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For solving linear problems stemming from PDEs with the FEM, such as the Poisson equation or the wave equation, it is customary to use the "simplest" numerical quadrature that exactly integrates the quantities that define the mass/stiffness matrices. More precisely, given shape functions $\phi_k$ for nodes $k$ on an element $K$, we use a quadrature rule that exactly integrates the expression

$$ \int_K \phi_i \, \phi_j \, \mathrm{d} x $$

for each combination of $i$ and $j$ for the mass matrix, and $$ \int_K \nabla \phi_i \, \nabla \phi \, \mathrm{d} x $$ for the stiffness matrix. It turns out that for other linear problems, such as linear elasticity, we can use the same quadrature rule because the components of the matrix are made up of constants times products of the basis function gradients.

However, for many non-linear problems, such as non-linear elasticity with a Neo-Hookean material model, or other models that involve transcendental functions, we cannot exactly evaluate the integrals with standard quadrature rules.

My understanding is that the rule-of-thumb in these cases is to use the same quadrature as you would use for linear problems; presumably the quadrature error due to inexact integration is small compared to other sources of error. Another criterion that seems very important is to make sure that we do not lose rank of the matrix. In other words, the rank of our approximate matrix should be the same as that of the matrix obtained through exact integration.

I am looking for literature that discusses the choice of quadrature in a non-linear but smooth standard FEM setting (ignoring discontinuous material parameters or non-standard discretizations). In particular, I have several questions whose answers are not directly obvious to me:

  1. Can we pose "minimum" requirements on a quadrature rule so that we do not lose rank of our matrix compared to exact integration? Is using the same quadrature as for the linear problem sufficient?
  2. Are there prominent examples where we need higher accuracy than the "linear problem quadrature"?
  3. Are there other concerns than the ones I listed here? (*)

Any advice, pointers or insights would be most welcome!

(*) In some special settings, it may be important to capture certain characteristics. For example, with a Neo-Hookean elastic model you may want to have enough quadrature points to reasonably ensure that the determinant of your deformation gradient remains positive throughout the element, in order to prevent local inversion of the element.

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  • $\begingroup$ I do not fully understand your question. Something what you always can do is to oversample or overintegrate your solution with a standard quadrature rule and project the solution back to your original space. This is often done to prevent geometrical aliasing when curved metrics are used, e.g., based on NURBS. $\endgroup$
    – ConvexHull
    Dec 16, 2021 at 17:20
  • $\begingroup$ The deformation gradient is computed at each quadrature point using values of displacement at the nodes. Therefore, it depends entirely on the interpolation functions over an element. I don't see how increasing quadrature points will prevent negative det(F) values. Do you have a reference where that issue has been explored? $\endgroup$
    – Biswajit Banerjee
    Dec 16, 2021 at 19:51
  • $\begingroup$ @BiswajitBanerjee Please define $F$. $\endgroup$
    – ConvexHull
    Dec 16, 2021 at 20:00
  • $\begingroup$ Moreover i don't understand your statement: "The deformation gradient is computed at each quadrature point using values of displacement at each node. Therefore, it depends entirely on the interpolation functions over an element." Your FEM solution is more than just the nodal values. $\endgroup$
    – ConvexHull
    Dec 16, 2021 at 20:44
  • $\begingroup$ @ConvexHull F = deformation gradient. I am addressing the part in italics. The accuracy of the stress solution (and the material model it comes from + the accuracy of the integral over the stress) should have no impact on the apparent local inversion of an element. Inversion is a due to the choice of the interpolation functions and, sometimes, a failure of the element to model the physical phenomenon correctly. $\endgroup$
    – Biswajit Banerjee
    Dec 16, 2021 at 21:09

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