We have the following system:
$\dot{x}=x-y-x(x^2+y^2)$
$\dot{y}=y+x-y(x^2+y^2)$
- Determine the equilibrium points
- Show that this system has a periodic solution. Use the following substitution $x=r\cos(\phi), y=r\sin(\phi)$.
- Give the explicit solution for this system. Also determine the period of this solution.
I found the following equilibrium point: $(x,y)=(0,0)$. I believe this is the only one. I don't know how to use this substitution to solve the problem.
$$x^2+y^2=r^2$$
Differentiating this gives you: $$2xx'+2yy'=2rr'$$
Now you substitute $x'$ and $y'$:
$$x(x-y-x(x^2+y^2))+y(y+x-y(x^2+y^2))=rr'$$
This gives us:
$$x^2-xy-x^2(x^2+y^2)+y^2+xy-y^2(x^2+y^2)=rr'$$
$$x^2-xy-x^2r^2+y^2+xy-y^2r^2=rr'$$
$$x^2-r^2(x^2+y^2)+y^2=rr'$$
$$r^2-r^4=rr'$$
$$r-r^3=r'$$
To determine angle use:
$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$
Using the quotient rule gives us:
$$\theta' = \dfrac{x y' - y x'}{r^2}=\dfrac{xy-y^2-xy(x^2+y^2)-xy-x^2+xy(x^2+y^2)}{r^2}$$
$$\theta'=-1$$
$$r'=r-r^3=r(1-r)(1+r)$$
check:
Polar coordinates differential equation
$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$
$$ \tan \theta = \dfrac{y}{x} \rightarrow \theta=arctan\frac{y}{x} $$ Using the quotient rule gives us:
$$\theta' = arctan(\frac{y}{x})=\dfrac{(\frac{y}{x})'}{1+(\frac{y}{x})^2} =\dfrac{x y' - y x'}{r^2}$$