Timeline for Solution to a PDE with constant data - what is the fault in my proof? [closed]

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Feb 2, 2015 at 6:23	history	closed	Michael Renardy Stefan Kohl♦ Denis Serre András Bátkai Ryan Budney		Not suitable for this site
Feb 1, 2015 at 20:48	comment	added	Terry Tao		In the degenerate case, the mean zero condition means that the domain of J consists solely of the zero function. This suggests that one take a closer look at your claim that "one can remove the mean value condition required on the test function".
Feb 1, 2015 at 18:14	comment	added	jamesC		@TerryTao Thanks for the comment. (On the non-degenerate case): my feeling is the domain of $J$ may not be well-defined. The function $y \mapsto \int_\Omega d(x,y) =: (Md)(y)$ is in fact such that $Md \in H^1(0,\infty)$ for $d \in H^1(C)$, hence $Md \in C^0([0,\infty))$. So if we take the continuous representative then we must have $Md(y) = 0$ for all $y$ including $y=0$. Could you please give another hint to what you meant?
Feb 1, 2015 at 2:03	comment	added	Terry Tao		Hint 1: You are getting an apparent contradiction even in the degenerate case when $\Omega$ is a zero-dimensional point and $C$ is a half-line (note that there is clearly no $H^1$ solution to $\Delta v = 0, Tv = 1$ in this case). Hint 2: check all of your claims about $J$ carefully.
Jan 31, 2015 at 19:37	comment	added	jamesC		Yes, it is zero Neumann BCs.
Jan 31, 2015 at 19:36	history	edited	jamesC	CC BY-SA 3.0	added 108 characters in body
Jan 31, 2015 at 19:28	review	Close votes
Feb 2, 2015 at 6:23
Jan 31, 2015 at 19:12	comment	added	Michael Renardy		It looks like you are missing a boundary condition on $\partial\Omega$.
Jan 31, 2015 at 18:39	history	edited	jamesC	CC BY-SA 3.0	added 178 characters in body
Jan 31, 2015 at 18:32	review	First posts
Jan 31, 2015 at 18:37
Jan 31, 2015 at 18:32	history	asked	jamesC	CC BY-SA 3.0