= Finite-dimensional continuation criterion
{title2=$T_+<\infty\ \Longrightarrow\ \text{escape from every compact subset of }U$}
Under the <Picard-Lindelöf theorem>'s hypotheses on $U\subset\mathbb R\times\mathbb R^d$, extend a <solution of a differential equation> as far as possible, to a maximal <interval> $(T_-,T_+)$. If $T_+<\infty$, the time-state trajectory $(t,y(t))$ must eventually leave every <compact subset> of $U$. Equivalently, confinement near $T_+$ to one <compact subset> of $U$ permits continuation past $T_+$. See https://www.mat.univie.ac.at/~gerald/ftp/book-ode/ode.pdf[Teschl, Section 2.6, Lemma 2.14 and Corollaries 2.15–2.16, pp. 52–53].
The domain restriction matters even for a bounded trajectory. Restrict $y'=1$, $y(0)=0$, to $U=\mathbb R\times(-1,1)$. Then $y(t)=t$ exists in this domain only for $-1<t<1$; at $t=1/2,3/4,9/10$ its values are $1/2,3/4,9/10$. As $t\uparrow1$, it approaches the excluded <boundary> point $(1,1)$ while remaining bounded. Its trajectory cannot fit into any <compact subset> of $U$ near that endpoint. This is escape from a finite domain, without <finite-time blowup> in norm. Enlarging the domain to $\mathbb R\times\mathbb R$ gives the same formula $y(t)=t$ for every real $t$.
Conversely, $y'=y$, $y(0)=1$, stays in the <compact set> $[0,T]\times[1,e^T]$ for every fixed $T>0$. Its growing <exponential function> is unbounded over the entire future, but has no <finite-time blowup>. At $T=2$, its largest value on $[0,2]$ is only $e^2\approx7.38905610$. The relevant question is boundedness before each finite endpoint, rather than one bound valid forever. The <tangent function> in the diagram under the <Picard-Lindelöf theorem> illustrates the other possibility: its norm escapes every bound before $T=\pi/2$.
This criterion is useful because an estimate can secure continuation without an explicit formula. In a <finite-dimensional vector space>, a bounded <closed ball> is <compact>; a bounded state trajectory therefore supplies the needed <compact set> when its time-state closure is inside the domain. That final proviso prevents mistaking a domain boundary for <finite-time blowup>.
A common sufficient estimate comes from global <Lipschitz continuity> in the state. Assume $U=I\times\mathbb R^d$ for an open time <interval> $I$, with $f$ continuous and a <Lipschitz constant> $L\geq0$ valid for all states on each compact time <interval> under consideration. For $[t_0,T]\subset I$, let $B=\sup_{t_0\leq s\leq T}\|f(s,0)\|$, finite by <continuity>. Then $\|f(t,y)\|\leq B+L\|y\|$, and the <Gronwall inequality> gives, for $t_0\leq t\leq T$ wherever the solution already exists,
$$
\|y(t)\|\leq\begin{cases}\|y_0\|e^{L(t-t_0)}+\dfrac{B}{L}\bigl(e^{L(t-t_0)}-1\bigr),&L>0,\\\|y_0\|+B(t-t_0),&L=0.\end{cases}
$$
For instance, $y'=y+1$, $y(0)=0$, has $L=B=1$ and the exact solution $y(t)=e^t-1$, attaining this estimate; at $t=1$ the value is $e-1\approx1.71828183$. Finite bounds of this kind give continuation throughout $I$, with the analogous estimate backwards in time. When $I=\mathbb R$, this gives existence for all real times. If $I$ has a finite endpoint, it gives no continuation beyond that prescribed domain. The underlying linear-growth criterion is <Ordinary Differential Equations and Dynamical Systems by Gerald Teschl>[Teschl, Theorem 2.17, p. 53]{ref}.
A brief explanation of the continuation mechanism suffices: within the protecting <compact set>, the <continuous function> $f$ is bounded, say by $M$. The <integral equation> yields $\|y(t)-y(s)\|\leq M|t-s|$ near the finite endpoint, so $y(t)$ has a limit $y_*$. The protecting <compact set> puts $(T_+,y_*)$ inside $U$. Restarting the <initial value problem> there and using local uniqueness joins the trajectories.
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