Minimizing the potential energy in a hyperelasticity problem

variational formulation
1

Hello,
I am currently using the FEniCS/DOLFINx package to simulate deformations on a mesh volume.
Following the tutorial, I am using the following equation to find u such as F(u)=0 :
F = \overrightarrow{grad} \overrightarrow{v} . P ~\overrightarrow{dx} - \overrightarrow{T} . \overrightarrow{v} \overrightarrow{ds}
with

  • P the first Piola Kirchhoff stress P = \frac{\partial \psi}{\partial F}
  • \overrightarrow{dx} and \overrightarrow{ds} the volume and surface element
  • \overrightarrow{T} a traction force applied on the surface.

Originally, the goal would be to minimize the potential energy, which writes itself as:

\Pi = \int_{\Omega} \psi (u) \,dx - \int_{\partial \Omega} T.u \,ds

My question is: what is the link between these two equations ? How to get to the first one starting from the second one ? I know that the Gateaux derivative is used in that case but I don’t know how to apply it to \Pi.

Thank you.

2

(Note: I apologize in advance for the ugly typesetting, but it seems that the usual way of including inline equations is now rendering them on separate lines, at least in Firefox, and I couldn’t find a workaround by Googling!)

If you’ve studied calculus of variations before, it may help to use the notation \delta\mathbf{u} for the test function instead of \mathbf{v}\text{ ,} then think about setting the Gateaux derivative of \Pi with respect to \mathbf{u} in the direction \delta\mathbf{u} to zero (by analogy to seeking the minimum of a univariate function by looking for an argument such that its ordinary derivative is zero):

d\Pi(\mathbf{u};\delta\mathbf{u}) = \delta\Pi = \int_\Omega \frac{\partial\psi}{\partial\mathbf{F}}:\delta \mathbf{F}\,d\Omega - \int_{\partial\Omega} \mathbf{T}\cdot\delta\mathbf{u}\,d\partial\Omega = 0\text{ ,}

where the variation \delta\mathbf{F} is also a Gateaux derivative with respect to displacement of \mathbf{F} = \operatorname{grad}\mathbf{u} + \mathbf{I}\text{ ,} i.e.,

\delta\mathbf{F} = d\mathbf{F}(\mathbf{u};\delta\mathbf{u}) = \operatorname{grad}(\delta\mathbf{u})\text{ .}

Making the substitutions \partial\psi/\partial\mathbf{F}\to\mathbf{P}~,~\delta\mathbf{F}\to\operatorname{grad}(\delta\mathbf{u})~,~\delta\mathbf{u}\to\mathbf{v}\text{ ,} you recover the weak form

\int_\Omega \operatorname{grad}\mathbf{v}:\mathbf{P}\,d\Omega - \int_{\partial\Omega}\mathbf{T}\cdot\mathbf{v}\,d\partial\Omega = 0\quad\forall\mathbf{v}\text{ .}

2 Likes
3

Thanks a lot for your help @kamensky, I understand it a lot better now !

I guess the volume and surface element somehow correspond to the linked integral in the code formulation.