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I am trying to prove that a positively asymptotically stable fixed point cannot also be classified as negatively stable.

I am employing the definition of positive (negative) stability as: A point p is said to be positively stablie if

(1) There exists $r>0$ such that when $|\zeta -p|< r$ the solution $x(t, \zeta )$ is defined for all $t \ge 0$ (or $t \le 0$ for negatively stability)

(2) Given $\epsilon \ge 0$ there exists $\delta \ge 0$ such that $|x(t, \zeta ) -p|< \epsilon $ for all $t \ge 0$ (or $t \le 0$ for negatively stability) when $|\zeta -p|< \delta $

Furthermore a point is said to be positively asymptotically stable when the additional condition is satisfied

(3) There exists $\gamma \ge 0$ such that $\lim_{t\to\infty}x(t,\zeta)=p$ whenever $| \zeta -p| < \gamma$

Do I prove this by contradiction by allowing p to be positively asymptotically stable and assume that it is also negatively stable and show the two are not compatible? I'm not entirely sure how to start the proof.

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Assume both, and take the $\epsilon$ in the definition of negatively stable smaller than the $\gamma$ in the definition of positively asymptotically stable. Then think of a solution which starts within distance $\gamma$ from the fixed point, but not within distance $\epsilon$. After a while, it will be within distance $\delta$. Now what happens if you start from that position and run time backwards?

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  • $\begingroup$ If we take $\epsilon$ <γ, then choose β such that ϵ<β<γ.Then when|ζ−p|<β, $x(t,\zeta )$is defined for all $t \ge 0$. As time increases given $\beta > \epsilon \ge 0 $ there exists $\delta(\beta) >0$ such that $|x(t,\zeta )-p|< \beta$ for all $t \ge 0$ when $|\zeta -p|< \delta(\beta)$. Now reversing the time will give the contradiction, but here I am unsure of the specifics. When time reverses from this point, then given $0 \le \epsilon < \gamma$ then we have found that $|x(t,\zeta )-p|> \epsilon$ for all $t \le 0$ when $|\zeta -p|< \delta(\beta)$. Is this the desired contradiction? $\endgroup$
    – user75514
    Feb 10, 2014 at 5:44
  • $\begingroup$ Yes, except that the last "all" should be "some". $\endgroup$ Feb 10, 2014 at 6:50

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