3
$\begingroup$

In regard to the stability analysis and control properties of the linear system $\dot{x}=Ax$.

Consider the solution $P$ of the continuous Lyapunov equation $AP+PA^T+Q=0$, where $A,Q,P \in {\mathbb{R}}^{n\times n}$, $A$ is a stable matrix and $Q$ is positive semidefinite.

Stability theory states that $P$ is positive definite.

If we were to modify the linear dynamics to $\dot{x}=(A-\alpha I) x$ where $\alpha >0$ then the new corresponding Lyapunov equation is $(A-\alpha I)P_m+P_m(A-\alpha I)^T+Q=0$ with solution $P_m$ (which is again positive definite).

The solution $P $ can also be found directly by $P=\int_0^\infty e^{At}Qe^{A^Tt}dt$ and similarly $P_m=\int_0^\infty e^{(A-\alpha I)t}Qe^{(A^T-\alpha I)t}dt=\int_0^\infty e^{-\alpha t} e^{At}Qe^{A^Tt}dt$.

Consequently, as $\forall t\geq 0$ we have $e^{-\alpha t}\leq {1}$, then $P_m \leq P$ but does a stronger bound exist?

For example a tighter bound of the form $P_m \leq f(\alpha ) P$.

$\endgroup$
2
$\begingroup$

Probably not without extra assumptions. If the eigenvalues of $A$ are fast, the integral for $P_m$ (in whose expression I believe a factor 2 is missing) will be weighted towards its values when $t$ is small, and $e^{-\alpha t}$ is close to one. So you can't have an expression that depends on $\alpha$ only.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.