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 $$u(t)=((p-1)(T_0-t))^{-\frac1{p-1}}\;\;\;\text{for }t<T_0\tag2$$ 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?

  • $\begingroup$ 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. $\endgroup$
    – AVK
    Aug 1, 2021 at 13:18
  • $\begingroup$ Is writing the variable limit over the arrow rather than under common? $\endgroup$ Aug 10, 2021 at 7:59
  • 2
    $\begingroup$ It is not the inverse function $f^{-1}(s)$, but the reciprocal $f(s)^{-1}$ in the integral criterion. $\endgroup$ Aug 10, 2021 at 7:59

1 Answer 1


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?).

  • $\begingroup$ 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? $\endgroup$
    – 0xbadf00d
    Aug 13, 2021 at 5:54
  • $\begingroup$ 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? $\endgroup$
    – Kore-N
    Aug 14, 2021 at 6:33

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