On Thu, Mar 3, 2011 at 2:45 AM, Fadoua El Moustaid <fad...@aims.ac.za> wrote:
>
> \frac{\partial b}{\partial t} = D \frac{\partila ^2 b}{\partial x^2}
> with initial condition
> b(x, t=0) = b0 constant for 0<x<L
> and I'm using as boundary conditions
> b(x=0, t) = 0 for t>0
> b(x=L, t) = 0 for t>0
Does FiPy work for you when solving these equations and boundary conditions?
> This means that once the particles reach the boundaries their density
> becomes zero and they do not belong to the particles in the free space
> anymore. In this case, for me to know how many particles got stuck at time
Sorry, I don't know anything about particles. The equation above has a
solution. Is your problem that the equation & BCs above are not being
solved correctly or that the equation & BCs do not describe the
physics correctly?
> T I just do
> b0 - \int_0^T b(x, ti) dti
> because there is now production or growth of the particles. Now my
> question is about how to measure the particles stuck to the boundaries in
Are you asking how to calculate the quantity "b0 - \int_0^T b(x, ti)
dti" as you calculate your problem using FiPy/python? I can explain
that.
> case of growth, which means if my equation is in the following form:
> \frac{\partial b}{\partial t} = D \frac{\partila ^2 b}{\partial x^2} +
> rb(1-b/k)
> where r and k are constants.
That is a different equation from above. Do you have a problem solving
it or posing it in FiPy?
--
Daniel Wheeler
