Having trouble understanding the concept of "mixing" in dynamical systems.

467 Views Asked by At

I'm trying to understand the concept of mixing in dynamical systems theory, especially when the system in question has a measure-preserving flow. Here's how the condition is expressed mathematically: If $\mu$ is the measure and $\phi$ is the flow, then for all subsets $A$ and $B$ of positive measure, $\lim_{t \rightarrow \infty}\mu(\phi^{t}(B) \cap A) = \mu(B) \times \mu(A)$.

Now suppose $B$ is an arbitrary set with measure greater than 0 and less than 1. If the flow is measure preserving, then for all $t$, $\mu(\phi^{t}(B)) = \mu(B)$. Pick $A = \lim_{t \rightarrow \infty}\phi^{t}(B)$. Then, $\mu(A) = \mu(B)$. It follows that $\lim_{t \rightarrow \infty}\mu(\phi^{t}(B) \cap A) = \mu(A) = \mu(B)$.

So if the dynamics is mixing, then we will have $\mu(B) = \mu(B) \times \mu(B)$. But this is only possible if $\mu(B)$ is 0 or 1, contradicting our initial assumption.

Isn't this a problem with the definition of mixing? Is the definition in my source wrong? Or am I doing something wrong?

2

There are 2 best solutions below

0
On

There seems to be suspicious reasoning as pointed out in the comments when you define $A$ and then conclude that since $\mu(A)=\mu(B)$ then $\lim_{t\rightarrow\infty}\mu(\phi^t(B)\cap A)=\mu(A)$.

Have you encountered ergodic theory? Your mixing condition is more formally referred to as "strong mixing" which implies ergodicity, in that $\phi_t$ must be ergodic. By definition a flow is ergodic when $A=\phi_{-t}(A)$ implies $\mu(A)=0,1$.

0
On

You may want to think about the notion of independent events in probability theory: two events are independent if $Pr(A \cap B) = Pr(A)\cdot Pr(B).$
So the definition you give says that in the large time limit, the events of a point being in $\phi^t(B)$ and in $A$ are independent. So however $A$ and $B$ are positioned with respect to one another, after a long time $t$, the position of $\phi^t(B)$ is completely independent of the position of $A$.

Intuitvely, the points in $B$ are being completely mixed throughout the set, independently of where they were originally positioned. Hence the term mixing.