> I understand that your description implies that you're comparing the
residuals for a timestep where both schemes have taken the same number of
Newton iterations. But nevertheless, I'm not quite sure that there's any
value in comparing these two quantities. They are both discrete residuals
Hi David,
One last point I don't understand right now is the magnitude of the
machine precision you talk about:
> You'll see that they all exhibit quadratic convergence, with only a
slight difference in the convergence history that can probably be
attributed to accumulated round-off errors.
Hi Jean-Paul,
> The indexing is correct - I've taken that part of the code from some
other codes that I've fully verified. But I thought it would still be a
useful exercise to re-implement the assembly routine for the other two
parameterisations. Maybe this would further convince you of its
Hi Jean-Paul,
first of all, thank you very much for your comprehensive answer. It's a
little bit unfortunate that there is no way for the deformed test gradient
evaluation using matrix-free. However, your code snippets and the
explanation are traceable and the required changes for the tangent
Hi David,
So as I'm sure that you know, if you want to assemble the linear system
for quasi-static non-linear (hyper)elasticity with a referential
DoFHandler (using no Eulerian mapping to transform the shape functions
to the spatial setting) AND without doing a similar transformation by
hand
Maybe as an edit: what I currently do looks the following way:
```
// Get gradient in reference frame
const Tensor<2, dim, VectorizedArrayType> grad_u = phi.get_gradient(q_point
);
// Compute deformation gradient
const Tensor<2, dim, VectorizedArrayType> F = Physics::Elasticity::
