Lyapunov Stability

Wikipedia · Lyapunov stability · CC BY-SA 4.0

Energy-based stability

A ball at the bottom of a bowl is stable: push it, and it rolls back. A ball on top of a hill is unstable: push it, and it rolls away. The Lyapunov function is the "height" function. If the system always moves downhill (V decreases), it converges to the bottom (equilibrium). You don't need to compute the trajectory. You just need to show that V keeps decreasing.

The conditions

A function V(x) is a Lyapunov function for the equilibrium x=0 if: (1) V(0) = 0, (2) V(x) is greater than 0 for x not equal to 0 (positive definite), and (3) dV/dt is less than or equal to 0 along trajectories (V is non-increasing). If dV/dt is strictly negative (not just non-increasing), the equilibrium is asymptotically stable: the state converges to zero.

LaSalle's invariance principle

What if dV/dt is only zero on some set, not everywhere? LaSalle's principle says: the system converges to the largest invariant set where dV/dt = 0. This handles cases where the Lyapunov function is not strictly decreasing everywhere but still guarantees convergence.

; Lyapunov analysis: x' = -x^3
; Candidate V(x) = x^2 / 2
; dV/dt = x * x' = x * (-x^3) = -x^4 <= 0
; V is positive definite, dV/dt is negative definite -> stable

(define dt 0.01)

(define (simulate x steps)
  (if (= steps 0)
    (begin (display "Done.") (newline))
    (let* ((x-dot (* -1.0 x x x))         ; x' = -x^3
           (V (* 0.5 x x))                 ; V(x) = x^2/2
           (dVdt (* -1.0 x x x x))         ; dV/dt = -x^4
           (new-x (+ x (* x-dot dt))))
      (if (= 0 (modulo steps 50))
        (begin
          (display "  x=") (display x)
          (display "  V=") (display V)
          (display "  dV/dt=") (display dVdt)
          (newline))
        'skip)
      (simulate new-x (- steps 1)))))

(display "Lyapunov stability: x'=-x^3, V=x^2/2") (newline)
(display "(V decreases at every step -> stable)") (newline)
(simulate 2.0 500)

# Lyapunov analysis: x' = -x^3
# V(x) = x^2/2, dV/dt = -x^4 < 0 (stable)
dt = 0.01
x = 2.0

print("Lyapunov stability: x'=-x^3, V=x^2/2")
print("(V decreases at every step -> stable)")
for step in range(500):
    V = 0.5 * x**2
    dVdt = -x**4
    if step % 50 == 0:
        print(f"  step={step:3d}  x={x:+.4f}  V={V:.6f}  dV/dt={dVdt:.6f}")
    x += (-x**3) * dt

print(f"  final   x={x:+.4f}  V={0.5*x**2:.6f}")