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Feb 23, 2018 at 16:38 review Close votes
Feb 24, 2018 at 18:09
Feb 23, 2018 at 13:40 comment added Hannes Look at the "time-expansion" for $E$, so $(\mathcal{E}u)(t) := E(u(t))$ with $E$ being the $W^{m,p}(\Omega) \to W^{m,p}(\mathbb{R}^n)$ extension operator.
Feb 23, 2018 at 13:25 history asked PeteAgor CC BY-SA 3.0