Ordinary Differential Equations/Locally linear

We will study autonomous systems $${\mathrm{x}}^{\prime }=\mathrm{f}(\mathrm{x}),$$ 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 $${\mathrm{x}}^{\prime }=\mathrm{A}\mathrm{x}+\mathrm{g}(\mathrm{x})$$ is called locally linear around a critical point ${\mathrm{x}}_0$ of $\mathrm{f}$ if $$\frac{\Vert \mathrm{g}(\mathrm{x})\Vert }{\Vert \mathrm{x}-{\mathrm{x}}_0\Vert }\to 0\text{ as }\mathrm{x}\to {\mathrm{x}}_0.$$

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.

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 $\mathrm{F}(x,y):=\left(\begin{array}{c}y\\ \\ -\gamma y-{\omega }^2\sin (x)\end{array}\right)$ around arbitrary critical point $(x_0,y_0)$: $$\begin{array}{cc}\mathrm{F}(x,y)& =\mathrm{F}(x_0,y_0)+{\mathrm{J}}_{\mathrm{F}}(x_0,y_0)(\genfrac{}{}{0ex}{}{x-x_0}{y-y_0})+o(\Vert (x-x_0,y-y_0)\Vert )\\ & =\left(\begin{array}{cc}0& 1\\ & \\ -{\omega }^2\cos (x_0)& -\gamma \end{array}\right)(\genfrac{}{}{0ex}{}{x-x_0}{y-y_0})+o(\Vert (x-x_0,y-y_0)\Vert ).\end{array}$$
3. Here ${\mathrm{J}}_{\mathrm{F}}(x_0,y_0)$ is the Jacobian matrix at $(x_0,y_0)$ which, for function $\mathrm{F}(x,y)=(\genfrac{}{}{0ex}{}{{\mathrm{F}}_1(x,y)}{{\mathrm{F}}_2(x,y)})$, is defined as:$$\begin{array}{c}{J}_{\mathrm{F}}(x_0,y_0):=\left(\begin{array}{cc}\frac{\mathrm{d}{\mathrm{F}}_1}{\mathrm{d}x}(x_0,y_0)& \frac{\mathrm{d}{\mathrm{F}}_1}{\mathrm{d}y}(x_0,y_0)\\ & \\ \frac{\mathrm{d}{\mathrm{F}}_2}{\mathrm{d}x}(x_0,y_0)& \frac{\mathrm{d}{\mathrm{F}}_2}{\mathrm{d}y}(x_0,y_0)\end{array}\right)\end{array}$$
4. The linearization around $(x_0,y_0)=(k\pi ,0)$ for an even integer $k$ is: $$\frac{\mathrm{d}}{\mathrm{d}t}(\genfrac{}{}{0ex}{}{x}{y})=\left(\begin{array}{cc}0& 1\\ & \\ -{\omega }^2& -\gamma \end{array}\right)(\genfrac{}{}{0ex}{}{x-k\pi }{y})+o(\Vert (x-k\pi ,y)\Vert ).$$
5. The eigenvalues of that matrix are: $${\lambda }_1,{\lambda }_2=\frac{-\gamma \pm \sqrt{{\gamma }^2-4{\omega }^2}}{2}.$$
6. 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.
7. If ${\gamma }^2-4{\omega }^2=0$, then the eigenvalues are repeated, real, and negative. Therefore, the critical points will be stable nodes.
8. 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.
9. The linearization around $(x_0,y_0)=(k\pi ,0)$ for odd integer $k$ is:$$\frac{\mathrm{d}}{\mathrm{d}t}(\genfrac{}{}{0ex}{}{x}{y})=\left(\begin{array}{cc}0& 1\\ & \\ {\omega }^2& -\gamma \end{array}\right)(\genfrac{}{}{0ex}{}{x-k\pi }{y})+o(\Vert (x-k\cdot \pi ,y)\Vert ).$$
10. The eigenvalues of that matrix are: $${\lambda }_1,{\lambda }_2=\frac{-\gamma \pm \sqrt{{\gamma }^2+4{\omega }^2}}{2}.$$
11. 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.