A dynamical system is governed by the equation $\frac{dx}{dt}=2\sqrt{1-x^2}$, $|x|\leq 1$. Then By equating $\frac{dx}{dt}$ to $0$ we get $1,-1$ are the fixed points. But how to check their stability? We can not check values of $\frac{dx}{dt}$ left to $-1$ and right to $1$.

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    $\begingroup$ You might want to declare that $-1$ is unstable and $1$ is stable since $x_0=-1+\epsilon$ with $\epsilon>0$ and small does not yield $x(t)-(-1)\to0$ while $x_0=1-\epsilon$ with $\epsilon>0$ and small yields $x(t)-1\to0$. $\endgroup$ – Did Sep 11 '18 at 10:00
  • $\begingroup$ would you please explain a bit $\endgroup$ – Anupam Sep 11 '18 at 13:24
  • $\begingroup$ would you please explain a bit $\endgroup$ – Anupam Sep 11 '18 at 13:24
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    $\begingroup$ I did. (Gaear Grimsrud, in Fargo).. $\endgroup$ – Did Sep 11 '18 at 13:36
  • $\begingroup$ @Anupam you can see this Question, some extensive discussion with Hans Lundmark is carried out, the main thing is $dx/dt $ is increasing over $[-1,1]$ and analysing the phase portrait, any other query do let me know and just to state Dynamical Systems Chaos theory chat room, hope this helps! $\endgroup$ – BAYMAX Sep 23 '18 at 13:27

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