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Feb 7, 2012 at 16:57 vote accept Willie Wong
Feb 7, 2012 at 16:56 answer added Willie Wong timeline score: 3
Feb 6, 2012 at 17:13 comment added Terry Tao The point is that the diamagnetic inequality argument shows that if there is a minimiser u that changes sign, then the diamagnetic inequality holds with equality for that solution. Admittedly this is not yet a contradiction if the solution only vanishes to second order or more at its sign changes, but I think one can work a bit more to eliminate this possibility (e.g. by dividing u into its positive part max(u,0) and its negative part min(u,0))
Feb 6, 2012 at 13:03 answer added Willie Wong timeline score: 2
Feb 4, 2012 at 10:05 comment added Willie Wong Isn't that the same sort of argument I described? (I may be misunderstanding you; please do say if that is the case.) (And sorry about the choice of name; I'll blame Lions for that (: ) Anyway, diamagnetic would say, in the same spirit of rearrangement, that if there exists any minimizer, there must exist a non-negative one. (There's the slight issue that for the focusing case where this is interesting the action is actually not minimized, just minimized over nontrivial solutions; but that's an aside.) Then if one has uniqueness this says the original solution is signed...
Feb 3, 2012 at 17:30 comment added Terry Tao Ah, I see. (I'd say that your G should be called something else, such as V, to avoid confusion.) For positivity, one can show from the diamagnetic inequality that S[|u|] <= |S[u]| if G is even, and a slight perturbation of that argument would show that minimisers are automatically signed, I think.
Feb 3, 2012 at 17:05 history edited Willie Wong CC BY-SA 3.0
clarification
Feb 3, 2012 at 17:02 comment added Willie Wong No, my main problem with GNN is that GN assumes that the solution is positive. A priori I don't know how to claim that the "action minimizer" must be function that doesn't change signs. Also, if I read it right, GNN works at the level of the equation $\triangle u = G'(u)$, and their $C^1$ assumption applies to $G'$, so by assuming only $C^1$ for $G$ we are still missing a bit.
Feb 3, 2012 at 16:53 comment added Terry Tao It seems to me that the Gidas-Ni-Nirenberg paper ams.org/mathscinet-getitem?mr=634248 only needs C^1 hypotheses on G to obtain spherical symmetry at least, provided of course that one can justify the Euler-Lagrange equation. The method (moving planes) is also likely to get radial decrease, and ODE uniqueness theorems should get the uniqueness. So is your main concern the justification of the Euler-Lagrange equations?
Feb 3, 2012 at 11:31 history asked Willie Wong CC BY-SA 3.0