# Criteria for finite time blow up for two simple ODEs

I've got two simple questions on criteria for a finite time blow up of solutions of two simple ODEs:

1: If $$u$$ is a solution of $$u'=f(u)\ge0$$, $$u(0)=u_0$$, how do we see that $$u$$ blows up in finite time (i.e. there is a $$T>0$$ s.t. $$|u(t)|\xrightarrow{t\to T-}\infty$$) if and only if $$\int_{u(0)}^\infty f^{-1}(s)\:{\rm d}s<\infty\tag1?$$

I've only got an idea for this if we additionally assume that $$f$$ is Lipschitz continuous. We then see that if $$u:I\to\mathbb R$$ is a solution of $$u'=f(u)$$ on some compact interval $$I:=[a,b]$$, then: If $$f(u(t_1))=0$$ for some $$t_1\in I$$, then $$u\equiv u(t_1)$$ on $$I$$ by uniqueness (for which we need the Lipschitz continuity of $$f$$). So, since $$f\ge0$$, we see that $$u$$ can only blow up if $$f>0$$ on $$[u(a),\infty)$$ ... But how do we need to proceed? And can we drop the Lipschitz assumption? And what's happening for general, possibly negative, $$f$$?

2: If $$p>1$$, the solution of $$u'=u^p$$ is given by $$u(t)=((p-1)(T_0-t))^{-\frac1{p-1}}\;\;\;\text{for }t for some $$T_0\in\mathbb R$$. We see that $$T_0=\frac1{(p-1)u_0^{p-1}}\tag3$$ and hence $$u(t)=\left(\frac{u_0^{p-1}}{1-(p-1)u_0^{p-1}t}\right)^{\frac1{p-1}}\tag4.$$ I think we should have $$u(t)\xrightarrow{t\to T_0-}\infty$$; at least if $$u_0>0$$. But what happens if $$u_0\le0$$? Does the solution then exists at all?

• Did you mean $f(u)>0$? If $f(u)\ge 0$, 1 is not true. For example, if $f(u_0)=0$, then $u_0$ is an equilibrium point and the solution $u(t)\equiv u_0$ does not blow up regardless of all other conditions.
– AVK
Aug 1, 2021 at 13:18
• Is writing the variable limit over the arrow rather than under common? Aug 10, 2021 at 7:59
• It is not the inverse function $f^{-1}(s)$, but the reciprocal $f(s)^{-1}$ in the integral criterion. Aug 10, 2021 at 7:59

Since $$f$$ is positive, $$u$$ is increasing, so we can see it as a timechanged version of the line $$t \mapsto t$$. Then blow up amounts to asking whether the time change reaches $$\infty$$ in finite time, which turns out to be the condition you require.
1. Let's make this rigorous and simple. We have that $$t = \int_{u_0}^{u(t)} \frac{1}{f(s)} ds$$ for all $$t < T_\mathrm{fin}$$ (tha latter being the blow-up time, possibly $$\infty$$). Taking the limit $$\lim_{t \to T_{\mathrm{fin}}}$$in the formula gives the result.
2. If $$u_0 < 0$$ we have that $$v= - u$$ solves $$v^\prime = (-1)^p v^p$$ and assume $$p \in \mathbb{N}$$ to have real valued solutions. Then if $$p$$ is even we have explosion as you explained. If $$p \in 2\mathbb{N}+1$$ the solution exists globally and converges to $$0$$ (can you see why?).
• Thank you for your answer! Do we need Lipschitz continuity of $f$? It seems so, since in order to derive the formula $t=\int_{u_0}^{u(t)}\frac1f$, we need to exclude the possibility that $f\circ u$ is constant. And one way to do this is assuming Lipschitz continuity, since then $(f\circ u u)(t_1)=0$ for some $t_1$ already implies $u\equiv u(t_1)$ by the usual uniqueness result for ODEs. So, do we need the Lipschitz continuity or is there another argumentation? Aug 13, 2021 at 5:54
• Of course you want $u$ to exist, so $f$ should satisfy some regularity assumption, such as local (but not global! Otherwise what's the point of asking for blow up?) Lipschitz continuity. Note that there is no $f \circ u$ in the formula so I don't know what you mean. But we should assume that $f(r)>0$ for $r \geq u_0$. If not there is no blow up: do you see why? Aug 14, 2021 at 6:33