Local asymptotically stability at a non derivable equilibrium point

Question

Given $\alpha , \beta > 0$ we define $$f(x)= \begin{cases} \alpha x & x \geq 0 \\ -\beta x & x <0 \\ \end{cases}$$

It´s obvious that $p=0$ is a fixed point of the system $x_{n+1}=f(x_n)$, I´m having problems to get the conditions for $\alpha , \beta > 0$ such p is an locally asymptotically stable point.

Well, I see clearly that they should be lower than $1$, but I cannot use the derivative criterium because $f$ is not derivable at $0$, how I can prove this formally? (probably with the definition, but I cannot see)

copper.hat · Accepted Answer · 2020-05-19T18:09:01.197

1

If $\beta>0$ then $f(x_n) \ge 0$ after the first iteration, so we can assume that $x_n \ge 0$, so we need only examine $\alpha$.

If $\alpha >1$ then if $x_0 = 1$ we have $x_n = \alpha^n$, which does not converge.

if $\alpha=1$ then if $x_0 = 1$ we have $x_n =1$, so clearly it does not converge to $0$.

If $\alpha \in [0,1)$ then we have $x_n = \alpha^n x_0$ and since $|\alpha|<1$ we see that $x_n \to 0$.

edited May 19 '20 at 18:09

answered May 19 '20 at 17:28

copper.hat

172,524

But f is lipchitz rank less 1 only if $\alpha , \beta$ are less than one, and in the beggining you do not know – Tt Nach00 May 19 '20 at 17:43
I did not read the question properly, I have updated my answer. – copper.hat May 19 '20 at 18:09
Nice, ok thankyou, now I notice this little detail, it works by definition. – Tt Nach00 May 19 '20 at 18:23

Local asymptotically stability at a non derivable equilibrium point

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