Timeline for Is a "global period" similar to a "local period"?

6 events

when toggle format	what		by	license	comment
Jun 16, 2021 at 14:13	history	edited	G. Panel	CC BY-SA 4.0	edited title
Jun 15, 2021 at 13:40	comment	added	G. Panel		For $\omega>0$, the planar vector field $x'=-\omega y$ and $y'=\omega x$ has for solution $(x,y)(t)=\mathcal{R}_{\omega t}(x_0,y_0)$ where $\mathcal{R}_{\theta}$ is the rotation matrix with angle $\theta$. So if $(x_0,y_0)\neq (0,0)$, then $T(x_0,y_0)=2\pi/\omega$, so $T$ has a continuous extension such that $T(0,0)=2\pi/\omega$. This does not depend on the modulus of the eigenvalues of $A_{(0,0)}$ (which is $\omega$).
Jun 15, 2021 at 1:29	comment	added	rpotrie		What do you mean? Typically (for instance, if $A$ has some eigenvalue with modulus >1) the period must go to infinity as the point approaches $E$, do you want the function $1/T$ rather than $T$? Still, for the linear vector.field in the plane $x'=-y$ and $y'=x$ this does not hold.
Jun 14, 2021 at 23:00	history	edited	G. Panel	CC BY-SA 4.0	deleted 37 characters in body
Jun 11, 2021 at 22:35	history	edited	G. Panel	CC BY-SA 4.0	added 10 characters in body
Jun 11, 2021 at 22:18	history	asked	G. Panel	CC BY-SA 4.0