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Consider the differential equation $ x'=f(x) $ where $ f(x)=\begin{cases} 0 & x = 0 \\[12pt] -x^3\sin\left( {\frac{1}{x}} \right) & x \ne 0 \end{cases} $

I have to study the equilibrium points. First, I've proved that $f(x) \in C^1(\mathbb{R})$. Then I've found that the equilibrium points are $x=0$ and $x_k = \frac{1}{k\pi}$, with $k \in \mathbb{Z} \setminus \{0\}$. The points $x_k$ can be easily classified (stable, asymptotically stable or unstable) because they are hyperbolic points ($f'(x_k) \ne 0$). What about $x=0$? I've read on Hale-Koçak that $x=0$ seems to be stable but not asymptotically stable.

I managed to prove that we cannot find a $\delta > 0$ s.t. $xf(x)<0$ for $0<\vert x\vert <\delta$: by a lemma on Hale-Koçak, this tells us the point is not asymptotically stable.

What about stability? I should prove that $ \exists \delta > 0, \, \vert x \vert < \delta \Rightarrow xf(x)\le 0 $

I don't manage to prove this. How would you do? Thanks for your help.

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    @Romeo : Please notice how I used the ampersand (&) in formatting the "cases" environment. This gets $y$ou proper alignment without fiddling with \quad.2012-06-19

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I add to my comment above.

To prove the stability of $\hat x=0$ you do not need to prove (you simply can't) that

$\exists \delta > 0, \, \vert x \vert < \delta \Rightarrow xf(x)\le 0$

Actually, this example is given in the book because it shows that the implication $\hat x$ is stable $\Rightarrow$ $(x-\hat x)f(x)\leq 0$ is not true (whereas the converse is true).

To prove the stability of $\hat x=0$ you just need to use the definition.

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    Yes, you are right: it cannot jump over other equilibrium points. Now it's clear. Thanks for your kind explanation, you've been exceptionally clear. Thank you very much.2012-06-20