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Nov 10, 2017 at 0:29 vote accept Z.Bingo
S Nov 3, 2017 at 3:09 history suggested jeq CC BY-SA 3.0
Corrected one typo.
Nov 3, 2017 at 2:41 review Suggested edits
S Nov 3, 2017 at 3:09
Nov 2, 2017 at 0:24 comment added Z.Bingo $x(t)$ is a real valued function since $x(t)$ has the constraint $m\le x(t)\le M$.
Nov 2, 2017 at 0:10 answer added Pietro Majer timeline score: 4
Oct 31, 2017 at 21:03 answer added Carlo Beenakker timeline score: 3
Oct 31, 2017 at 15:40 comment added Christian Remling Sometimes the minimum value is zero, when you can solve $x'=-x$ within your constraints.
Oct 31, 2017 at 15:33 comment added Pietro Majer is $x$ real valued? then what is $\|\dot x(t)+x(t)\|^2$?
Oct 31, 2017 at 13:14 review First posts
Oct 31, 2017 at 13:18
Oct 31, 2017 at 13:12 history asked Z.Bingo CC BY-SA 3.0