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Consider a discrete dynamical system $x_{k+1} = f(x_k)$, where $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$, sufficiently smooth, and let $C \subseteq \mathbb{R}^2$ be an invariant, closed curve in the phase space.

By Jordan's theorem, $C$ gives rise to two more connected regions, the interior $C_{\text{int}}$ and the exterior $C_\text{ext}$ of $C$.

If the system were continuous, then $C_{\text{int}}$ and $C_\text{ext}$ would be invariant regions, as the phase spaces of a continuous system can be shown to be a disjoint union of the orbits of the system - which means, that in particular the orbits cannot cross $C$.

But if the system is, as stated, discrete, this argument does not apply anymore, as it would be a priori conceivable that between two discrete time steps a jump across $C$ occurs (though perhaps some really simple argument that I just fail to see prevents this). Is there any other way to show that in the discrete case $C_\text{int}$ and $C_\text{ext}$ are invariant? Or does perhaps a counterexample exist?

temo
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    Your point is valid. For instance, this may happen when we discretize a nonlinear continuous-time system using a simple Euler scheme. However, if you consider the continuous-time system $\dot{x}=f(x)$, the set of equilibrium point is defined by $f(x)=0$, which would correspond to $C_{\mathrm{int}}$. On the other hand, the set of equilibrium points of the same function but considered in discrete-time is described by $x=f(x)$, which is different from $C_{\mathrm{int}}$. So, it would be important to clarify the context of this question. – KBS May 06 '22 at 00:49
  • @KBS Why would $f(x)=0$ correspond to all of $C_\text{int}$ in the continuous case? I think there are various examples where a limit cycle arises that has a single stable point inside it, which would invalidate this claim. Yes, in discrete time the stable points are given by solving $f(x)=x$, but I don't see how that relates to $C_\text{int}$ in any immediate way. – temo May 09 '22 at 23:30

2 Answers2

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This is a counter-example over the punctured plane.

Counter-example. [$f$ smooth over $\mathbb{R}^2\setminus \left\{0\right\}$] Consider the family of maps or discrete-time dynamical systems (indexed by $\alpha>0$) $f_{\alpha}\,:\mathbb{R}^{2}\setminus \left\{0\right\}\longrightarrow \mathbb{R}^2$ given by $f_{\alpha}(x,y)=\left(\frac{x}{\|(x,y)\|^{\alpha}},\frac{y}{\|(x,y)\|^{\alpha}}\right)$, where $\|\cdot \|$ is the Euclidean norm. Remark that $\|f_{\alpha}(x,y)\|=\|(x,y)\|^{1-\alpha}$. The unit circle $S$ is invariant w.r.t. any map $f_{\alpha}$ in the family. If $\alpha>1$, the dynamical system $f_{\alpha}$ contracts: if $(x_k,y_k)$ lies outside the unit circle (i.e., $\|(x_k,y_k)\|>1$), then $\|f_{\alpha}(x_k,y_k)\|=\|(x_k,y_k)\|^{1-\alpha}<1$, i.e., $f_{\alpha}(x_k,y_k)$ lies inside the unit circle. In other words, $S_{{\sf ext}}$ is not invariant w.r.t. any $f_{\alpha}$ with $\alpha>1$. On the other hand, if $\alpha<1$, the dynamical system $f_{\alpha}$ expands and $S_{{\sf int}}$ is not invariant w.r.t. any $f_{\alpha}$ with $\alpha<1$.

  • What is $f_{\alpha}(0,0)$? – Jacob Manaker May 06 '22 at 18:37
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    If we insist in having $f_{\alpha}$ smooth over the whole $\mathbb{R}^{2}$, we can simply redefine $f_{\alpha}$ to be constant over some (small) ball around the origin and there is no problem in terms of the dynamics (especially if $\alpha<1$) as it expands, i.e., it will get out of the interior. I will edit to reflect that. Thank you. – Augusto Santos May 06 '22 at 18:46
  • I made a quick hacky numerical implementation of your example. I must say your example is actually pretty interesting, as $C$ turns out to be attracting and the iterates jump around $C$, converging to it (so while you still gave a counterexample to my claim, it seems that what you had in mind, i.e. that $S_\text{int}$ being invariant for $\alpha >0$, does not hold either). [...] – temo May 10 '22 at 00:55
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    Here is how it looks like for $\alpha = 1.6$ and starting value $(40,40)$ and 200 iterations (the starting value itself is no displayed, only the next value, which almost lies at the origin of the circle, in the next iteration, because the numerator is so small, it shoots outside again, to the upper right corner, then is inside again and so on): https://ibb.co/1KJvHfM – temo May 10 '22 at 00:55
  • That is pretty interesting! Regarding having in mind that $S_{{\sf int}}$ was invariant for $\alpha>1$ or $S_{{\sf ext}}$ was invariant for $\alpha<1$: I thought about it at the time and was not certain, I would be just not surprised of the possibility (that's why I referred to these regimes quite wrongly as 'contracting' and 'expanding'). I was not certain that the unit circle was a (possibly global) attractor either (just that the origin could not be). It is quite interesting that the system accumulates onto C by peeping in and out of it. Thank you for sharing this numerical result! – Augusto Santos May 10 '22 at 05:27
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You state that $f$ is sufficiently smooth. However, the essential topological property, which you do not address, is whether or not the function $f$ is injective. This makes all the difference.

The time $t$ flow of a vectorfield (without finite time blowup) is a diffeomorphism. In particular it is injective and this prevent flowlines from crossing, whence the phenomena you mention for that case. In the case of a continuous map $f$, if the image of the interior contains points in the interior as well as in the exterior then being continuous it also contains points on the separating closed curve $C$ so it cannot be injective (Jordan Curve Theorem). In this case you can say very little about invariant sets.

Instead, let us assume that $f$ is injective and continuous and maps the Jordan curve $C$ surjectively onto itself. Then suddenly you can say a lot about invariant domains. I claim that each of $C_{\rm int}$ and $C_{\rm ext}$ are both invariant domains and moreover, $C_{\rm int}$ must be completely invariant, i.e. $f(C_{\rm int})=C_{\rm int}$. Note that we only assumed injectivity and continuity of $f$.

Proof: Invariance of Domain implies that $f$ is an open map of the Euclidean plane, i.e. a homeomorphism onto its image $I=f({\Bbb R}^2)$. This implies that $f$ maps a simply connected domain in the plane onto a simply connected domain. $D=C\cup C_{\rm int}$ is simply connected. Thus, any closed curve $\gamma$ may be continuously deformed within $D$ into a point. Thus, $f(D)$ is also contractible and this is not possible if $f$ maps $C_{\rm int}$ into $C_{\rm ext}$ and preserves $C$ (since $C$ would not be contractible within the image). Thus $f(D)\subset D$. And since $f(D)$ is simply connected and contains $C$ it can not omit any points in $D$, so we must have $f(D)=D$. Then $f$ must also map the exterior into itself, although not necessarily surjectively.

The above is related to the topology of the plane. On e.g. the punctured plane or on the sphere you may have diffeos exchanging interior and exterior.

H. H. Rugh
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  • Indeed: i) in continuous time the smoothness of the vector field engenders flows that are diffeomorphisms (as maps from the initial conditions to time $t$ state) and that's the critical attribute for uniqueness; whereas ii) in discrete-time the diffeomorphism property should be imposed in the discrete-time flow to preserve the invariance of $C_{{\sf int}}$. That is the main dichotomy. Quite interesting. – Augusto Santos May 09 '22 at 17:09
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    In fact, topology and continuity are the driving forces here. Being differentiable is not so important although in practice you often use differentiability and the lack of critical points to show injectivity locally. – H. H. Rugh May 09 '22 at 17:28
  • Great answer, thanks. But I have some unclarities: "If the image of the interior contains points in the interior as well as in the exterior then be continuity it also contains points on the separating closed curve so it cannot be injective" Could you prove this more formally? It seems that this contradicts Augusto Santo's example, that is injective and continuous, yet crossing $C$ is still possible. – temo May 09 '22 at 23:34
  • Also, is the fact that $f$ maps $C$ onto itself surjectively really necessary? Would it suffices to simply assume $f$ maps $C$ onto itself (intuitively I would assume that that means there is some more complicated flow on $C$, perhaps including stable point, but I'm not sure) – temo May 09 '22 at 23:36
  • Indeed, it is injective when $\alpha\neq 1$ (I missed that). It would be great if Hans Rugh could have a look at the counter-example in my answer. Thank you. – Augusto Santos May 09 '22 at 23:57
  • I have added some clarification. I am not sure what your problem is with the first claim, i.e. that if the image of the interior contains points in the interior and in the exterior then must also contain points on the boundary so it can not be injective. This is just the Jordan curve theorem. To @AugustoSantos: The technical remark you make is actually a topological obstruction. You can not find a continuous map that maps the unit disk to the exterior and preserves the boundary (try it out in details). So you counter-example does not work on the non punctured plane. – H. H. Rugh May 10 '22 at 06:16
  • I see... there must lie a technical problem with the technical remark then. The counter-example only works over the punctured plane. I have edited my answer to reflect that. Thank you. – Augusto Santos May 10 '22 at 06:24
  • I resolved the issue regarding Augusto's example, but the following point regarding your first claim I'm still unsure about: Consider the following explicit proof of your first claim: Assume $x,y$ are in the interior but that $f(x)$ is in the interior, while $f(y)$ is in the exterior. Since the interior is path connected by the Jordan curve theorem, the image of the interior will, by continuity, also be path connected. Thus a path $P$ between $x,y$ will be mapped to a path between $f(x),f(y)$. [...] – temo May 10 '22 at 21:11
  • [...] This path has to intersect $C$ at some point $z$, which is the image of some point on $P$, that lies in the interior. Now you claim that $f$ therefore cannot be injective, but I can't see why that is true; why should also a point on $C$, different form $z$, exist that is mapped to $z$? Couldn't $f$ restricted to $C$ be the identity? Then one wouldn't obtain a contradiction. – temo May 10 '22 at 21:16
  • When $f$ is globally injective and $f(C)=C$, then every point in $C$ is already the image of some (possibly other) point in $C$. So no points in the complement of $C$ should map to $C$. Does this settle your question? – H. H. Rugh May 10 '22 at 21:55