On Fri, Oct 19, 2012 at 11:03 AM, Wei Shao - CITG <[email protected]> wrote: > In order to solve Richards equation: > > $C(h)\frac{{\partial h}}{{\partial t}} = \frac{\partial }{{\partial > z}}\left[ {K(h)\left( {\frac{{\partial h}}{{\partial z}} + 1} \right)} > \right] + Q$ > > > > I try to define the convection term by K/C/h. Is that a correct expression?
Hi Kris, I am a little confused about the question. There is nothing wrong with "K / C / h" as an expression per se. It doesn't seem to be correct to divide the equation through by C and then take the C inside the derivatives. Is this in fact what you did? Or am I missing some steps? > > But when I change the function to this form: Do you mean the "function" or the "equation"? Little confused by that. > > Req = (TransientTerm(coeff=1.0) == > > DiffusionTerm(coeff=K(h)/C(h)) + > > VanLeerConvectionTerm (coeff=K(h)/C(h)/h)) The above doesn't seem to resemble the latex version of the equation. Is Q just 0? I can't seem to derive the FiPy equation from the latex equation given above. > > The FiPy report me mistake. > > How can I define the convection term of Richards equation, if not possible, > also let me know. I think you can. You may know better, but at first glance I would treat the transient part of the equation as a source and a transient term and avoid dividing through by C. Rearrange the convection part of the equation to be $\left( K \right)_z = K' h_z = \left( K' h \right)_z - h K'' h_z$ Also you might want to construct a problem where C is just a constant to start with just so that you can test the convection and diffusion without worrying about the issues with C(h). Anyway, that is how I would approach it. Good luck! -- Daniel Wheeler _______________________________________________ fipy mailing list [email protected] http://www.ctcms.nist.gov/fipy [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ]
