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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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    $\begingroup$ Mmmmm.... For $x \neq 0$, $$x f(x) = -x^4 \sin \frac{1}{x}.$$ Do you believe this expression may have constant sign? $\endgroup$
    – Siminore
    Commented Jun 19, 2012 at 17:07
  • $\begingroup$ Well, I agree: you are right. But how would you prove stability? I do not know other useful theorems that helps with this. Maybe we should use directly the definition. What do you think? Thanks. $\endgroup$
    – Romeo
    Commented Jun 19, 2012 at 17:16
  • $\begingroup$ The equilibrium $\hat x=0$ is not isolated therefore it cannot be asymptotically stable. Since the state space is one dimensional and any neighborhood of $\hat x=0$ contains other equilibria, hence $\hat x=0$ is neutrally stable. $\endgroup$
    – Artem
    Commented Jun 19, 2012 at 17:23
  • $\begingroup$ @Romeo : Please notice how I used the ampersand (&) in formatting the "cases" environment. This gets you proper alignment without fiddling with \quad. $\endgroup$ Commented Jun 19, 2012 at 17:42

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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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  • $\begingroup$ First, I thank you for your answer. Let's take $\varepsilon \gt 0$. I have to show that there exists $\delta > 0$ s.t. for every $\vert x_0 \vert < \delta$ I have $\vert x(t,x_0) \vert < \varepsilon$ for every $t \ge 0$. I take $k$ s.t. $x_k=\frac{1}{k\pi}$ is asymptotically stable and $k \gt \frac{1}{\varepsilon \pi}$: now, I think it's enough to apply the definition of asymptotically stability to the point $x_k$ with $\varepsilon_1=\varepsilon-\frac{1}{k\pi}$. But I still have some doubts, I'm not sure about this. What do you think? $\endgroup$
    – Romeo
    Commented Jun 20, 2012 at 10:52
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    $\begingroup$ @Romeo Given any $\varepsilon$ you need to find a $\delta$. Find the smallest $k$ such that $1/(k\pi)<\varepsilon$ and put $\delta=1/(k\pi)$. Any orbit with the initial point $|x_0|<\delta$ will stay in a neighborhood of $\hat x=0$ and hence $|x(t;x_0)|<\varepsilon$. Hence the Lyapunov (neutral) stability. $\endgroup$
    – Artem
    Commented Jun 20, 2012 at 11:35
  • $\begingroup$ Sorry, but I've one more question: you have written "Any orbit with the initial point $\vert x_0 \vert < \delta$ will stay in a neighborhood of $\hat{x}=0$". Why is this true? I'd expect something like "Any orbit with the initial point $\vert x_0 -1/k\pi \vert < \delta $ will stay in a neighborhood of $1/(k\pi)$". I can't understand why we can take the initial point $\vert x_0 \vert <\delta$. Thanks. $\endgroup$
    – Romeo
    Commented Jun 20, 2012 at 11:53
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    $\begingroup$ @Romeo This is definition. Given any $\varepsilon$ you need to find a $\delta$ so that any orbit starting in an $\delta$-neighborhood of $\hat x=0$ will stay in its $\varepsilon$-neighborhood. Take, for example, $\varepsilon=0.01$. Then I claim that any orbit with $x_0<1/(32 \pi)$ will be such that $|x(t;x_0)|<\varepsilon$, just because it cannot jump over other equilibrium points. But this is exactly the definition of Lyapunov stability. $\endgroup$
    – Artem
    Commented Jun 20, 2012 at 12:02
  • $\begingroup$ 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. $\endgroup$
    – Romeo
    Commented Jun 20, 2012 at 12:11

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