# Do the maximal intervals of an ODE solution vary continuously?

Consider the following Cauchy problem (which we will denote by IVP$_{x_0}$ ) $$\dot {x} = F (x)$$ $$x(0) = x_0,$$ where $F:U \subset \mathbb {R}^n \rightarrow \mathbb {R}^n$ is a $\mathcal {C}^{\infty}$ function, and $U$ is an open set.

Let $\varphi (t,x)$ be the flow of the above differential equation, i. e. $\varphi$ satisfies the conditions:

• $\frac {d \varphi }{dt}(t,x) = F (\varphi (t,x))$.

• $\varphi (0,x) = x$, $\forall$ $x$ $\in$ $U$.

• $\varphi (t,x)$ is the maximal solution of the IVP$_{x}$.

For each $y$ $\in$ $U$ fixed, we can associate this value to an open interval $\left(- \omega_{y}^ {(1)}, \omega_{y}^{(2)}\right)$ corresponding to the domain of the maximal solution of the IVP$_y$, i. e, $\left(- \omega^{(1)}_{y}, \omega_{y}^{(2 )}\right)$ is the domain of $g (t) = \varphi (t,y).$

Now, I would like to know if

\begin{align*}\Psi : U &\rightarrow \mathbb {R} \\ x &\mapsto \min \left\{\omega_{x}^{(1)} , \omega_{x}^{(2)},1\right\} \end{align*} is a continuous function. Does anyone know how I prove this?

EDIT: I found an easy counter-example:

Define $U := \left(-\frac{1}{4},\frac{1}{4}\right)\times \left(-\frac{1}{4},\frac{1}{4}\right) \setminus \{0,0\}$. Consider the function

\begin{align*}F: U& \rightarrow \mathbb{R}^2\\ u& \mapsto (1,0), \end{align*}

then trivially $\Psi$ is not continuous, this example is particularly interesting because $F$ has no singular points.

Simplest example: in the line consider $F$ with a single zero such that all other solutions nonglobal. Say $F(x)=x^2$.
The best that you can say is that $$\left\{\left(x,\omega_x^{(1)},\omega_x^{(2)}\right):x\in\mathbb R^n\right\}$$ is an open set (see for example Hale's ODE book).