Consider the system
\begin{align}
\dot{x}&= -y+x(1-2x^2-3y^2)\\
\dot{y}&=x+y(1-2x^2-3y^2)
\end{align}
(a)Find the equilibrium points and determine their stability
(b)Rewrite the equations in polar coordinates
My Attempt:
(a) As far as I know, an equilibrium point will be a $(x,y)$ coordinate so that the Right Hand Side of the above two equations is equal to zero. But when I try to compute them directly, the calculation becomes very messy. So I would like to know whether there is a better way of doing that.
And for the stability.. should I have to consider the linearization of the system and obtain their Eigen values? (to do that I will first need the equilibrium points anyway)
(b)By using $x=r\cos\theta$ and $y=\sin\theta$, We get
\begin{align}
\dot{x}&= \dot{r}\cos\theta-r\sin\theta\times\dot{\theta}\\
\dot{y}&=\dot{r}\sin\theta+r\cos\theta.\times\dot{\theta}
\end{align}
Thus
\begin{align}
\dot{r}&= r-2r^3-r^3\sin^2\theta\\
\dot{\theta}&=1
\end{align}
I would like to know whether there is a way to get rid of that $\sin^2\theta$ term
Appreciate your help
Best Answer
We have the system
$$\begin{align} \dot{x}&= -y+x(1-2x^2-3y^2)\\ \dot{y}&=x+y(1-2x^2-3y^2) \end{align}$$
We can solve $x' = y' = 0$ and find a single critical point of
$$(x, y) = (0, 0)$$
The Jacobian is
$$J(x, y) = \begin{bmatrix} \dfrac{\partial x'}{\partial x}& \dfrac{\partial x'}{\partial y}\\ \dfrac{\partial y'}{\partial x}& \dfrac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} -6 x^2-3 y^2+1 & -6 x y-1 \\ 1-4 x y & -2 x^2-9 y^2+1 \end{bmatrix}$$
Evaluating the Jacobian at the critical point, $J(0,0)$, we have the eigenvalues
$$\lambda_{1, 2} = 1 \pm i$$
This gives us a phase portrait (unstable spiral since we have a positive real part)
To convert to polar, using $x = r \cos \theta, y = r \sin \theta$, we have
$$\begin{align}r' &= \dfrac{xx' +yy'}{r} = \dfrac{1}{2} r \left(r^2 \cos (2 t)-5 r^2+2\right) \\\theta' &= \dfrac{y' x -x' y}{r^2} = 1 \end{align}$$