There is a better approach in @lee2013smooth page 314.
(I have to fix the notation)
What follows is from @warner prop 1.48:
Proposition
Let $X$ be a vector field on a manifold $M$. For each $p \in M$, there exist $a(p),b(p) \in \mathbb{R} \cup \{\pm \infty\}$ and a differentiable curve
$$ \gamma_p: (a(p), b(p)) \longrightarrow M $$such that:
1. $0 \in (a(p), b(p))$ and $\gamma_p(0)=p$
2. $\gamma_p$ is an integral curve of $X$ (i.e. $d(\gamma_p)_s(\frac{d}{dt})=X(\gamma_p(s))$)
3. Uniqueness: If $\mu:(c,d) \longrightarrow M$ is another curve satisfying the above conditions, then $(c,d) \subseteq (a(p), b(p))$ and $\gamma_p|_{(c,d)}=\mu$.
4. For each $t\in \mathbb{R}$, we define a transformation $X_t$ with domain $D_t=\{q \in M: t \in (a(q),b(q))\}$ by the expression $X_t(q)=\gamma_q(t)$, so that for each $p\in M$, there exists an open neighborhood $V$ and an $\epsilon >0$ such that the map
$$ (t,q)\longrightarrow X_t(q) $$is defined on $(\epsilon, \epsilon)\times V$. This transformation is called the flow of $X$ and is usually denoted by $\phi_X(t,q).$
5. $D_t$ is open for each $t$.
6. $\bigcup_{t>0} D_t =M$.
7. $X_t: D_t\longrightarrow D_{-t}$ is a diffeomorphism with inverse $X_{-t}$.
8. Let $s,t \in \mathbb{R}$. The domain of $X_s \circ X_t$ is contained in $D_{t+s}$ (but in general, it is not equal to $D_{t+s}$). It is equal to $D_{t+s}$ if $s$ and $t$ have the same sign. Moreover, in the domain of $X_s \circ X_t$, we have $X_s \circ X_t=X_{t+s}$. Therefore, we have a local group of transformations.
See flow of the sum of two vector fields
Proposition
Let $X$ be a vector field, and $k$ a constant. Let's denote $\phi_X$ and $\phi_{kX}$the flow of $X$ and $kX$ respectively. It is satisfied that:
$$ \phi_{kX}(t,q)=\phi_{X}(kt,q). $$Proof
We only have to check that $\hat{\phi}(t,q)=\phi_{X}(kt,q)$ satisfies 1. and 2. for the vector field $kX$, and then we will have the result. In fact:
1. $\hat{\phi}(0,q)=\phi_{X}(0,q)=q$.
2. $\hat{\phi}(t,q)$ are integral curves of $kX$:
$$ \frac{d}{dt}\hat{\phi}(t,q)=\frac{d}{dt}\phi_{X}(kt,q)=k X_{\phi_{X}(kt,q)} $$$\blacksquare$
Proposition
Let $X$ be a vector field and $\phi_t(p)$ its flow. For every $t$ the map $\phi_t(-)$ is a diffeomorphism, so we can consider its Jacobian
$$ J(t,x):=det(D_x \phi_t(x)). $$It is satisfied that
$$ \frac{d}{dt}|_{t=0} J(t,x)=\mbox{div}X (x) $$where $\mbox{div}$ is the divergence.
Moreover
$$ \frac{d}{dt} J(t,x)=J(t,x) \mbox{div}X (\phi_t(x)). $$Proof
(See this answer in Mathstackexchange)
It is satisfied that
$$ \frac{d}{dt} \phi_t(x)=X(\phi_t(x)) $$and then, taking derivatives with respect to the coordinates $x_1,\ldots,x_n$ we obtain the matrix equation
$$ \frac{d}{dt} d(\phi_t)_x=dX_{\phi_t(x)} d(\phi_t)_x. $$We can consider the linear system given by
$$ \dot{y}=dX_{\phi_t(x)} y,\quad y: \mathbb R\to \mathbb R^n, $$and then $d(\phi_t)_x$ is a matrix whose columns are solutions to it. We now apply Liouville's formula to this linear system, obtaining
$$ J(t,x)=J(t_0,x)e^{\int_{t_0}^t \mbox{tr}dX_{\phi_s(x)}ds} $$and deriving with respect to $t$:
$$ \frac{d}{dt} J(t,x)=J(t,x) \mbox{div}X (\phi_t(x)). $$$\blacksquare$
See xournal_155 page 5 for relation with Jacobi last multipliers.
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Author of the notes: Antonio J. Pan-Collantes
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