Thanks for the clarification, Jean-Paul. I'm glad to hear that my syntax is
correct.
I am indeed using an extractor, e.g. here's a line from my code:
const Tensor<2, dim> gradv = fe_values[velocity].gradient(j, quad);
I'm having some trouble finding an example of the S[i][j] syntax in either
tutorial. Part of step 44 has the following for the geometric stress
contribution:
data.cell_matrix(i, j) += grad_Nx[i][component_i] * tau
* grad_Nx[j][component_j] * JxW;
which is at least using the same operators to read the components; but it's
not clear to me that this maps to the $S_{i,j}$ notation we discussed. It
seems they have to do some work to map to the correct indices, i.e. with
const unsigned int component_i = fe.system_to_component_index
<https://dealii.org/developer/doxygen/deal.II/classFiniteElement.html#a27220a135402b96c7e6eecbb04acda56>
(i).first;
For now I'm satisfied with having an experience user confirm that my syntax
is correct; but I might want to formulate a test, unless a clear example
already exists somewhere. I should be able to write something down for a
simple PDE with a simple basis. I get the feeling that I'd just be testing
a well established deal.II code behavior. Maybe I'm missing something
obvious in the information you already provided.
I'm going to describe my specific problem a bit more in case there's an all
around better way to do this: This is coming from the incompressible
Navier-Stokes momentum equation with the term $(\mathbf{u} \cdot \nabla)
\mathbf{u}$. The nonlinear variational form uses the operator
$c(\mathbf{w};mathbf{z},mathbf{v} = \int{[(\mathbf{w} \cdot \nabla)
\mathbf{z}] \cdot \mathbf{v} d\Omega}_\Omega =
\sum{\sum{\int{w_j(\partial_j z_i)v_i d\Omega}_\Omega}_{j=1}^n}_{i=1}^n$
When I implement this with higher level library, FEniCS, I don't have to
write the summation; but I think I have to write the summation in deal.II;
hence I need to access $\partial_j z_i$, which I'm doing with the S[i][j]
syntax that we discussed. I actually implement something similar to the
integrand of this operator, for a given quadrature point and degree of
freedom, with the lambda function:
auto c = [](
const Tensor<1, dim> _w,
const Tensor<2, dim> _gradz,
const Tensor<1, dim> _v)
{
double sum = 0.;
for (unsigned int i = 0; i < dim ; ++i)
{
for (unsigned int j = 0; j < dim; ++j)
{
sum += _w[j]*_gradz[i][j]*_v[i];
}
}
return sum;
};
Thanks for the help!
P.S. What's the best way to communicate math equations on the mailing list?
I'm just typing what I think is roughly the correct LaTeX syntax, but this
seems a bit clunky.
On Wednesday, April 5, 2017 at 3:04:04 PM UTC+2, Jean-Paul Pelteret wrote:
>
> Dear Alex,
>
> What you've stated is correct. The tensor component $S_{ij}$ is exactly
> component S[i][j] of a Tensor<2,dim> S if "S" represents $S_{ij} =
> \partial v_{i} \partial x_{j}$.
>
> As for an example, both steps 18
> <https://dealii.org/developer/doxygen/deal.II/step_18.html#TopLevelupdate_quadrature_point_history>
>
> and 44
> <https://dealii.org/developer/doxygen/deal.II/step_44.html#Solidupdate_qph_incremental>
>
> both extract gradient of the vector-valued solution field. In step-18, the
> gradient tensor is constructed manually after a the call to
> fe_values.get_function_gradients
> <https://dealii.org/developer/doxygen/deal.II/classFEValuesBase.html#a69c63f8be0311c8970253276ac7df138>
>
> (incremental_displacement,
> displacement_increment_grads);
> while in step-44 an extractor is used as well in order to produce $\Grad
> \mathbf{u}$ (u representing the displacement) at each quadrature point
> directly:
> scratch.fe_values_ref[u_fe].get_function_gradients(scratch.
> solution_total,
> scratch.
> solution_grads_u_total);
>
> Depending on whether or not you use an extractor to perform this task, the
> data structure in which the result is placed (here,
> displacement_increment_grads
> and scratch.solution_grads_u_total) is slightly different. When using an
> extractor you will end up with the vector (each entry representing data at
> a quadrature point) of rank-2 tensors because it is known that this is the
> gradient of a vector field - this would be exactly "S" if the extractor
> pertains to the velocity field. In contrast, the FEValues object knows
> nothing about the solved fields if no extractor is used, so you'd have to
> reconstruct "S" manually, in a similar way to that which is done in step-18.
>
> I hope that this helps clarify things a little bit.
>
> Best regards,
> Jean-Paul
>
>
> On Wednesday, April 5, 2017 at 1:10:00 PM UTC+2, Alex Zimmerman wrote:
>>
>> I am debugging a code where part of my weak form requires indexing into
>> the Tensor<2,dim> that results from getting the gradient of a Tensor<1,dim>
>> (in this case, a velocity vector).
>>
>> I found the convention for the index notation here:
>> http://dealii.org/8.4.1/doxygen/deal.II/classFEValuesViews_1_1Vector.html
>>
>> The important part of that documentation for me is that "The gradient of
>> a vector $d_k, 0 <= k < dim$ is defined as $S_{ij} = \frac{\partial
>> d_i}{\partial x_j}, 0 <= i,j < dim$.
>>
>> What is the exact syntax for getting component $S_{ij}$ in my code? For a
>> Tensor<2,dim> S
>>
>> I am currently assuming that I want
>> S[i][j]
>>
>> But I am not 100% sure that don't have this backwards. I have reviewed
>> the documentation about how these tensors are rolled/unrolled; but I'm
>> having trouble devising a test to ensure that I am understanding this
>> correctly, and I'm guessing that this is something quite obvious to most
>> users.
>>
>> Also it would help if someone could point me toward an existing
>> tutorial/example that accesses components of the gradient of a physical
>> vector. I had no luck finding one.
>>
>> Thanks,
>>
>> Alex
>>
>
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