On an $n$-dimensional differentiable manifold $\mathcal{M}$, select a smooth tangent vector field $v : \mathcal{M} \to T\mathcal{M}$. Let $\varphi : U \to \mathbb{R}^n$ be a chart on an open subset $U \subseteq \mathcal{M}$. A curve $\gamma : \mathbb{R} \to U$ is an integral curve of $v$ if $\ \frac{d}{dt}\big{(}\varphi \circ \gamma\big{)}(t) = v \circ \gamma(t)\ \ \forall t \in \mathbb{R}$. The intuition is that, as you follow the curve, the tangent vector to the curve where you currently are must match what $v$ specifies for where you currently are.
Denote $\gamma_x$ as the unique integral curve to $v$ with $\gamma_x(0)=x$. Then the flow $\Phi : \mathcal{M} \times \mathbb{R} \to \mathcal{M}$ of $v$ can be defined as $\Phi(x,t) = \gamma_x(t)$. The intuition is that $\Phi(x,t)$ selects the integral curve initialized at $x$, follows it for a duration $t$, and tells you where on the manifold it landed. It is essentially an assemblage of integral curves with a "selector" argument.
Because $\gamma_x$ is an integral curve, it of course satisfies $\ \frac{d}{dt}\big{(}\varphi \circ \gamma_x\big{)}(0) = v \circ \gamma_x(0) = v(x)$. This leads me to an alternative definition for the flow of $v$ on $\mathcal{M}$ as the map $\Phi : \mathcal{M} \times \mathbb{R} \to \mathcal{M}$ such that,
$$
\frac{d}{dt}\big{(}\varphi_x \circ \Phi(x,t)\big{)}\big{|}_{t=0} = v(x)\ \ \ \forall x\in\mathcal{M} \tag{1}
$$
where $\varphi_x$ is a chart on $\mathcal{M}$ with $x$ in its domain. What I like about this perspective is that there is one condition for each $x \in \mathcal{M}$, rather than one condition for each $t \in \mathbb{R}$ along every integral curve. Though not proved here, it is well known that integral curves cannot cross and thus these conditions appear one-to-one.
The usual equation $\ \frac{d}{dt}\big{(}\varphi \circ \gamma\big{)} = v \circ \gamma\ $ is a general autonomous ODE. For a Euclidean manifold with an identity chart we write $\dot{\gamma} = v(\gamma)$. I am trying to interpret Eq.1 in a similar light as $\ \dot{\Phi}(x,0) = v(x)\ $. But something is weird about that, because it seems to be trivially satisfied by $\Phi(x,t) = tv(x)$.
Does this mean that Eq.1 is somehow not a sufficient condition for $\Phi$ to be the flow? Am I just misinterpreting the equation? Or is there a deeper flaw in my reasoning above?
Best Answer
$(1)$ does not uniquely determine the map $\Phi$, as your counterexample suggests. Indeed, there are a great varienty of distinct maps $\mathcal{M}\times\mathbb{R}\to\mathcal{M}$ whose derivatives all agree on $\mathcal{M}\times\{0\}$. The missing piece of information is that the ODE determined by a fixed vector fields is autonomous, in that it satisfies $$ \Phi(x,t_1)=\Phi(y,t_2)\implies\frac{\partial\Phi}{\partial t}(x,t_1)=\frac{\partial\Phi}{\partial t}(y,t_2) $$ Put another way, the vector field $\left.\frac{\partial\Phi}{\partial t}\right|_{t=t_0}$ does not depend on $t_0$. One can show that this condition is equivalent to $\Phi$ being an $\mathbb{R}$-action, satisfying $$ \Phi(\Phi(x,t_1),t_2)=\Phi(x,t_1+t_2) $$ Flows are typically defined as maps having this property. In particular, given a vector field $v$ there is a unique flow satisfying $(1)$.
To describe a greater variety of maps $\mathcal{M}\times\mathbb{R}\to\mathcal{M}$, one often generalizes to time-dependent flows which need not be autonomous, and are generated by time-dependent vector fields. However, as the nomenclature suggests, to specify such a map one must specify its derivatives for all $t\in\mathbb{R}$.