11.1 Lyapunov’s direct method
Definition 11.1.1. A continuous function \(V\) on a neighbourhood \(D\) of \(0\) is positive definite if \(V(0) = 0\) and \(V(x) > 0\) for \(x \neq 0\). A \(C^1\) positive definite \(V\) with \(\dot V \leq 0\) on \(D\) is a Lyapunov function; if \(\dot V < 0\) on \(D \setminus \{0\}\) it is a strict Lyapunov function.
Theorem 11.1.2 (Lyapunov). If there is a Lyapunov function, the equilibrium \(0\) is stable. If there is a strict Lyapunov function, \(0\) is asymptotically stable.
Proof. Stability. Given \(\varepsilon > 0\) with \(\bar B(0, \varepsilon ) \subset D\), let \(m = \min \{V(x) : \|x\| = \varepsilon \} > 0\), a minimum of a positive continuous function on a compact sphere. By continuity of \(V\) at \(0\) there is \(\delta < \varepsilon \) with \(V(x) < m\) for \(\|x\| < \delta \). A solution starting there has \(V(\phi _t(x_0)) \leq V(x_0) < m\) for \(t \geq 0\), because \(\dot V \leq 0\), so it can never reach the sphere \(\|x\| = \varepsilon \), where \(V \geq m\): it stays in the ball.
Asymptotic stability. With \(\delta \) as above, \(V(\phi _t(x_0))\) is non-increasing and bounded below, so it tends to a limit \(c \geq 0\). Suppose \(c > 0\). By continuity of \(V\) at \(0\) there is \(\eta > 0\) with \(V < c\) on \(B(0, \eta )\), so the solution stays in the compact shell \(\eta \leq \|x\| \leq \varepsilon \), on which the continuous function \(\dot V\) is negative, with maximum \(-\gamma < 0\). Then \(V(\phi _t(x_0)) \leq V(x_0) - \gamma t \to -\infty \), which is impossible. So \(c = 0\), and since \(V\) is positive definite, \(\phi _t(x_0) \to 0\) (a solution staying away from \(0\) would keep \(V\) away from \(0\), by the same compactness argument). □
The theorem has a clear geometric meaning. The level sets \(\{V = c\}\) near a positive definite minimum are closed surfaces around \(0\), and \(\dot V \leq 0\) says that solutions cross them inwards or along them, never outwards. Finding \(V\) is an art; physical energy, quadratic forms, and sums of squares of the state are the usual first candidates.
Theorem 11.1.3 (Chetaev’s instability theorem). If there is a \(C^1\) function \(V\) with \(V(0) = 0\), taking positive values at points arbitrarily close to \(0\), and with \(\dot V > 0\) wherever \(V > 0\) in a neighbourhood of \(0\), then \(0\) is unstable.
Proof. Fix \(r > 0\) with \(\bar B(0, r)\) in the neighbourhood, and take \(x_0\) arbitrarily close to \(0\) with \(V(x_0) = a > 0\). While the solution stays in \(\bar B(0, r)\), \(V\) increases along it, so it remains in the compact set \(K = \{x \in \bar B(0, r) : V(x) \geq a\}\), which does not contain \(0\), and on which \(\dot V \geq \gamma > 0\). Then \(V(\phi _t(x_0)) \geq a + \gamma t\), which is unbounded, while \(V\) is bounded on the ball; so the solution must leave \(\bar B(0, r)\). Since \(x_0\) was arbitrarily close to \(0\), the equilibrium is unstable. □
Example 11.1.4. For \(x' = -y - x^3\), \(y' = x - y^3\), the linearisation is a centre, and Theorem 10.2.2 is silent. With \(V = x^2 + y^2\), \(\dot V = 2x(-y - x^3) + 2y(x - y^3) = -2(x^4 + y^4) < 0\) for \((x, y) \neq 0\), so \(V\) is a strict Lyapunov function on the whole plane, and the origin is asymptotically stable. With the signs of the cubic terms reversed, \(\dot V = 2(x^4 + y^4) > 0\) and Chetaev’s theorem gives instability.
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