On Feb 20, 2015, at 12:08 PM, Daniel Wheeler <[email protected]> wrote:
> On Fri, Feb 20, 2015 at 6:39 AM, Ronghai Wu <[email protected]> wrote: >> Thanks, I worked it out. But still have three questions: >> (1) The 4th-order is split by "psi = d2fdphi2(phi-phiold) + dfdphi - >> epsilon**2*laplace phi". I do not understand why we need >> d2fdphi2(phi-phiold)? > > Where do you see this? Do you have a link to this? It's in http://www.ctcms.nist.gov/fipy/examples/cahnHilliard/generated/examples.cahnHilliard.mesh2DCoupled.html that Ronghai was originally asking about. _______________________________________________ fipy mailing list [email protected] http://www.ctcms.nist.gov/fipy [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ]
