TITLE: How to solve a linearised system around a critical point
QUESTION [1 upvotes]: Considering a nonlinear autonomous system:
$\frac{d}{dt} \begin{bmatrix} x(t) \\ y(t)  \end{bmatrix} = \begin{bmatrix} F(x,y) \\ G(x,y) \end{bmatrix}$
Suppose $(x,y)=(a,b)$ is a critical point of the system and we want to linearise around this point:
Setting $z_1 = x - a, z_2 = y-b$ we obtain:
$\frac{d}{dt} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} F_x(a,b) & F_y(a,b) \\ G_x(a,b) & G_y(a,b) \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}$
This is where I have confusion, can I simply solve the linearised system for $z_1, z_2$ then if I want a solution for $x(t),y(t)$ I can simply just add $\begin{bmatrix} a \\ b \end{bmatrix} $ on the rhs of the solution for $ \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}$?
Furthermore, the critical point of a system is a particular solution of the linearised system around that point? Appreciate the help.

REPLY [1 votes]: In order to solve the linearized system at the equilibrium points you need to solve:
$$\frac{d}{dt} \begin{bmatrix} x-a \\ y-b \end{bmatrix} = \begin{bmatrix} F_x(a,b) & F_y(a,b) \\ G_x(a,b) & G_y(a,b) \end{bmatrix} \begin{bmatrix} x-a \\ y-b \end{bmatrix}$$
$$\implies \frac{d}{dt} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} F_x(a,b) & F_y(a,b) \\ G_x(a,b) & G_y(a,b) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}-\begin{bmatrix} F_x(a,b) & F_y(a,b) \\ G_x(a,b) & G_y(a,b) \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}$$
This is an inhomogeneous linear differential equation with constant coefficients. The solution vector $\boldsymbol{x}$ is given as the linear superposition of the homogeneous solution $\boldsymbol{x}_{\text{h}}$ of the system  
$$\dot{\boldsymbol{x}}_h = \begin{bmatrix} F_x(a,b) & F_y(a,b) \\ G_x(a,b) & G_y(a,b) \end{bmatrix} \boldsymbol{x}_h$$
and the particular solution $\boldsymbol{x}_{\text{p}}=\begin{bmatrix}a,b\end{bmatrix}^T$ (verify by plugging into the differential equation). 
Hence, the solution is given as
$$\boldsymbol{x}(t)=\boldsymbol{x}_{\text{h}}(t)+[a,b]^T.$$
As $\boldsymbol{x}_{\text{h}}$ is nothing but the solution for your system in $z_i$ you were right in assuming that you could simply add $[a,b]^T$ to the solution to obtain the solution for $\boldsymbol{x}(t)$.
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