Timeline for Given a fixed convex domain $\Omega$ in 3D, for what value $c$ the function $f(c) := \int_{\partial \Omega} |x-c| d \sigma_x$ gets its minimum?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2016 at 1:37 | comment | added | Michael Renardy | Actually, I think it is clear that the minimum is unique, since $|x-c|$ is a convex function of $c$, and, therefore, so is $f(c)$. | |
| Nov 4, 2016 at 0:09 | comment | added | Alexandre Eremenko | It is not even clear that the critical point (or absolute minimum) is unique, the thing one would like to know before solving for $c$. | |
| Nov 4, 2016 at 0:09 | comment | added | student | @Alexandre Eremenko, you are right. Thank you for your comment and time! | |
| Nov 4, 2016 at 0:08 | comment | added | Alexandre Eremenko | Of course this is difficult to solve for $c$, but this equation indicates that this is not the center as you conjectured. | |
| Nov 4, 2016 at 0:00 | comment | added | student | @Alexandre Eremenko, we can always try to find critical point. Here using $\nabla |x-c|=\frac{c-x}{|x-c|}$, I get $\int_{\partial \Omega}\frac{c-x}{|x-c|} d\sigma_x=0$ , but I cannot solve for the $c$. I agree that the critical number may not be at the center, but I believe it should be a point related to the geometry of $\Omega$. Like when $\Omega$ is a ball, one can really prove $c$ is the center of the ball. | |
| Nov 3, 2016 at 23:43 | comment | added | Alexandre Eremenko | It is highly unlikely that it will be in the center. Hint: differentiate and write that all partial derivatives at c are equal to 0. | |
| Nov 3, 2016 at 21:38 | history | asked | student | CC BY-SA 3.0 |