I want to find the equilibrium solutions and determine their stability.
$(1)\left\{\begin{matrix} \dot{x}=-y-x(1-\sqrt{x^2+y^2})^2\\ \dot{y}=x-y(1-\sqrt{x^2+y^2})^2 \end{matrix}\right.$
I also want to check the behavior of the solutions of $(1)$ when $t \to \infty$. There are five possible answers.
there exists exactly one equilibrium solution and it is asymptotically stable. Furthermore, $\forall$ solution $(x,y)$ of $(1)$ we have that $\lim x(t)=x_0$ and $\lim y(t)=y_0$ where $(x_0,y_0)$ equilibrium solution.
there exists exactly one equilibrium solution and it is asymptotically stable. Furthermore, $\forall$ solution $(x,y)$ of $(1)$ we have that $\lim x(t) \neq x_0$ and $\lim y(t) \neq y_0$ where $(x_0,y_0)$ equilibrium solution.
$\exists$ exactly one equilibrium solution and it is stable but not asymptotically stable. Also $\forall$ solution $(x,y) \neq (x_0, y_0)$ of $(1)$ we have that $\lim x(t) \neq x_0$ and $\lim y(t) \neq y_0$ where $(x_0,y_0)$ is the equilibrium solution.
$\exists$ exactly one equilibrium solution and it is stable but not asymptotically stable. Also $\forall$ solution $(x,y) \neq (x_0, y_0)$ of $(1)$ with $x^2(0)+y^2(0)>1$ we have that $x^2(t)+y^2(t) \geq 1$.
$\exists$ exactly one equilibrium solution and it is unstable. $\forall$ other solution $(x,y)$ of $(1)$ we have that $\lim (x^2(t)+y^2(t))=1$.
I have thought the following so far.
In order to find the equilibrium solutions, we set $\dot{x}=0$ and $\dot{y}=0$.
$\dot{x}=0 \Rightarrow -y-x (1-\sqrt{x^2+y^2})^2=0 (\star)$
and
$\dot{y}=0 \Rightarrow x-y (1-\sqrt{x^2+y^2})^2=0 \Rightarrow x=y(1-\sqrt{x^2+y^2})^2$
$(\star): -y-y(1-\sqrt{x^2+y^2})^4=0 \Rightarrow y=0 \text{ or } 1+(1-\sqrt{x^2+y^2})^4=0, \text{ which is rejected}$.
So $y=0$ and $x=0$.
So $(0,0)$ is the only equilibrium solution.
If we set $f_1(x,y)=-y-x(1-\sqrt{x^2+y^2})^2$ and $f_2(x)=x-y(1-\sqrt{x^2+y^2})^2$, then we have
$\frac{\partial{f}}{\partial{x}}=-(1-\sqrt{x^2+y^2})^2+\frac{2x^2(1-\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}$.
Then we have
$$\frac{\partial{f_1}}{\partial{x}}(0,0)=-1 \\ \frac{\partial{f_1}}{\partial{y}}(0,0)=-1 \\ \frac{\partial{f_2}}{\partial{x}}(0,0)=1 \\ \frac{\partial{f_2}}{\partial{y}}(0,0)=-1$$
and thus the Jacobi matrix at $(0,0)$ gets the following form:
$J=\begin{pmatrix} -1 & -1\\ 1 & -1 \end{pmatrix}$, right?
Then we get that the eigenvalues are these ones: $-1 \pm i$ and so $Re(\lambda_1)=Re(\lambda_2)=-1<0$ which implies that the equilibrium is asymptotically unstable.
So we have that there is exactly one equilibrium solution and it is asymptotically unstable, right? When $(x,t)$ is any other solution of the problem, what can we say about the limits $\lim x(t)$ and $\lim y(t)$ ?