---
title: "Non-Linear System"
type: concept
related: [Function, Second Order Linear Differential Equation, Non Linear Systems]
source: https://www.jemoka.com/posts/kbhnon_linear_systems/
confidence: high
status: active
---

&ldquo;Chaotic Dynamics&rdquo; 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 &ldquo;linearization&rdquo; 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&rsquo;t actually perform the linearization technique (raising diagonalized \(A\) to powers) highlighted here.
So we need something new.
Sudden Review of Vector Functions Let&rsquo;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&rsquo;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 &ldquo;slope&rdquo; of this function to eliminate the independent variable wholesale&mdash;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 &ldquo;stable&rdquo; because, for each \(\epsilon &gt;0\), \(\exists \delta &gt;0\), such that:
\begin{equation} |x_0-x*| &lt; \delta \implies |x(t)-x*| &lt; \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, \(\exists \delta &gt;0\) such that:
\begin{equation} |x-x*| &lt; \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.
