Re: diffusion PDE
- To: mathgroup at smc.vnet.net
- Subject: [mg130993] Re: diffusion PDE
- From: Roland Franzius <roland.franzius at uos.de>
- Date: Sun, 2 Jun 2013 00:27:33 -0400 (EDT)
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Am 01.06.2013 12:05, schrieb sabrinacasanova at gmail.com: > 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] > I've written the system a bit more readable. There seems to be no problem exept a lacking boundary condition at energymin and the fact, that the solution seems to be independent of time. The start condition seems to be already a solution, make a test of it applying the operator pde on the start condition. chi = 1.; d0 = 3. 10^27; n = 300.; norm = 4. 3.14 1.8/(3 10^10); B = 100. Sqrt[n/10000.]; Dif = chi*d0/Sqrt[B/3]; tau0 = 2.*10^5/(n/300.)* 3.10^7; gam = 0.5; radiusmin = 0.00001; radiusrad = 20.; radiusmax = 40.; timemin = 0.; timemax = 1000 (40.*3.18 10^18)^2/(6.*Dif); energymin = 1.; energymax = 100000000000.; energycut = 1000000; pde = D[y[time, radius, energy], time] - Dif *energy^gam *D[y[time, radius, energy], radius, radius] - Dif* energy^gam/radius * 2* D[y[time, radius, energy], radius] + y[time, radius, energy]/tau0 + energy D[y[time, radius, energy], energy]/tau0 == 0 startcond = y[timemin, radius, energy] == norm/ energy^2.7 Exp[-energy/energycut] rmaxcond = y[time, radiusmax, energy] == norm/energy^2.7* Exp[-energy/energycut] rmincond = Derivative[0, 1, 0][y][time, radiusmin, energy] == 0 energymaxcond = y[time, radius, energymax] == 0 energymincond = y[time, radius, energymin] == startcond[[2]] /. energy -> energymin s = NDSolve[{pde, startcond, rmincond, rmaxcond, energymincond, energymaxcond}, y, {radius, radiusmin, radiusmax}, {energy, energymin, energymax}, {time, timemin, timemax}, MaxStepFraction -> {1/1000, 1/40, 1/40}] Plot3D[y[time, radiusmax, energy] /. s, {time, timemin, timemax}, {energy, energymin, energymax/10}, PlotRange -> All, PlotPoints -> 200, AxesLabel -> {"time", "energy", "y"}] -- Roland Franzius