For some non-linear function, we can use its first Jacobian to create a linear system. Then, we can use that system to write the first order Taylor:
\begin{equation} y’ = \nabla F(crit)y \end{equation}
where \(crit\) are critical points.
Phase Portrait stability
if all \(Re[\lambda] < 0\) of \(\qty(\nabla F)(p)\) then \(p\) is considered stable—that is, points initially near \(p\) will exponentially approach \(p\)
if at least one \(Re[\lambda] > 0\) of \(\qty(\nabla F)(p)\) then \(p\) is considered unstable—that is, points initially near \(p\) will go somewhere else
if all \(Re[\lambda] \leq 0\) and at least one \(\lambda\) is pure imaginary of \(\qty(\nabla F)(p)\), then there are no conclusions and \(p\) is considered marginal
If there are no purely imaginary values, then the solution paths of the ODE look like that of \(y’ = (\nambla F)(p) y\).
Worked Example
Let’s Lotha-Volterra Prey-Predictor Equation again as an example
\begin{equation} \begin{cases} x_1’ = 2x_1-x_1x_2 \\ x_2’ = x_1x_2 - 3x_2 \end{cases} \end{equation}
we can stare at this (and factor \(x\) out) to understand that there are only two stationary points:
\begin{equation} (x_1,x_2) = (0,0), (3,2) \end{equation}
Let’s analyze this function for linearilzation.
Let’s write this expression in terms of the linear and non linear parts
\begin{equation} \begin{cases} x’ = \mqty(2 & 0 \\ 0 & -3) \mqty(x_1 \\ x_2) + \mqty(-x_1x_2 \\ x_1 x_2) \end{cases} \end{equation}
Near \((0,0)\)
You will note that the right non-linear parts becomes very small near \((0,0)\), meaning we can analyze this in terms of a normal phase portrait.
Near \((3,2)\)
We can translate this down:
Let:
\begin{equation} y = x - \mqty(3 \\2) \end{equation}
meaning:
\begin{equation} y’ = x’ = F\qty(y+\mqty(3 \\ 2)) \end{equation}
we can use a Taylor expansion to get:
\begin{equation} y’ = x’ = F\qty(y + \mqty(3\\2)) + \qty(\nabla F)y + \dots \end{equation}
Recall that \(F\) is given as:
\begin{equation} \mqty(2x_1 - x_1x_2 \\ x_1x_2-3x_2) \end{equation}
meaning:
\begin{equation} \nabla \mqty(2x_1 - x_1x_2 \\ x_1x_2-3x_2) = \mqty(2-x_2 & -x_1 \\ x_2 & x_1-3) \end{equation}
plugging in \((3, 2)\) obtains:
\begin{equation} y’ = \mqty(0 & -3 \\ 2 & 0) y \end{equation}
which we can analyze in the usual manners.