Why infinitesimal generator of the autonomous differential equation $\frac{dx(t)}{dt}=\textbf{b}(x(t))$ is gradient operator?

Question

Let $\mathscr{X}=C(\mathbb{R}^n,\mathbb{R}), b\in C^{1}(\mathbb{R}^n,\mathbb{R}^n)$, consider $$\frac{dx(t)}{dt}=\textbf{b}(x(t)),$$ where $x(0)=\xi$ is a autonomous differential equation.

For $\xi \in \mathbb{R}^n $, $\exists x(t,\xi)$, $0\leq t<\infty$, s.t. $x(t)\in C^{1}(\mathbb{R}^1,\mathbb{R}^n)$. Define operator $$T(t): f(\xi) \rightarrow f(x(t,\xi)).$$

My questions:

1. what does $f$ mean? I try to show $T(t)T(s)=T(s+t)$, but $(T(s+t)f)(\xi)=f(x(s+t,\xi))=?$

2. How to prove: if $A$ is the infinitesimal generator, then $C_{c}^{1}(\mathbb{R}^n,\mathbb{R}^1)\subset D(A).$ And as $f \in C_{c}^{1}(\mathbb{R}^n,\mathbb{R}^1)$, $$ (Af)(x)=\sum_{i=1}^n b^{i}(x)\frac{\partial f(x)}{\partial x_{i}},$$ where $\textbf{b}(x)=(b^{1}(x),..,b^{n}(x))$.

Answer 1

By the chain rule you get $$\frac{df(x(t))}{dt}=f'(x(t))\dot x(t)=f'(x(t))b(x(t)),$$ and if you transpose that, you get $$\frac{df(x(t))}{dt}=b(x(t))^Tf'(x(t))^T=b(x(t))^T\nabla f(x(t)).$$

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By the way, your notation does not make sense. $f(ξ)$ is a function value, a point, or a real number in this case. What you can to is define $$ T(t):C^1(R^n,R)\to C^1(R^n,R),~~ [T(t)(f)](ξ)=f(x(t;ξ)) $$

Answer 2

Because I am too lazy to look in the literature for a proof of the group property, and because it is, however, much too long for a comment, I am giving it here as an answer.

Assume that the $C^1$ vector field $\mathbf{b}$ is forward complete: For every $\xi \in \mathbb{R}^n$ the domain of the (necessarily) unique nonextendible solution of the IVP $$ \begin{cases} \displaystyle \frac{dx(t)}{dt} = \mathbf{b}(x(t)) \\[0.5em] x(0) = \xi \end{cases} $$ contains $[0, \infty)$. Denote such a solution by $\varphi(\cdot, \xi)$.

We want to show that $$ \tag{$*$} \varphi(t+s,\xi) = \varphi(t, \varphi(s, \xi)) \quad \text{ for all }s, t \ge 0, \ \xi \in \mathbb{R}^n. $$ Fix $s \ge 0$ and $\xi \in \mathbb{R}^n$. Consider the mapping $$ [0, \infty) \ni t \mapsto \varphi(t+s,\xi) \in \mathbb{R}^n, $$ as well as the mapping $$ [0, \infty) \ni t \mapsto \varphi(t, \varphi(s, \xi)) \in \mathbb{R}^n. $$ They both satisfy the IVP $$ \begin{cases} \displaystyle \frac{dx(t)}{dt} = \mathbf{b}(x(t)) \\[0.5em] x(0) = \varphi(s, \xi). \end{cases} $$ By uniqueness, they must be the same mapping, which concludes the proof of $(*)$.

Now, for any (even not necessarily continuous) function $f \colon \mathbb{R}^n \to \mathbb{R}^1$ the property $$ T(t) (T(s) f) = T(t + s) f \quad \text{ for all }s, t \ge 0 $$ follows directly from $(*)$.