Given a smooth vector field we know what it means for a curve to be an integral curve of . We even know how to find them by starting at a point and solving differential equations as far out as we can. For every , let be the maximal open interval containing on which we can define the integral curve with .
Now, I say that there is a unique open set and a unique smooth map such that — the set cuts out the interval from the copy of at — and further for all . This is called the “maximal flow” of .
Since there is some integral curve through each point , we can see that . Further, it should be immediately apparent that is also a local flow. What needs to be proven is that is open, and that is smooth.
Given a , let be the collection of for which there is a neighborhood of contained in on which is differentiable. We will show that is nonempty, open, and closed in , meaning that it must be the whole interval.
Nonemptiness is obvious, since it just means that is contained in some local flow, which we showed last time. Openness also follows directly from the definition of .
As for closedness, let be any point in , the closure of . We know there exists some local flow with and . Now pick an close enough to so that and — this is possible since is in the closure of and is continuous. Then choose an interval around so that for each . And finally the continuity of at tells us that there is a neighborhood of so that .
Now, is defined and differentiable on , showing that . Indeed, if and , then and , so is defined. The curve is an integral curve of , and it equals at . Uniqueness tells us that is defined, and is thus differentiable at .