Using polar coordinates to find the critical points

Question

I have written the following DE system $$\dot{x} = x+y-x(x^2+y^2)\\ \dot{y} = -x+3y-y(x^2+y^2) $$ as follows in Polar form: $$\dot{r} = -2r\cos^2\theta+r(3-r^2) $$ $$\dot{\theta} = 2\sin\theta\cos\theta-1$$ What I wonder is could I also find all the critical points of this system by setting $\dot{\theta}$ and $\dot{r}$ equal to zero? Since these two systems are actually equivalent to each other this seemed like a logical thing to me and at the same time finding the critical points through setting $\dot{x}$ and $\dot{y}$ equal to zero seems like much more tedious in this state.

Solving for $\dot{\theta}=0$ leads to $2\sin\theta\cos\theta = 1$ which is the same as $\sin2\theta=1$ leading to $\theta = \frac{\pi}{4}$ or $\theta = -\frac{3\pi}{4}$. Substituting these within $\dot{r}$ leads to: $$-2r\cos^2\theta+r(3-r^2) = -2r(\pm\frac{\sqrt{2}}{2})^2 +r(3-r^2)=r(2-r^2)$$ So $\dot{r}=0$ if $r=0$ or $r^2=2$. This leads us to the following critical points: $(0,0)$, $(\sqrt{2},\sqrt{2})$, $(-\sqrt{2},-\sqrt{2})$. However what bugs me is that entering these points into the non-polar system does not lead to $\dot{x}=\dot{y}=0$ for $(\sqrt{2},\sqrt{2})$, $(-\sqrt{2},-\sqrt{2})$.

What am I missing here? Is the whole thought of trying to find critical points through putting $\dot{\theta}$ and $\dot{r}$ equal to zero unwise? Any help or hint with this will be much appreciated.

Answer 1

If $r=\sqrt 2$ and $\theta=\pi/4$ you get $x=y=\sqrt 2\frac 1{\sqrt2}=1$. Then $(1,1)$ is a critical point. Similarly, you can check the other solution $(-1,-1)$.

Answer 2

It is not that difficult to proceed from $\dot x=0=\dot y$. Multiply the first with $y$, the second equation with $x$ and subtract. Then $$ 0=xy+y^2+x^2-3xy=(x-y)^2 $$ has solutions only for $x=y$. Insert that in either equation to get $$ 0=2x -x(x^2+x^2)=2x(1-x^2) $$ to conclude that the values taken are $x=y\in\{-1,0,1\}$.