Yes! Thank you very much, I was indeed missing the mathematical derivation.
I better checked my computations and I actually need:
*div([I + grad(u)]^{-T})* instead of *div([I + grad(u)]^{-1})* but
following your steps that leads to a similar result:
*div([I + grad(u)]^{-T}) = - Finv [grad(grad(u)^T)] Finv*
Then I guess that using get_function_hessians() on *u* is what I need to
assemble correctly my weak form.
Thank you again!
Il giorno venerdì 4 aprile 2025 alle 17:49:05 UTC+2 Wolfgang Bangerth ha
scritto:
> 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.
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: bang...@colostate.edu
> www: http://www.math.colostate.edu/~bangerth/
>
>
>
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