# 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.
* First we find the critical points. From the previous section we have:$(k\pi ,0){\text{ for any integer }}k.$
* 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 $(x_{0},y_{0})$: ${\begin{aligned}\mathrm {F} (x,y)&=\mathrm {F} (x_{0},y_{0})+\mathrm {J} _{\mathrm {F} }(x_{0},y_{0}){\binom {x-x_{0}}{y-y_{0}}}+o(\left\|(x-x_{0},y-y_{0})\right\|)\\&={\bigl{\begin{smallmatrix}0&1\\&\\-\omega ^{2}\cos(x_{0})&-\gamma \end{smallmatrix}}{\bigr )}{\binom {x-x_{0}}{y-y_{0}}}+o(\left\|(x-x_{0},y-y_{0})\right\|).\end{aligned}}$
* 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:${\begin{aligned}J_{\mathrm {F} }(x_{0},y_{0}):={\bigl{\begin{smallmatrix}{\frac {\mathrm {d} \mathrm {F} _{1}}{\mathrm {d} x}}(x_{0},y_{0})&{\frac {\mathrm {d} \mathrm {F} _{1}}{\mathrm {d} y}}(x_{0},y_{0})\\&\\{\frac {\mathrm {d} \mathrm {F} _{2}}{\mathrm {d} x}}(x_{0},y_{0})&{\frac {\mathrm {d} \mathrm {F} _{2}}{\mathrm {d} y}}(x_{0},y_{0})\end{smallmatrix}}{\bigr )}\end{aligned}}$
* The linearization around $(x_{0},y_{0})=(k\pi ,0)$ for an even integer $k$ is: ${\frac {\mathrm {d} }{\mathrm {d} t}}{\binom {x}{y}}={\bigl{\begin{smallmatrix}0&1\\&\\-\omega ^{2}&-\gamma \end{smallmatrix}}{\bigr )}{\binom {x-k\pi }{y}}+o(\left\|(x-k\pi ,y)\right\|).$
* The eigenvalues of that matrix are: $\lambda _{1},\,\lambda _{2}={\frac {-\gamma \pm {\sqrt {\gamma ^{2}-4\omega ^{2}}}}{2}}.$
* 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.
* If $\gamma ^{2}-4\omega ^{2}=0$, then the eigenvalues are repeated, real, and negative. Therefore, the critical points will be stable nodes.
* 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.
* The linearization around $(x_{0},y_{0})=(k\pi ,0)$ for odd integer $k$ is:${\frac {\mathrm {d} }{\mathrm {d} t}}{\binom {x}{y}}={\bigl{\begin{smallmatrix}0&1\\&\\\omega ^{2}&-\gamma \end{smallmatrix}}{\bigr )}{\binom {x-k\pi }{y}}+o(\left\|(x-k\cdot \pi ,y)\right\|).$
* The eigenvalues of that matrix are: $\lambda _{1},\,\lambda _{2}={\frac {-\gamma \pm {\sqrt {\gamma ^{2}+4\omega ^{2}}}}{2}}.$
* 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.