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Mar 9, 2020 at 17:02 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Nov 8, 2014 at 19:26 answer added gerw timeline score: 1
Sep 28, 2014 at 2:26 comment added Michael Renardy $H^1(\Omega)^*$ is not a space of distributions in $\Omega$. For instance, the functional on the left hand side of your equation is in $H^1(\Omega)^*$. This defeats the objective of separating the inhomogeneous term in the PDE from the boundary condition.
Sep 27, 2014 at 14:35 comment added Joonas Ilmavirta It might be just a matter of convenience: it is easier to work with an $L^2$ function than a more general distribution.
Sep 27, 2014 at 14:21 history edited riem CC BY-SA 3.0
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Sep 27, 2014 at 14:13 history asked riem CC BY-SA 3.0