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[Alessio Quaglino <alessio@xxxxxx>]

As far as I know the gradient is an opperation on scalar fields giving a vector, while the jacobian is on vector fields giving a tensor. There's no assumption on the dimension of the vector (which you have with the divergence). I then guess page 42 of ffc documentation must be corrected, since there is stated that grad() applied to vectors gives the jacobian matrix.

Alessio


________________________________________
Da: Garth N. Wells [gnw20@xxxxxxxxx]
Inviato: venerdì 28 novembre 2008 10.25
A: Alessio Quaglino
Cc: ffc-dev@xxxxxxxxxx
Oggetto: Re: [FFC-dev] R:  bug with triangles?

Alessio Quaglino wrote:
> Thank you for the reply. So you mean that such limitation in present
> only for triangles, while for tets you can take gradients of
> n-vectors?

It's not a limitation and it applies to all spatial dimensions. The
gradient is an operator that is only defined for  tensors of dimension n
in a space of dimension n. The gradient of a function with three
components in two dimensions is not defined.

Then I don't understand, also because the workaround I
> used in the code I posted treats grad(phi) and grad(b) as 3x2
> matrices and converts them to vectors, and I don't get any "index out
> of range" error.
>

We should add a test that would give an error for your second piece of
code because the gradient isn't defined for the type of function that
it's being given.

You can still take derivatives of your function with three components,
but you'll have to to it using .dx(0), etc or by defining you own
grad-like operator.

Garth

> Alessio
>
>
> ________________________________________ Da: Garth N. Wells
> [gnw20@xxxxxxxxx] Inviato: venerdì 28 novembre 2008 9.41 A: Alessio
> Quaglino Cc: ffc-dev@xxxxxxxxxx Oggetto: Re: [FFC-dev] bug with
> triangles?
>
> Alessio Quaglino wrote:
>> I have the following code:
>>
>> q = 2 EL1 = VectorElement("Lagrange", "triangle", q, 3) EL2  =
>> VectorElement("Lagrange", "triangle", q - 1, 3) Mh = EL1 + EL2
>> (psi, phi) = TestFunctions(Mh) (b, w) = TrialFunctions(Mh) k = dot(
>> grad(phi),grad(b) )
>>
>> which gives me the error:
>>
>> k = dot( grad(phi),grad(b) ) File
>> "/Library/Python/2.5/site-packages/ffc/compiler/language/operators.py",
>> line 102, in dot form = form + v[i][j]*w[i][j] IndexError: list
>> index out of range
>>
>> I don't think I hven't understood what I supposed to give to ffc,
>> since if I define the following I get the desired behavior:
>>
>> def vect(a): return [ a[i][j] for j in range(2) for i in range(3) ]
>>  gphi = vect( grad(phi) ) gb = vect( grad(b) ) k = dot( gphi,gb )
>>
>> Is there a problem in handling (n x d) matrices with 2D meshes?
>
> The problem is that FFC doesn't know how to take the gradient of a
> vector function of length 3 on a 2D element (triangle). The below
> works
>
> q = 2 EL1 = VectorElement("Lagrange", "triangle", q, 2) EL2  =
> VectorElement("Lagrange", "triangle", q - 1, 2) Mh = EL1 + EL2 (psi,
> phi) = TestFunctions(Mh) (b, w) = TrialFunctions(Mh) k = dot(
> grad(phi),grad(b) )
>
> The '2' at the end of the element definition is redundant in this
> case since FFC will default to 2 for 2D elements.
>
> Garth
>
>> Alessio _______________________________________________ FFC-dev
>> mailing list FFC-dev@xxxxxxxxxx
>> http://www.fenics.org/mailman/listinfo/ffc-dev
>
>
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