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Aug 29, 2013 at 17:03 comment added George Lowther ...as long as you also have boundary conditions specified at $x=\pm1$
Aug 29, 2013 at 16:54 comment added George Lowther You can treat this as the generator for a markov process in $(x,t)$ (introducing a separate random variable for the time parameter), from which you can see that there will be a unique solution with the piecewise initial conditions.
Aug 29, 2013 at 2:10 answer added user39289 timeline score: 0
Aug 16, 2013 at 18:48 answer added Bob Terrell timeline score: 2
Aug 14, 2013 at 18:20 review First posts
Aug 14, 2013 at 18:21
Aug 14, 2013 at 18:02 history asked vojta havlíček CC BY-SA 3.0