Given the planar linear system $ x' = Ax $ with $ A $ as below, find a symmetric matrix $ P $ such that $ V(x) = x^T P x $ is a Lyapunov function whose time derivative along solutions satisfies $ \dot{V} = -x^T Q x $ with $ Q $ the identity matrix.\[A = \begin{pmatrix} -1 & 0 \\ 2 & -1 \end{pmatrix}\]Explanation: - The problem involves determining a symmetric matrix $ P $ for the Lyapunov function $ V(x) $.- The matrix $ A $ is given as $ \begin{pmatrix} -1 & 0 \\ 2 & -1 \end{pmatrix} $.- $ Q $ is specified as the identity matrix.- The objective is to ensure that the derivative $ \dot{V} $ is expressed in the form $ -x^T Q x $.

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Given the planar linear system ' = A.x with A as below, find a symmetric matrix P such that V (x) = x" P is a Lyapunov function whose time derivative along solutions satisfies V = –xTQx with Q the identity matri -1 A = 2.

Given the planar linear system ' = A.x with A as below, find a symmetric matrix P such that V (x) = x" P is a Lyapunov function whose time derivative along solutions satisfies V = –xTQx with Q the identity matri -1 A = 2.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Given the planar linear system $ x' = Ax $ with $ A $ as below, find a symmetric matrix $ P $ such that $ V(x) = x^T P x $ is a Lyapunov function whose time derivative along solutions satisfies $ \dot{V} = -x^T Q x $ with $ Q $ the identity matrix.

\[
A = \begin{pmatrix} -1 & 0 \\ 2 & -1 \end{pmatrix}
\]

Explanation:
- The problem involves determining a symmetric matrix $ P $ for the Lyapunov function $ V(x) $.
- The matrix $ A $ is given as $ \begin{pmatrix} -1 & 0 \\ 2 & -1 \end{pmatrix} $.
- $ Q $ is specified as the identity matrix.
- The objective is to ensure that the derivative $ \dot{V} $ is expressed in the form $ -x^T Q x $.

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