Timeline for A problem on real valued functions in $\mathbb{R}^2$ with least variation
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 24, 2016 at 10:07 | comment | added | Rajesh D | in two dimensions, our curve divides domain into 2 regions, inside and outside. I am looking at this type of generalization. | |
| Feb 24, 2016 at 10:06 | comment | added | Rajesh D | i don't agree with your generalization to 1 dimension. Here you are taking two points $0$ and $1$ as bounadries, but according to me, there is only one point which acts as boundary, for exaple $0$ and divides domain into two regions, left of $0$ and right of $0$. Example is Heavside step function. (This solution is unique). en.wikipedia.org/wiki/Heaviside_step_function | |
| Feb 24, 2016 at 9:17 | history | edited | Dirk | CC BY-SA 3.0 |
added 21 characters in body
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| Feb 22, 2016 at 13:40 | comment | added | Dirk | May work… Anyway, all the best for you undertakings. | |
| Feb 22, 2016 at 13:38 | comment | added | Rajesh D | Ok, then additionally, I go as $J \in BV(0,L)$ | |
| Feb 22, 2016 at 12:59 | comment | added | Dirk | Hmm, this may get complicated as sets of 1-dimensional Hausdorff measure can get quite nasty. Also this assumption would not even imply boundedness of $f$… I think the framework is most natural, if you assume that $f$ is $L^1$ on the boundary. Usually, these spaces of differentiable functions are not well suited for variational problems. The framework of weak derivatives is much more handy here. | |
| Feb 22, 2016 at 12:46 | comment | added | Rajesh D | Thanks for the response and answer. Would it make sense to relax regularity of $f$ as $$f \in C^{\infty}(\Omega \setminus R)$$ where the 1-dimensional Hausdorff measure of $R$ is zero. | |
| Feb 22, 2016 at 10:35 | history | answered | Dirk | CC BY-SA 3.0 |