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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?

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  • $\begingroup$ 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
    Commented Jul 3, 2014 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
    Commented Jul 3, 2014 at 21:12
  • $\begingroup$ @ChristianRemling you're right of course, I read the question in a hurry. $\endgroup$
    – Vidit Nanda
    Commented Jul 3, 2014 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
    Commented Jul 4, 2014 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
    Commented Jul 4, 2014 at 4:44

1 Answer 1

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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 $(0,0)$.

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  • $\begingroup$ This answer works! But I wonder what happens if we insist that $U$ must be convex? $\endgroup$
    – Darren Ong
    Commented Jul 4, 2014 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
    Commented Jul 4, 2014 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
    Commented Jul 4, 2014 at 17:12
  • $\begingroup$ @RobertIsrael: My comment may have been misleading. I am not claiming that nonzero eigenvalues is a sufficient condition. $\endgroup$
    – user39719
    Commented Jul 4, 2014 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
    Commented Jul 4, 2014 at 21:45

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