Discrete Time Dynamical Systems
We'll go over systems in which time is discrete rather than continuous. These systems are called difference equations, @recursion-relations, iterated maps, or simply maps. These can be interesting; even in one dimension they can exhibit oscillations and chaos.
A difference equation has the general form
$$\vec{x_{n+1}} = \vec{f(x_n, n)}, $$
where $n$ is a time-like variable, $\vec{f}$ and $\vec{x_n}$ may be vectors, $n = 0, 1, 2, \dots = \mathbb{N}_0.$
Solutions to difference equations are sequences $(x_0, x_1, ..., x_n).$ If $x_0$ is the initial condition, then
$$\begin{aligned} x_1 & = f(x_0, 0) \\ x_2 & = f(x_1, 0) \\ & \vdots \\ x_n & = f(x_{n-1}, 0) \\ \end{aligned} $$
So the solution is $(x_1, x_2, \dots, x_n).$ Long term dynamics are given by moving along this sequence as $n \to \infty.$
We will consider ourselves, for now, with @autonomous-systems where there is no explicit dependence on $n,$ i.e.
$$ x_{n+1} = f(x_n). $$
So $x_n \in \mathbb{R}^N$ and $f : \mathbb{R}^N \to \mathbb{R}^N. $
Example: $x_{n+1} = - x_n.$ If we start with $x_0,$ then $x_1 = -x_0, x_2 = -x_1 = -(-x_0) = x_0, dots$ so the solution is the sequence
$$ x = (x_0, -x_0, x_0, -x_0, \dots). $$
Orbits and Fixed Points
A sequence $(x_0, x_1, x_2, \dots)$ is called the orbit (or solution) starting from $x_0.$ This is analogous to a trajectory in a continuous time dynamical system.
A fixed point in a DTDS is a point whose further iteration does not change, i.e. $x_{n+1} = x_n.$ Fixed points satisfy
$$ x^* = f(x^*).$$
Stability of Linear Maps
Consider
$$ x_{n+1} = \lambda x_n, \quad \lambda \in \mathbb{R}. $$
Starting with $x_0,$ $x_1 = \lambda x_0, x_2 = \lambda x_1 = \lambda ( \lambda x_0) = \lambda^2 x_0, \dots,$ so our general solution is
$$ x_n = \lambda^n x_0. $$
Fixed points satisfy $x^* = f(x^*) \implies x^* = \lambda x^* \implies x^*(\lambda - 1) = 0.$ Therefore, $x^* = 0$ is always a fixed point, and if $lambda = 1,$ then every $x^* \in \mathbb{R}$ is a fixed point, i.e. the map is just $x_{n+1} = x_n.$
Given that $x^* = 0$ is always a fixed point of a linear map, we'll consider different cases of the value of $\lambda$ and how it impacts the stability of $x^*.$
If $\lambda > 1,$ then $\lambda^n \to \infty$ as $n \to \infty.$ Thus, $x_n$ grows without bound and $x^* = 0$ is unstable.
If $0 < \lambda < 1,$ then $\lambda^n \to 0$ as $n \to \infty.$ Thus, $x_n \to 0$ and keeps the same sign. Hence, $x^* = 0$ is stable.
If $\lambda = 0,$ then $x_n = 0$ for all $n \geq 1.$ Hence $x^* = 0$ is stable.
If $-1 < \lambda < 0,$ then $|\lambda|^n \to 0$ as $n \to \infty,$ but $x_n$ alternates sign (oscillates.) Hence, $x^* = 0$ is stable.
If $\lambda < -1,$ then $|\lambda|^n \to \infty$ as $n \to \infty,$ and $x_n$ oscillates with increasing magnitude. Hence $x^* = 0$ is unstable.
If $\lambda = -1,$ then $x_n = (-1)^n x_0,$ so the solution alternates between $x_0$ and $-x_0.$ The fixed point $x^* = 0$ is unstable.
If $\lambda = 1,$ then $x_n = x_0$ for all $n,$ so every point is a stable fixed point.
In summary, if $|\lambda| < 1,$ then $x^* = 0$ is stable. if $|\lambda| > 1,$ then $x^* = 0$ is unstable.