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1

By using the output of ListContourPlot3D and the new Mathematica 10.0 feature DiscretizeGraphics, one can nicely generate a meshed contour region which is suitable for NIntegrate. We can show this for the above example for energy contours from 1.0 to 2.0 with a step width of 0.1: Monitor[Table[ e = ListInterpolation[data, {{-1, 1}, {-1, 1}, {-1, 1}}]; f ...


3

First note that, $$ \oint_{E(\vec{x})=\epsilon}\frac{dS}{|\nabla E(\vec{x})|} =\oint_{E(\vec{x})=\epsilon}\frac{\nabla E(\vec{x})\cdot d\vec{S}}{|\nabla E(\vec{x})|^2} = \int_{E(\vec{x})\leq\epsilon} \nabla\cdot\left( \frac{\nabla E(\vec{x})}{|\nabla E(\vec{x})|^2}\right) dV$$ By now, the volume integral can be evaluated by Integrate[f ...


1

The calculation seems to work, producing the curve, Plot[Sol[x], {x, -.45, 0}] Finding zeroes of Sol requires a modification of the expression in the Question, FindRoot[Sol[x], {x, -0.45}, Evaluated -> False] (* {x -> -0.325184} *) It is not possible to comment on why this might not agree with another calculation without seeing that other ...


2

It may be overkill, but here's a version with everything rounded that gets rid of the extraneous Null. Clear[yy]; dx = 1./100; dt = 1./10000; yy[x0_, t0_] := With[{x = Round[x0, dx], t = Round[t0, dt]}, Which[ x == 0, yy[0, t] = 0, x == 1, yy[1, t] = 0, t == 0, yy[x, 0] = Exp[-1000 (x - .3)^2]] ] Table[With[{ x = Round[x, dx], t = ...



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