Consider $$ \dot x=0,\qquad \dot y =1 \qquad (\star) $$
Rectify the induced direction field is to find a diffeomorphism $(\tau,u,v) =\phi(t,x,y)$ such that $\frac{d u}{d \tau}=0=\frac{d v}{d\tau}$.
- Why does $\phi$ modify $t$ in $\tau$? I can't find an example where we have $\tau\neq t$.
In the Polar coordinates (where $x$ is the radius and $y$ is the polar angle), this differential equation defines a direction field whose line through the point $(x,y)$ has orthogonal vector $(x,y)$.
In the Cartesian coordinates, this differential equation defines a direction field whose lines are vertical at every point.
Here take $\tau=t,u=x,v=v-t$ and we have $\dot u=0=\dot v$.
This rectified field is parallel to the t axis for both coordinate system.
So we see that $\phi$ (so the rectification) doesn't depend on the coordinate systems.
But we won't have the same result if $\tau \neq t$, in that case, the rectified direction field won't be the same with different coordinate systems ?
Why a vector field is not rectifiable near a periodic orbit ?