On Mon, Sep 15, 2008 at 02:54:25PM +0100, Garth N. Wells wrote:
Anders Logg wrote:
On Mon, Sep 15, 2008 at 02:45:58PM +0100, Garth N. Wells wrote:
Anders Logg wrote:
On Mon, Sep 15, 2008 at 11:21:59AM +0100, Garth N. Wells wrote:
DOLFIN wrote:
One or more new changesets pushed to the primary dolfin repository.
A short summary of the last three changesets is included below.
changeset: 4754:23602808c60413cb8faffca818e7a8c04527d3ec
tag: tip
user: Anders Logg <logg@xxxxxxxxx>
date: Sun Sep 14 19:46:40 2008 +0200
files: demo/pde/elasticity/python/demo.py
description:
Use symmetric gradient in variational form in elasticity demo
In particular reason for this? It is simpler (and still correct) to use
the gradient.
Garth
To make the form and the matrix symmetric. I showed the demo to a
friend (in computational mechanics) and he insisted that we replace
grad(v) by epsilon(v).
He's wrong :). When you take the inner product A : B, where B is
symmetric, the inner product only 'sees' the symmetric part of A. It's a
classic exercise and you can prove it by writing out the indices.
Garth
I don't understand how that makes it wrong. My point (and maybe his)
is that since B is symmetric, we may replace A by its symmetric part
(as you say easy to see by writing out the indices). So, both
epsilon(v) and grad(v) are correct.
They are both correct (I never said they they aren't). Quoting your
earlier email
"I showed the demo to a friend (in computational mechanics) and he
insisted that we replace grad(v) by epsilon(v)."
which is not correct.
Yes, it is correct. He did insist that we should replace grad(v) by
epsilon(v)! ;-)
We can use grad(v) *or* epsilon(v). Either way,
the matrix will be symmetric. grad(v) leads to a slightly small .h file
and I would expect is easier for FFC.
Garth
Yes, and I realize now that the matrix will still be symmetric if we
use grad(v). So the reason is something different. Maybe related to
boundary conditions as suggested by Shawn.
