have some problem with Neumann boundary conditions over simple rectangle. Here my code

currentTime := 0;
bounds := 4;
timeLimit := 50;
rect = Rectangle[{-bounds, -bounds}, {+bounds, +bounds}]
bsf = Interpolation@Flatten[Table[{{x, y}, 0.5 + 0.2 * RandomReal[{-1, 1}]}, {x, -bounds, +bounds}, {y, -bounds, +bounds}], 1];
initial = {f[x, y, 0] == bsf[x, y], D[f[-bounds, -bounds, t], t] == D[f[bounds, -bounds, t], t] ==D[f[-bounds, bounds, t], t] == D[f[bounds, bounds, t], t] == 0}
equation := With[{d = 0.1, a = 0.8, b = 0.005, c = 0.0005, g = 0.000005,    tm = 0.3, tmc = 0.2}, D[f[x, y, t], t] == d*Laplacian[a*(tm - tmc)*f[x, y, t] - (b*f[x, y, t]^3) + (c*f[x, y, t]^5) - g*Laplacian[f[x, y, t], {x, y}], {x, y}]]
ProgressIndicator[Dynamic[currentTime], {0, timeLimit}]
result = NDSolve[{equation, initial}, f[x, y, t], {x, y} \[Element] rect, {t, 0.1, timeLimit}, EvaluationMonitor :> (currentTime = t;), Method -> {"MethodOfLines", Method -> {"FixedStep", "StepSize" -> .5, 
  Method -> "ImplicitRungeKutta"}, "SpatialDiscretization" -> {"TensorProductGrid", "MinPoints" -> 50, "MaxPoints" -> 100}}]
plots = Table[DensityPlot[f[x, y, t] /. result, {x, y} \[Element] rect], {t,0,timeLimit, 0.5}];

There are error "Boundary conditions is not specified on a single edge of the boundary of the computational domain", but i already specified conditions. What's wrong ?

Thanks for attention !

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    – Michael E2
    Jan 10 '16 at 23:08

What's wrong: at least the boundary condition. They are defined only in 4 points, (-bounds,-bounds), (bounds,-bounds), etc.

You need to define them on the edge, that is

f[x,bounds,0] == Sin[x]

for example. Redefine your initial variable and try again, there may be other errors.

  • $\begingroup$ I defined boundaries on edges initial = {f[x, y, 0] == bsf[x, y], D[f[x, -bounds, t], t] == D[f[x, +bounds, t], t] == D[f[-bounds, y, t], t] == D[f[bounds, y, t], t] == 0} but still have same error. May be exist more.. correct way to define Neumann boundaries over rectangle ? $\endgroup$
    – K. Lonhus
    Jan 11 '16 at 9:29
  • $\begingroup$ Your derivative is with respect to time, so that's not a neumann condition, you need to derivate with respect to space. Have you looked at the reference $\endgroup$ Jan 11 '16 at 13:42

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