Draw phase portrait of this system

Question

Consider the system: $$ \begin{cases} x'=xy\\ y'= -x^2. \end{cases} $$ I find that for this system, the line $x=0$ are a line of fixed points. I wonder how to draw the phase portrait for this system.

Answer 1

Change to polar coordinates: $$ x(t)=R(t)\cos(\varphi(t))\qquad y(t)=R(t)\sin(\varphi(t)) $$ $$ x'(t)=R'(t)\cos(\varphi(t))-R(t) \varphi '(t) \sin (\varphi (t)) $$ $$ y'(t)=R'(t)\sin(\varphi(t))+R(t) \varphi '(t) \cos (\varphi (t)) $$ Then: $$ R'(t)\cos(\varphi(t))-R(t) \varphi '(t) \sin (\varphi (t)) =R(t)^2\cos(\varphi(t))\sin(\varphi(t)) $$ $$ R'(t)\sin(\varphi(t))+R(t) \varphi '(t) \cos (\varphi (t)) =-R(t)^2\cos(\varphi(t))^2 $$ This is a linear equation for $R'(t)$ and $\varphi'(t)$, solving it yields: $$ \begin{cases} R'(t)=0\\ \varphi'(t)=-R(t)\cos(\varphi(t))\\ \end{cases}\Rightarrow \begin{cases} R'=0\\ \varphi'=-R\cos(\varphi)\\ \end{cases} $$ Thus $R(t)=c$, so the curves should be circles.

Using the following Mathematica code helps:

StreamPlot[{x y, -x^2}, {x, -2, 2}, {y, -2, 2}]

(StreamPlot reference)