Non-Linear System — Jemoka Knowledge Base

“Chaotic Dynamics” Because the word is sadly nonlinear. motivating non-linearity \begin{equation} \dv t \mqty(x \ y) = f\left(\mqty(x\y)\right) \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.