Timeline for How to find the minimum of the integral?
Current License: CC BY-SA 3.0
15 events
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| Nov 1, 2017 at 22:07 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Nov 1, 2017 at 22:01 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Nov 1, 2017 at 15:49 | comment | added | Carlo Beenakker | good catch; for fixed $m,M$, I'm not able to construct such a regular solution for times larger than about $2T^*$. | |
| Nov 1, 2017 at 14:12 | comment | added | Pietro Majer | Ops, sorry, the last line is $I(y)=(1-e^{-2T})/{\bf 2}=1/{\bf 2}+o(1)$, still less than $\log 2=0.69..$ | |
| Nov 1, 2017 at 12:47 | comment | added | Pietro Majer | For instance, take e.g. $m:=1$ and $M:=\cosh T$, so $T^*=\log \cosh T$. Then $x(t):=\max(Me^{-t},1)$ has functional level $I(x)=T-\log \cosh T=\log2-\log(1+e^{-2T})=\log2+o(1)$ for $T\to\infty$; and $y(t):=\cosh(t-T)=(e^Te^{-t})/2+(e^{-T} e^t)/2$ is admissible for the constraint and has the same level of $(e^{-T} e^t)/2$ namely $I(y)=(1-e^{-2T})/4=1/4+o(1)$. | |
| Nov 1, 2017 at 12:47 | comment | added | Pietro Majer | I think one can prove that the minimizer is a decreasing solution of $\ddot x=x$ until it possibly reaches the value $m$ at $t=T^*\le T$, and then it is constant, but I suspect that the minimizer is always regular and $T^*=T$ , so it is the $y=ae^t+be^{-t}$ with minimum $|a|$ that fits within the constraints $m\le y(t)\le M$ for $0\le t\le T$. | |
| Nov 1, 2017 at 7:30 | comment | added | Carlo Beenakker | it does not, I have (hopefully) accounted for this. | |
| Nov 1, 2017 at 7:29 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Oct 31, 2017 at 22:24 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Oct 31, 2017 at 22:15 | comment | added | Pietro Majer | Since $x(t)=ae^t+be^{-t}$ is not necessarily monotone, it is not clear to me why $m\le x_0\le M$ and $m\le x_T\le M$ should imply $m\le x(t)\le M$ for all $t$... | |
| Oct 31, 2017 at 21:52 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Oct 31, 2017 at 21:43 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Oct 31, 2017 at 21:37 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Oct 31, 2017 at 21:09 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Oct 31, 2017 at 21:03 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |