# Equilibrium points and stability

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?

• What is this $R$ in your definition? – Vincent Dec 14 '19 at 16:26
• 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 $B$. – Hans Lundmark Dec 14 '19 at 16:42
• 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.) – Lutz Lehmann Dec 14 '19 at 19:20
• 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 – LearningProb Dec 15 '19 at 10:37
• How can you find $\delta$ if $x^* = B$ so that your conditions are hold for arbitrary $\epsilon$? – Pasha Dec 20 '19 at 18:34