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[Martin Sandve Alnæs <question152914@xxxxxxxxxxxxxxxxxxxxx>]

Question #152914 on DOLFIN changed:
https://answers.launchpad.net/dolfin/+question/152914

Martin Sandve Alnæs posted a new comment:
Kristian, can you check if this seems right? Is det(f) == f for scalar f
useful?

In [3]: V = FiniteElement("CG", interval, 1)

In [4]: f = Coefficient(V)

In [5]: grad(f)
Out[5]: SpatialDerivative(Coefficient(FiniteElement('Lagrange',
Cell('interval', 1, Space(1)), 1), 0), MultiIndex((FixedIndex(0),), {0: 1}))

In [6]: grad(f) == f.dx(0)
Out[6]: True

In [7]: div(f) == f.dx(0)
Out[7]: True

In [10]: inner(f,f) == f*f
Out[10]: True

In [11]: dot(f,f) == f*f
Out[11]: True

In [13]: det(f) == f
Out[13]: True

Martin


On 15 April 2011 12:30, Kristian Ølgaard <k.b.oelgaard@xxxxxxxxx> wrote:

> On 15 April 2011 10:53, Martin Sandve Alnæs <martinal@xxxxxxxxx> wrote:
> > Change grad(T) to T.dx(0), i.e. derivative in a single direction.
> > Maybe we should make grad work like .dx(0) in 1D.
>
> I think that would be a good idea.
> When testing stuff in 1D (which is intended for 2D and 3D) I find it
> annoying to change back and forth between grad() and .dx(0).
>
> Kristian
>
> > Martin
> >
> > On 15 April 2011 10:46, Bento <question152914@xxxxxxxxxxxxxxxxxxxxx>
> wrote:
> >>
> >> New question #152914 on DOLFIN:
> >> https://answers.launchpad.net/dolfin/+question/152914
> >>
> >> Hi,
> >>
> >> I want to solve a PDE on a one-dimensional domain. The equation involves
> a
> >> first derivative. How do I express this in 1D? Here is (basically) my
> >> equation:
> >>
> >> dT/dx + d^2T/dx^2 = 0
> >>
> >> and weak form
> >>
> >> a = (grad(T)*v - inner(grad(T), grad(v)))*dx
> >>
> >> with testfunction v and trialfunction T. When I use grad(T), I get a
> >> "shape mismatch in sum" error. The mesh is Interval(100, 0, 10)
> >>
> >> --
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