My gut reaction is that the second implementation is probably better, as it should be possible to make things more implicit and coupled, even though you're "wasting" calculation of PDE1 over most of the 2D domain where you don't care about it.
> On Jul 14, 2016, at 11:18 AM, Guyer, Jonathan E. Dr. (Fed) > <[email protected]> wrote: > >> What’s the best way to implement this problem in FiPy? > > Don't! > > > Assuming you won't take that advise, I've posted a couple of attempts at this > problem at: > > https://gist.github.com/guyer/bb199559c00f6047d466daa18554d83d > > > >> On Jul 9, 2016, at 1:37 PM, Gopalakrishnan, Krishnakumar >> <[email protected]> wrote: >> >> Hi, >> >> **We’re sorry, the previous posting omitted the complete definition of the >> boundary condition for the 2nd PDE, and also a couple of typos are fixed >> below. >> >> We have a system of equations, wherein the BC of one PDE couples with the >> source term of another PDE. >> >> We have a regular 2D unit grid in x and y. >> >> There are two PDEs to be solved: >> a) The first PDE (elliptic diffusion problem) is defined only at y = 1, >> acting along the x-axis (i.e. it acts in the x-direction and only along the >> top of the cartesian grid). This x-axis is discretized with a fixed >> grid-spacing, generating a finite number of nodes. Let this set of >> nodes/co-ordinates be represented by ‘X’. >> b) The second PDE is a time-varying diffusion problem. This is defined >> only along the y-axis, but for all x-nodes (i.e. for all ‘X’), where the 1st >> PDE is being solved. >> >> Mathematically (plain-text version) : >> >> PDE1: divergence(S * grad(a)) = f(w,a) >> BC1: a = 0 , at x = 0 (dirichelet) >> BC2: d_a/dx = 1 at x = 1 (Neumann) >> >> D = faceVariable(rank=2, value = 1.0) # Rank 2 tensor for anisotropic >> diffusion (i.e. allowed only along the y-axis) >> >> PDE2: partial_B/partial_t = divergence( [[[0, 0], [0, D]]] * grad(B) ) >> BC1: B = 0, at y = 0 at all ‘X’ (dirichelet), i.e. along the bottom face >> BC2: d_B/dy = g(w,a) (Neumann) at y =1, i.e. along the top face of the grid >> >> f and g are linear functions in ‘w(x,y,t)’ and ‘a(x,y,t)’. B is defined in >> the 2D grid as ‘B(x,y,t)’. >> >> Importantly, w = B at y = 1, for all ‘X’; i.e., the BC2 of the 2nd PDE >> couples with the Implicit Source Term of the 1st PDE along the top face of >> the Cartesian mesh. >> >> What’s the best way to implement this problem in FiPy? >> >> Best Regards, >> >> Krishna >> >> >> PS: Since the problem is mathematically harder to express in plain-text, >> here is a HTML formatted version of the PDEs and BCs. >> PDE1: <image002.png> >> BC1: <image004.png> (Dirichelet) >> BC2: <image006.png> (Neumann) >> >> D = faceVariable(mesh=pos_p2d_mesh, rank=2, value = 1.0) # Rank 2 >> tensor for anisotropic diffusion (i.e. allowed only along the y-axis) >> >> PDE2: <image008.png> = <image010.png> >> BC1 is: <image012.png> i.e. along the bottom face of the grid >> BC2 is: <image014.png> (Neumann) at y =1, i.e. along the top face of the >> grid >> >> f and g are linear functions in <image015.png> and<image016.png>. >> <image018.png> is defined in the 2D grid as <image035.png> >> >> Importantly, <image036.png> i.e., the BC2 of the 2nd PDE couples with the >> Implicit Source Term of the 1st PDE along the top face of the Cartesian mesh. >> >> <image017.png><image024.png><image026.png><image023.png><image025.png><image022.png><image011.png><image019.png><image020.png>_______________________________________________ >> fipy mailing list >> [email protected] >> http://www.ctcms.nist.gov/fipy >> [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ] > > > _______________________________________________ > fipy mailing list > [email protected] > http://www.ctcms.nist.gov/fipy > [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ] _______________________________________________ fipy mailing list [email protected] http://www.ctcms.nist.gov/fipy [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ]
