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Let $\dot{x}(t) \in F(x(t))$ be a differential inclusion, with $F: \mathbb{R}^n \rightrightarrows \mathbb{R}^n$ an uppersemicontinuous, convex and compact valued set-valued map.

On Wikipedia it is said that such an inclusion always admits a local solution (i.e. on an interval $[0, \epsilon)$ ), and if such a solution does not blow up, then we can extend it to a global solution (i.e. on $[0, +\infty)$ ). Such a result seems quite intuitive, and I believe has a simple proof.

Is there any reference where such a result is explicitly stated? I have only managed to found some reference which impose conditions on $F$.

Any help would be appreciated.

EDIT: Actually, looking at the proof of (local) existence in Aubin-Cellina, I am no longer sure that this is true. However, I am not able to construct a counterexample.

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  • $\begingroup$ All you need is that that for every $x_0$ there exists a continuous $f$ such that $f(x)\in F(x)$ for $x$ in a neighborhood of $x_0$. Then you can apply Peano's existence theorem for odrdinary differential equation $x'=f(x)$. $\endgroup$
    – Alexandre Eremenko
    Commented Jan 6, 2022 at 14:37
  • $\begingroup$ @AlexandreEremenko I believe not any usc map has such an $f$. For instance, $F(x) = \textrm{sign}(x)$ and $F(0) = [-1, 1]$ do not. Though the differential inclusion admits a solution. $\endgroup$
    – J. Doe
    Commented Jan 6, 2022 at 16:35
  • $\begingroup$ And what is the solution in your case? $x(t)=|t|?$ But it is not differentiable at $0$, so you have to explain what do you exactly mean by a solution. $\endgroup$
    – Alexandre Eremenko
    Commented Jan 6, 2022 at 17:06
  • $\begingroup$ $x(t)$ is said to be a solution of a differential inclusion, if it absolutely continuous, and the inclusion is satisfied for almost every $t$. For this example a solution can be $x(t) = 0$ up to some $t_h$ and then $x(t) = t-t_h$. $\endgroup$
    – J. Doe
    Commented Jan 7, 2022 at 15:20

1 Answer 1

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Consider the initial value $x(t_0)=x_0$. Let $B$ be the closed ball around $x_0$ with radius $1$. Let $\rho$ denote a retraction onto $B$. Then $G=F\circ\rho$ satisfies $G|_B=F|_B$ and is upper semi-continuous with nonempty closed convex values. Moreover, the range of $G$ is contained in the compact set $F(B)$, and so $G$ is bounded.

Now standard results imply the existence of a local solution of $x'\in G(x)$, $x(t_0)=x_0$. Since $x$ is in particular continuous at $t_0$, it follows that on some interval $[t_0-\varepsilon,t_0+\varepsilon]$ the solution $x$ assumes only values in $B$, and so actually $x'\in F(x)$ on that subinterval.

This proves the local existence result.

The extension result is not so easy to formulate in the absence of uniqueness. One version (if one has a-priori bounds for all possible solutions) follows in a standard way by Hausdorff's maximality principle (or Zorn's lemma).

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