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Let assume that the dynamical system $x_{t+1}=f(x_t)$ (with $f(0)=0$) is globally unstable (which means from any initial condition $x\neq0$, the states go to infinity). Can we prove that $x_{t+1}=f(x_t)+y_t$ (where $y_t$ is norm bounded $||y_t||<c$ and is not a function of $x_t$. $y_t$ also changes its vlue over time with some unknown random distribution. So its value is not fixed over time) is unstable for some $x_0 \neq 0$ with $||x_0||\leq c$ for any $c>0$?

Amir Khazraei
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  • Given $a,$ if you define $y_t=a-f(a)$ for all $t$ then when $x_0=a$ your recurrence is constant, and hence does not converge to infinity. So it is not globally unstable. – Thomas Andrews Sep 22 '21 at 23:56
  • Thank you for your comment, but I think in your example $y_0$ is a function of $x_0$ and this will be also the case for the following time steps. – Amir Khazraei Sep 23 '21 at 00:01
  • It is not a function of $x_0.$ It is a dynamic sequence, and it determines one $x_0$ which does not converge to infinity. You are not going to be able to define “not a function of $x_t$“ rigorously. – Thomas Andrews Sep 23 '21 at 00:02
  • By having $x_0=a$, we have $y_0=a-f(a)=x_0-f(x_0)$. This will result in $x_1=f(a)+a-f(a)=a$ and your $y_1=a-f(a)=x_1-f(x_1)$. But you are right about the assumption and let assume that y_t is a bounded random sequence. – Amir Khazraei Sep 23 '21 at 00:08
  • $f(x)=2x$ has your property. But $y_t=c$ for any constant $c\neq 0$ has some fixed point. – Thomas Andrews Sep 23 '21 at 00:10
  • A constant is a random sequence, it is just a random sequence with one value. You are going to need to be more specific about what you mean by “$y_t$ is a random bounded sequence.” – Thomas Andrews Sep 23 '21 at 00:12
  • Yes, you are right.But in my system $y_t$ changes its vlue over time with some unknown random distribution. So its value is not fixed over time. – Amir Khazraei Sep 23 '21 at 00:17
  • The you are going to need to be specific in the question. Nothing in your question suggests there will be a random process involved, and nothing in your question suggests that we can’t use a constant $y_t.$ It is up to you to ask a complete question. – Thomas Andrews Sep 23 '21 at 00:19

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