Phase portrait of non linear system

Question

I have a following system:

$x'(t)=\sin x \cos x$

I know that stationary points are $x_s=\frac{k \pi}{2}$ where $k \in \mathbb{Z}$

What will lineared system look like and how to draw its phase portrait by hand?

Answer 1

Just compute the Jacobian of the system and you get $$\text{D}f(x)=\cos(x)^2-\sin(x)^2$$

The linearized systems around your fixed points $x_s$ reads as $\xi'(t)=\text{D}f(x_s)\xi$ where

$$\text{D}f(x_s)=\text{D}f(k\pi/2)=\begin{cases} 1 &,k=0 \\ 1 &, k \text{ even} \\ -1 &,k \text{ odd} \end{cases}$$

Notice that your linearized systems at $x_s=0$ and $x_s=k\pi/2$ for $k$ even reads $\xi'(t)=\xi$ and its solution can be seen immediately, it is $\xi(t)=\xi_0e^t$ which means for increasing $t$ you move away from your fixed point.

At the other fixed points you get $\xi'(t)=-\xi$ with the solution $\xi(t)=\xi_0e^{-t}$ and you see that for increasing $t$ you move towards zero. Notice that in the linearized system your fixed point is transferred to the origin.

Sketching the phase portrait gives you: