Ljapunov's theorem - Aetherwisp

Ljapunov's theorem gives a method for determining whether an equilibrium point of an ODE is stable or not.

There is also a corollary, called the Lagrange-Dirichlet theorem, that applies to conservative systems which states that isolated minima of the potential energy are always stable equilibrium points. An isolated minimum is a minimum that is a single point (a non-isolated minimum would be an entire interval).

> We know that $f(x)$ can be described by the derivative of a [[potential energy]] $V(x)$ as $f(x)=-V'(x)/m$, where $m$ is the [[mass]]. If the potential energy has an isolated minimum in $x^{*}\in \mathbb{R}$, then $\mathbf{c}=(x^{*},0)\in \mathbb{R}^{2}$ is a stable equilibrium point in the future. > > **Proof.** Let $x^{*}$ be an isolated minimum for $V$. Then, $V'(x^{*})=0$ and so $f(x^{*})=0$. Thus, $\mathbf{c}=(x^{*},0)$ is an equilibrium point. To see that it is also stable, consider the total mechanical [[energy]] > $$E=\frac{1}{2}mv^{2}+V(x)

as a dynamical variable. This quantity satisfies Ljapunov's theorem since

$E$ has an isolated minimum in $(x^{*},0)$ because the kinetic energy is zero if $v=0$ and $V(x^{*})$ is an isolated minimum.

Mechanical energy is a constant of motion, so $\mathcal{L}_{f}E=0$ .

Since it verifies the theorem, it is a stable in the future.

Examples#

> Our potential energy is the usual $V(x)=m\omega ^{2}x^{2}/2$. The ODE is $\ddot{x}=-\omega ^{2}x-2\mu \dot{x}$, where $\mu>0$ is the dampening constant. The vector field is > $$f=\begin{pmatrix} > v \\ > -\omega ^{2}x-2\mu v > \end{pmatrix}

To have $f=\begin{pmatrix}0 \\ 0\end{pmatrix}$ we just need $\mathbf{c}=(x,v)=(0,0)$ , which is our equilibrium point. We just need to apply Ljapunov's theorem in a neighborhood of $(0,0)$ using the total energy. We already know that $E$ is a constant of motion, so we just need to check that the Lie derivative is always non-positive:

> This verifies the theorem, so $\mathbf{c}$ is stable in the future.