Define $h(x)=-y(-x)$. Then $$h'(x)=y'(-x)=f(y(-x))=f(-h(x))=f(h(x)),$$ where we have used the eveness of $f$. Now, notice that $h(0)=-y(0)=0=y(0)$. As the solution should be unique you obtain that $y$ is odd.

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There is a little problem in the formulation of this question.

In order to even wonder whether $\alpha=\beta$, a further piece of information is necessary: That the solution is defined in a maximum possible domain.

Clearly, if the flux function (here $f$) satisfies conditions which guarantee uniqueness (which here it does as $f$ is $C^1$, and thus locally uniformly Lipschitz continuous), then it is possible to talk about a "solution defined in a maximum open interval". This is obtained as the set-theoretic union of all solutions of this particular IVP; these solutions do not have the same domain, but each two of them agree in their common domain due to uniqueness. (Note that a function, in Set Theory, is defined as a set on ordered pairs.) If uniqueness is not enjoyed by an IVP, then we simply talk about a "solution defined in a maximal interval", and apparently its definition requires the use of Zorn's lemma.

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