Ordinary Differential Equations/Locally linear

We will study autonomous systems $$ \mathrm{x}' = \mathrm{f} (\mathrm{x}),$$ where the components of $$\mathrm{f}$$ are $$C^{1}$$ functions so that we are able to Taylor expand them to first order. A system of the form $$ \mathrm{x}'=\mathrm{A}\mathrm{x}+\mathrm{g}(\mathrm{x})$$ is called locally linear around a critical point $$\mathrm{x}_{0}$$ of $$\mathrm{f}$$ if $$ \frac{\left\|\mathrm{g}(\mathrm{x})\right\|}{\left\|\mathrm{x}-\mathrm{x}_{0}\right\|}\to 0\text{ as }\mathrm{x}\to\mathrm{x}_{0}. $$

Example presenting the method
We study the damped oscillating pendulum system: $$\frac{dx}{dt}=y,\,\frac{dy}{dt}=-\gamma y-\omega^{2}\sin(x), $$ where $$\gamma$$ is called the damping constant and as in the spring problem it is responsible for removing energy. \end{smallmatrix}\bigr)\binom{x-k\pi}{y}+o(\left\|(x-k\pi,y)\right\|). $$
 * 1) First we find the critical points. From the previous section we have:$$ (k\pi,0)\text{ for any integer }k.$$
 * 2) Second we Taylor expand the RHS of the system  $$\mathrm{F}(x,y):=\bigl(\begin{smallmatrix}y\\\\-\gamma y-\omega^{2}\sin(x)\end{smallmatrix}\bigr)$$ around arbitrary critical point $$(x_{0},y_{0})$$: $$\begin{align}\mathrm{F}(x,y)&=\mathrm{F}(x_{0},y_{0})+\mathrm{J}_{\mathrm{F}}(x_{0},y_{0})\binom{x-x_{0}}{y-y_{0}}+o(\left\|(x-x_{0},y-y_{0})\right\|)\\&=\bigl(\begin{smallmatrix} 0&1 \\  &\\ -\omega^{2}\cos(x_{0})&-\gamma \end{smallmatrix}\bigr)\binom{x-x_{0}}{y-y_{0}}+o(\left\|(x-x_{0},y-y_{0})\right\|).\end{align}$$
 * 3) Here $$\mathrm{J}_{\mathrm{F}}(x_{0},y_{0})$$ is the Jacobian matrix at $$(x_{0},y_{0})$$ which, for function $$\mathrm{F}(x,y)=\binom{\mathrm{F}_{1}(x,y)}{\mathrm{F}_{2}(x,y)}$$, is defined as:$$ \begin{align}J_{\mathrm{F}}(x_{0},y_{0}):=\bigl(\begin{smallmatrix}\frac{\mathrm{d}\mathrm{F}_{1}}{\mathrm{d}x}(x_{0},y_{0}) & \frac{\mathrm{d}\mathrm{F}_{1}}{\mathrm{d}y}(x_{0},y_{0}) \\&\\\frac{\mathrm{d}\mathrm{F}_{2}}{\mathrm{d}x}(x_{0},y_{0}) & \frac{\mathrm{d}\mathrm{F}_{2}}{\mathrm{d}y}(x_{0},y_{0})\end{smallmatrix}\bigr)\end{align}$$
 * 4) The linearization around $$(x_{0},y_{0})=(k\pi,0)$$ for an even integer $$k$$ is: $$ \frac{\mathrm{d}}{\mathrm{d}t}\binom{x}{y}=\bigl(\begin{smallmatrix} 0&1 \\  &\\ -\omega^{2}&-\gamma
 * 1) The eigenvalues of that matrix are: $$ \lambda_{1},\,\lambda_{2}=\frac{-\gamma\pm \sqrt{\gamma^{2}-4\omega^{2}}}{2}.$$
 * 2) If $$\gamma^{2}-4\omega^{2}>0$$, then the eigenvalues are real, distinct, and negative. Therefore, the critical points will be stable nodes.We observe that the basins of attractions for each even-integer critical points are well-separated.
 * 3) If $$\gamma^{2}-4\omega^{2}=0$$, then the eigenvalues are repeated, real, and negative. Therefore, the critical points will be stable nodes.
 * 4) If $$\gamma^{2}-4\omega^{2}<0$$, then the eigenvalues are complex with negative real part. Therefore, the critical points will be stable spiral sinks.
 * 5) The linearization around $$(x_{0},y_{0})=(k\pi,0)$$ for odd integer $$k$$ is:$$ \frac{\mathrm{d}}{\mathrm{d}t}\binom{x}{y}=\bigl(\begin{smallmatrix} 0&1 \\  &\\ \omega^{2}&-\gamma \end{smallmatrix}\bigr)\binom{x-k\pi}{y}+o(\left\|(x-k\cdot \pi,y)\right\|). $$
 * 6) The eigenvalues of that matrix are: $$ \lambda_{1},\,\lambda_{2}=\frac{-\gamma\pm \sqrt{\gamma^{2}+4\omega^{2}}}{2}.$$
 * 7) Therefore, it has one negative eigenvalue $$\lambda_{1}<0$$ and one positive eigenvalue $$\lambda_{2}>0$$, and so the critical points will be unstable saddle points.