Timeline for Reference request: Optimal controls can be assumed to take values in a convex set
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 6, 2021 at 21:05 | comment | added | Abhishek Halder | I will leave the formalization but mention the key ideas. In the set-valued integral I wrote, the integrand is a set parameterized by a choice of $s$ in $[0,t]$. The set $U$ is fixed. Different choices of $s$ in the integrand result in different linear transforms of the set $U$. If you uniformly discretize $[0,t]$, then that integral is the limit of the Minkowski sum of these linearly transformed sets, where the limit is w.r.t. discretization length. So Lyapunov theorem is saying that repeated Minkowski sums eventually convexify. You can view it as the continuum version of Shapley-Folkman. | |
| Oct 3, 2021 at 6:39 | comment | added | Nate River | Thank you! Can you possibly elaborate a litle more on how Lyapunov’s theorem implies the set valued integral is convex? As stated it applies to a single function integrated over all possible Borel sets, while your version is over different points in $U$. It is not immediately obvious to me how to go from my version to yours. | |
| Oct 3, 2021 at 6:12 | comment | added | Abhishek Halder | The reason why the Proposition follows from the Theorem is this: the set $\mathcal{R}_{U}$ can be written as the translation of a set-valued integral given by $\exp(tA)x_{0} + \int_{0}^{t}\exp(sA)B U\:ds$. The plus sign denotes a Minkowski sum. The set valued integral, by Lyapunov's theorem, is a convex set and compact. From here, you can also show that the set-valued integral is invariant under the closure of convexification of $U$. | |
| Oct 3, 2021 at 3:13 | history | edited | Nate River | CC BY-SA 4.0 |
deleted 4 characters in body
|
| Oct 3, 2021 at 3:08 | comment | added | Nate River | @PiyushGrover It refers to the second part of the proposition that states that $\mathcal R_U$ is equal to $\mathcal R_{\overline {\text{co U}}}$. | |
| Oct 3, 2021 at 3:06 | comment | added | Piyush Grover | Title of your post is misleading.Your post says that the $\text{reachable sets}$ form a convex set when control is in a compact set. | |
| Oct 3, 2021 at 3:00 | history | asked | Nate River | CC BY-SA 4.0 |