I have designed a matrix ($A \in \Bbb R^{n\times n}$) that will converges the vector $x\in\Bbb R^{n\times 1}$ to its equilibrium point $ x^* \ne 0$. If i do not translate $x$ as $x'=x-x^*$, then
$ \lim_{k \to \infty} x(k+1) = Ax(k) \to x^* \neq 0.$
this (perhaps) means that not all the eigenvalues of $A$ are within the unit circle and at least one of the eigenvalues is on unit circle as $x(k) \not \to 0$?
This means that without the translation $(x'=x-x^*)$, we cannot represent the asymptotic stability of the system, where $x^* \neq 0$ ?
DETAIL: In my case, the equilibrium point of the system is not known exactly and decided dynamically, so I cannot use the translation ($x'=x-x^{*}$) and the matrix $A$ designed without the translation has a single eigenvalue at unit circle. how can i show the asymptotic stability without using translation and one of the eigenvalue at unit circle.
Best Answer
If the operating point is decided dynamically, it all depends on the dynamics how the operating point is moved around. I see two possibilities:
If you can write the dynamics of the operating point as a system of ODEs you can just include these in your original system. Then you check stability of this new system.
If you have no ODEs for the operating point you have to treat its movement as an exogenous input, so you get a time varying system which you can check for stability.
If you post a concrete example, it might be possible to give more specific help.