Is time one flow in R^2 of vector field rln(r)d/dr a C^1 mapping of R^2? Given vector field in $\mathbb{R}^2$ defined by $rln(r)d/dr$ is it flow defining $C^1$ automorphism of $\mathbb{R}^2$? By $r$ we denote the radial coordinate on $\mathbb{R}^2$.
 A: The particles move radially with a speed given by
$$\dot r=r\>\log r\ .$$
It follows that $\dot r/ r=\log r$ or $(\log r)^\cdot=\log r$. Therefore
$\log\bigl(r(t)\bigr)=C\>e^t$ for some real $C$, which implies
$\log r(t)=\log r(0)\> e^t$, whence
$$r(1)=\bigl(r(0)\bigr)^e\ .$$
Note that the function $z\mapsto z^e$ maps ${\mathbb R}_{\geq0}$ bijectively onto itself.
It follows that the time-$1$-map $\Phi:\ (x,y)\mapsto (u,v)$ can be written as
$$u(x,y)=x(x^2+y^2)^{(e-1)/2},\quad v(x,y)=y(x^2+y^2)^{(e-1)/2}\ .$$
It is easy to see that $\Phi:\ {\mathbb R}^2\to{\mathbb R}^2$ is bijective and that $\Phi$ is differentiable even at the origin. Since $d\Phi(0,0)=0$  the map $\Phi$ fails to be a diffeomorphism, however.
A: Note that the vector field, I assume, is defined to be $0$ at the origin and, as such, is continuous (not differentiable). I also assume you mean the flow for fixed time $t$ when you ask if you get a diffeomorphism. And, yes, for every $t\ge 0$ I believe you do get a $C^1$ map, because we're looking at the flow $f(r) = r^{e^t}$. However, for $t>0$ the derivative vanishes at $r=0$.
