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I am studying the dynamical system on a discrete standard map

$$x_{n+1} = f(x_n, y_n)$$ $$y_{n+1} = g(x_n, y_n)$$

First of all, could anyone explain the difference between the stationary point and the fixed point? As stated in some book, for the points which satisfying $f(x_0, y_0)=0$ and $g(x_0, y_0)=0$, that point is the steady point. But in other book, they called it fixed point. So they are the same thing?

But my question is if I write down the Jacobian as $J$, so what's the eigenvalue of $J$ really mean? Is the eigenvalue related to the angular frequency of the period-n orbits (in some material, they give the frequencies based on the eigenvalues. Thanks.

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I will concatenate $x$ and $y$ and work with a single state-transition equation

$$x_{k+1} = f (x_k)$$

where $f : \mathbb{R}^n \to \mathbb{R}^n$. Given a state $x$, function $f$ gives you the next state $f (x)$. It's an infinite state machine! Suppose that $f (\bar{x}) = \bar{x}$ for some isolated point $\bar{x} \in \mathbb{R}^n$. We say that $\bar{x}$ is a fixed point, because if $x_0 = \bar{x}$, then $x_1 = f (x_0) = f (\bar{x}) = \bar{x}$, and $x_2 = f (x_1) = f(\bar{x}) = \bar{x}$, and so and so on. In other words, if the system starts at $x_0 = \bar{x}$, it stays there for all $k \geq 0$. Hence, the word "fixed" in "fixed point".

The Taylor series expansion of $f$ around $\bar{x}$ is

$$f (x) = f (\bar{x}) + (D f) (\bar{x}) (x - \bar{x}) + \text{H.O.T}$$

where $(D f)$ is the Jacobian matrix-valued function, and "H.O.T." stands for "higher-order terms". In a sufficiently small neighborhood of $\bar{x}$, we can neglect the higher-order terms, and thus

$$x_{k+1} = f (x_k) \approx f (\bar{x}) + (D f) (\bar{x}) (x_k - \bar{x})$$

and, since $f (\bar{x}) = \bar{x}$, we obtain $x_{k+1} - \bar{x} \approx (D f) (\bar{x}) (x_k - \bar{x})$. Let $e_k := x_{k} - \bar{x}$ be the error vector that measures the deviation from $\bar{x}$, and let $A := (D f) (\bar{x})$. We finally obtain the error dynamics $e_{k+1} \approx A \, e_k$, which yields $e_k \approx A^k \, e_0$. In other words, in a sufficiently small neighborhood of $\bar{x}$, the magnitude of the eigenvalues of matrix $A$ will tell us whether the error vector will converge to the origin or diverge, or equivalently, whether $x_k$ will converge to the fixed point $\bar{x}$ or diverge. If the former is the case, we say that the fixed point $\bar{x}$ is locally stable, whereas if the latter is the case we say that the fixed point $\bar{x}$ is unstable.

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Thanks Rod. The explanation of "fixed points" helps to clarify my question. As for the Jacobian, I now understand the math but if the system is a pendulum-like physical system. If I work out eigenvalues of the Jacobian and find that it is real, does it mean that the system is not even in oscillation (because in the text, they said the frequency is given by the complex eigenvalues)? It is strange because the system is really oscillating ... –  Lee Sep 9 '12 at 3:32
    
Instead of "magnitude of eigenvalues" I would say "the sign of real parts". .. @Lee are you sure your computation is correct? Pendulum should not have real eigenvaluea unless it's overdamped. –  user31373 Sep 9 '12 at 4:06
    
@ Lee: If you are modeling a pendulum, you should have a differential equation, not a discrete-time dynamical system! Take a look at chapter 1 of Sontag's book (available online): math.rutgers.edu/~sontag/mct.html –  Rod Carvalho Sep 9 '12 at 4:27
    
@ LVK: Do note that this is a discrete-time dynamical system and, therefore, it's not the sign of the real part of the eigenvalues that counts, but their magnitude: if the eigenvalues are inside the unit circle of the complex plane, then we have stability. –  Rod Carvalho Sep 9 '12 at 4:29
    
@ROD. Thanks for the feedback. I go back to check the text and I think I put a wrong statement. The one I read is the kicked rotor not pendulum, and the kicking is not continuous, so the map is discrete. –  Lee Sep 9 '12 at 7:49
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