A local solution can be extended as long as its graph remains inside a region where the vector field is regular. A maximal solution records the largest interval on which this continuation is possible.
Gluing and maximal intervals
Theorem. Let \(f\) be locally Lipschitz in \(x\) on an open set \(D\subset\mathbb R\times\mathbb R^d\). Through every \((t_0,x_0)\in D\) there is a unique maximal solution \[ x:(\alpha,\beta)\to\mathbb R^d. \] Every other solution through the same point is its restriction.
Proof. Take the union of all local solution intervals through \((t_0,x_0)\). Uniqueness makes solutions agree on overlaps, so they glue to one solution on the union, which is an interval. Any proper extension would belong to the union, contradicting maximality.
Continuation criterion
Theorem. If \(\beta<\infty\) is the right endpoint of a maximal solution, then its graph leaves every compact subset of \(D\) as \(t\uparrow\beta\). In particular, if \([t_0,\beta]\times\mathbb R^d\subset D\), then \[ \|x(t)\|\to\infty \] along some sequence \(t\uparrow\beta\).
Proof. If a sequence \((t_n,x(t_n))\), \(t_n\uparrow\beta\), remained in a compact \(K\Subset D\), boundedness of \(f\) there would make \(x\) uniformly Lipschitz near \(\beta\). It would have a limit \(x_\beta\), and local existence from \((\beta,x_\beta)\) would extend the solution, a contradiction.
Linear-growth global existence
Theorem. If \[ \|f(t,x)\|\le a(t)+b(t)\|x\| \] with \(a,b\) locally integrable and \(f\) locally Lipschitz in \(x\), then every solution is global on the time interval where \(a,b\) are integrable.
Proof. The integral equation and Gronwall give, for \(t\ge t_0\), \[ \|x(t)\| \le \left(\|x_0\|+\int_{t_0}^ta(s)\,\mathrm{d}s\right) \exp\left(\int_{t_0}^tb(s)\,\mathrm{d}s\right). \] Thus the solution remains bounded on every finite subinterval. The continuation criterion rules out finite-time blow-up.
Finite-time blow-up
Example. For \[ \dot x=x^p,\qquad x(0)=x_0>0,\quad p>1, \] separation gives \[ x(t)= \left(x_0^{1-p}-(p-1)t\right)^{-1/(p-1)}. \] The maximal right endpoint is \[ \beta=\frac{x_0^{1-p}}{p-1}, \] and \(x(t)\to\infty\) as \(t\uparrow\beta\).
Differential inequalities for blow-up
If a nonnegative quantity \(y(t)\) satisfies \[ y'(t)\ge c y(t)^p,\qquad p>1, \] then comparison with the explicit scalar solution forces blow-up no later than \(y(0)^{1-p}/\{c(p-1)\}\). Conversely, dissipative estimates such as \[ y'(t)\le C(1+y(t)) \] preclude finite-time blow-up. For PDEs, energy or norm inequalities play the same continuation role.
Escape to the boundary
Failure of continuation need not mean norm blow-up if the vector field's domain has a boundary. For \(\dot x=1\) on \(D=(0,1)\), the solution exits at \(x=1\) in finite time while remaining bounded. The correct criterion is compact escape from \(D\), not divergence in a chosen coordinate.