# The Unapologetic Mathematician

## 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