Timeline for Existence of a global solution to a differential inclusion that does not blow up
Current License: CC BY-SA 4.0
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| Jun 5 at 5:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
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| Jan 16, 2022 at 23:31 | answer | added | Martin Väth | timeline score: 1 | |
| Jan 7, 2022 at 15:20 | comment | added | J. Doe | $x(t)$ 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 $x(t) = t-t_h$. | |
| S Jan 6, 2022 at 21:05 | history | suggested | Buzz | CC BY-SA 4.0 |
Fixed parens
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| Jan 6, 2022 at 20:35 | review | Suggested edits | |||
| S Jan 6, 2022 at 21:05 | |||||
| Jan 6, 2022 at 17:06 | comment | added | Alexandre Eremenko | 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. | |
| Jan 6, 2022 at 16:37 | history | edited | J. Doe | CC BY-SA 4.0 |
added 172 characters in body
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| Jan 6, 2022 at 16:35 | comment | added | J. Doe | @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. | |
| Jan 6, 2022 at 14:37 | comment | added | Alexandre Eremenko | 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 $x'=f(x)$. | |
| Jan 6, 2022 at 12:20 | history | asked | J. Doe | CC BY-SA 4.0 |