1
$\begingroup$

I am trying to formalize the following proof on Perko's Differential Equations and Dynamical Systems, which says that a periodic orbit has index +1.

enter image description here

My only problem is trying to prove that the map $g$ is continuous on $T$. Geometrically it seems obvious, and I am trying to prove it using limits, but I do not get anything. The following figure (also from Perso's book) is a nice illustration:

enter image description here

$\endgroup$
0
$\begingroup$

All follows from the definition of the vector field $\bf u$. For example, on the line $s=t$ you have $$ \lim_{t\to s^+}\frac{x(t)-x(s)}{\|x(t)-x(s)\|}=\lim_{t\to s^+}\frac{\frac{x(t)-x(s)}{t-s}}{\left\|\frac{x(t)-x(s)}{t-s}\right\|}=\frac{x'(s)}{\|x'(s)\|}={\bf u}(x(s)). $$

The proof and figures are copied from Coddington and Levinson's book, really the best book there is on this particular topic.

$\endgroup$
  • $\begingroup$ If I am not mistaken, you are only approaching the line $s=t$ from "above", but I think we need to compute that limit aproaching from everywhere (for example on a diagonal) $\endgroup$ – user203327 Mar 23 '16 at 20:53
  • $\begingroup$ Of course, but on the diagonal it is trivial. $\endgroup$ – John B Mar 23 '16 at 20:57
  • $\begingroup$ I mean on something like, on the point $(1/2,1/2)$, the line $y=-1/4+3/2*x$ aproaching $(1/2,1/2)$ from the right $\endgroup$ – user203327 Mar 23 '16 at 21:09
  • $\begingroup$ That is a very particular case of my answer. $\endgroup$ – John B Mar 23 '16 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.