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Mar 10, 2017 at 9:42 history edited CommunityBot
replaced http://dl.dropbox.com/ with https://dl.dropbox.com/
Apr 9, 2013 at 14:49 answer added Loïc Teyssier timeline score: 2
Nov 21, 2010 at 4:51 answer added partition_of_unity timeline score: 3
Nov 21, 2010 at 4:48 history edited partition_of_unity CC BY-SA 2.5
Moving update into an answer to not lose the original question
Nov 21, 2010 at 3:08 history edited partition_of_unity CC BY-SA 2.5
Highlighted the main questions.
Nov 21, 2010 at 3:00 history edited partition_of_unity CC BY-SA 2.5
added images and enhanced the question.
Nov 19, 2010 at 20:45 answer added Willie Wong timeline score: 5
Nov 12, 2010 at 22:36 comment added Will Jagy Let $\theta$ be the angle around the boundary, from $0$ to $2 \pi$ as usual. Let your boundary function be $f(\theta) = \sin ( 3 \theta)$ and make a new graphic, and make sure that one of the level sets depicted is $ u_f = 0.$
Nov 12, 2010 at 12:50 comment added partition_of_unity That's really interesting, is there a theorem that explains this? Is there a way to choose the boundary data in such a way to make the level sets not arcs? maybe to make the boundaries of the level sets to be only a Lipschitz curve?
Nov 12, 2010 at 3:40 history edited partition_of_unity
Added tag
Nov 12, 2010 at 3:00 comment added Will Jagy The thing to concentrate on is that in the interior, connected components of level sets of a harmonic function are either nice arcs, as your graphic shows, or like the zero set of the imaginary part of $$ ( x + i y)^n $$ for integer $ n \geq 2.$ If you picked exactly the right level in your graphic you could see such a point where multiple arcs intersect.
Nov 12, 2010 at 2:21 history edited partition_of_unity CC BY-SA 2.5
Added image, sorry can't seem to embed.
Nov 12, 2010 at 1:26 history asked partition_of_unity CC BY-SA 2.5