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Dynamical system $f(x)=x^4\sin(1/x)$ (for x not equal to zero) and $f(x)=0$ (for x=0). How to find equilibrium points, and determining the stability of each equilibrium point?

I found the equilibrium points x=0 and $x=1/kπ$, where k is integer.

$f'(x)=4x^3\sin(1/x)-x^4\cos(1/x)$.

When $k$ is 'even', $f'(x=1/kπ)=1/π^4$ which is positive, therefore not asymptotically stable.

And when $k$ is 'odd', $f'(x=1/kπ)=-1/π^4$ which is negative, therefore asymptotically stable.

But $f'(x=0)=0$, which is not hyperbolic. And this is the part that I can not solve.

Spiral
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  • Apply the definition of stability directly and consider that 0 is a limit point of stable equilibria. – Tobias Nov 10 '16 at 20:31
  • Will it eventually be unstable at x=0 (that is my guess)? Can you explain how to use the delta-epsilon definition to prove this? – Spiral Nov 10 '16 at 23:23

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Hint: Consider a small perturbation $x(t)$ of the zero solution, say with $x(0)=\epsilon \neq 0$. This solution will be caught in between two of the other equilibrium points (or perhaps exactly at one of them), and this lets you permit what happens to the solution $x(t)$ in the long run.

Hans Lundmark
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  • This is the very first question of Dynamical system that I am trying to solve. The examples in the lecture notes are not as complicated as this one. – Spiral Nov 10 '16 at 22:01
  • Thanks for the hint. I am still not sure how to continue (This is the very first question of Dynamical system that I am trying to solve. The examples in our lecture notes are not as complicated as this one). I am guessing that at very close to x=0 it is not stable because there will be always an odd and an even K which makes the f'(1/kπ) once negative and once positive respectively. So at one side of x=0 the arrow is toward the x=0, and at another side the arrow is away from x=0. Hence it will be unstable at x=0. Please correct me if I am wrong. – Spiral Nov 10 '16 at 22:16
  • No, you have infinitely many equilibrium points that “pile up” as you approach the origin. In between these, the arrows alternatingly go left and right (since $x^4 \sin(1/x)$ oscillates between positive and negative values). – Hans Lundmark Nov 11 '16 at 06:46