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 $C^1$ ?



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


