Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 5 results · all verified · 1 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 4 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

SUPG, residual stabilization and variational multiscale methods — Examples

1 · Prerequisites

SUPG, residual stabilization and variational multiscale methods

2 · Summary

These examples separate what each stabilization term controls. The two-cell boundary-layer calculation shows exact nodal fitting with a large within-cell error; the one-degree-of-freedom load example shows how an oversized parameter can damp the discrete solution. The periodic Q1/Q1 example gives an exact pressure mode in the divergence-coupling kernel, which a PSPG pressure-gradient term detects; the next example proves that grad-div alone leaves that kernel unchanged. The final VMS example attains the residual-closure defect bound exactly.

The checkerboard nodal pattern on the $2 \times 2$ periodic Q1 mesh is shown below. Opposite edges are identified periodically; the corner and edge signs therefore agree under those identifications. On each cell the bilinear interpolant is $σ_{i j} (1 - 2 ξ) (1 - 2 η)$ , as verified algebraically in its item.

\begin{tikzpicture}[scale=1.2]
\draw[step=1,gray!70] (0,0) grid (2,2);
\node at (0,0) {$+$};
\node at (1,0) {$-$};
\node at (2,0) {$+$};
\node at (0,1) {$-$};
\node at (1,1) {$+$};
\node at (2,1) {$-$};
\node at (0,2) {$+$};
\node at (1,2) {$-$};
\node at (2,2) {$+$};
\draw[<->,dashed] (-0.25,0.2) -- (-0.25,1.8);
\draw[<->,dashed] (0.2,2.25) -- (1.8,2.25);
\node[left] at (-0.25,1) {$y$-periodic};
\node[above] at (1,2.25) {$x$-periodic};
\end{tikzpicture}

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passOpen item page →

A periodic Q1/Q1 checkerboard pressure mode is invisible to divergence

Example

Take the periodic square with a $2 \times 2$ unit-square Q1 mesh. At the grid node $(i, j)$ , for $i, j \in {0, 1, 2}$ , set $q_{i j} = (- 1)^{i + j}$ and identify opposite boundaries periodically. Assume $τ_{K} > 0$ 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 $2 \times 2$ Q1 mesh, equal-order Q1 velocity and pressure, strictly positive cell parameters $τ_{K}$ , 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 $\sum_{K} τ_{K} (\nabla q_{h}, R_{M, K} (u_{h}, p_{h}))_{K}$ ; for the Stokes specialization and $u_{h} = 0$ , $p_{h} = q_{h}$ , $f = 0$ , the residual is $\nabla q_{h}$ .

Verification

1.1givenconstructalgebra

On a cell $K_{i j}$ with local coordinates $(ξ, η) \in [0, 1]^{2}$ , the corner values are $σ_{i j} (1, - 1, - 1, 1)$ with $σ_{i j} = (- 1)^{i + j}$ . Q1 interpolation therefore gives $q_{h} ∣_{K_{i j}} = σ_{i j} (1 - 2 ξ) (1 - 2 η)$ . 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 $v_{1}$ , $\partial_{x} v_{1}$ on a cell is affine in $η$ and independent of $ξ$ . Since $\int_{0}^{1} (1 - 2 ξ) d ξ = 0$ , $\int_{K_{i j}} q_{h} \partial_{x} v_{1} = 0$ . Similarly, $\partial_{y} v_{2}$ is independent of $η$ and $\int_{0}^{1} (1 - 2 η) d η = 0$ , so $\int_{K_{i j}} q_{h} \partial_{y} v_{2} = 0$ . Summing cells gives $(q_{h}, \nabla \cdot v_{h}) = 0$ for every Q1 velocity.

3.1F1step 1.1step 2.1givenalgebra∎

On each unit cell, $\int_{K} ∣ \nabla q_{h} ∣^{2} = 4 / 3 + 4 / 3 = 8 / 3$ . Thus $\sum_{K} τ_{K} ∥ \nabla q_{h} ∥_{0, K}^{2} = (8 / 3) \sum_{K} τ_{K} > 0$ 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.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passOpen item page →

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.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A two-scale VMS closure attains its Schur defect bound

Example

Let each resolved and unresolved space be $R$ with the absolute-value norm (and give its dual the corresponding dual norm). Take the block matrix $(\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array})$ , load $(1, 0)$ , and closure inverse $T = 1 / 4$ . The exact and modeled coarse solutions differ by exactly the defect bound in Finite-dimensional VMS residual closure has an explicit Schur defect bound.

Facts & Assumptions

Given: The two-block VMS decomposition and approximate closure, with scalar block entries $A_{c c} = A_{f f} = 2$ , $A_{c f} = A_{f c} = 1$ , load $F_{c} = 1$ , $F_{f} = 0$ , and $T = 1 / 4$ .

[F1]

Finite-dimensional resolved–unresolved VMS split and residual closure: if $G = A_{f f}^{- 1}$ , the exact coarse Schur system is $(A_{c c} - A_{c f} G A_{f c}) u_{c} = F_{c} - A_{c f} G F_{f}$ and the modeled system replaces $G$ by $T$ .

[F2]

Finite-dimensional VMS residual closure has an explicit Schur defect bound: the coarse error is bounded by $∥ {\tilde{S}}^{- 1} ∥ ∥ A_{c f} ∥ ∥ T - G ∥ ∣ F_{f} - A_{f c} u_{c} ∣$ .

Verification

1.1F1givenalgebra

Here $G = 1 / 2$ , so the exact Schur coefficient is $2 - 1 / 2 = 3 / 2$ and the exact coarse solution is $u_{c} = 2 / 3$ . The exact unresolved component is $u_{f} = G (F_{f} - A_{f c} u_{c}) = - 1 / 3$ .

2.1F1step 1.1algebra

Replacing $G$ by $T = 1 / 4$ gives the approximate Schur coefficient $2 - 1 / 4 = 7 / 4$ , hence ${\tilde{u}}_{c} = 4 / 7$ , and ${\tilde{u}}_{f} = T (F_{f} - A_{f c} {\tilde{u}}_{c}) = - 1 / 7$ . The coarse error is $∣ 4 / 7 - 2 / 3 ∣ = 2 / 21$ .

3.1F1F2step 1.1step 2.1algebra∎

The bound in [F2] equals $(4 / 7) \cdot 1 \cdot (1 / 4) \cdot (2 / 3) = 2 / 21$ , because $∥ {\tilde{S}}^{- 1} ∥ = 4 / 7$ , $∣ A_{c f} ∣ = 1$ , $∣ T - G ∣ = 1 / 4$ , and $∣ F_{f} - A_{f c} u_{c} ∣ = 2 / 3$ . Thus the defect bound is attained exactly in this finite example.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Overlarge streamline stabilization suppresses a two-cell solution

Example

Use $ϵ = b = 1$ , $f = 1$ , homogeneous endpoint values, and one interior P1 degree of freedom on the two-cell mesh $h = 1 / 2$ . Replace the exact parameter $τ$ by the deliberately oversized value $\tilde{τ} = M τ$ , $M > 0$ . The discrete midpoint value tends to zero as $M \to \infty$ , while the exact midpoint value is fixed and positive.

Facts & Assumptions

Given: The one-dimensional SUPG form, the two-cell homogeneous P1 space, $ϵ = b = 1$ , $f = 1$ , and the modified parameter $\tilde{τ} = M τ$ .

[F1]

One-dimensional stationary convection–diffusion model and uniform P1 space: the continuous problem is $- u^{''} + u^{'} = 1$ on $(0, 1)$ with $u (0) = u (1) = 0$ .

[F2]

One-dimensional residual SUPG form with the exact cell parameter: the stabilized diffusion coefficient in the P1 bilinear form is $ϵ + \tilde{τ} b^{2}$ , and the stabilization load is $\sum_{K} \tilde{τ} (f, b ϕ^{'})_{K}$ .

[F3]

The one-dimensional SUPG parameter is positive and bounded: the reference parameter $τ$ is strictly positive.

Verification

1.1F1givenalgebra

The function $u (x) = x - (e^{x} - 1) / (e - 1)$ satisfies the endpoint conditions. Its derivatives are $u^{'} = 1 - e^{x} / (e - 1)$ and $u^{''} = - e^{x} / (e - 1)$ , so $- u^{''} + u^{'} = 1$ . Hence $u (1 / 2) = u_{*} : = 1 / 2 - 1 / (\sqrt{e} + 1) > 0$ .

1.2F2givenalgebra

Let $ϕ$ be the midpoint hat function and $U$ its coefficient. Its derivative is $2$ on the left cell and $- 2$ on the right. Thus $\int_{0}^{1} ∣ ϕ^{'} ∣^{2} d x = 4$ , $\int_{0}^{1} ϕ^{'} ϕ d x = \frac{1}{2} [ϕ^{2}]_{0}^{1} = 0$ , and $\int_{0}^{1} ϕ d x = 1 / 2$ . Also $\int_{0}^{1} ϕ^{'} d x = ϕ (1) - ϕ (0) = 0$ , so the stabilization load vanishes. The single discrete equation is therefore $4 (1 + M τ) U = 1 / 2$ , so $U = 1 / [8 (1 + M τ)]$ .

2.1F3step 1.1step 1.2algebra∎

Since $τ > 0$ by [F3], $U \to 0$ as $M \to \infty$ , whereas the exact value $u_{*} > 0$ from step 1.1 is fixed. In particular, every $M$ satisfying $M τ > 1 / (4 u_{*}) - 1$ gives $U < u_{*} / 2$ . Oversizing the parameter can therefore smear this discrete solution.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

A two-cell SUPG mesh misses the interior shape of an outflow layer

Example

Take $b > 0$ , $ϵ > 0$ , $f = 0$ , and endpoint data $u (0) = 0$ , $u (1) = 1$ . On the two-cell mesh $h = 1 / 2$ , the exact-parameter P1 SUPG solution has the exact nodal value at $x = 1 / 2$ . Nevertheless, at a fixed point inside the outflow cell its error tends to $0.98$ as $ϵ ↓ 0$ . The example separates nodal fitting from resolution of the layer inside an element.

Facts & Assumptions

Given: The model on $(0, 1)$ with $b > 0$ , $ϵ > 0$ , $f = 0$ , endpoint values $0, 1$ , two equal cells, and the exact parameter $τ$ from the one-dimensional SUPG form.

[F1]

One-dimensional stationary convection–diffusion model and uniform P1 space: the equation is $- ϵ u^{''} + b u^{'} = f$ , the two-cell P1 trial function is continuous with the prescribed endpoint values, and $P e = b h / (2 ϵ)$ for $b > 0$ .

[F2]

One-dimensional residual SUPG form with the exact cell parameter: $a_{h} (w_{h}, v_{h}) = ϵ (w_{h}^{'}, v_{h}^{'}) + b (w_{h}^{'}, v_{h}) + τ b^{2} (w_{h}^{'}, v_{h}^{'})$ , with the corresponding residual-weighted load, and $τ = h (2 b)^{- 1} (\coth (P e) - {P e}^{- 1})$ for $b > 0$ .

[F3]

The one-dimensional SUPG parameter is positive and bounded: the reference parameter $τ$ is strictly positive.

Verification

1.1F1givenalgebra

Put $r = b / ϵ$ . The function $u (x) = (e^{r x} - 1) / (e^{r} - 1)$ has endpoint values $0, 1$ , and $- ϵ u^{''} + b u^{'} = (- ϵ r^{2} + b r) e^{r x} / (e^{r} - 1) = 0$ . Hence it is the exact solution.

2.1F1F2F3step 1.1algebra

Let $ϕ$ be the P1 hat function at $x = h = 1 / 2$ , let $χ_{h} \in S_{h} (0, 1)$ have nodal values $(0, 0, 1)$ , and write $u_{h} = U ϕ + χ_{h}$ . Put $D = ϵ + τ b^{2} > 0$ by [F3]. Since $ϕ^{'} = 1 / h$ on the left cell and $- 1 / h$ on the right, $\int_{0}^{1} (ϕ^{'})^{2} d x = 2 / h$ and $\int_{0}^{1} ϕ^{'} ϕ d x = \frac{1}{2} [ϕ^{2}]_{0}^{1} = 0$ , so $a_{h} (ϕ, ϕ) = 2 D / h$ . Also $χ_{h}^{'} = 0$ on the left and $1 / h$ on the right, giving $D \int χ_{h}^{'} ϕ^{'} = - D / h$ and $b \int χ_{h}^{'} ϕ = b / h \int_{h}^{1} ϕ = b / 2$ . Thus $a_{h} (χ_{h}, ϕ) = - D / h + b / 2$ and the one interior equation is $D (2 U - 1) / h + b / 2 = 0$ . By [F2] and the Péclet formula in [F1], $D = ϵ + (b h / 2) (\coth (P e) - 1 / P e) = (b h / 2) \coth (P e)$ , so $U = (1 - \tanh (P e)) / 2 = 1 / (e^{2 P e} + 1)$ . The exact value $u (h) = (e^{b h / ϵ} - 1) / (e^{2 b h / ϵ} - 1)$ is the same number because $2 P e = b h / ϵ$ .

3.1step 1.1step 2.1algebra∎

At $x = 0.99$ , the piecewise-linear value is $u_{h} (0.99) = 0.98 + 0.02 U$ , which tends to $0.98$ as $ϵ ↓ 0$ . The exact value is $(e^{0.99 b / ϵ} - 1) / (e^{b / ϵ} - 1)$ , which tends to zero because $b > 0$ and $0.99 < 1$ . Therefore the pointwise error tends to $0.98$ . The exact midpoint node does not resolve the exponential layer inside the final cell.

SUPG, residual stabilization and variational multiscale methods — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A periodic Q1/Q1 checkerboard pressure mode is invisible to divergence

Example

Facts & Assumptions

Verification

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

Example

Facts & Assumptions

Verification

A two-scale VMS closure attains its Schur defect bound

Example

Facts & Assumptions

Verification

Overlarge streamline stabilization suppresses a two-cell solution

Example

Facts & Assumptions

Verification

A two-cell SUPG mesh misses the interior shape of an outflow layer

Example

Facts & Assumptions

Verification

Sources