# Ordinary Differential Equations/Locally linear

We will study autonomous systems

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

## Example presenting the method

We study the damped oscillating pendulum system:

${\displaystyle {\frac {dx}{dt}}=y,\,{\frac {dy}{dt}}=-\gamma y-\omega ^{2}\sin(x),}$

where ${\displaystyle \gamma }$  is called the damping constant and as in the spring problem it is responsible for removing energy.
1. First we find the critical points. From the previous section we have:
${\displaystyle (k\pi ,0){\text{ for any integer }}k.}$

2. Second we Taylor expand the RHS of the system ${\displaystyle \mathrm {F} (x,y):={\bigl (}{\begin{smallmatrix}y\\\\-\gamma y-\omega ^{2}\sin(x)\end{smallmatrix}}{\bigr )}}$  around arbitrary critical point ${\displaystyle (x_{0},y_{0})}$ :
{\displaystyle {\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}}}

3. Here ${\displaystyle \mathrm {J} _{\mathrm {F} }(x_{0},y_{0})}$  is the Jacobian matrix at ${\displaystyle (x_{0},y_{0})}$  which, for function ${\displaystyle \mathrm {F} (x,y)={\binom {\mathrm {F} _{1}(x,y)}{\mathrm {F} _{2}(x,y)}}}$ , is defined as:
{\displaystyle {\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}}}

4. The linearization around ${\displaystyle (x_{0},y_{0})=(k\pi ,0)}$  for an even integer ${\displaystyle k}$  is:
${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\binom {x}{y}}={\bigl (}{\begin{smallmatrix}0&1\\&\\-\omega ^{2}&-\gamma \end{smallmatrix}}{\bigr )}{\binom {x-k\pi }{y}}+o(\left\|(x-k\pi ,y)\right\|).}$

5. The eigenvalues of that matrix are:
${\displaystyle \lambda _{1},\,\lambda _{2}={\frac {-\gamma \pm {\sqrt {\gamma ^{2}-4\omega ^{2}}}}{2}}.}$

6. If ${\displaystyle \gamma ^{2}-4\omega ^{2}>0}$ , 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.
7. If ${\displaystyle \gamma ^{2}-4\omega ^{2}=0}$ , then the eigenvalues are repeated, real, and negative. Therefore, the critical points will be stable nodes.
8. If ${\displaystyle \gamma ^{2}-4\omega ^{2}<0}$ , then the eigenvalues are complex with negative real part. Therefore, the critical points will be stable spiral sinks.
9. The linearization around ${\displaystyle (x_{0},y_{0})=(k\pi ,0)}$  for odd integer ${\displaystyle k}$  is:
${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\binom {x}{y}}={\bigl (}{\begin{smallmatrix}0&1\\&\\\omega ^{2}&-\gamma \end{smallmatrix}}{\bigr )}{\binom {x-k\pi }{y}}+o(\left\|(x-k\cdot \pi ,y)\right\|).}$

10. The eigenvalues of that matrix are:
${\displaystyle \lambda _{1},\,\lambda _{2}={\frac {-\gamma \pm {\sqrt {\gamma ^{2}+4\omega ^{2}}}}{2}}.}$

11. Therefore, it has one negative eigenvalue ${\displaystyle \lambda _{1}<0}$  and one positive eigenvalue ${\displaystyle \lambda _{2}>0}$ , and so the critical points will be unstable saddle points.