linearilzation

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 \left(\nabla F\right)(p) then p is considered stable—that is, points initially near p will exponentially approach p
if at least one Re[\lambda] > 0 of \left(\nabla F\right)(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 \left(\nabla F\right)(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\left(y+\mqty(3 \\ 2)\right) \end{equation}

we can use a Taylor expansion to get:

\begin{equation} y’ = x’ = F\left(y + \mqty(3\\2)\right) + \left(\nabla F\right)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.