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I am trying to get an intuition for (i) why a chaotic "driver" system can be predicted by a "response" system in the first place, and (ii) why is feedback delay within the latter a necessary condition for this. Here are the relevant differential equations for one such case (the Rössler system):

$\dot x_1 = -x_2 - x_3$

$\dot x_2 = x_1 + ax_2$

$\dot x_3 = b + x_3(x_1 - c)$

$\dot y_1 = -y_2 - y_3 + k(x_1 - y_{1,\tau})$

$\dot y_2 = y_1 + ay_2$

$\dot y_3 = b + y_3(y_1 - c)$

The behavior of $x$ is independent of $y$, so we call it the "driver" system in this context; and $y$ receives input from $x$, so it is a "response" system. $a,b,c$ are constants, $k$ is the coupling constant. The delayed feedback is $y_{1,\tau} \equiv y_1(t-\tau)$. Here is a simulation of this from Stepp & Turvey (2011) (solid = driver, dashed = response system):

enter image description here

How is it possible for the $y$ response system to anticipate $x$ if the latter is chaotic? And what role is feedback delay playing? Stepp & Turvey say that

The effect of the coupling term $k(x - y_{1,\tau})$ is to minimize the difference between the state of $x$ at the current time, and the state of $y$ at a past time. If this difference is successfully minimized, then the difference between the present state of $y$ and future state of $x$ is also minimized. The effect of this minimization is the synchronization of $y$ with the future of $x$.

But I have no intuition why the coupling term earns its name - what is it doing to "couple" $x$ and $y$?

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It seems to me that there is a confusion. Here, "chaotic" does not mean "unpredictable". So, it is totally possible to anticipate the behavior of the driver system.

To understand this, first note that without the coupling, the models for $x$ and $y$ are the same. So, if the initial conditions are the same, both systems will have the same trajectories.

Now, with the coupling term, one can see that the response system compares the past value of its state $y_1(t-\tau)$ and $x_1(t)$, which is the current value for $x_1$. So, this can be understood as the fact that we want to anticipate the behavior of the system by $\tau$ seconds since we have $y_1(t)\approx x_1(t+\tau)$ when the difference $y_1(t-\tau)-x_1(t)$ is small, which is the goal here as explained in the paragraph you last quoted.

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  • $\begingroup$ That makes sense, thank you. I have a follow-up question: why does the sign of $k$ have to be positive for this synchronization to happen over time? $\endgroup$ Commented Feb 27, 2022 at 22:11
  • $\begingroup$ You need to be positive for stability here. In fact, the factor of $y_1(t-\tau)$ should be negative. Otherwise, you get an unstable behavior. You can observe via simulations. $\endgroup$ Commented Feb 27, 2022 at 22:33
  • $\begingroup$ @LutzLehmann Yes thanks. I have corrected it. $\endgroup$ Commented Feb 28, 2022 at 14:17

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