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diffusion PDE

  • To: mathgroup at smc.vnet.net
  • Subject: [mg130985] diffusion PDE
  • From: sabrinacasanova at gmail.com
  • Date: Sat, 1 Jun 2013 06:26:08 -0400 (EDT)
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Hello,


In the notebook below you can find  PDE which represents a particle diffusion equation
with energy dependence and not only time/spatial dependences. The
particles should diffuse in a energy dependent way inside, lets say, a
sphere. There is a spherical symmetry, so only the radius of this
sphere is important. I have an initial condition and a boundary
condition on the spatial border of the sphere. My question is whether Mathematica
is able to solve such PDEs at all. If yes, do you know of somebody
who has already done it ?
Best

Sabrina

--------------------------------

chi = 1.
d0 = 3. Power[10., 27]
n = 300.

norm = 4. 3.14 1.8/( 3 Power[10., 10.])
Evaluate [norm]
B = 100. Power[n/10000., 0.5]
Dif = chi*d0*Power[B/3., -0.5]
tau0 = 2.*Power[10., 5]*Power[(n/300.), -1] 3. Power[10., 7]
gam = 0.5

x2min = 0.00001
x2rad = 20.
x2max = 40.
x1min = 0.
x1max = Power[(40.*3.18 Power[10., 18.]), 2]/(6.*Dif)
x3min = 1.
x3max = 100000000000.
x3cut = 1000000

pde = D[y[x1, x2, x3], x1] - 
   Dif Power[x3, gam] D[y[x1, x2, x3], {x2, 2}] - 
   Dif Power[x3, gam] Power[x2, -1] 2 D[y[x1, x2, x3], x2] + 
   y[x1, x2, x3]/tau0 + x3  D[y[x1, x2, x3], x3]/tau0 == 0


s = NDSolve[{pde, 
   y[x1min, x2, x3] == 
    norm Power[x3, -2.7] Exp[-x3/x3cut] UnitStep[x2 - x2max] UnitStep[
      x3max - x3], 
   y[x1, x2max, x3] == 
    norm Power[x3, -2.7] Exp[-x3/x3cut] UnitStep[x3max - x3], 
   Derivative[0, 1, 0][y][x1, x2min, x3] == 0, 
   y[x1, x2, x3max] == 0 }, 
  y, {x1, x1min, x1max}, {x2, x2min, x2max}, {x3, x3min, x3max}]

Plot3D[Evaluate[y[x1, x2max, x3] /. s], {x1, x1min, x1max}, {x3, 
  x3min, x3max}, PlotRange -> All]



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