On 4/4/25 09:24, Sclah wrote:
- take the gradient of the previous solution
/fe_values[components].get_function_gradients(old_sol, local_old_grad);/
- for every quadrature point "q" compute the deformation gradient (F =
Identity + grad(u))
/Tensor<2, dim> F = Identity + local_old_grad(q);/
/Tensor<2, dim> F_inv = invert(F);/
Now I'd like to compute the divergence of /F_inv/, is there a way to do that?
I was thinking about automatic differentiation but maybe there is a more
straightforward way
You need to apply the chain rule. If I understand correctly, you need
div ( [I + grad(u)]^{-1} )
= trace grad ( [I + grad(u)]^{-1} )
which is of the form
grad ( A^{-1} )
and which you can compute by observing that on the one hand
grad ( A A^{-1} ) = grad I = 0
and on the other hand by using the product rule
grad ( A A^{-1} ) = (grad A) A^{-1} + A (grad A^{-1})
and so
grad A^{-1} = - A^{-1} (grad A) A^{-1}
with an appropriate choice of contraction over indices. So that gives you
something (you'll have to check the details) of the form
grad ( [I + grad(u)]^{-1} )
= - Finv grad (I + grad(u)) Finv
= - Finv [grad^2(u))] Finv
Best
W.
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