## Classification of Critical Points

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.
The linearisation matrix
$\left( \begin{array}{cc} \frac{\partial F}{\partial x}|_{(x_0, y_0)} & \frac{\partial F}{\partial y}|_{(x_0, y_0)} \\ \frac{\partial G}{\partial x}|_{(x_0, y_0)} & \frac{\partial G}{\partial y}|_{(x_0, y_0)} \end{array} \right)$
has eigenavlues
$\lambda_1 , \: \lambda_2$
and corresponding eigenvectors
$\mathbf{v}_1, \: \mathbf{v}_2$
.
If
$\lambda_1 \: \lambda_2 \lt 0$
then the point
$(x_0, y_0)$
is stable. Any initial point on the eigenline (a line at starting at the point
$(x_0, y_0)$
in the direction of an eigenvector) will tend to the point
$(x_0, y_0)$
along that eigenline. Any initial point not on an eigenline will tend to the eigenline will the eigenline with the corresponding most negative eigenvalue, then towards the point
$(x_0, y_0)$
. This is a stable node.
If
$\lambda_1 \: \lambda_2 \gt 0$
then the point
$(x_0, y_0)$
is unstable. Any initial point on the eigenline will tend to move away from the point
$(x_0, y_0)$
along that eigenline. Any initial point not on an eigenline will tend to the eigenline will the eigenline with the corresponding most positive eigenvalue, then away from the point
$(x_0, y_0)$
. This is an unstable node.
If
$\lambda_1 \gt 0, \: \lambda_2 \,t 0$
then the point
$(x_0, y_0)$
is unstable. Any initial point not on the eigenline corresponding to
$\lambda_2$
will tend to the point
$(x_0, y_0)$
.Any other initial point on the eigenline corresponding to
$\lambda_1$
will tend to move away from the point
$(x_0, y_0)$
along that eigenline, or will tend to that eigenline away from the point
$(x_0, y_0)$
If the eigenvalues are complex
$\lambda_1=a+bi, \: \lambda_2 =a-bi$
then if
$a \lt 0$
the point
$(x_0, y_0)$
is stable (a stable spiral) and unstable (an unstable spiral)if
$a \gt 0$
. If
$a = 0$
then any point will move in a circle about the point
$(x_0, y_0)$
. The point
$(x_0, y_0)$
is a centre.
If there is only one eigenvalue
$\lambda$
and only one eigenvector
$\mathbf{v}$
then the point
$(x_0,y_0)$
is a stable degenerate node if
$\lambda \lt 0$
and an unstable degenerate node if
$\lambda \gt 0$
.