Difference quotient for solutions of ODE and Liouville equation

Question

Suppose that $\Phi$ is the solution of $$\begin{cases} \frac{d}{dt}\Phi(x,t) = f(\Phi(x,t),t) \quad t >0 \\ \Phi(x,0) = x \quad x \in \mathbb{R}^N \end{cases}$$

How does one prove that $$\tilde \Phi(x,y,t) = \left(\Phi(x,t), \frac{\Phi(x + r y,t) - \Phi(x,t)}{r} \right)$$ is the flow of the ODE with $$\tilde{f}_r(x,y,t) = \left(f(x,t), \frac{f(x+r y,t) - f(x,t)}{r} \right)$$ as a vector field?

Also, in an answer to Prove that the flow of a divergence-free vector field is measure preserving, it was proved that if $\mu_t = (\Phi(\cdot,t))_{\sharp} \mu$ denote the image of the measure $\mu$ by the flow of $f$, then the family of measures $\{\mu_t\}_{t\in \mathbb R}$ satisfies Liouville equation $$ \begin{cases} \partial_t \mu_t + \operatorname{div\,} (f \mu_t) = 0 \\ \mu_0 = \mu \end{cases} $$ in the sense of distributions.

What PDE does $\tilde\mu_t = (\tilde\Phi_t)_{\sharp} \mu$ solve?

Answer 1

Question 1: Denoting $U:= \Phi(x,t)$ and $\displaystyle V:={\Phi(x + r y,t) - \Phi(x,t)\over r}$ the components of $\tilde \Phi(x,y,t)$ , we have $U+rV=\Phi(x + r y,t) $, and $$\displaystyle\frac{f(U+r V,t) - f(U,t)}{r} = \frac{f(\Phi(x + r y,t),t) - f(\Phi(x,t),t)}{r} .$$ So

$$\partial_t\tilde \Phi(x,y,t)= $$$$= \left(\partial_t \Phi(x,t) , \frac{\partial_t\Phi(x + r y,t) - \partial_t\Phi(x,t) }{r} \right)$$ $$= \left(f(\Phi(x,t),t), \frac{f(\Phi(x + r y,t),t) - f(\Phi(x,t),t)}{r} \right)$$ $$= \left(f(U,t), \frac{f(U+r V,t) - f(U,t)}{r} \right)$$$$=\tilde{f}_r(U,V,t) $$ $$=\tilde{f}_r(\tilde \Phi(x,y,t),t).$$

Question 2: as to $\tilde\mu_t$, the same result of the quoted link apply in particular to the flow $\tilde\Phi$, therefore you still have the Liouville equation $\partial_t {\tilde\mu}_t + \operatorname{div\,} ({\tilde f} {\tilde\mu_t) }= 0$