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For a dynamical system governed by the equation $$\frac{dx}{dt} = 2\sqrt{1-x^2}$$

where $|x|\leq 1$. I want to compute its stable points. The question was asked in one of the entrance exam for PG course.

My approach I am completely new to this topic dynamical system. However, going through web searches I found the following way to solve this problem. Kindly correct me if I am wrong.

Let us take $f(x) = 2\sqrt{1-x^2}$. Then its equilibrium points are given by setting $f(x) = 0$ which are $x^* = -1, 1$. Further, if $f′(x^*)<0$, the equilibrium $x(t)=x^*$ is stable, and if $f′(x^*)>0$, the equilibrium $x(t)=x^*$ is unstable.

Since in this case $f^\prime(x) = \frac{-4x}{\sqrt{1-x^2}}$, we see that $f^\prime(x^*)$ goes to infinity so the points $x^* = -1, 1$ are unstable equilibrium points.

Thanks in advance

Theo Bendit
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Srijan
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1 Answers1

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Well, one goes to $-\infty$ and the other goes to $+\infty$, so you can expect the former to be stable and the latter to be unstable. In point of fact, notice that any solution $x(t)$ is strictly increasing as long as $x(t)\in(-1,1)$, therefore $1$ is asymptotically stable and $-1$ is unstable.