Lyapunov's Direct Method

Lyapunov's direct (or second) method determines stability without solving the differential equation, by finding an appropriate energy-like function.

Statement

Theorem7.7Lyapunov stability theorem (complete)

Let $\mathbf{x}_0$ be an equilibrium of $\mathbf{x}' = \mathbf{F}(\mathbf{x})$ and let $V: D \to \mathbb{R}$ be a $C^1$ function on a neighborhood $D$ of $\mathbf{x}_0$ with $V(\mathbf{x}_0) = 0$ .

If $V$ is positive definite and $\dot{V} \leq 0$ (negative semi-definite), then $\mathbf{x}_0$ is stable.
If $V$ is positive definite and $\dot{V}$ is negative definite, then $\mathbf{x}_0$ is asymptotically stable.
If $V$ is positive definite, $\dot{V}$ is negative semi-definite, and the only solution of $\mathbf{x}' = \mathbf{F}(\mathbf{x})$ that stays in $\{\dot{V} = 0\}$ is $\mathbf{x}(t) \equiv \mathbf{x}_0$ , then $\mathbf{x}_0$ is asymptotically stable (LaSalle).

Proof of Part 1 (Stability)

Proof

Given $\varepsilon > 0$ , let $B_\varepsilon = \{\|\mathbf{x} - \mathbf{x}_0\| < \varepsilon\} \subset D$ . Since $V$ is positive definite and continuous, $m = \min_{\|\mathbf{x}-\mathbf{x}_0\|=\varepsilon} V(\mathbf{x}) > 0$ .

Since $V$ is continuous and $V(\mathbf{x}_0) = 0$ , there exists $\delta > 0$ such that $V(\mathbf{x}) < m$ for $\|\mathbf{x}-\mathbf{x}_0\| < \delta$ .

For $\|\mathbf{x}(0) - \mathbf{x}_0\| < \delta$ : $V(\mathbf{x}(0)) < m$ . Since $\dot{V} \leq 0$ , $V(\mathbf{x}(t)) \leq V(\mathbf{x}(0)) < m$ for all $t \geq 0$ .

If $\mathbf{x}(t)$ ever reached $\|\mathbf{x}(t) - \mathbf{x}_0\| = \varepsilon$ , then $V(\mathbf{x}(t)) \geq m$ , contradicting $V(\mathbf{x}(t)) < m$ . Therefore $\|\mathbf{x}(t) - \mathbf{x}_0\| < \varepsilon$ for all $t \geq 0$ . $\blacksquare$

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Instability Theorem

Theorem7.8Chetaev's instability theorem

Let $V: D \to \mathbb{R}$ be $C^1$ with $V(\mathbf{x}_0) = 0$ . If in every neighborhood of $\mathbf{x}_0$ there exists a point where $V > 0$ , and $\dot{V} > 0$ on the set $\{V > 0\} \cap D$ , then $\mathbf{x}_0$ is unstable.

ExampleInstability via Chetaev

For $x' = y + x(x^2+y^2)$ , $y' = -x + y(x^2+y^2)$ : try $V = x^2 + y^2$ .

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

Since $V > 0$ near the origin and $\dot{V} > 0$ , the origin is unstable.

RemarkComparison with linearization

Lyapunov's method and linearization are complementary:

Linearization is easy when eigenvalues have nonzero real parts.
Lyapunov functions handle non-hyperbolic cases and provide stability regions.
For nonlinear systems, Lyapunov functions can prove global stability, which linearization cannot.