[petsc-users] Elasticity tensor in ex52

[Matthew Knepley]

On Sat, Apr 19, 2014 at 5:25 PM, Miguel Angel Salazar de Troya <
salazardetroya at gmail.com> wrote:

> Thanks for your response. Now I understand a bit better your
> implementation, but I am still confused. Is the g3 function equivalent to
> the f_{1,1} function in the matrix in equation (3) in this paper:
> http://arxiv.org/pdf/1309.1204v2.pdf ? If so, why would it have terms
> that depend on the trial fields? I think this is what is confusing me.
>

Yes, it is. It has no terms that depend on the trial fields. It is just
indexed by the number of components in that field. It is
a continuum thing, which has nothing to do with the discretization. That is
why it acts pointwise.

   Matt


> On Sat, Apr 19, 2014 at 11:35 AM, Matthew Knepley <knepley at gmail.com>wrote:
>
>> On Fri, Apr 18, 2014 at 1:23 PM, Miguel Angel Salazar de Troya <
>> salazardetroya at gmail.com> wrote:
>>
>>> Hello everybody.
>>>
>>> First, I am taking this example from the petsc-dev version, I am not
>>> sure if I should have posted this in another mail-list, if so, my
>>> apologies.
>>>
>>> In this example, for the elasticity case, function g3 is built as:
>>>
>>> void g3_elas(const PetscScalar u[], const PetscScalar gradU[], const
>>> PetscScalar a[], const PetscScalar gradA[], const PetscReal x[],
>>> PetscScalar g3[])
>>> {
>>>   const PetscInt dim   = spatialDim;
>>>   const PetscInt Ncomp = spatialDim;
>>>   PetscInt       compI, d;
>>>
>>>   for (compI = 0; compI < Ncomp; ++compI) {
>>>     for (d = 0; d < dim; ++d) {
>>>       g3[((compI*Ncomp+compI)*dim+d)*dim+d] = 1.0;
>>>     }
>>>   }
>>> }
>>>
>>> Therefore, a fourth-order tensor is represented as a vector. I was
>>> checking the indices for different iterator values, and they do not seem to
>>> match the vectorization that I have in mind. For a two dimensional case,
>>> the indices for which the value is set as 1 are:
>>>
>>> compI = 0 , d = 0      ----->     index = 0
>>> compI = 0 , d = 1      ----->     index = 3
>>> compI = 1 , d = 0      ----->     index = 12
>>> compI = 1 , d = 1      ----->     index = 15
>>>
>>> The values for the first and last seem correct to me, but they other two
>>> are confusing me. I see that this elasticity tensor (which is the
>>> derivative of the gradient by itself in this case) would be a four by four
>>> identity matrix in its matrix representation, so the indices in between
>>> would be 5 and 10 instead of 3 and 12, if we put one column on top of each
>>> other.
>>>
>>
>> I have read this a few times, but I cannot understand that you are
>> asking. The simplest thing I can
>> respond is that we are indexing a row-major array, using the indices:
>>
>>   g3[ic, id, jc, jd]
>>
>> where ic indexes the components of the trial field, id indexes the
>> derivative components,
>> jc indexes the basis field components, and jd its derivative components.
>>
>>    Matt
>>
>>
>>> I guess my question is then, how did you vectorize the fourth order
>>> tensor?
>>>
>>> Thanks in advance
>>> Miguel
>>>
>>> --
>>> *Miguel Angel Salazar de Troya*
>>> Graduate Research Assistant
>>> Department of Mechanical Science and Engineering
>>> University of Illinois at Urbana-Champaign
>>> (217) 550-2360
>>> salaza11 at illinois.edu
>>>
>>>
>>
>>
>> --
>> 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
>>
>
>
>
> --
> *Miguel Angel Salazar de Troya*
> Graduate Research Assistant
> Department of Mechanical Science and Engineering
> University of Illinois at Urbana-Champaign
> (217) 550-2360
> salaza11 at illinois.edu
>
>


-- 
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
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