Hi Daniel, Thanks for your reply :-)
I started analyzing the data, as I need to use the result as a step in another
computation.
What I want is n(theta, r) where r is fixed. I can obtain this value by calling
the variable "n" with a list of coordinates. I do this by adding the following
code:
def radial_value(th,r,order=1):
x = r*np.sin(th)
y = r*np.cos(th)
return n((x, y), order=order)
and I plot
plot(theta, radial_value(theta, 1.5))
here, 1.5 is arbitrary. I need the value at different radii depending on other
parameters.
Around theta=0 and theta=pi, which are close to the left of my domain
(cylindrical radius=0), I expect zero flux (-> flat function). However, the
slope is non-negligible. I attach a plot of n(theta,r=1.5), for sigma=9 and
v=-0.01).
The result doesn't change much if I shift the mesh with
+ ((1e-5,),(0.,))
Could it be the the multiplication of the cell volume and face area by the
radius cause of bad handling of the zero flux BC on the left of the domain with
radius=0.
In that case, I couldn't solve satisfactorily my axisymmetric problem, is this
right?
Thanks,
Pierre
On Wed, Feb 20, 2013 at 09:32:33AM -0500, Daniel Wheeler wrote:
> On Sun, Feb 17, 2013 at 9:11 AM, Pierre de Buyl <pdeb...@ulb.ac.be> wrote:
>
> > Dear FiPy users and developers,
> >
> > Thanks for this software! The way FiPy expresses a PDF a an equality
> > (double
> > equal sign) in which one enters the different terms really motivated me to
> > give
> > it a try (the CISE paper is nice, too).
> >
>
> Hi Pierre,
>
> Great to have you on board and thanks for the feedback.
>
>
> > Thanks for any comment on the method.
> >
>
> There don't seem to be any glaring issues, it all seems reasonable to me.
> Nice work.
>
> Cheers,
>
> Daniel Wheeler
>
> --
> Daniel Wheeler
<<attachment: adv_diff_sigma9_radial.png>>
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