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]