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An equilibrium point $x^* ∈X$ for an ODE or a DE is stable (S) if $$∀ε > 0, ∃δ>0$$ s.t. $t_0 ∈ T$, $x_0 ∈ X$, $\|x_0−x^*\|_{I\!R^n} < δ =⇒ \|x(t;t_0,x_0)−x^*\|_{I\!R^n} ≤ ε$ $∀t ∈ T$

If it is not stable it is unstable (U).

If this is the definition, why do we say that B is unstable? In the end we can easily pick $$δ,ε>0 ~~s.t.~~ t_0 ∈ T, x_0 ∈ X, ||x_0−x^*||_{I\!R^n} < δ =⇒ \|x(t;t_0,x_0)−x^*\|_{I\!R^n} ≤ ε~~∀t ∈ T$$

Both from the right where it is also A.S. and from the left. I do not understand why B in the phase diagram below is unstable if the definition of eq point is as I stated above. Can you help me?enter image description here

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  • $\begingroup$ What is this $R$ in your definition? $\endgroup$ – Vincent Dec 14 '19 at 16:26
  • $\begingroup$ If I understand correctly what you're asking, it doesn't work from the left if $\varepsilon$ is less than the distance from $A$ to $B$. $\endgroup$ – Hans Lundmark Dec 14 '19 at 16:42
  • $\begingroup$ Somehow you are missing the definition of what $T$ is. Also, I would expect that only the forward dynamic is interesting for the definition, that is, $t\ge t_0$. (Also look at the first "Related" link, which atm is math.stackexchange.com/questions/686673, as it is a concretization of your sketch.) $\endgroup$ – Lutz Lehmann Dec 14 '19 at 19:20
  • $\begingroup$ Yes, it is as you said. I know that B is unstable but it is not vlear ti me why this is true, using the definition $\endgroup$ – LearningProb Dec 15 '19 at 10:37
  • $\begingroup$ How can you find $\delta$ if $x^* = B$ so that your conditions are hold for arbitrary $\epsilon$? $\endgroup$ – Pasha Dec 20 '19 at 18:34

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