Timeline for Euler-Lagrange equations for minimizer of energy with indicator function
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Jul 25 at 9:01 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 23 at 6:33 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
generalization
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| Jul 23 at 6:00 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 22 at 21:01 | comment | added | Carlo Beenakker | @DanieleTampieri You are right, I have fixed it. The factor of two is gone. | |
| Jul 22 at 21:00 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
corrected factor of due from missing indicator term
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| Jul 22 at 18:01 | comment | added | Daniele Tampieri | @CarloBeenakker possibly the problem is caused by the fact that the regions $V=\Omega\cap\{u>0\}$ and its boundary $S$ are independent from the functional variation $\epsilon(x)$ even in the expression of $J(u+\epsilon)$ and this cannot be true, in this particular case. | |
| Jul 22 at 17:48 | comment | added | Daniele Tampieri | @CarloBeenakker I'm a little bit confused too, as the equation I got is consistent with equation (0.1) of the paper of Alt and Caffarelli with $Q^2=1$ which, however, makes no difference at all. The path I followed is expressing the characteristic function $\chi_{\{u>0\}}$ in a form more suitable for the calculation of the functional derivative. | |
| Jul 22 at 17:11 | comment | added | Carlo Beenakker | I notice a factor of two difference between the boundary term I have above, $2|\nabla u|^2=1$, and in the answer of Daniele Tampieri, $|\nabla u|=1$. I'm unsure why. The expression above is consistent with Eq. (2.8) of this source. | |
| Jul 22 at 17:05 | comment | added | Carlo Beenakker | boundary term: consider first a one-dimensional integration, with $u(x)>0$ for $x<0$ and $u(x)<0$ for $x>0$; then $$\int \frac{d}{du}\chi_{u>0}dx=\int \delta(u)dx=\int |u'(x)|^{-1}\delta(x)dx=|u'(0)|^{-1}.$$ I used that the derivative of the indicator function is a delta function, and $\delta(f(x))=|f'(x)|^{-1}\delta(x-x_0)$ if $f(x)$ vanishes at $x=x_0$. In the $d$-dimensional case a $d-1$ dimensional surface integral remains, $$\int \frac{d}{du}\chi_{u>0}dx=\int_S |n\cdot\nabla u|^{-1}ds=-\int_S (n\cdot\nabla u)^{-1}ds.$$ | |
| Jul 22 at 16:57 | comment | added | BBB | Thank you. How did you get the boundary term in the second line of your first displayed equation? | |
| Jul 22 at 16:40 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
rewrote the final expression in order to make contact with the other answer
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| Jul 22 at 15:29 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
more general case.
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| Jul 22 at 13:02 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 22 at 12:38 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 22 at 12:15 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 22 at 12:10 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 22 at 12:02 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 22 at 11:56 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 22 at 11:51 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |