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ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck pass
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A periodic Q1/Q1 checkerboard pressure mode is invisible to divergence

Example

Take the periodic square with a 2×2 unit-square Q1 mesh. At the grid node (i,j), for i,j∈{0,1,2}, set qij=(−1)i+j and identify opposite boundaries periodically. Assume τK>0 on every cell. These nodal values define a nonzero mean-zero Q1 pressure. It is invisible to the ordinary velocity-divergence coupling, although its PSPG pressure-gradient term is strictly positive.

Facts & Assumptions

Given: The periodic 2×2 Q1 mesh, equal-order Q1 velocity and pressure, strictly positive cell parameters τK, and the PSPG residual term from Residual PSPG and grad-div terms for equal-order flow spaces.

[F1]

Residual PSPG and grad-div terms for equal-order flow spaces: PSPG adds ∑KτK(∇qh,RM,K(uh,ph))K; for the Stokes specialization and uh=0, ph=qh, f=0, the residual is ∇qh.

Verification

1.1givenconstructalgebra

On a cell Kij with local coordinates (ξ,η)∈[0,1]2, the corner values are σij(1,−1,−1,1) with σij=(−1)i+j. Q1 interpolation therefore gives qh∣Kij=σij(1−2ξ)(1−2η). Its integral on every cell is zero, so its global mean is zero, and it is nonzero at every mesh node.

2.1step 1.1algebra

For any Q1 velocity component v1, ∂xv1 on a cell is affine in η and independent of ξ. Since ∫01(1−2ξ) dξ=0, ∫Kijqh∂xv1=0. Similarly, ∂yv2 is independent of η and ∫01(1−2η) dη=0, so ∫Kijqh∂yv2=0. Summing cells gives (qh,∇⋅vh)=0 for every Q1 velocity.

3.1F1step 1.1step 2.1givenalgebra∎

On each unit cell, ∫K∣∇qh∣2=4/3+4/3=8/3. Thus ∑KτK∥∇qh∥0,K2=(8/3)∑KτK>0 by the strict positivity in Given. By [F1], the PSPG term detects this mode, while the unstabilized divergence equation does not. This finite mode check proves neither a mesh-uniform inf-sup bound nor general PSPG accuracy.

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