A bounded field whose solution leaves the domain in finite time: $x^{\mathrm{\prime }}=v$, $v^{\mathrm{\prime }}=0$ on $(0,\mathrm{\infty })\times (-2,2)$

The curve $z(t)=(1-t,-1)$ is defined for $t<1$ and lies in $\mathbb{R}\times U$ there because $1-t>0$ and $-1\in (-2,2)$; it is differentiable with $z^{\mathrm{\prime }}(t)=(-1,0)=F(z(t))$ for every $t<1$, and $z(0)=(1,-1)$. So it is a solution on $(-\mathrm{\infty },1)$.

Maximality: suppose a solution $\zeta$ through $(1,-1)$ were defined on an interval containing some $t_0>1$. By [L1] it agrees with $z$ on the overlap of their intervals and is continuous at $t=1$, so $\zeta (1)={\lim }_{t\uparrow 1}(1-t,-1)=(0,-1)$, i.e. its position is $0$ at time $1$; but then $(1,\zeta (1))=(1,(0,-1))$ does not lie in $\mathbb{R}\times U$, contradicting the domain condition for a solution on that interval. Hence no solution through the datum is defined at $t_0>1$, while $z$ is defined at every $t<1$; the maximal interval is exactly $(-\mathrm{\infty },1)$. In particular the maximal solution escapes the domain at the finite time $1$ with $\mathrm{\parallel }F(x,v)\mathrm{\parallel }=\mathrm{\mid }v\mathrm{\mid }<2$ bounded on $U$, so boundedness of the field alone does not guarantee a global solution.