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Solve Boltzmann eq in 2D with NDSolve and the method of lines in less time?

Ralph Trenkler

Ralph Trenkler, Universität Hamburg

Posted 8 years ago

I tried to solve the Boltzmann equation in 2 spatial dimensions with NDSolve and the method of lines. In the first call of NDSolve in the attached Mathematica notebook the computation time jumps from 3 minutes for MinPoints -> 19 to more than 5 hours for MinPoints -> 20. What is the reason for this? What can I do to limit the computation time? pde = D[f[t, x, y, kx, ky], t] + (v.Grad[f[t, x, y, kx, ky], {x, y}]) + 1/hbar(F.Grad[f[t, x, y, kx, ky], {kx, ky}]) == 0 ics = {f[0, x, y, kx, ky] == f0[hbar^2((kx - kxShift)^2 + ky^2)/(2m), mu, T], f[t, -L/2, y, kx, ky] == f0[hbar^2((kx - kxShift)^2 + ky^2)/(2m), mu, T], f[t, L/2, y, kx, ky] == f0[hbar^2((kx - kxShift)^2 + ky^2)/(2m), mu, T], f[t, x, -L/2, kx, ky] == f0[hbar^2((kx - kxShift)^2 + ky^2)/(2m), mu, T], f[t, x, L/2, kx, ky] == f0[hbar^2((kx - kxShift)^2 + ky^2)/(2m), mu, T]} parameters = {hbar -> 1.05458866410^-34, e -> 1.6021889210^-19, kB -> 1.38066210^-23, T -> 300., mu -> 2.34810^-18, m -> 9.10938310^-31, kxShift -> 8.010^9, xE -> 0, yE -> 0, tau -> 0.3} borders = {L -> 40.0, kmax -> 4.10^10} (sol1 = NDSolve[Union[{pde}, ics] /. parameters /. borders, f, {t, 0, 1}, {x, -L/2, L/2} /. borders, {y, -L/2, L/2} /. borders, {kx, -kmax, kmax} /. borders, {ky, -kmax, kmax} /. borders, Method -> {"MethodOfLines", "TemporalVariable" -> t, "SpatialDiscretization" -> {"TensorProductGrid", "MinPoints" -> 19}}, WorkingPrecision -> MachinePrecision, PrecisionGoal -> 4, AccuracyGoal -> 12]) // AbsoluteTiming Attachments:

I tried to solve the Boltzmann equation in 2 spatial dimensions with NDSolve and the method of lines. In the first call of NDSolve in the attached Mathematica notebook the computation time jumps from 3 minutes for MinPoints -> 19 to more than 5 hours for MinPoints -> 20. What is the reason for this? What can I do to limit the computation time?

pde = D[f[t, x, y, kx, ky], 
    t] + (v.Grad[f[t, x, y, kx, ky], {x, y}]) + 
   1/hbar*(F.Grad[f[t, x, y, kx, ky], {kx, ky}]) == 0

ics = {f[0, x, y, kx, ky] == 
   f0[hbar^2*((kx - kxShift)^2 + ky^2)/(2*m), mu, T],
  f[t, -L/2, y, kx, ky] == 
   f0[hbar^2*((kx - kxShift)^2 + ky^2)/(2*m), mu, T],
  f[t, L/2, y, kx, ky] == 
   f0[hbar^2*((kx - kxShift)^2 + ky^2)/(2*m), mu, T],
  f[t, x, -L/2, kx, ky] == 
   f0[hbar^2*((kx - kxShift)^2 + ky^2)/(2*m), mu, T],
  f[t, x, L/2, kx, ky] == 
   f0[hbar^2*((kx - kxShift)^2 + ky^2)/(2*m), mu, T]}

parameters = {hbar -> 1.054588664*10^-34, e -> 1.60218892*10^-19, 
  kB -> 1.380662*10^-23, T -> 300., mu -> 2.348*10^-18, 
  m -> 9.109383*10^-31, kxShift -> 8.0*10^9, xE -> 0, yE -> 0, 
  tau -> 0.3}

borders = {L -> 40.0, kmax -> 4.*10^10}

(sol1 = NDSolve[Union[{pde}, ics] /. parameters /. borders, 
    f, {t, 0, 1}, {x, -L/2, L/2} /. borders, {y, -L/2, L/2} /. 
     borders, {kx, -kmax, kmax} /. borders, {ky, -kmax, kmax} /. 
     borders, 
    Method -> {"MethodOfLines", "TemporalVariable" -> t, 
      "SpatialDiscretization" -> {"TensorProductGrid", 
        "MinPoints" -> 19}}, WorkingPrecision -> MachinePrecision, 
    PrecisionGoal -> 4, AccuracyGoal -> 12]) // AbsoluteTiming

POSTED BY: Ralph Trenkler

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Sam Carrettie

Sam Carrettie, Freelancer

Posted 8 years ago

Yes, exactly. @Ralph Trenkler, how do you think your `hbar -> 1.054588664*10^-34` competes with `WorkingPrecision -> MachinePrecision` ? To add to Frank's comments, you should specify as many parameters as you can in terms of whole not decimal numbers to keep it simple - tracking of precision. Physicist usually do this with physical unit rescaling - going into unit-less space - dividing by hbar - etc. And that would land you approximately in parameter values order of 1.