Consider a map $\Phi: [0,T] \times \mathbb{R}^d \to \mathbb{R}^d$, and assume that there exists a set $U \subset \mathbb{R}^d$ such that $\Phi\rvert_{[0,T] \times U}$ is $L$-Lipschitz. It is well known that, in general, there exists a $L$-Lipschitz extension $\tilde{\Phi}$ to the whole space with the same Lipschitz constant, this is the Kirszbraum theorem. What I wonder is the following: assume $\Phi$ is not a general map, but the flow of some regular vector field $b$, and let's say that the above condition on $\operatorname{Lip}(\Phi\rvert_{[0,T] \times U})$ holds. The question is: is there an extension $\tilde{\Phi}$ such that $\tilde{\Phi}|_{[0,T] \times U} = \Phi_{[0,T] \times U}$ and $\tilde{\Phi}$ is still the flow of some other velocity field $\tilde{b}$? Or is there a counterexample/simple reason why this cannot be in general? I'd be ok with less (I guess it's less, maybe it's not). In fact, what I'd need is an extension $\tilde{\Phi}$ such that it admits an inverse $\tilde{\Phi} ^{-1}$ and $\tilde{\Phi}$ has the semigroup property, i.e. $\tilde{\Phi} (s+ t , x)= \tilde{\Phi}(s, \tilde{\Phi} (t,x))$.
P.S. I would be happy even if the Lipshitz constant does not stay the same but can be controlled by some multiple of $L$ (but not $\exp(L)$, this would be too much), something like what happens if instead of the Kirszbraum extension you consider a McShane extension componentwise.
EDIT: I think in general this is false. Consider the situation where instead of $\mathbb{R}^d$ you have the ball $B_1$ (let’s say that the velocity which gives the flow is 0 at the boundary). Then assume that $U=B_1 \setminus B_\varepsilon$, and that the flow acts by squeezing the set $U$ in, say, $B_1 \setminus B_{1/2}$. Since I still want a flow, it should be invertible, thus the image of $B_\varepsilon$ should be $B_{1/2}$ and I cannot have a uniform bound on the Lipschitz constant of the new flow. Maybe something better can be said if $div(b)\in L^\infty$, but I think an example like this can still be constructed with a little more work.
