# Dependence of the blow-up time of existence of an ODE with respect to initial condition.

Let $V$ be a smooth vector field on $\mathbb{R}^n$. Assume that the maximal solution to the Cauchy problem $x'=V(x), x(0)=x_0$ exist only for $t\in [0,T)$, where $T$ is finite, denote this time by $T(x_0)$. Is $T$ continuous with respect to $x_0$ ? Is it $C^1$ ?

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No, but you can say it is lower semicontiuous, even wrto the initial time (that is, the optimistic situation: perturbing a little the initial data the existence is ensured almost up to $T$ , and could even be much greater) .
Precisely, given a Banach space $E$, an open set $\Omega\subset \mathbb{R}\times E$ and $f:\Omega\subset E\times \mathbb{R}\rightarrow E$ in the Cauchy-Lipshitz-Picard hypoteses, for any $(t_0,x_0)\in \Omega$ the Cauchy problem $$u(t_0)=x_0$$ for the ODE $$\dot u =f( t, u(t))$$
admits a maximal solution defined in an interval $\big(\tau_*(t_0,x_0), \tau^*(t_0,x_0)\big)\subset\mathbb{R}$, where $$\tau _ *:\Omega\to [-\infty,0 )$$ is upper semicontinuous and $$\tau ^ *:\Omega\to (0,+\infty]$$ is lower semicontinuous. This amount to saying that: the domain of the "general solution" $\xi:\Xi\subset \Omega\times\mathbb{R}\rightarrow E$ defined as $\xi(t_0,x_0,t):=u(t)$ with the solution $u(t)$ of the above Cauchy problem, that is the set $$\Xi:=\{ (s,x,t)\in \Omega\times\mathbb{R} \, :\, \tau _ * (s,x) < t < \tau ^ *(s,x) \}$$ (that is the zone between the graph of $\tau _ *$ and the graph of $\tau ^ *$), is an open set.
Interesting, do you know of an example where a sequence of initial conditions $x_i$ with $\tau^*(x_i)=+\infty$ converges to an $x$ with $\tau(x)<+\infty$ ? I have some intuition with nested periodic orbits which turn faster and faster as you get to the "outside", and which finaly go to a finite-time blow-up orbit, but I don't see how to make it rigorous. – Thomas Richard Nov 28 '12 at 10:47
To the first question: one (trivial) example is: just remove a point (e.g. $\Omega:=\mathbb{R}^2\setminus\{(1,1)\}$, and $f(t,x)=1$, so that the solution with $u(0)=0$ ends at time $1$, whereas with $u(0)=\epsilon$ they last forever). But clearly there are examples of scalar ODE's with $f:\mathbb{R}^2\to\mathbb{R}$, maybe not immediate to wirte down. But there is no obstruction to having one solution blowing up at $T=1$, e.g. $u(t)=1/(1-t)$ for $t <1$ and below its graph, solutions defined for all times. – Pietro Majer Nov 28 '12 at 11:33