4
$\begingroup$

Let $U$ be an open subset of $\mathbb{R}^2$. For my purposes, we can assume that $U$ is just a rectangle. I have an infinitely differentiable map $M:U\to U$ that has a unique fixed point $p$ in $U$. Furthermore, the Jacobian of $M$ at $p$ only has eigenvalues with absolute value less than $1$, so I know that $p$ is a globally attracting fixed point in some neighborhood of $p$.

I also know that the Jacobian of $M$ taken at every other point in $U$ also only has eigenvalues with absolute value less than $1$. Does this imply that the $M$-orbit of every point in $U$ must converge to $p$?

EDIT: What happens if we insist that the $U$ must be convex?

$\endgroup$
  • $\begingroup$ You will probably need an additional condition to ensure that the orbits don't run off to the boundary of $U$. Note that your condition on the Jacobian makes $M$ a contraction mapping, so the Banach fixed point theorem applies if $U$ was closed but not otherwise. Maybe you want to find some closed $U' \subset U$ containing $p$ which gets sent to itself by $M$. $\endgroup$ – Vidit Nanda Jul 3 '14 at 20:51
  • $\begingroup$ @ViditNanda: I don't understand the statement about $M$ being a contraction. After all, we only have information about the eigenvalues. Say $M(x,y)=(10y,0)$ on $U=\mathbb R^2$ (add $\epsilon$ times the identity if you want a unique fixed point). (hi Darren!) $\endgroup$ – Christian Remling Jul 3 '14 at 21:12
  • $\begingroup$ @ChristianRemling you're right of course, I read the question in a hurry. $\endgroup$ – Vidit Nanda Jul 3 '14 at 23:13
  • $\begingroup$ What you do have is $|\det(J)| < 1$ everywhere, implying that $M$ shrinks areas. Thus there is no $A \subseteq U$ of Lebesgue measure $> 0$ such that $M(A) = A$. In particular, $\bigcap_{n=1}^\infty M^n(U)$ has measure $0$. On the other hand, $M^n(U)$ is also connected. This would seem to make it difficult to avoid having just the one fixed point... $\endgroup$ – Robert Israel Jul 4 '14 at 4:38
  • $\begingroup$ @RobertIsrael: That can't be all there is to it. $M(x,y)=(x/2,y)$ shrinks areas and has lots of fixed points. $\endgroup$ – Christian Remling Jul 4 '14 at 4:44
4
$\begingroup$

No consider $U=\{(x,y)\in \mathbb{R}^2| |xy|<1/4\}$, $M(x,y)=(y^2,x^2)$ then $(0,0)$ is the only fixpoint but the orbit of $(1,0)$ does not converge to $(0,0)$.

$\endgroup$
  • $\begingroup$ This answer works! But I wonder what happens if we insist that $U$ must be convex? $\endgroup$ – Darren Ong Jul 4 '14 at 15:58
  • $\begingroup$ @DarrenOng: Darren, sorry I was a bit quick in my email. It isn't the convexity that breaks it. This answer is a bit pathological, because at $(1, 0)$, the differential of $M$ has only zero eigenvalues. I would suggest that in your question you also state that the eigenvalues are nonzero. $\endgroup$ – user39719 Jul 4 '14 at 16:34
  • $\begingroup$ You could perturb this slightly and avoid zero eigenvalues: try $M(x,y) = (y^2+\epsilon x(x-1), x^2 + \epsilon y(y-1))$. $\endgroup$ – Robert Israel Jul 4 '14 at 17:12
  • $\begingroup$ @RobertIsrael: My comment may have been misleading. I am not claiming that nonzero eigenvalues is a sufficient condition. $\endgroup$ – user39719 Jul 4 '14 at 17:31
  • $\begingroup$ The shape of $U$ really doesn't have much to do with it, as we can always start out with this $M$ near the fixed point and the periodic orbit and then extend in a pretty much arbitrary way (as long as we keep $J$ small). $\endgroup$ – Christian Remling Jul 4 '14 at 21:45

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.