a Lyapunov function
/ lee-ah-POO-nawf /
Imagine a ball rolling in a bowl. You do not need to solve its equations of motion to know it ends up at the bottom — you just notice that its height (its energy) never increases as it rolls, so it must keep sinking toward the lowest point. A Lyapunov function is exactly this idea made into a tool: a single quantity, shaped like a bowl around an equilibrium, that the system can only keep flat or push downhill, which forces the trajectory to stay near, or fall into, the bottom.
Formally, near an equilibrium x* (placed at the origin for convenience) a Lyapunov function V(x) is a scalar function that is positive everywhere nearby except zero at x* itself — that is what 'bowl shaped' means (positive definite). The crucial test is on its rate of change along trajectories: V' = (gradient of V) dotted with f(x), the derivative of V as you ride a solution. If V' is less than or equal to zero (V never increases), the equilibrium is Lyapunov stable; if V' is strictly negative away from x* (V always strictly decreases), the equilibrium is asymptotically stable. You compute V' directly from the equations, without ever solving them.
The power here is that you certify stability with no formula for the solution — priceless for nonlinear systems that cannot be solved. The honest difficulties are two. First, finding a working V is an art: there is no universal recipe, though physical energy is often a superb first guess. Second, failing to find one proves nothing — a Lyapunov function that works confirms stability, but the absence of one you happened to try says nothing about instability.
For x' = -x^3, try V(x) = x^2; then V' = 2x · (-x^3) = -2x^4, which is strictly negative for x ≠ 0. So the origin is asymptotically stable — even though the linearization gives a zero eigenvalue and tells you nothing.
V is a bowl (positive definite) and V' is strictly downhill, certifying asymptotic stability without solving the equation.
A Lyapunov function is sufficient, not necessary: failing to find one does not prove instability. And it must be positive definite — merely positive somewhere is not enough; the bowl has to sit cleanly above zero around the equilibrium.