Phase Portrait Using Polar Coordinates

Question

I converted a system to polar coordinates and got: $$r'=r^2 \sin \theta \\ \theta'=r^2\cos\theta $$ Now I have to graph the phase portrait near the fixed point at (0,0) and don't know where to begin. Usually, in cartesian coordinates, I would find the Jacobian, eigenvalues, eigenvectors, and the plotting is simple. Can anyone give me a hint?

Answer 1

You can divide both equations to obtain

$$\dfrac{dr}{d\theta}=\tan\theta \implies dr = \tan \theta\,d\theta$$ $$\implies r = -\ln |\cos \theta| + c.$$

Using the initial condition $(r_0,\theta_0)$ we can find the constant

$$r_0=-\ln |\cos \theta_0|+c.$$

This implies $$r-r_0=-\ln \left|\dfrac{\cos \theta}{\cos \theta_0} \right|$$ $$r = r_0-\ln \left|\dfrac{\cos \theta}{\cos \theta_0} \right|.$$