Breakdown versus trapping: the blow-up of $x^{\mathrm{\prime }}=x^2$ and the global trapped solution of $x^{\mathrm{\prime }}=-x^2$

The quotient rule [L1] gives $x_{+}^{\mathrm{\prime }}(t)=(1-t{)}^{-2}=x_{+}(t{)}^2=F_{+}(t,x_{+}(t))$ on $[0,1)$ and $x_{+}(0)=1$, and $x_{-}^{\mathrm{\prime }}(t)=-(1+t{)}^{-2}=-x_{-}(t{)}^2=F_{-}(t,x_{-}(t))$ on $[0,\mathrm{\infty })$ and $x_{-}(0)=1$.

Local Lipschitz continuity of $F_{\pm }$: by [L1], $G_{\pm }$ are differentiable and continuous, with $G_{\pm }^{\mathrm{\prime }}(z)=\pm 2z$; consequently $F_{\pm }$ are continuous, since $(t,z)\to (t_0,z_0)$ implies $z\to z_0$ and $F_{\pm }(t,z)=G_{\pm }(z)$. Here a compact time-state cylinder means $C=T\times [a,b]$, where $T\subseteq \mathbb{R}$ is a compact interval and $a\le b$; this includes the one-dimensional closed-ball cylinders used in [L2] of Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it. Put $S:=[a,b]$. Set $R:=\max \{\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid }\}$, so $\mathrm{\mid }{G}_{\pm }^{\mathrm{\prime }}(z)\mathrm{\mid }\le 2R$ on $S$. For $(t,z),(t,w)\in C$ with the same time and $z\mathrm{\ne }w$, the mean value inequality [L3] on the interval between $z$ and $w$ gives $\mathrm{\mid }{F}_{\pm }(t,z)-F_{\pm }(t,w)\mathrm{\mid }=\mathrm{\mid }{G}_{\pm }(z)-G_{\pm }(w)\mathrm{\mid }\le 2R\mathrm{\mid }z-w\mathrm{\mid }$, since that interval lies in $S$; for $z=w$ the inequality is immediate. If $C$ is empty it holds with $L=0$. This verifies the local Lipschitz condition of Local Lipschitz continuity in the state variable, locally uniform in time and parameters on every such cylinder.

Non-extendability in (i): suppose a solution $\tilde{x}$ of the IVP $x^{\mathrm{\prime }}(t)=F_{+}(t,x(t))$, $x(0)=1$ existed on $[0,1+\epsilon )$ for some $\epsilon >0$. Both $\tilde{x}$ and $x_{+}$ solve the same IVP on the common interval $[0,1)$ containing $0$, and $F_{+}$ is continuous and locally Lipschitz in the state variable by step 1.2. The uniqueness clause of Picard-Lindelöf (Picard-Lindelöf local existence and uniqueness for first-order systems) gives $\tilde{x}=x_{+}$ on $[0,1)$, so $\tilde{x}(t)=1\mathrm{/}(1-t)\to +\mathrm{\infty }$ as $t\uparrow 1$. But $\tilde{x}$ is continuous [L1] on the compact interval $[0,1]$, so [L2] bounds $\mathrm{\mid }\tilde{x}\mathrm{\mid }$ there, a contradiction. Hence no extension past $b=1$ exists.

Escape in (i): let $K\subseteq D$ be compact and $\eta >0$. Put ${\eta }_0:=\min \{\eta ,1\mathrm{/}2\}>0$, so $(1-{\eta }_0,1)\subseteq (1-\eta ,1)$ and all times in the smaller interval lie in $[0,1)$. By [L2], $K$ is bounded, so there is $R>0$ with $\mathrm{\mid }z\mathrm{\mid }\le R$ for every $(t,z)\in K$. Choose $t\in (1-{\eta }_0,1)$ with $\frac{1}{1-t}>R$, which is possible since $\frac{1}{1-t}\to +\mathrm{\infty }$; then $x_{+}(t)>R$, so $(t,x_{+}(t))\mathrm{\notin }K$. This is the escape asserted by A non-extendable solution leaves every compact subset of the domain near its finite endpoint; an autonomous solution trapped in a compact subset of its state domain cannot break down there for this solution.

Trapping in (ii): $K_0=[0,1]\subseteq \mathbb{R}={\mathbb{R}}^1$ is closed and bounded, hence compact by [L2], and $0<x_{-}(t)\le 1$ for every $t\ge 0$, so $x_{-}(t)\in K_0$ throughout its domain. For any finite $b>0$, its restriction to $[0,b)$ is a solution trapped in $K_0$ near $b$; clause 2 of A non-extendable solution leaves every compact subset of the domain near its finite endpoint; an autonomous solution trapped in a compact subset of its state domain cannot break down there therefore extends that restriction past $b$. Thus no finite right endpoint can be a nonextendable endpoint for this solution, and the displayed formula gives its global extension.