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Sep 29, 2018 at 9:44 history edited Martin Sleziak
Removed the deprecated (abstract-algebra) tag - see the tag info: https://mathoverflow.net/tags/abstract-algebra/info (if there are some other suitable tags, choose them instead.)
Sep 29, 2018 at 1:22 vote accept Saj_Eda
Sep 28, 2018 at 21:10 answer added Christopher A. Wong timeline score: 3
Sep 27, 2018 at 23:54 comment added Saj_Eda Yes, bounded in $C(0,T;H^2(0,L))$ norm.
Sep 27, 2018 at 23:30 comment added Nate Eldredge Can you explain precisely what "bounded" means? Do you mean bounded in $C(0,T; H^2(0,L))$ norm, or something else?
Sep 27, 2018 at 23:07 history edited Saj_Eda CC BY-SA 4.0
added 6 characters in body
Sep 27, 2018 at 22:41 history asked Saj_Eda CC BY-SA 4.0