Cool Facts about Control

(Under Construction)

Consider the differentiable function \(f:\mathbb{R}^n \rightarrow \mathbb{R}^n\). Let \(x^*\in\mathbb{R}^n\) be an isolated equilibrium point. Because \(x^*\) is isolated, there exists \(\varepsilon >0\) such that for the ball \(B_{\varepsilon}(x^*)\) centered at \(x^*\) there are no other equilibrium points. Let \(S_{\varepsilon}(x^*)\) be the boundary of \(B_{\varepsilon}(x^*)\). Then we can give the well-defined map:

\[ f_{\varepsilon} : S_{\varepsilon}(x^*) \rightarrow S^{n-1} : x \mapsto \frac{f(x)}{||f(x)||}. \]

We define the index of the isolated equilibrium point \(x^*\) to be the topological degree of \(f_{\varepsilon}\) around \(x^*\). Since \(S_{\varepsilon}(x^*)\) and \(S^{n-1}\) are both compact smooth manifolds and \(f_{\varepsilon}\) is differentiable on its domain, we can use a particular definition of degree which is rather straight-forward to compute with respect to some regular value \(y \in S^{n-1}\)

\[ \text{Ind}_{x^*}(f) = \text{deg} (f_{\varepsilon}) = \sum_{x\in f^{-1}(y)} \text{sign} \left( \det \left( \frac{\partial f_{\varepsilon}}{\partial x}\Big\rvert_{x} \right) \right) \]

Note that because \(S_{\varepsilon}(x^*)\) is compact and \(y\) is a regular value, the cardinality of \(f^{-1}(y)\) is finite. The index of a map around an equilibrium point can be useful since it provides a necessary condition for local asymptotic stability with the following statement.

Theorem: Consider the autonomous system

\[ \dot x = f(x) \]

in dimension \(n\neq 4\) with isolated equilibrium point \(x^*\). Then \(x^*\) is locally asymptotically stable only if the index of \(f\) at \(x^*\) is \((-1)^n\), that is

\[ \text{Ind}_{x^*}(f) = (-1)^n. \]

This can be a useful condition to check for nonlinear systems as opposed to the sufficient Lyapunov test. However, the index is preserved under homotopy, which is a rather weak notion of equivalance. There are plenty of linear systems which are homotopic to one another, but have different stability properties.

The proof that I learned requires \(n\neq 4\) since it is not known if sublevel sets of a smooth Lyapunov function are diffeomorphic to a \(3\) dimensional disk (they are atleast homoemorphic). This may seem like a funny intricacy because, due to Perelman's proof of the Poincare conjecture, Wilson's theorem says that super level sets of smooth Lyapunov functions are diffeomorphic to \(S^{n-1}\) for all dimensions. One would think that the interior of such \(n-1\) sphere would look a lot like a disk. This condition can be removed by considering homoemorphism of trajectories and maybe I will take the time to prove this soon.