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Aizawa attractor creates a closed surface. Let's ignore the transient state in the beginning. I am wondering how to calculate its surface? I even don't know where to start.

The dynamic model of Aizawa attractor is \begin{align} \begin{cases} \dot{x} = (z-b)x-dy\\ \dot{y} = dx+(z-b)y \\ \dot{z} = c+az-\frac13 z^3-(x^2+y^2) (1+e z)+f z x^3\\ \end{cases} \end{align}

Though, it is preferred that the general form of problem is solved, a special case of the parameters is

\begin{align} &\quad~a=0.95\\ &\quad~b=0.7\\ &\quad~c=0.6\\ &\quad~d=3.5\\ &\quad~e=0.25\\ &\quad~f=0.1 \end{align}

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  • $\begingroup$ If you don't write down any equation whatsoever it is hard to answer here $\endgroup$
    – Giuseppe Negro
    Apr 26, 2018 at 15:10
  • $\begingroup$ @GiuseppeNegro, thank you very much for this comment. $\endgroup$
    – Katrine
    Apr 26, 2018 at 23:11
  • $\begingroup$ You are welcome but I didn't do anything! :-) Anyway, what do you exactly want to compute? The equations describing the attractor, or just its surface area? I suspect that the equations are very complicated. $\endgroup$
    – Giuseppe Negro
    Apr 27, 2018 at 8:28
  • $\begingroup$ P.S.: The video is beautiful. $\endgroup$
    – Giuseppe Negro
    Apr 27, 2018 at 8:37
  • $\begingroup$ @GiuseppeNegro, I look for an analytical solution if possible. $\endgroup$
    – Katrine
    Apr 27, 2018 at 9:10

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