MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $P_m \leq f(\alpha ) P$.

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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