Using polar coordinates to sketch phase portrait

Question

I am trying to use polar coordiantes to sketch the phase portrait of the following system,

$$x'=xy-x^2y+y^3$$ $$ y'=y^2+x^3-xy^2$$

I know the formula for polar coordinates is the following,

$$r'=\frac{xx'+yy'}{r}$$ $$\theta'=\frac{y'x-yx'}{r^2}$$

But how do I change my system into these polar coordinates and then draw the phase portait?

Answer 1

\begin{align*} r' &= \frac{xx'+yy'}{r} \\ &= \frac{x(xy-x^2y+y^3)+y(y^2+x^3-xy^2)}{r} \\ &= \frac{x^2y-x^3y+xy^3+y^3+x^3y-xy^3}{r} \\ &= \frac{y(x^2+y^2)}{r} \\ &= r^2\sin \theta \\ \theta' &= \frac{xy'-xy'}{r} \\ &= \frac{x(y^2+x^3-xy^2)-y(xy-x^2y+y^3)}{r^2} \\ &= \frac{xy^2+x^4-x^2y^2-xy^2+x^2y^2-y^4}{r^2} \\ &= \frac{(x^2+y^2)(x^2-y^2)}{r^2} \\ &= r^2(\cos^2 \theta-\sin^2 \theta) \\ &= r^2\cos 2\theta \end{align*}

Note that the vector field is $$ \begin{align*} \mathbf{A}(r,\theta) &= r' \mathbf{e}_r+r\theta' \mathbf{e}_{\theta} \\ &= r^2\sin \theta \, \mathbf{e}_r+r^3\cos 2\theta \, \mathbf{e}_{\theta} \end{align*}$$