Criteria for finite time blow up for two simple ODEs

Question

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?

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?).