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Assume there is a discrete standard map

$$ \begin{cases} x_{n+1} = x_n + y_n,\\[3mm] y_{n+1} = ay_n + b\sin(x_{n+1}) \end{cases} $$

for solving the fixed points, we have $$ (x_{fm}, y_{fm}) = \left(\arcsin(-2(1-c)p\pi/b), -2p\pi\right) $$

where $p$ is interger

The fundamental period-1 fixed point is $(x_{f0}, y_{f0}) = (0, 0)$, right? When we say it is fixed point, we actually mean any input argument will map into itself, right? Ok, then there is a question. If I write the Jacobian of the map and solve the eigenvalue, for some a and b, I might get a complex eigenvalue $A\exp(\pm iB)$, which B tells the frequency of the orbit? But if I have a period-1 orbit, what can I tell the frequency of the orbit from the eigenvalue? Also, if the eigenvalue is complex, so does it mean that the system might be damping and decay into something else instead of staying at the fixed point? If that's true, so why it is still called a fixed point?

What I am interesting is what happen if I focus on period-n orbit, if I want to find the eigenvalue, I should use that period-n fixed point, right? In that case, what's the different in dynamics for having the system start from period-1 or period-n fixed point? Or I should ask is there any different between the eigenvalues when considering period-1 or period-n fixed points? Will either one decay faster than the other? why?


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A fixed point is just a point $x_0$ with $f(x_0)=x_0$. The derivative (Jacobian) tells us the behaviour of points sufficiently close to $x_0$. In your case eigenvalues $A e^{\pm iB}$ mean that the Jacobian essentlially looks like a rotation by $B$ and a scaling by $A$. If $0\le A<1$ than $x_0$ is an attracting fix point, if $A>1$ it is repelling. However, $x_0$ itself will always stay fixed.

Period-n orbits are just fixed points of $n$ iterates of $f$. The Jacobian of a point of $f^n$ is simply the product aof all jacobians of $f$ along the orbit of the point.

Hope that helps clarifying.

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Thanks Hagen. Let's focus on the period-1 orbit. If I have eigenvalues $Ae^{\pm iB}$ and $A>1$. According to your explanation, if a particle initially stay at the fixed point, because of repelling (A>1), it moves spirally away the fixed point and eventually outside the island? If $0<A<1$, if the particle start it motion anywhere within the island, it will move towards the fixed point? If A=1, the particle will be moving around a circular orbit forever? So my question is for the first case (A>0), what's B? it is not a circular orbit, does B still mean oscillation frequency? Why – Lee Sep 9 '12 at 19:32

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