dolfin team mailing list archive

["Matthew Knepley" <knepley@xxxxxxxxx>]

On Feb 11, 2008 4:02 PM, Matthew Knepley <knepley@xxxxxxxxx> wrote:
>
> On Feb 11, 2008 1:57 PM, Matthew Knepley <knepley@xxxxxxxxx> wrote:
> > On Feb 11, 2008 10:27 AM, Anders Logg <logg@xxxxxxxxx> wrote:
> > >
> > > On Mon, Feb 11, 2008 at 08:05:57AM -0600, Matthew Knepley wrote:
> > > > On Feb 11, 2008 5:05 AM, Anders Logg <logg@xxxxxxxxx> wrote:
> > > > > On Mon, Feb 11, 2008 at 11:54:14AM +0100, Kristen Kaasbjerg wrote:
> > > > > > Hi dolfin users
> > > > > >
> > > > > > I am trying to use dolfin for the following simple
> > > > > > electrostatic problem:
> > > > > > - finding the electrostatic potential in a domain composed
> > > > > >   of two subdomains with different dielectric constants.
> > > > > >   This amounts to solving Laplace equation with appropriate
> > > > > >    BC's.
> > > > > >
> > > > > > One thing that I cannot figure out how to do is to tell dolfin
> > > > > > that I have 2 subdomains and on the internal boundary between
> > > > > > these two domains, the normal derivative of the potential has
> > > > > > a discontinuity given by the difference between the dielectric
> > > > > > constants.
> > > > > > Could anyone give me a hint of how to approach this problem
> > > > > > with dolfin ? I am relatively unexperienced to FEM so a nice
> > > > > > reference would also be welcome.
> > > > > >
> > > > > > Regards
> > > > > > Kristen
> > > > >
> > > > > The easiest thing to do (if there is just a coefficient in your
> > > > > problem that is discontinuous) is to just define a Function for the
> > > > > dielectric constant and make it discontinuous. Then just plug it in to
> > > > > your equation.
> > > > >
> > > > > Look in the demos (under src/demo/pde) for how to define a function.
> > > > >
> > > > > For example, if some parameter is 0 or 1 depending on whether x is
> > > > > below or above 0.5, then just do something like this in eval():
> > > > >
> > > > >   if (x[0] > 0.5)
> > > > >     return 1.0;
> > > > >   else
> > > > >     return 0.0;
> > > >
> > > > This is not going to give the correct answer unless you do the boundary integral
> > > > that comes from the integration by parts. Or am I missing something?
> > > >
> > > >    Matt
> > >
> > > I don't know. Maybe I'm ignorant but if you have -div(a*grad(u)) = f,
> > > then it looks to me that you may just go ahead and integrate by parts
> > > even if the coefficient a is discontinuous. grad(u) is already
> > > discontinuous (for the discretization) so I don't see the difference.
> >
> > I do not agree. There is a jump term along the boundary. You can get
> > some solution without it, which looks alright, it is just wrong.
>
> How about this example. You can tell me what is wrong:
>
>   Take a 1-D domain from -1 to 1 which is divided at 0 into
> \epsilon_0 and \epsilon_1. Then we have something like
>
>   \int^1_{-1} v div (\epsilon grad u)
>
>  = \int^0_{-1} v div (\epsilon_0 grad u) + \int^1_0 v div (\epsilon_1 grad u)
>
>  = -\int^0_{-1} grad v \epsilon_0 grad u + \epsilon_0 v grad u |_0
>      -\int^1_0 grad v \epsilon_1 grad u - \epsilon_1 v grad u |_0
>
>  = -\int^1_{-1} \epsilon grad v grad u - (\epsilon_1 - \epsilon_0) v grad u |_0
>
> Dmitry is now of the opinion that this is wrong based on distribution theory
> arguments. However, I was sure that physically we get a layer of induced
> charge at a dielectric jump proportional to the jump times the field. I will try
> and look it up in Jackson.

So, it appears that I had the wrong interpretation of our variational
form. I did
remember correctly from Jackson that this problem is equivalent (in a
distributional sense) to

  (\epsilon_1 - \epsilon_0) \delta(0) + \epsilon \Delta u

However, when we go to the variational form, the integration by parts with the
piecewise constant \epsilon exactly cancels the delta term. This is
the result we
get when mollifying \epsilon. My derivation above interprets the
original integral
as just some Lebegue integral, and not as a derivative in the
distributional sense.
Is this how you understand it?

    Matt

>    Matt
>
>
> >    Matt
> >
> > > I tried to put in a discontinuous coefficient in the DOLFIN Poisson
> > > demo and the solution looks fine.
> > >
> > > --
> > >
> > > Anders
> > > _______________________________________________
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> > > http://www.fenics.org/mailman/listinfo/dolfin-dev
> > >
> >
> >
> >
> > --
> > What most experimenters take for granted before they begin their
> > experiments is infinitely more interesting than any results to which
> > their experiments lead.
> > -- Norbert Wiener
> >
>
>
>
> --
> What most experimenters take for granted before they begin their
> experiments is infinitely more interesting than any results to which
> their experiments lead.
> -- Norbert Wiener
>



-- 
What most experimenters take for granted before they begin their
experiments is infinitely more interesting than any results to which
their experiments lead.
-- Norbert Wiener