Timeline for Variational problem with constraint
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 3, 2022 at 5:54 | comment | added | Varun Vejalla | The problem comes down to finding $g(y):[0,2\pi]\to[-\pi/2,\pi/2]$ such that $4\int_0^{2\pi}\cos(g(y))dy$ is maximized subject to the constraint $4\int_0^{2\pi}\cos(y)g(y)dy=T$. The region $D$ would then be $D=\{(x,y)\mid \frac{\pi}{2}-g(y)\le |x-\pi|\le \pi, 0\le y \le 2\pi\}$. I believe $g(y)=\arcsin(\lambda \cos(y))$, obtained using the technique here, is the maximizing function, although I'm not sure I did it correctly. | |
| Aug 2, 2022 at 20:42 | comment | added | Varun Vejalla | Just a small contribution, but you have that $\int_{D^c}\cos(x)\mathrm{d}x \mathrm{d}y=\int_0^{2\pi}\int_0^{2\pi}\cos(x)\mathrm{d}x \mathrm{d}y-\int_D \cos(x)\mathrm{d}x \mathrm{d}y=-\int_D \cos(x)\mathrm{d}x \mathrm{d}y$, so the problem becomes maximizing $2\int_D \cos(x)\mathrm{d}x \mathrm{d}y$ subject to $2\int_D \cos(y)\mathrm{d}x \mathrm{d}y>T$. | |
| Aug 2, 2022 at 20:31 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Changed the symbol for set complement.
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| Aug 2, 2022 at 12:32 | history | edited | gmvh | CC BY-SA 4.0 |
Fixed obvious typo in formula
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| S Aug 2, 2022 at 12:06 | review | First questions | |||
| Aug 2, 2022 at 12:32 | |||||
| S Aug 2, 2022 at 12:06 | history | asked | TryingToLearn | CC BY-SA 4.0 |