The Unapologetic Mathematician

Mathematics for the interested outsider

The Maximal Flow of a Vector Field

Given a smooth vector field X\in\mathfrak{X}M we know what it means for a curve c to be an integral curve of X. We even know how to find them by starting at a point p and solving differential equations as far out as we can. For every p\in M, let I_p be the maximal open interval containing 0 on which we can define the integral curve \Phi_p with \Phi_p(0)=p.

Now, I say that there is a unique open set W\subseteq\mathbb{R}\times M and a unique smooth map \Phi:W\to M such that W\cap(\mathbb{R}\times\{p\})=I_p\times\{p\} — the set W cuts out the interval I_p from the copy of \mathbb{R} at p — and further \Phi(t,p)=\phi_p(t) for all (t,p)\in W. This is called the “maximal flow” of X.

Since there is some integral curve through each point p\in M, we can see that \{0\}\times M\subseteq W. Further, it should be immediately apparent that \Phi is also a local flow. What needs to be proven is that W is open, and that \Phi is smooth.

Given a p\in M, let I\subseteq I_p be the collection of t for which there is a neighborhood of (t,p) contained in W on which \Phi is differentiable. We will show that I is nonempty, open, and closed in I_p, meaning that it must be the whole interval.

Nonemptiness is obvious, since it just means that p is contained in some local flow, which we showed last time. Openness also follows directly from the definition of I.

As for closedness, let t_0 be any point in \bar{I}, the closure of I. We know there exists some local flow \Phi':I'\times V'\to M with 0\in I' and \Phi_p(t_0)\in V'. Now pick an t_1\in I close enough to t_0 so that t_0-t_1\in I' and \Phi_p(t_1)\in V' — this is possible since t_0 is in the closure of I and \Phi_p is continuous. Then choose an interval I_0 around t_0 so that t-t_1\in I' for each t\in I_0. And finally the continuity of \Phi at (t_1,p) tells us that there is a neighborhood V of p so that \Phi(t_1\times V)\subseteq V'.

Now, \Phi is defined and differentiable on I_0\times V, showing that t_0\in I. Indeed, if t\in I_0 and q\in V, then t-t_1\in I' and \Phi(t_1,q)\in V', so \Phi'(t-t_1,\Phi(t_1,q)) is defined. The curve s\mapsto\Phi'(s-t_1,\Phi(t_1,q)) is an integral curve of X, and it equals \Phi(t_1,q) at t_1. Uniqueness tells us that \Phi(t,q)=\Phi'(t-t_1,\Phi(t_1,q)) is defined, and \Phi is thus differentiable at (t,q).


May 30, 2011 - Posted by | Differential Topology, Topology


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