PLS HELP ME. I have a problem with this exercise given by the professor for home. It's about Lyapunov equation and autonomous systems. Here it is:
Prove that if the state of equilibrium $x^*=0$ $(x^*\in\mathbb{R}^n)$ of the system:
$x(k+1)=e^A x(k)$ with $A\in \mathbb{R}^{n\times n}$
is asymptotically stable then even the equilibrium state $x^{**}=0$ of the system:
$\dot{x}(t)=Ax(t)$
is asymptotically stable.
I thought I could apply the step invariant transformation for an autonomous system and since it should preserve the response it should be asymptotically stable. But i'm not sure... Hope someone of you can help me, thanks in advice!
I'm not sure what tools you have available, but here is an approach using the definition of asymptotic stability.
Consider trajectories corresponding to the common initial state $x(0) = x_0$. For any $k = 0,1,2,3,\dots$ and $t \in [k,k+1]$, we have $$ \begin{align} \left\|e^{tA} x_0\right\| &= \left\|e^{(t-k)A} \cdot e^{kA} \cdot x_0\right\| \\ & \leq \|e^{(t-k)A}\| \cdot \left\|e^{kA} \cdot x_0\right\| \\ & \leq e^{\|(t-k)A\|} \cdot \left\|e^{kA} \cdot x_0\right\| \\ & \leq e^{\|A\|} \cdot \left\|e^{kA} \cdot x_0\right\| = e^{\|A\|} \cdot \|x(k)\|. \end{align} $$ In other words, we have $0 \leq \|e^{tA}x_0\| \leq e^{\|A\|} \cdot \|x(\lfloor t\rfloor )\|$. If $\lim_{k \to \infty}\|x(k)\| = 0$, then it follows from the squeeze theorem that $\lim_{t \to \infty} \|e^{tA}x_0\| = 0$, which is to say that the continuous time system is asymptotically stable.