mathematics//dynamical systems//Lyapunov function

How to prove that a system loses imbalance without solving every trajectory. A **Lyapunov function** assigns a scalar to the state in order to study stability. It introduces no new force and need not coincide with physical energy.

How to prove that a system loses imbalance without solving every trajectory. A Lyapunov function assigns a scalar to the state in order to study stability. It introduces no new force and need not coincide with physical energy.

For an equilibrium at the origin, a typical candidate satisfies V(0)=0\mathcal V(0)=0V(0)=0 and V(x)>0\mathcal V(x)>0V(x)>0 away from it. Its derivative along the model is

V˙=∇VTf(x).\dot{\mathcal V}=\nabla\mathcal V^{\mathsf T}f(x).V˙=∇VTf(x).

Under the conditions of the corresponding theorem, a non-positive derivative proves stability and a strictly negative one adds convergence. If they only hold in a region, the conclusion only holds there. On an unbounded domain one must also rule out escape and ensure future existence.

A proof for connected tanks. In a closed passive network (nodal capacity), Mh˙=−LhM\dot h=-LhMh˙=−Lh with MMM diagonal positive and LLL symmetric positive semidefinite. Let h∗h_*h∗​ be the common level compatible with the initial volume and e=h−h∗1e=h-h_*\mathbf1e=h−h∗​1. Choose V(e)=12eTMe\mathcal V(e)=\tfrac12e^{\mathsf T}MeV(e)=21​eTMe. Then

V˙=eTMe˙=−eTLe≤0.\dot{\mathcal V}=e^{\mathsf T}M\dot e=-e^{\mathsf T}Le\le0.V˙=eTMe˙=−eTLe≤0.

If the graph is connected, the derivative vanishes only at constant states, and within the subspace of conserved volume the only constant state with 1TMe=0\mathbf1^{\mathsf T}Me=01TMe=0 is e=0e=0e=0. An invariance argument completes the convergence. This does not prove a return to a level fixed independently of the volume: it proves consensus toward the level that conservation allows.

Passivity and dissipation are not the same claim. With an input bbb, the algebraic identity

Mh˙=−Lh+b,ddt12hTMh=−hTLh+hTbM\dot h=-Lh+b,\qquad \frac{d}{dt}\tfrac12h^{\mathsf T}Mh=-h^{\mathsf T}Lh+h^{\mathsf T}bMh˙=−Lh+b,dtd​21​hTMh=−hTLh+hTb

shows storage, dissipation and supply. This model has a passivity relation for the pair (b,h)(b,h)(b,h). Reading it as physical energy or power needs the right factors and units; in gravity-driven hydraulics ρg\rho gρg can be included.

Passivity does not always mean absence of oscillation. An ideal oscillator conserves energy and oscillates. The absence of oscillatory modes in this particular network comes from its first-order structure and from M−1LM^{-1}LM−1L being similar to a symmetric positive semidefinite matrix.

The candidate here measures a quadratic imbalance; the general notion of stability it certifies is the one in equilibrium and stability.