$y^{\mathrm{\prime }}=y^2$, $y(0)=1$, has maximal solution $y(t)=(1-t{)}^{-1}$ on $(-\mathrm{\infty },1)$

Differentiation gives $((1-t{)}^{-1}{)}^{\mathrm{\prime }}=(1-t{)}^{-2}=y(t{)}^2$ and $y(0)=1$. On the connected component of a solution's nonzero set containing $0$, [L2] gives $(1\mathrm{/}y{)}^{\mathrm{\prime }}=-1$, so integration forces $1\mathrm{/}y=1-t$. If that component had a finite boundary $c$ inside the solution interval, continuity would give $y(t)\to 0$ there and hence $\mathrm{\mid }1\mathrm{/}y(t)\mathrm{\mid }\to +\mathrm{\infty }$, while the identity gives $1\mathrm{/}y(t)\to 1-c\in \mathbb{R}$, a contradiction. Thus the component is the whole solution interval.

On every compact state interval $\mathrm{\mid }y\mathrm{\mid },\mathrm{\mid }z\mathrm{\mid }\le R$, one has $\mathrm{\mid }{y}^2-z^2\mathrm{\mid }\le 2R\mathrm{\mid }y-z\mathrm{\mid }$, so the polynomial field satisfies the local state-Lipschitz hypothesis of [L3]. The formula is defined on $(-\mathrm{\infty },1)$ and tends to $+\mathrm{\infty }$ as $t\uparrow 1$; no finite value permits continuation through $1$, while the formula continues indefinitely to the left. Thus it is the unique maximal solution from [L3], consistently with the compact-escape conclusion [L1].