Timeline for Calculus of variations when functional involves inverse of the function
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 27, 2016 at 18:25 | vote | accept | StevenMurray | ||
| May 27, 2016 at 18:23 | comment | added | StevenMurray | Thanks everyone. I think I have solved it, and it doesn't require the inverse after all. @Robert 's answer led me to the solution by showing that I could construct the integral from a to b rather than some function of $u$ or its inverse. Once in that space, it is not too hard. | |
| May 22, 2016 at 7:46 | comment | added | Peter Kravchuk | One could try introducing the inverse via a Lagrange multiplier, i.e. add a constraint $f(u(x))-x=0$, where $f$ will be the inverse. This is perhaps in the end equivalent to Robert's suggestion | |
| May 21, 2016 at 22:21 | answer | added | Robert Bryant | timeline score: 4 | |
| May 21, 2016 at 19:31 | history | edited | StevenMurray | CC BY-SA 3.0 |
added 1550 characters in body
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| May 21, 2016 at 19:17 | comment | added | StevenMurray | Yes, a very specific application -- I guess it may turn out that I have constructed the problem incorrectly, but I don't think so. I'll edit the question with more details as to the nature of the problem | |
| May 21, 2016 at 18:30 | comment | added | Andreas Rüdinger | For me the question does not look very "natural". The functional inverse of $u: [a,b] \to \mathbb{R}$ is not meant to be integrated over the same domain as $u$. Do you have an application in mind? | |
| May 21, 2016 at 18:29 | comment | added | StevenMurray | Yeah, that's why I thought it might actually be impossible in a general sense. But you never know what some clever mathematician has come up with... | |
| May 21, 2016 at 17:42 | comment | added | Siminore | It looks hard, since you need to invert $u + \varepsilon \varphi$... | |
| May 21, 2016 at 17:10 | comment | added | StevenMurray | Ah, good point. I meant the former. | |
| May 21, 2016 at 12:22 | comment | added | Igor Khavkine | By "inverse" $u^{-1}$, do you mean $u(u^{-1}(x)) = x$ or $u(x) u^{-1}(x) = 1$? | |
| May 21, 2016 at 11:14 | review | First posts | |||
| May 21, 2016 at 12:23 | |||||
| May 21, 2016 at 11:13 | history | asked | StevenMurray | CC BY-SA 3.0 |