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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 $\lim y(t)$ ?

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    $\begingroup$ Your stated there are four possible answer yet you have five bullet points. Wassup with that? Cheers! $\endgroup$ – Robert Lewis Sep 14 '18 at 0:20
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    $\begingroup$ @RobertLewis lol - copy and paste error! the enemy of a developer! $\endgroup$ – Chinny84 Sep 14 '18 at 0:57
  • $\begingroup$ I changed it... Do you maybe have an idea? $\endgroup$ – Evinda Sep 14 '18 at 6:31
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    $\begingroup$ Your post has something backwards; if $Re(\lambda_1), Re(\lambda_2) = -1 < 0$, the equilibrium point is asymptotically stable, not unstable. Because the then $e^{\lambda_i t} \to 0$ as $t \to \infty$. Also, check your exponents in $(\star)$. Cheers! $\endgroup$ – Robert Lewis Sep 14 '18 at 16:38
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    $\begingroup$ And I do have an idea; will post shortly. $\endgroup$ – Robert Lewis Sep 14 '18 at 16:43

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