## Linearising a System

Suppose we have a system of coupled differential equations
$\dot{x}=F(x,y)$

$\dot{y}=G(x,y)$

This system may be very difficult or impossible to solve, but we can find the nature of the solution about a point
$(x_0,y_0)$
by linearising the system about the point
$(x_0,y_0)$

$\dot{x}=\frac{\partial F}{\partial x}|_{(x_0, y_0)}(x-x_0) +\frac{\partial F}{\partial y}|_{(x_0, y_0)}(y-y_0) + higher \: order \: terms$

$\dot{y}=\frac{\partial G}{\partial x}|_{(x_0, y_0)}(x-x_0) +\frac{\partial G}{\partial y}|_{(x_0, y_0)}(y-y_0) + higher \: order \: terms$

We can define
$x'=x-x_0, \: y'=y-y_0$
and the system becomes, in matrix form:
$\begin{pmatrix}\dot{x}'\\ \dot{y}'\end{pmatrix} = \left( \begin{array}{cc} \frac{\partial F}{\partial x}|_{(x_0, y_0)} & \frac{\partial F}{\partial y}|_{(x_0, y_0)} \\ \frac{\partial G}{\partial x}|_{(x_0, y_0)} & \frac{\partial G}{\partial y}|_{(x_0, y_0)} \end{array} \right) \begin{pmatrix}x'\\ y'\end{pmatrix}$

We are especially interested in the point(s)
$(x_0,y_0)$
that satisfy
$\dot{x}=\dot{y}=0$
. These are called critical points.