# integrability and area-preservation property of dynamical systems

Suppose we have a map defined on a plane, $x_{1}=f(x_{0})$, where $x \in \mathbb{R}^{2}$. Assume it is integable: there exists a function $I$ of the phase space variable $x$ such that $I(x)=I(f(x))$. I do not assume global integrability in the sense of Liouville-Arnold - just that the phase space is foliated by invariant curves of the type $I(x)=c$, where $c$ is constant.

Must this map be necessarily area-preserving?

I do not know either how to prove this, or come up with a counter example. Similarly, for maps defined for $\mathbb{R}^{n}$...

• Comment to the question (v1): The condition $I(x)=I(f(x))$ is trivially satisfied for any $f$ if $I$ is a constant function. – Qmechanic Oct 31 '15 at 0:59
• Of course, but my question concerns the existence of a nontrivial integral @Qmechanic – Alex Oct 31 '15 at 17:00
• Here's another counterexample. Let me identify the plane $\mathbb{R}^{2}\cong \mathbb{C}$. Assume $f:\mathbb{C}\to \mathbb{C}$ is of the form $f(z)=f_0(z)+i{\rm Im}(z)$, where $f_0:\mathbb{C}\to \mathbb{R}$ is a real-valued function chosen such that $f$ is not area-preserving, e.g. $f_0(z)=2{\rm Re}(z)$. Let $I:\mathbb{C}\to \mathbb{R}$ be the imaginary part $I(z):={\rm Im}(z)=I(f(z))$. – Qmechanic Oct 31 '15 at 19:00
• It looks to me that your example can just be easily given in $\mathbb{R}^{2}$, without identification with $\mathbb{C}$, if we put $x_{1}=2x_{0},\quad y_{1}=y_{0}$, right? But then the variables are decoupled and we essentially have two 1-dimensional maps with trivial dynamics... @Qmechanic – Alex Nov 2 '15 at 0:37
• $\uparrow$ Yes. – Qmechanic Nov 2 '15 at 0:40

How about, e.g., the Hénon map with $a=1.25$ and $b=0.3$? It has an attractor, so it's not area-preserving, and this attractor is periodic, so the system is integrable.
Unless our definitions don't match, any integral map in $\mathbb{R}^n$ that has an attractor is a counter-example to integrability and area-preservation being linked.