“Chaotic Dynamics” Because the word is sadly nonlinear.
motivating non-linearity
\begin{equation} \dv t \mqty(x \\ y) = f\qty(\mqty(x\\y)) \end{equation}
This function is a function from \(f: \mathbb{R}^{2}\to \mathbb{R}^{2}\). All the work on Second-Order Linear Differential Equations, has told us that the above system can serve as a “linearization” of a second order differential equation that looks like the follows:
\begin{equation} \dv t \mqty(x \\y) = A \mqty(x \\ y) +b \end{equation}
Actually going about deriving a solution to this requires powers of \(A\) to commute. If \(A\) has a independent variable in it, or if its a time-varying function \(A(t)\), you can’t actually perform the linearization technique (raising diagonalized \(A\) to powers) highlighted here.
So we need something new.
Sudden Review of Vector Functions
Let’s take some function:
\begin{equation} f: \mathbb{R}^{2} \to \mathbb{R}^{2} \end{equation}
It will output a vector:
\begin{equation} f(x,y) = \mqty(f_1(x,y)\\ f_{2}(x,y)) \end{equation}
Solving Non-Linear Systems, actually
Let’s take a non-linear system:
\begin{equation} \begin{cases} \dv{x}{t} = F(x,y) \\ \dv{y}{t} = G(x,y) \end{cases} \end{equation}
Overarching Idea: To actually solve this, we go about taking a Taylor Series (i.e. linearize) the functions next to its critical points. Then, we use an epsilon-delta proof to show that the linearization next to those critical points are a good approximation.
So! Let us begin.
Let \((x*,y*)\) be a critical point of \(F\). Naturally, \(d 0=0\), so it is also a critical point of \(G\).
So we have:
\begin{equation} F(x*,y*)=G(x*,y*) = 0 \end{equation}
Now, we will begin building the “slope” of this function to eliminate the independent variable wholesale—by dividing:
\begin{equation} \dv{y}{x} = \dv{y}{t} / \dv{x}{t} = \frac{G(x,y)}{F(x,y)} \end{equation}
a divergence into epsilon delta proof
stable
A critical point is considered “stable” because, for each \(\epsilon >0\), \(\exists \delta >0\), such that:
\begin{equation} |x_0-x*| < \delta \implies |x(t)-x*| < \epsilon \end{equation}
asymptotically stable
For every trajectory that begins close to the critical point, it will end up at the critical point as time increases. That is, \(\exists \delta >0\) such that:
\begin{equation} |x-x*| < \delta \implies \lim_{t \to \infty } x(t)=x* \end{equation}
This is essentially epsilon delta, but the limit traces out the entire process descending so the critical point is stable through the whole descend.