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.

Solutions to difference equations are sequences $(x_0,x_1,...,x_n).$ If $x_0$ is the initial condition, then

So the solution is $(x_1,x_2,\dots ,x_n).$ Long term dynamics are given by moving along this sequence as $n\to \mathrm{\infty }.$

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 fixed point in a DTDS is a point whose further iteration does not change, i.e. $x_{n+1}=x_n.$ Fixed points satisfy

$$x^{\ast }=f(x^{\ast }).$$

Stability of Linear Maps

Consider

$$x_{n+1}=\lambda x_n,\lambda \in \mathbb{R}.$$

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 $x^{\ast }=f(x^{\ast })⟹x^{\ast }=\lambda x^{\ast }⟹x^{\ast }(\lambda -1)=0.$ Therefore, $x^{\ast }=0$ is always a fixed point, and if $lambda=1,$ then every $x^{\ast }\in \mathbb{R}$ is a fixed point, i.e. the map is just $x_{n+1}=x_n.$

Given that $x^{\ast }=0$ 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 $x^{\ast }.$

If then ${\lambda }^n\to \mathrm{\infty }$ as $n\to \mathrm{\infty }.$ Thus, $x_n$ grows without bound and $x^{\ast }=0$ is unstable.

If then ${\lambda }^n\to 0$ as $n\to \mathrm{\infty }.$ Thus, $x_n\to 0$ and keeps the same sign. Hence, $x^{\ast }=0$ is stable.

If $\lambda =0,$ then $x_n=0$ for all $n\ge 1.$ Hence $x^{\ast }=0$ is stable.

If then ${\left|\lambda \right|}^n\to 0$ as $n\to \mathrm{\infty },$ but $x_n$ alternates sign (oscillates.) Hence, $x^{\ast }=0$ is stable.

If then ${\left|\lambda \right|}^n\to \mathrm{\infty }$ as $n\to \mathrm{\infty },$ and $x_n$ oscillates with increasing magnitude. Hence $x^{\ast }=0$ 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 $x^{\ast }=0$ is unstable.

If $\lambda =1,$ then $x_n=x_0$ for all $n,$ so every point is a stable fixed point.

In summary, if then $x^{\ast }=0$ is stable. if then $x^{\ast }=0$ is unstable.