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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asked Nov 28, 2012 at 9:56
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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.
answered Nov 28, 2012 at 10:31
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$\begingroup$ 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. $\endgroup$ Commented Nov 28, 2012 at 10:47
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$\begingroup$ By the way, do you have a reference for this theorem ? $\endgroup$ Commented Nov 28, 2012 at 10:48
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$\begingroup$ 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. $\endgroup$ Commented Nov 28, 2012 at 11:33
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$\begingroup$ Reference: the somehow geometric way I told it is perhaps written e.g. in Cartan's book (Calcul Differentiel) or in books on Dynamical Systems (maybe Arnold's). But actually most textbooks on ODE should have it, maybe stated in other forms. It's part of the property of continuous dependence by initial data. $\endgroup$ Commented Nov 28, 2012 at 11:39
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$\begingroup$ Ok, I think I see why it follows from the continuity of the flow of an ODE now. Thanks ! $\endgroup$ Commented Nov 28, 2012 at 12:54