Timeline for Functional minimization problem
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Apr 21, 2015 at 9:47 | vote | accept | Mandrill | ||
| Apr 21, 2015 at 9:38 | answer | added | Pietro Majer | timeline score: 11 | |
| Apr 21, 2015 at 7:24 | comment | added | Mandrill | No, because I don't know how to do it. Tried to understand how to use Lagrange multipliers with Euler-Lagrange method but got stuck, I am not a mathematician and I am not that smart. | |
| Apr 21, 2015 at 7:16 | comment | added | Benjamin | Ah, sorry. I didn't spot that. Have you used a Lagrange multiplyer to impose the constraint? | |
| Apr 21, 2015 at 7:09 | comment | added | Mandrill | @Benjamin the monotonic is from f'(x)<=0. This equation is the drag from a revolution solid with height/radio = k. This equation was my attempt to calculate drag using gas kinetic theory but long time ago now I know it is only valid at extremely low gas densities (free path >> height and free path >> radius). If the surface does a zig zag then there would be multiple colisions that the equation does not account for (but the equation is fine if the generation function is monotonic). | |
| Apr 21, 2015 at 7:08 | comment | added | Fan Zheng | I missed the condition $f'\le 0$ ( reading on my phone but not rotating to the right direction). But still I think there is no smooth minimizer. I'll write it up later | |
| Apr 21, 2015 at 7:06 | comment | added | Benjamin | Also, which exact algorithm converged. There are many subtleties to minmising a functional numerically and often one can't make a conclusion about global optima from numerics. Are you really looking for a minimiser or a stationary curve as this is what the EL equations find. | |
| Apr 21, 2015 at 6:58 | comment | added | Fan Zheng | By the way, it also has no smooth maximizer: You can make f drop sharply to 0 sharply and remain 0 thereafter to make the integral arbitrarily close to 1/2. Again no smooth function will make it exactly 1/2. Actually, the EL equation is singular at 0, so there is even no smooth critical point of the functional. | |
| Apr 21, 2015 at 6:56 | comment | added | Benjamin | The integrand is positive and the functional can be made arbitrarily close to zero as you point out. Thus, by a squeezing argument, if there is smooth minimiser. then it is assigned 0 by the functional. However, there clearly isn't one. I don't understand your comment. Where is this additional "monotonic restriction" coming from. Are you asking to impose it? | |
| Apr 21, 2015 at 6:54 | comment | added | Mandrill | @ Fan Zheng Sorry but I don't agree with your conclusion. There is a monotonic restriction (the functional does not evaluate correctly a non monotonic function) and minimizing doesn't mean being exactly 0. | |
| Apr 21, 2015 at 6:49 | comment | added | Fan Zheng | You got the point. By making your function more and more Zigzagging, you can make the integral arbitrarily close to 0. Yet no smooth function will make it exactly 0, so the answer is no. | |
| Apr 21, 2015 at 6:46 | comment | added | Mandrill | @ Fan Zheng I did, but it doesn't work (at least not the standard way to use it) because a "zigzag" (going up and down with a small x variation) function would make a very high $f'(x)$ and that would minimize the functional. I don't know how to introduce a monotonic restriction in the Euler-Lagrange method. | |
| Apr 21, 2015 at 6:45 | review | Close votes | |||
| Apr 21, 2015 at 9:06 | |||||
| Apr 21, 2015 at 6:45 | history | edited | Federico Poloni | CC BY-SA 3.0 |
better title
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| Apr 21, 2015 at 6:41 | history | edited | Mandrill | CC BY-SA 3.0 |
edited body
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| Apr 21, 2015 at 6:40 | comment | added | Fan Zheng | Have you tried Euler-La grange equation? | |
| Apr 21, 2015 at 6:38 | history | edited | Mandrill | CC BY-SA 3.0 |
added 92 characters in body
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| Apr 21, 2015 at 6:34 | comment | added | Mandrill | $k^2$ is a constant, a form factor. | |
| Apr 21, 2015 at 6:34 | comment | added | Nate Eldredge | @RyanBudney: Maybe $k$ is supposed to be fixed? | |
| Apr 21, 2015 at 6:28 | comment | added | Ryan Budney | I can't tell what the problem is. If you let $k$ get arbitrarily large, your integrals tend to zero. So there would appear to be no minimum. | |
| Apr 21, 2015 at 6:26 | history | edited | Amritanshu Prasad | CC BY-SA 3.0 |
added 4 characters in body
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| Apr 21, 2015 at 6:25 | review | First posts | |||
| Apr 21, 2015 at 7:27 | |||||
| Apr 21, 2015 at 6:21 | history | asked | Mandrill | CC BY-SA 3.0 |