“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.