Lagrange-Dirichlet theorem

The Lagrange-Dirichlet theorem gives a condition to find stable equilibrium points in mechanical systems.

> is the [[kinetic energy]] as given by the [[Kinetic energy|kinetic matrix]] $\mathrm{a}$ and $V$ is a velocity-independent [[Potential]]. Then, if $V$ has a strict minimum, that minimum is a stable [[equilibrium point]]. > [!quote]- Proof > Let $\mathbf{q}^{*}$ be the minimum of $V$. Then it is a [[Punto critico|stationary point]] of $V$: > $$\frac{ \partial V }{ \partial q_{i} } (\mathbf{q}^{*})=0,\quad \forall\ i=1,\ldots,n

This guarantees that it is an equilibrium point. To prove stability, we use Ljapunov's theorem with E(q,q˙)=T(q,q˙)−V(q)E(\mathbf{q},\dot{\mathbf{q}})=T(\mathbf{q},\dot{\mathbf{q}})-V(\mathbf{q}) as a Ljapunov function. Then

  • In a neighborhood of c=(q∗,0)\mathbf{c}=(\mathbf{q}^{*},0) we must have E>V(q∗)E>V(\mathbf{q}^{*}).
  • E(q,q˙)E(\mathbf{q},\dot{\mathbf{q}}) is a constant of motion, so LfE=0\mathcal{L}_{f}E=0, where ff is the Vector field of the system (as in x˙=f(x)\dot{\mathbf{x}}=f(\mathbf{x})).

This satisfies Ljapunov's theorem, so it is stable.

Some notes:

  • The theorem can also be proven to be valid for velocity dependent potentials V(q,q˙)=V0(q)+V1(q,q˙)V(\mathbf{q},\dot{\mathbf{q}})=V_{0}(\mathbf{q})+V_{1}(\mathbf{q},\dot{\mathbf{q}}). In that case, the Ljapunov function must be E=T−V0E=T-V_{0} instead.
  • A lesser form of the theorem is also valid in nonconservative systems, but the stability is only guaranteed in the future.