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One staggered Fourier mode is removed exactly by pressure correction

Statement

Let $N \geq 2$ be even, $d \geq 1$ , $h = 1 / N$ , and use the staggered operators from Periodic staggered-grid velocity and pressure spaces. Put $q_{j} = (- 1)^{j_{1}}$ , so $q \in Q_{h}$ , and let $ν, k > 0$ . Start one step of Backward-Euler incremental pressure correction on a periodic staggered grid from $u^{n} = 0$ , $p^{n} = 0$ with $f^{n + 1} = k^{- 1} G_{h} q + ν A_{h} G_{h} q .$ Then the unique predictor is ${\tilde{u}}^{n + 1} = G_{h} q$ , the pressure increment is $ϕ^{n + 1} = k^{- 1} q$ , and the corrected velocity is exactly $u^{n + 1} = 0$ . The pressure update is $p^{n + 1} = k^{- 1} q$ .

Facts & Assumptions

Given: The periodic staggered grid, even $N$ , the alternating cell pressure $q_{j} = (- 1)^{j_{1}}$ , positive $k, ν$ , and the stated one-step data.

[F1]

The staggered operators satisfy $Δ_{h} = D_{h} G_{h}$ and $A_{h}$ is the negative componentwise periodic Laplacian (Periodic staggered-grid velocity and pressure spaces).

[F2]

The predictor, pressure-increment equation and velocity/pressure updates are those in Backward-Euler incremental pressure correction on a periodic staggered grid.

[F3]

Both solves are unique, the pressure increment is mean-zero, and the correction is exactly the discrete Helmholtz projection (Well-definedness of the periodic pressure-correction solves, Mass-orthogonal periodic discrete Helmholtz projection and exact incompressibility).

Proof

1.1F1givenalgebra

Since $N$ is even, $q$ has zero mean. The staggered gradient is not zero: $(G_{h} q)_{1, j} = - 2 q_{j} / h$ . Also $Δ_{h} q = - 4 q / h^{2}$ , so [F1] gives $D_{h} G_{h} q = Δ_{h} q$ . Periodic summation shows both $G_{h} q$ and $A_{h} G_{h} q$ have zero component means, hence the force is admissible.

2.1F1F2F3step 1.1algebra

Substituting ${\tilde{u}}^{n + 1} = G_{h} q$ and $u^{n} = p^{n} = 0$ into the predictor equation [F2] gives $k^{- 1} G_{h} q + ν A_{h} G_{h} q = f^{n + 1}$ , so uniqueness [F3] identifies it as the predictor. For the pressure equation, $Δ_{h} (k^{- 1} q) = k^{- 1} D_{h} G_{h} q = k^{- 1} D_{h} {\tilde{u}}^{n + 1};$ both sides have mean zero and [F3] gives $ϕ^{n + 1} = k^{- 1} q$ .

3.1F2step 2.1algebra∎

The velocity update [F2] now gives $u^{n + 1} = G_{h} q - k G_{h} (k^{- 1} q) = 0$ , while the pressure update gives $p^{n + 1} = k^{- 1} q$ . This is exact one-step removal of the selected mode.

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Sources

J. L. Guermond, P. Minev, and J. Shen, An overview of projection methods for incompressible flows (standard reference, not scraped)
NPTEL, Lecture 5.4: Methods based on Coupling of Pressure and Velocity (standard reference, not scraped)