about Jacobian and eigenvalues 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.
 A: 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.
