Just to make sure that I chose a reasonable approach:
//create FEValues object whenever I need it, that is,
'number_of_cells*number_of_qps_per_cell' times
FEValues<2> fe_values(dof_handler.get_fe(),
Quadrature<2>(my_point_in_ref_coords),
update_gradients);
fe_values.reinit(cell);
//evaluate, say, the gradient of the second shape function in real coords
at "my_point_in_ref_coords"
Tensor<1,2> shape_gradient_real_second = fe_values.shape_grad(1, 0);
Without measuring the time but based on your intuition - the approach using
FEPointEvaluation would not be much faster, would it?
Best
Simon
Am Do., 7. Juli 2022 um 03:45 Uhr schrieb Wolfgang Bangerth <
[email protected]>:
> On 7/5/22 09:28, Simon Wiesheier wrote:
> >
> > What is the approach to achieve this?
> > My only idea is to create a new FEValues object for each quadrature
> point and
> > pass an appropriate Quadrature object. But seems to be very inefficient.
>
> This is basically what the VectorTools::point_value() and
> VectorTools::point_gradient() functions do. You might want to read through
> their implementations to see how this is done.
>
> The only other approach is with classes like FEPointEvaluation. If you
> read
> through their implementation, you might be able to understand how they can
> also be used to evaluate just a single shape function. In the end, the
> values
> and gradients of a single shape function correspond to the values and
> gradients of a finite element field that corresponds to a solution vector
> with
> one 1.0 and the rest of the elements set to zero.
>
> Best
> W.
>
>
> --
> ------------------------------------------------------------------------
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> www: http://www.math.colostate.edu/~bangerth/
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