# lyapunov stability of autonomous system

Consider a discrete-time system of the form

$$x(k+1) = Ax(k)$$

where $$x \in \mathbb R^{n \times 1}$$ and $$A \in \mathbb R^{n \times n}$$. This system is (globally) asymptotically stable if the eigenvalues of $$A$$ are inside the unit circle. If the eigenvalues are inside the unit circle then the (only) equilibrium point "must" be the origin; $$x \to x_e=0$$ as $$k \to \infty$$ (where $$x_e$$ denotes the equilibrium point). The states $$x(k)$$ will converge to zero.

What if $$x_e \neq 0$$? In that case, one of the eigenvalue will lie on the unit circle; $$x \to x_e \neq 0$$, as $$k \to \infty$$. Please correct me if I am wrong.

In this case, the equilibrium point will change with initial points, $$x_0$$ $$(x_0 \in R^{n \times 1} )$$, that is the initial point $$x_{0,1}$$ will lead to $$x_{e,1, }$$, while $$x_{0,2} \to x_{e,2}$$ and so on..., (also $$x_0=0, \to x_e \to 0$$).

So we can say that the set of initial values converges to a set of equilibrium points. Since every initial point always converges to a particular equilibrium point the system should be "asymptotically" stable?

In other words, how can I prove the asymptotic stability when there is no fixed equilibrium point (and depends upon the initial values) and an eigenvalue at unit circle.

• If one of the eigenvalue lies on the unit circle itself, there's no convergence. Hence, there is no equilibrium point. – Rodrigo de Azevedo Oct 14 '19 at 20:30

It can be noted that all but one eigenvalue inside the unit circle and one on the unit circle isn't enough for $$x(k)\to x^*\neq0$$, as $$k\to\infty$$. For example $$x(k+1)=-x(k)$$ keeps changing sign for a nonzero initial condition. A similar thing can also be shown with complex conjugate eigenvalues with nonzero imaginary part. Your statement only hold when the eigenvalue on the unit circle is equal to one.
In order for an equilibrium point to be asymptotically stable all neighboring solutions should eventually converge to the same equilibrium point, which isn't the case. For example for $$x(k+1)=x(k)$$ only the solution starting at $$x(0)=a\in\mathbb{R}$$ "converges" to $$x^*=a$$.
• however you said that for asymptotic stability there should be a single equilibrium point. but in my case the equilibrium point depends upon the initial states (e.g, ${x_o}^T1=x^{*T}1$ and $Cx^*=constant$, where $C$ is constant diagonal matrix). The matrix $A$ always converges the $x_o$ to the corresponding equilibrium point. – fas Oct 15 '19 at 6:08
Following the comments of @Kwin i have slightly redesigned the Matrix $$A$$, to be applied on the state-errors, instead of states itself, that makes $$0$$ as a single equilibrium point for the system. the resulting matrix has eigenvalues inside the unit circle, :)