# Filippov's differential inclusion

Most theorems regarding solutions of differential equations are ill-equipped to deal with properly discontinuous ones, of the form

\begin{equation} y(t) = F(y(t), t) \end{equation}where $F$ may discontinuous in both $t$ and $y(t)$. We can easily construct such equations with well-defined solutions, such as, on $\mathbb{R} \setminus \{ 0 \}$,

\begin{equation} \dot{y}(t) = \mathrm{sgn}(y(t)) \end{equation}Our solutions will be some mixture of being locally $y(t) = t + C$, $y(t) = -t + C$ and $y(t) = C$. $y(t) = 0$ is a global continuous solution, as well as generally any solution of the form $y(t) = \max(t + c, 0)$ or $y(t) = \min(- t - c, 0)$. we can already see that this doesn't have unique solutions for some initial condition on $y$, as can be expected.

## Set-valued maps

The basic idea behind the Filippov differential inclusion is to replace the notion of a differential equation by a set relation, of the form

\begin{equation} y(t) \in \mathcal{F}[F](y(t), t) \end{equation}where $\mathcal{F}$ maps a function $ f : \mathbb{R}^n \to \mathbb{R}^m$ to a function mapping reals to a set of real numbers, $\mathbb{R}^n \to \mathcal{P}(\mathbb{R}^m)$. This set-valued function will essentially give us the values in between its left and right limit in most cases.

First, some properties we will require from set-valued maps. We will define continuity in two different ways.

- For a set-valued function $f$, given a neighbourhood $U_f$ of $f(x)$, there is a neighbourhood $U_x$ such that $f(U_x) \subset U_f$

$\mathcal{F}$ is given by the essential convex hull of the function :

## The Filippov differential inclusion

where $\mathcal{F}$ maps a function $\mathbb{R}^n \to \mathbb{R}^m$ to a function mapping reals to a set of real numbers, $\mathbb{R}^n \to \mathcal{P}(\mathbb{R}^m)$. This set-valued function will essentially give us the values in between its left and right limit in most cases. $\mathcal{F}$ is given by the essential convex hull of the funciton :

\begin{equation} \mathcal{F}[f](x) = \bigcap_{\delta > 0} \bigcap_{\mu(S) = 0} \overline{\mathrm{co}}[f(B(x, \delta) \setminus S)] \end{equation}We are taking a ball $B$ centered around the point considered $x$, of radius $\delta$,

...

An important example is the case of the step function, $\Theta(x)$, defined by

\begin{equation} \Theta(x) = \begin{cases} 0 & x \leq 0\\ 1 & x > 0 \end{cases} \end{equation}From the expression of the function, we can see that any image of a subset of $\mathbb{R}$ can only be one of four sets : $\varnothing$, $\{ 0 \}$, $\{ 1 \}$ or $\{ 0, 1 \}$. As we are considering intersections, an important thing to note here is that we only need the smallest set defined by $\overline{\mathrm{co}}$.

For any $x < 0$, any open ball around $x$ will map either to $\{ 0 \}$ or $\{ 0, 1 \}$. As removing any set of measure zero will still leave at least one point in this domain (and in fact many more), we still have that. In fact, if our open ball is such that the image is $\{ 0, 1 \}$, then our open ball has some overlap with $(-\infty, 0)$, of cardinality $\mathfrak{c}$, and removing any set of measure zero will therefore still leave such a cardinality below $0$. The convex closure of these two sets is therefore $\{ 0 \}$ and $[0, 1]$, and their intersection is simply $\{ 0 \}$. A similar proof shows that for $x > 0$, $\mathfrak{F}[\Theta](x) = \{ 1 \}$.

At $x = 0$, on the other hand, using the same method we used concerning the overlap, the image of our open ball minus some negligible set will always be $\{ 0, 1 \}$, and therefore $\mathfrak{F}[\Theta](0) = [0, 1]$.

## Properties of the mapping

Last updated :

*2020-03-18 15:03:40*