Differential equation and vector field flow In Marsden "Foundations of Mechanics" it is stated that given a vector field $X(x)$ with flow $F_t(x)$, $x \in R^n$ being the "spatial" coordinates, then $f(x,t)=g(F_t(x))$ is a solution of the following:
$\frac{\partial f}{\partial t}(x,t)=\nabla f(x,t)•X(x), f(x,0)=g(x)$
However, by direct derivation I can only arrive at $\frac{\partial f}{\partial t}(x,t)=\nabla f(F_t(x),0)•X(x)$, i.e., the point of evaluation of the gradient of $f$ is not $(x,t)$ but $(F_t(x),0)$
I'm sure I'm overlooking a stupid detail but I'm not able to find it.
How can this apparent easy statement be probed? (it is straightforward using Lie derivative but I wanted to check it by direct substitution)
 A: Compute the LHS and RHS separately. The LHS is
\begin{equation}
\frac{\partial f}{\partial t}(x,t)= \frac{\partial}{\partial t} (g(F_t(x)))\overset{\text{chain rule}}{=} Dg(F_t(x))\frac{d}{dt}F_t(x) = Dg(F_t(x)) X(F_t(x));
\end{equation}
while on the RHS,
\begin{equation}
\partial_x f(x,t)X(x) =  \partial_x (g(F_t(x))) X(x) = Dg(F_t(x)) DF_t(x) X(x).
\end{equation}
So we need to show $DF_t(x) X(x) = X(F_t(x))$. To that end, note
\begin{equation}
F_t(F_s(x))=F_{t+s}(x)=F_s(F_t(x)).
\end{equation}
So
\begin{equation}
\frac{d}{ds}F_t(F_s(x))=\frac{d}{ds}F_s(F_t(x)).
\end{equation}
The LHS is
\begin{equation}
\frac{d}{ds}F_t(F_s(x))=DF_t(F_s(x))X(F_s(x));
\end{equation}
while the RHS is
\begin{equation}
\frac{d}{ds}F_s(F_t(x))=X(F_s(F_t(x)));
\end{equation}
Evaluation at $s=0$ then shows, $DF_t(x)X(x) = X(F_t(x))$ as needed.$\square$
Remark: Background info normally kept secret: I "cheated" and worked backwards, unravelling the proof of Abraham Marsden Ratiu - Manifolds, Tensor Analysis and Applications prop 4.2.11. My notes of that proof:
\begin{equation}
\frac{\partial f}{\partial t} = \frac{d}{dt}F^*_t g
\overset{\text{prop 4.2.10}}{=}F^*_t\mathcal{L}_X g
\overset{\text{props 4.2.7(i),4.2.4}}{=}\mathcal{L}_X(F^*_t g) = X[f].
\end{equation}
$DF_t(x)X(x) = X(F_t(x))$ is $\phi$-relatedness in local coordinates, $X\sim_\phi X$, where $\phi = F_t$, while $\phi\circ F_s = F_s\circ\phi$ is
$F_t(F_s(x))=F_{t+s}(x)=F_s(F_t(x)),$ essentially the last couple of sentences of prop 4.2.4, reproduced below for convenience if you don't have the book.
AMR-Man Tensor A App- Prop 4.2.4:
Let $\phi : M → N$ be a $C^r$-mapping of manifolds, $X \in \mathfrak{X}^r(M)$ and $Y\in\mathfrak{X}^r(N)$. Let $F^X_t$ and
$F^Y_t$ denote the flows of $X$ and $Y$ respectively. Then $X∼_ϕ Y$ iff $\phi\circ F^X_t = F^Y_t\circ \phi$. In particular,
if $\phi$ is a diffeomorphism, then the equality $Y = \phi_∗ X$ holds iff the flow of $Y$ is $\phi\circ F^X_t\circ\phi^{−1}$. In particular,$(F^X_s)_∗X = X$.
