most recent 30 from http://mathoverflow.net 2013-05-19T04:46:50Z http://mathoverflow.net/feeds/question/89818 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89818/stronger-bound-for-a-modified-lyapunov-equation unknown (yahoo) 2012-02-29T01:16:59Z 2012-08-21T05:22:01Z

<p>In regard to the stability analysis and control properties of the linear system $\dot{x}=Ax$.</p> <p>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.</p> <p>Stability theory states that $P$ is positive definite.</p> <p>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).</p> <p>The solution $P $ can also be found directly by $P=\int_0^\infty e^{At}Qe^{A^Tt}dt$ and similarly $P_m=\int_0^\infty e^{(A-\alpha I)t}Qe^{(A^T-\alpha I)t}dt=\int_0^\infty e^{-\alpha t} e^{At}Qe^{A^Tt}dt$.</p> <p>Consequently, as $\forall t\geq 0$ we have $e^{-\alpha t}\leq {1}$, then $P_m \leq P$ but does a stronger bound exist?</p> <p>For example a tighter bound of the form $P_m \leq f(\alpha ) P$.</p>

http://mathoverflow.net/questions/89818/stronger-bound-for-a-modified-lyapunov-equation/91659#91659 Pait 2012-03-19T19:37:49Z 2012-03-19T19:37:49Z

<p>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.</p>