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Grad-div alone does not remove the Q1 checkerboard pressure kernel

Example

On the same periodic Q1/Q1 mesh and with the same checkerboard pressure from A periodic Q1/Q1 checkerboard pressure mode is invisible to divergence, add any grad-div coefficient $γ \geq 0$ to the velocity equation but omit PSPG. The pressure mode remains a nonzero null vector for every $γ$ .

Facts & Assumptions

Given: The periodic Q1/Q1 Stokes form, grad-div parameter $γ \geq 0$ , no PSPG term, and the checkerboard pressure mode.

[F1]

Residual PSPG and grad-div terms for equal-order flow spaces: grad-div adds $γ (\nabla \cdot u_{h}, \nabla \cdot v_{h})$ to the velocity block; the unstabilized mixed terms are $- (p_{h}, \nabla \cdot v_{h}) + (q_{h}, \nabla \cdot u_{h})$ .

[F2]

A periodic Q1/Q1 checkerboard pressure mode is invisible to divergence: the nonzero mean-zero pressure $q_{h}$ satisfies $(q_{h}, \nabla \cdot v_{h}) = 0$ for every Q1 velocity $v_{h}$ .

Verification

1.1F1F2given

Set $u_{h} = 0$ and $p_{h} = q_{h}$ . Against an arbitrary velocity test $v_{h}$ , the pressure contribution is $- (q_{h}, \nabla \cdot v_{h}) = 0$ by [F2], and the grad-div term is zero because $u_{h} = 0$ . Against an arbitrary pressure test $r_{h}$ , the continuity contribution is $(r_{h}, \nabla \cdot u_{h}) = 0$ .

2.1step 1.1algebra∎

Thus the homogeneous mixed operator maps the nonzero pair $(0, q_{h})$ to zero for every $γ \geq 0$ . This example shows that grad-div alone leaves a nonzero kernel for this equal-order pairing; it makes no pressure-uniqueness claim after adding another stabilization or changing the spaces.

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Sources

Felipe Galarce and Douglas R. Q. Pacheco, Fully consistent lowest-order finite element methods for generalised Stokes flows with variable viscosity, arXiv:2411.02100v1 (standard reference, not scraped)
Volker John, Numerical Methods for Convection-Dominated Problems, lecture notes, Chapter 5 (standard reference, not scraped)