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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ locally Lipschitz. Does the initial value problem $y'=f(y)$, $y(0)=y_0$ have a unique solution on $\mathbb{R}$ if $f\left(\frac{y_0}{2}\right)=f(2 y_0)=0$?

From Picard Lindelöf I know the existence and uniqueness of a solution on every open intervall around $0$. I don't know how to use the conditions on $f$.

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The only problem that could prevent global existence of the solution would be that $y$ goes off to $\pm \infty$ at some finite value of the independent variable. But in your case, what would happen as $y$ approached $y_0/2$ or $2 y_0$?

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  • $\begingroup$ I only know that the derivative of $y$ is $0$ at those points. Is y bounded by $y_0/2$ and $2y_0$? I don't see why. $\endgroup$ – Julian May 31 '12 at 21:19
  • $\begingroup$ I guess I don't know how to use that x is a independent variable. $\endgroup$ – Julian May 31 '12 at 21:23
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    $\begingroup$ Ok, I got it. The integral curves cannot cross each other and so the solution is indeed bounded by $y_0/2$ and $y_0$ $\endgroup$ – Julian May 31 '12 at 21:58

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