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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).

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Please read over your question again. I think your first improper integral equation should be an inequality. Also, you misspelled separable. – KConrad Aug 28 '10 at 15:33
The spelling error is corrected but I do indeed want to say that integral diverges. It is saying that if x does to infinity then $T$ must also go to infinity. – Dorian Aug 28 '10 at 15:42
Think about $f(x) = x^2$ versus $f(x) = x^{1/2}$ to get the divergence/convergence ideas right. – Dorian Aug 28 '10 at 15:43
Ok I will make a modification. – Dorian Aug 28 '10 at 16:24
The condition you're talking about (with the additional constraints mentioned in Willie Wong's answer) is called Osgood’s condition and known since 1898. See Walter's Ordinary Differential Equations (1998 ed., Springer GTM 182), p. 147. – Fizz Jan 29 '15 at 14:16
up vote 4 down vote accepted

(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 $x(s) = \infty$.

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Note that by horizontally translating $f$, the second example can be made so that your modified criteria that $x_0 > 0$ is satisfied. – Willie Wong Aug 28 '10 at 16:34
If you strengthen your assumption to $f(x) > 0$, allowing no stationary solutions, then the blow-up criterion boils down to the integral equality I wrote down at the bottom of my answer. – Willie Wong Aug 28 '10 at 16:38

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