Global Stability and Advanced Lyapunov Methods

Beyond local stability, Lyapunov methods can establish global stability and provide quantitative bounds on convergence rates.

Global Asymptotic Stability

Definition7.5Global asymptotic stability

An equilibrium $\mathbf{x}_0$ is globally asymptotically stable (GAS) if it is stable and $\mathbf{x}(t) \to \mathbf{x}_0$ as $t \to \infty$ for all initial conditions in $\mathbb{R}^n$ (not just those near $\mathbf{x}_0$ ).

Theorem7.5Global Lyapunov theorem

If there exists a $C^1$ function $V: \mathbb{R}^n \to \mathbb{R}$ such that:

$V(\mathbf{x}_0) = 0$ and $V(\mathbf{x}) > 0$ for $\mathbf{x} \neq \mathbf{x}_0$ .
$V(\mathbf{x}) \to \infty$ as $\|\mathbf{x}\| \to \infty$ (radially unbounded).
$\dot{V}(\mathbf{x}) < 0$ for $\mathbf{x} \neq \mathbf{x}_0$ .

Then $\mathbf{x}_0$ is globally asymptotically stable.

ExampleGlobal stability of a nonlinear system

For $x' = -x^3$ , $y' = -y$ : let $V = x^4/4 + y^2/2$ .

$\dot{V} = x^3(-x^3) + y(-y) = -x^6 - y^2 < 0$ for $(x,y) \neq (0,0)$ .

$V$ is radially unbounded since $V \to \infty$ as $\|(x,y)\| \to \infty$ . Therefore $(0,0)$ is GAS.

Constructing Lyapunov Functions

RemarkStrategies for finding Lyapunov functions

Quadratic forms: For linear systems $\mathbf{x}' = A\mathbf{x}$ , try $V = \mathbf{x}^T P\mathbf{x}$ where $P$ is positive definite. Then $\dot{V} = \mathbf{x}^T(A^TP + PA)\mathbf{x}$ . The Lyapunov equation $A^TP + PA = -Q$ (with $Q > 0$ ) has a unique solution $P > 0$ iff $A$ is stable.
Energy functions: For mechanical systems $m\ddot{q} = -\nabla U(q) - \beta\dot{q}$ , the total energy $E = \frac{1}{2}m|\dot{q}|^2 + U(q)$ satisfies $\dot{E} = -\beta|\dot{q}|^2 \leq 0$ .
Variable gradient method: Assume $\nabla V = M(\mathbf{x})\mathbf{x}$ for a matrix $M$ and determine $M$ from integrability conditions.

ExampleSolving the Lyapunov equation

For $A = \begin{pmatrix}-1&2\\0&-3\end{pmatrix}$ , solve $A^TP + PA = -I$ :

$\begin{pmatrix}-1&0\\2&-3\end{pmatrix}\begin{pmatrix}p_{11}&p_{12}\\p_{12}&p_{22}\end{pmatrix} + \begin{pmatrix}p_{11}&p_{12}\\p_{12}&p_{22}\end{pmatrix}\begin{pmatrix}-1&2\\0&-3\end{pmatrix} = -I.$

Solving: $p_{11} = 5/6$ , $p_{12} = 1/4$ , $p_{22} = 1/6$ . Check $P > 0$ : $\det P > 0$ and trace $> 0$ .

Barbashin-Krasovskii Theorem

Theorem7.6Barbashin-Krasovskii theorem

If $V$ satisfies $V > 0$ , $V$ radially unbounded, and $\dot{V} \leq 0$ , and if the largest invariant set in $\{\dot{V} = 0\}$ is $\{\mathbf{x}_0\}$ , then $\mathbf{x}_0$ is globally asymptotically stable.

ExampleDamped pendulum

$\theta'' + b\theta' + \sin\theta = 0$ ( $b > 0$ ). Energy: $V = \frac{1}{2}\dot\theta^2 + (1-\cos\theta) \geq 0$ , with $V = 0$ only at $(\theta, \dot\theta) = (0, 0)$ .

$\dot{V} = \dot\theta(\ddot\theta + \sin\theta) = \dot\theta(-b\dot\theta) = -b\dot\theta^2 \leq 0$ .

$\{\dot{V} = 0\} = \{\dot\theta = 0\}$ . On this set, $\ddot\theta = -\sin\theta$ , which is zero only at $\theta = n\pi$ . The only invariant subset near $\theta = 0$ is $\{(0,0)\}$ . By Barbashin-Krasovskii, the downward equilibrium is (locally) asymptotically stable.