2014-06-20 16:24 GMT+02:00 Daniel Wheeler <daniel.wheel...@gmail.com>:
> On Thu, Jun 19, 2014 at 11:53 AM, Benny Malengier > <benny.maleng...@gmail.com> wrote: > > Hi, > > > > another question. This time on SourceTerm use. I'm checking an article > with > > diffusion and reaction. So some species diffuse, others only react > (waiting > > for diffused species to start). If there is an example with this > floating on > > the internet, do point it my way! > > Not that I'm aware of. > > > You then have a coupled set of equations, of which one of the non > diffusive > > species will have equation like > > > > eqn3 = TransientTerm(var=P0) == ImplicitSourceTerm(-k3*M1, var=P0) + > > ImplicitSourceTerm(-k5*M2, var=P0) > > > > > > eqn4 = TransientTerm(var=P1) == ImplicitSourceTerm(k3*P0, var=M1) > > > > > > M1 will have another equation that has a diffusionterm. > > > > In eqn3 it is normal to use ImplicitSourceTerm as the rhs has -k3 M1 P0. > > Yes, but the right splitting for "-k3 * M1 * P0" is probably > > k3 * M1 * P0 - ImplicitSourceTerm(k3 * M1, var=P0) - > ImplicitSourceTerm(k3 * P0, var=M1) > > > In eqn4, we could actually write > > > > > > eqn4 = TransientTerm(var=P1) == k3*P0*M1 > > > > > > or > > > > > > eqn4 = TransientTerm(var=P1) == ImplicitSourceTerm(k3*M1, var=P0) > > > > > > In light that all will be coupled in one equation to solve, will this > lead > > to different solutions? > > It shouldn't assuming that there isn't an instability. Eventually, > they should converge to the same result, but not at the the same rate. > > > Or will fipy internally all give this the same > > matrix formulation? > > It certainly won't give the same matrix formulation for different > implicit / explicit choices. User needs to be aware of what's going > on. > > > Also, in eqn3, the first term in rhs is actually -k3 M1 P0. It seems > nicest > > if somehow it is guaranteed that values as in eqn4 for this term would be > > used. Will this be the case in this formulation? > > Don't understand the question fully. Do you mean that it would be nice > to be using the latest value of "M1"? This won't happen as "M1" is > explicit in Equation 3. Either "P0" is explicit or "M1" will be > explicit in Equation 3 (or both). There is no way to avoid this. > > > Perhaps there is a way to do this nicer and introduce a helper variable, > > U=M1*P0, so as to have only in eqn3 and eqn4 ImplicitSourceTerm(-k3, > var=U) > > and ImplicitSourceTerm(k3, var=U) respectively. If so, how to go about > to do > > this. If I do ImplicitSourceTerm(-k3, var=M1*P0) I get immediately the > error > > > > fipy.terms.SolutionVariableNumberError: Different number of solution > > variables and equations. > > The helper variable won't buy you anything. Use a Taylor expansion to > make the correct explicit / implicit choice > > S = Sc + var1 * dS / dvar1 + var2 * dS / dvar2 > > if S is the source term. The other choice would be to do full Newton > iteration. > > > So, I would need an equation for U, would that be > > > > eqnhelper = ImplicitSourceTerm(1, var=U) == M1*P0 > > > > Or am I making things too complicated now. > > This doesn't help since you still have an explicit representation for > M1 and P0 in one of the equations. > Thanks for the pointers, I'll try some things. My FiPy knowledge is slowly getting back. My main question seems to have come down on how intelligent the coupling mechanism in fipy is as opposed to using the old sweep and updating values when finished. If I understand you correctly, only the var option of ImplicitSourceTerm comes into play ( http://www.ctcms.nist.gov/fipy/examples/diffusion/generated/examples.diffusion.coupled.html ) so the other var present are explicit, also after coupling. Seems the logical thing. So, next question would be, can I use the nicer new notation suitable for coupling, and nevertheless sweep the coupled equation? But sweep requires a var option, so that seems not possible on the coupled equation, but perhaps is still possible on the seperate equations (would be nice if the coupled equation var could be passed in some way ...). So I could envision, doing one solve on the coupled equation, then doing sweep equation per equation, and continuing that till low residuals are obtained. If no sweeping is used, I would expect it best to take equal parts explicit and implicit. So not as you say k3 * M1 * P0 - ImplicitSourceTerm(k3 * M1, var=P0) - ImplicitSourceTerm(k3 * P0, var=M1) but - 1/2 *ImplicitSourceTerm(k3 * M1, var=P0) - 1/2*ImplicitSourceTerm(k3 * P0, var=M1) or if a fully explicit term is used, then 1/3 of three terms. If sweeping of equations written like that is supported, then sweeping this should go to full implicit. Anyway, my code is working already, so I can try different formulations out myself and see how much they result in different solutions. In case you wonder what I try to reproduce, see figures and equation in http://rspb.royalsocietypublishing.org/content/271/1548/1565 It's nice that after some hours with fipy you have a fully working script :-). But then the nagging question of how correct everything is pops up off course, and you need to put in the effort of verifying and checking... Greetings Benny > > Cheers, > > Daniel > > > -- > Daniel Wheeler > _______________________________________________ > fipy mailing list > fipy@nist.gov > http://www.ctcms.nist.gov/fipy > [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ] >
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