Hi I'm trying to understand the most general conditions under which I can conclude finite time blow up of an ODE of the form $\dot{x} = f(x)$ with initial condition $x\_0 > 0$ and $f(x) \geq 0$ for all $x \geq 0$. 


If I re-write this in a separable way so that $dt = \frac{dx}{f(x)}$ then I want to determine if there is some finite time $T$ for which $x(T) = \infty$. Ok so it is clear to me that if $\int\_{x\_0}^{\infty} \frac{dx}{f(x)} = \infty$ then there is **no finite time blowup** for the initial condition $x\_0$ (since we would need some finite $T$ for which $x(T)=\infty$ and this precludes that). I'm confused however if this is a necessary condition. In other words, if I know that $\int\_{x\_0}^{\infty} \frac{dx}{f(x)} < \infty$ does this necessarily tell me that I **DO** get finite time blowup? This is my question. It's not entirely clear to me why this should *prove* finite time blowup.


Two examples to keep in mind in all of this are $f(x) = x^2$ (finite time blowup) and $f(x) = x^{1/2}$ (no finite time blowup).



Criterion for finite time blowup of an ODE
(This used to be a comment, but I think it deserves to be an answer, after mulling over it a bit.)


I don't think your criteria are quite correct. Some counterexamples:


Let $f(x) = - x^2$, and $x(0) = -1$. This ODE blows up in finite time toward $-\infty$. But $\int\_{-1}^\infty dx / f(x) $ diverges due to the singularity at $x = 0$. 


Similarly, for any $f(x)\geq 0$ such that $f(0) = 0$, for any initial value $x(0) < 0$ we must have $x(0) \leq x(t) \leq 0$ for any $t > 0$. Hence we have global existence (no blow up). But if we take $f(x) = \sqrt{|x|}$ if $|x| \leq 1$ and $f(x) = x^2$ if $|x| > 1$, then $1/f(x)$ is integrable, and in particular $\int\_{x\_0}^\infty dx/f(x) < \infty$ for any $x\_0$. 


One key point used in your examples $f(x) = x^2$ and $f(x) = \sqrt{x}$ with initial data positive, is that there are no stationary points. And so for any positive initial datum the evolution eventually goes toward $x\to \infty$, and so the question of blowup reduces to a question of how fast that happens. And for that the integral test is a good one. 


Another way of saying this is that you wanted to use the equality


$$ \int\_0^s dt = \int\_{x(0)}^{x(s)} \frac{dx}{f(x)} $$


but you falsely presupposed that the end state is necessarily $x(s) = \infty$. 


