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).
Best Answer
(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 $x(s) = \infty$.