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](https://mathoverflow.net/questions/327781/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?
> 
> 
> 



Difference quotient for solutions of ODE and Liouville equation
*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 $\partial\_t {\tilde\mu}\_t + \operatorname{div\,} ({\tilde f} {\tilde\mu\_t) }= 0$


