Consider ODE $x'=x^2$ on $R$. Find its solution with initial value $x_0$. Are the solutions defined for all the time or do they escape to infinity in finite time?
The solution seems to be $x(t)=1/(1/x_0-t)$ by separation of variables, but what should we say to conclude "escaping to infinity in finite time"? $x^2$ is not globally Lipschitz or something else? I am confused about this part.
Thanks!
Best Answer
If you consider $x_0>0$, $\lim_{x \to (1/x_0)^-} = +\infty$. That is, you have blowup in finite time: it only takes time $1/x_0$ for the solution to blow up. By contrast $x=e^t$ does blow up, but it takes infinite time to do so. In light of Picard-Lindelof, this is only possible because the right side is not globally Lipschitz in $x$.