Amine - The n is effectively a power, or how many times the operator $\nabla\cdot\Gamma\nabla$ should be applied. The index i runs from 1..n to denote that each operator gets its own coefficient.
If n=1, then apply the operator $\nabla\cdot\Gamma_1\nabla$ to the argument $\phi$ and you get the "normal" 2nd-order diffusion. If n=2, then apply the operator $\nabla\cdot\Gamma_1\nabla$ to the argument consisting of the operator ($\nabla\cdot\Gamma_2\nabla$ applied to the argument $\phi$) and you get a 4th-order expression that comes up in Cahn-Hilliard problems. If n=3, then apply the operator $\nabla\cdot\Gamma_1\nabla$ to the argument consisting of the operator ($\nabla\cdot\Gamma_2\nabla$ applied to the argument ($\nabla\cdot\Gamma_3\nabla$ applied to the argument $\phi$)) and you get a 6th-order expression that comes up in phase field crystal. Repeat as necessary. If you don't have terms like this in your equations, don't worry about it. Even if you do, you're probably better off doing operator splitting, e.g., instead of \nabla\cdot(\Gamma_1\nabla[\nabla\cdot\Gamma_2\nabla\phi]) do \nabla\cdot(\Gamma_1\nabla\psi) and \psi = \nabla\cdot\Gamma_2\nabla\phi - Jon > On Jan 22, 2020, at 6:22 AM, A A <[email protected]> wrote: > > Here's what I mean about the diffusion term: > > <Untitled.png> > _______________________________________________ fipy mailing list [email protected] http://www.ctcms.nist.gov/fipy [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ]
