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✓ 8 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 7 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

SUPG, residual stabilization and variational multiscale methods

1 · Prerequisites

2 · Summary

This page keeps the homogeneous stationary one-dimensional P1 branch and its exact cell parameter τ=h2∣b∣(coth⁡(Pe)−Pe−1) separate from a multidimensional P1 branch with c−12∇⋅b≥ω>0 and an explicit small-parameter condition. The multidimensional result proves consistency, coercivity in a streamline norm, discrete solvability, and a mesh-dependent C2 error estimate. It does not extend the one-dimensional optimal parameter to general meshes.

PSPG, grad-div, and VMS are treated by separate definitions and arguments. The P1 Stokes PSPG result proves coercivity only in a norm that contains the pressure-gradient stabilization; it does not assert a mesh-uniform pressure estimate. The VMS result is finite-dimensional Schur elimination with a quantified residual-closure defect, not a universal PDE turbulence claim.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

One-dimensional stationary convection–diffusion model and uniform P1 space

Definition

Fix an integer N≥1, put h=1/N, and let xj=jh for j=0,…,N. The cells are Kj=[xj−1,xj]. For g0,g1∈R, let Sh(g0,g1) be the finite-dimensional affine space of continuous functions on [0,1] that are affine on each cell and have endpoint values g0,g1. Write Vh=Sh(0,0) for the homogeneous test and trial vector space; explicitly, Sh(g0,g1)=ℓ+Vh with ℓ(x)=(1−x)g0+xg1. “P1” means polynomial degree at most one on each interval.

Let ϵ>0, b∈R∖{0}, and f∈C([0,1]). The stationary boundary-value problem is to find u∈C2([0,1]) such that −ϵu′′+bu′=f on (0,1), with prescribed endpoint values u(0)=g0 and u(1)=g1. Its dimensionless cell Péclet number is Pe=∣b∣h/(2ϵ)>0. The stability theorem on this page uses the homogeneous branch g0=g1=0; the general endpoint notation also permits the boundary-layer example.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Stationary multidimensional convection–diffusion–reaction P1 model

Definition

Fix d≥2 and a bounded connected polyhedral open set Ω⊂Rd. Let {Th} be a shape-regular family of finite matching affine simplicial meshes in the sense of Matching affine simplicial meshes and shape regularity: one finite constant σ satisfies hK/ρK≤σ for every cell of every mesh. Put h=max⁡K∈ThhK. Let Vh be the space of continuous functions on Ω‾ that are affine on every cell and vanish on ∂Ω.

Let ϵ>0, b∈C1(Ω‾;Rd), c∈C(Ω‾), and f∈C(Ω‾). Define B:=∥b∥L∞(Ω) and C:=∥c∥L∞(Ω). Assume a fixed ω>0 satisfies c−12∇⋅b≥ω on Ω. The boundary condition for this branch is u=0 on the whole boundary, including the inflow part {x∈∂Ω:b(x)⋅n(x)<0}. A classical exact solution is a function u∈C2(Ω‾) satisfying −ϵΔu+b⋅∇u+cu=f in Ω and this boundary condition.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Residual PSPG and grad-div terms for equal-order flow spaces

Definition

Let d≥1 and Ω⊂Rd have a finite cell partition T. Let Vh be a finite-dimensional vector space of continuous vector-valued functions, each cellwise polynomial, and let Qh be a finite-dimensional scalar space of continuous cellwise polynomials with zero mean. Require all velocity components, their cellwise first derivatives and cellwise Laplacians, and all pressures and their cellwise first derivatives to belong to L2(Ω). Boundary conditions are stated with each application; the Stokes theorem below uses zero velocity boundary values, and the periodic example below uses periodic identification. The pair is equal order when velocity components and pressure use the same polynomial degree on each cell. On a cell K, derivatives in residuals are classical derivatives restricted to the cell interior. For well-defined cell integrals, take σ∈L∞(Ω), β∈L∞(Ω;Rd), f∈L2(Ω;Rd), and parameters τK≥0, γ≥0.

For viscosity ν>0, reaction coefficient σ≥0, advective field β, force f, and trial pair (uh,ph)∈Vh×Qh, define the strong momentum residual RM,K(uh,ph):=σuh+β⋅∇uh−νΔKuh+∇ph−f. For test pairs (vh,qh), the full residual PSPG mixed form is Bh((uh,ph),(vh,qh)):=ν(∇uh,∇vh)Ω+(β⋅∇uh+σuh,vh)Ω−(ph,∇⋅vh)Ω+(qh,∇⋅uh)Ω+∑KτK(∇qh,σuh+β⋅∇uh−νΔKuh+∇ph)K, with load (f,vh)Ω+∑KτK(∇qh,f)K. The grad-div addend in velocity tests vh is γ∑K(∇⋅uh,∇⋅vh)K with γ≥0. This definition alone asserts no inf-sup condition, pressure bound, energy estimate, or convergence result.

For piecewise affine velocities, ΔKuh=0 on each cell. Therefore the standard P1 PSPG residual omits the cellwise viscous Laplacian of the exact velocity; this fact does not justify a pressure-accuracy or convergence claim. Such conclusions require separate consistency and stabilization analysis.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite-dimensional resolved–unresolved VMS split and residual closure

Definition

Let Xc and Xf be finite-dimensional real vector spaces and set X=Xc⊕Xf; elements are written (uc,uf), with uc the resolved component and uf the unresolved component. For a real vector space Y, let Y∗ be its space of real linear functionals. Let Acc:Xc→Xc∗, Acf:Xf→Xc∗, Afc:Xc→Xf∗, and Aff:Xf→Xf∗ be linear maps, with Aff invertible, and let Fc∈Xc∗ and Ff∈Xf∗. The linear equation on X has block form (AccAcfAfcAff)(ucuf)=(FcFf). The fine residual for a resolved vector vc is rf(vc):=Ff−Afcvc. Exact elimination gives uf=Aff−1rf(uc) and the Schur equation (Acc−AcfAff−1Afc)uc=Fc−AcfAff−1Ff.

A residual closure selects a linear map T:Xf∗→Xf to approximate the exact inverse Aff−1:Xf∗→Xf. It defines A~cc:=Acc−AcfTAfc, F~c:=Fc−AcfTFf, and the modeled fine component u~f:=Trf(u~c), where u~c solves A~ccu~c=F~c. This is a finite-dimensional algebraic closure; an application to a PDE requires its own residual, spaces, invertibility, and approximation-defect analysis.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

One-dimensional residual SUPG form with the exact cell parameter

Definition

Use the uniform mesh, coefficients, and spaces of One-dimensional stationary convection–diffusion model and uniform P1 space. On every cell set τ=h2∣b∣(coth⁡(Pe)−1Pe),coth⁡s:=cosh⁡ssinh⁡s. For w∈Sh(g0,g1)+C2([0,1]) and vh∈Vh, define ah(w,vh):=ϵ∫01w′vh′ dx+b∫01w′vh dx+∑K∈Thτ∫K(−ϵw′′+bw′)(bvh′) dx, Fh(vh):=∫01fvh dx+∑K∈Thτ∫Kf(bvh′) dx. Second derivatives are taken classically on each cell. The strong equation residual is RK(w):=−ϵw′′+bw′−f. For P1 trial functions, wh′′=0 on each cell, so the cell term is τ∫K(bwh′)(bvh′). On the smooth consistency extension it retains −ϵw′′; thus the full residual pairing is τ∫K(RK(w)+f)(bvh′), while the load contains the matching f term. The SUPG equation is ah(wh,vh)=Fh(vh) and adds the same residual-weighted streamline test perturbation to the differential equation and its load.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Multidimensional P1 SUPG residual form and parameter scaling

Definition

Use the model and notation of Stationary multidimensional convection–diffusion–reaction P1 model. Define sb(K)=hK/∥b∥L∞(K) when that norm is positive and sb(K)=+∞ otherwise; define sc(K)=ω/∥c∥L∞(K)2 when that norm is positive and sc(K)=+∞ otherwise. For each cell set δK:=θmin⁡{hK2ϵ,sb(K),sc(K)},0<θ≤14. The minimum is finite and positive because its first entry is finite and positive. This scaling gives δK∥c∥L∞(K)2≤ω/4.

For a trial function w that is the sum of a cellwise smooth function and a member of Vh, and vh∈Vh, define the element residual RK(w):=−ϵΔKw+b⋅∇w+cw, and ah(w,vh):=ϵ(∇w,∇vh)Ω+(b⋅∇w+cw,vh)Ω+∑K∈ThδK(RK(w),b⋅∇vh)K, Fh(vh):=(f,vh)Ω+∑K∈ThδK(f,b⋅∇vh)K. Here (g1,g2)D:=∫Dg1g2 dx and ΔK is the classical Laplacian restricted to the interior of K. For wh∈Vh, ΔKwh=0 because wh is affine on K. No stability claim is part of this definition.

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One-dimensional P1 interpolation has an L2 error bound

Statement

Let 0=x0<⋯<xN=1 be a finite interval partition with maximum cell length h, and let u∈C2([0,1]). Let Ihu be the continuous piecewise-linear interpolant with the same nodal values as u. Then ∥u−Ihu∥L2(0,1)≤h2∥u′′∥L∞(0,1).

Facts & Assumptions

Given: A finite interval partition of [0,1], its maximum cell length h, and u∈C2([0,1]).

[F1]

By the definition of Ihu in the Statement, on each interval cell it is affine and agrees with u at both endpoints.

Proof

1.1F1givenconstruct

Fix a cell K=[a,b] of length hK, and let sK=(u(b)−u(a))/hK be the slope of Ihu on K. By the fundamental theorem of calculus, sK=hK−1∫abu′(t) dt. For every x∈K, the same theorem gives ∣u′(x)−u′(t)∣≤hK∥u′′∥∞, so averaging in t yields ∣(u−Ihu)′(x)∣=∣u′(x)−sK∣≤hK∥u′′∥∞.

2.1step 1.1givenalgebra

Since (u−Ihu)(a)=0, integrating the derivative bound from step 1.1 over [a,x] gives ∣u(x)−Ihu(x)∣≤hK2∥u′′∥∞ for every x∈K. Therefore ∥u−Ihu∥L2(K)2≤hK5∥u′′∥∞2.

3.1step 2.1givenalgebra∎

The cell interiors are disjoint, their lengths sum to 1, and hK≤h, so ∑KhK5≤h4∑KhK=h4. Summing step 2.1 over the cells and taking nonnegative square roots proves the estimate.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

Shape-regular affine P1 interpolation estimates for C2 functions

Statement

Use the fixed dimension d≥2 and the finite shape-regular affine simplicial mesh family from Stationary multidimensional convection–diffusion–reaction P1 model and Matching affine simplicial meshes and shape regularity. Let u∈C2(Ω‾), and on each simplex K let IKu=∑i=0du(ai)λi be its nodal affine interpolant, where ai are the vertices and λi the barycentric coordinates. Set Ihu∣K=IKu and M2:=∣Ω∣1/2∥D2u∥L∞(Ω), using the Euclidean matrix norm for D2u. There is a constant CI depending only on d and the shape-regularity bound such that ∥u−Ihu∥L2(Ω)≤CIh2M2,∥∇(u−Ihu)∥L2(Ω)≤CIhM2. If u=0 on ∂Ω, then Ihu=0 on every boundary facet.

Facts & Assumptions

Given: The finite matching affine simplicial mesh family, its shape-regularity bound σ, and u∈C2(Ω‾).

[F1]

Matching affine simplicial meshes and shape regularity: each cell is a nondegenerate affine simplex, its barycentric coordinates are affine, and its inradius ρK satisfies hK/ρK≤σ.

[F2]

Stationary multidimensional convection–diffusion–reaction P1 model: h=max⁡KhK, the mesh is finite, and ∑K∣K∣=∣Ω∣.

Proof

1.1F1givenconstruct

Fix a cell K and a vertex a0. Taylor's formula along the segment from a0 to x∈K gives u(x)=pK(x)+rK(x) with pK(x)=u(a0)+∇u(a0)⋅(x−a0), ∣rK(x)∣≤12hK2∥D2u∥L∞(K), and ∣∇rK(x)∣≤hK∥D2u∥L∞(K).

2.1F1step 1.1algebra

The nodal interpolant reproduces pK, so u−IKu=rK−∑irK(ai)λi. On K, 0≤λi≤1 and ∑iλi=1. The altitude from any vertex to its opposite facet is at least ρK, hence ∣∇λi∣≤ρK−1≤σhK−1 by [F1]. Substitution of the bounds in step 1.1 gives ∣u−IKu∣≤C(d)hK2∥D2u∥L∞(K) and ∣∇(u−IKu)∣≤C(d,σ)hK∥D2u∥L∞(K).

3.1F2step 2.1algebra

Squaring the two pointwise bounds, integrating on each cell, summing over the finite mesh, using hK≤h and ∑K∣K∣=∣Ω∣, and taking square roots proves both global estimates with M2=∣Ω∣1/2∥D2u∥∞.

4.1F1givenstep 2.1∎

On a shared interior facet, the affine traces from its two incident cells agree at all facet vertices, so they agree everywhere on that facet; thus the cell interpolants assemble to a continuous Ihu. If u vanishes on a boundary facet, all its vertices there have zero nodal values, so the affine trace also vanishes there. This proves both global conformity and the boundary claim.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

P1 PSPG Stokes form is coercive in its stabilization norm

Statement

Let d≥2, let Ω⊂Rd be a bounded connected polyhedral open set, and let a finite conforming nondegenerate simplicial mesh have connected cell adjacency. Let Vh be continuous vector-valued P1 functions vanishing on the boundary, and let Qh be continuous scalar P1 functions with zero mean. Let f∈L2(Ω;Rd), ν>0, τK>0, and γ≥0. For the stationary Stokes specialization of Residual PSPG and grad-div terms for equal-order flow spaces, define Bh((uh,ph),(vh,qh)):=ν(∇uh,∇vh)Ω−(ph,∇⋅vh)Ω+(qh,∇⋅uh)Ω+γ(∇⋅uh,∇⋅vh)Ω+∑KτK(∇qh,−νΔKuh+∇ph)K. For the residual load, set Fh(vh,qh):=(f,vh)Ω+∑KτK(∇qh,f)K. Define ∥(vh,qh)∥h2:=ν∥∇vh∥22+γ∥∇⋅vh∥22+∑KτK∥∇qh∥0,K2. Then this is a norm, Bh(zh,zh)=∥zh∥h2, and the discrete problem Bh(zh,⋅)=Fh(⋅) has a unique solution satisfying ∥zh∥h≤∥Fh∥h,∗, where ∥Fh∥h,∗:=sup⁡yh≠0∣Fh(yh)∣/∥yh∥h. The estimate controls pressure in the mesh-dependent gradient stabilization norm; it does not assert a mesh-uniform L2 pressure bound or convergence.

Facts & Assumptions

Given: A connected domain and connected conforming cell adjacency, continuous equal-order P1 velocity and pressure spaces, zero velocity boundary values, mean-zero pressure, ν>0, τK>0, γ≥0, and the Stokes residual form.

[F1]

Residual PSPG and grad-div terms for equal-order flow spaces: PSPG adds the element momentum residual tested against ∇qh, while grad-div adds γ(∇⋅uh,∇⋅vh); both are definitions and supply no stability claim by themselves.

[A1]

On each cell, uh is affine, so ΔKuh=0. The cell mesh is connected, the pressure is continuous, and its mean is zero.

Proof

1.1F1A1givenalgebra

Setting (vh,qh)=(uh,ph) in the displayed form cancels the mixed terms −(ph,∇⋅uh)+(ph,∇⋅uh). By [A1], the PSPG residual term becomes ∑KτK∥∇ph∥0,K2. The remaining terms are exactly ν∥∇uh∥22+γ∥∇⋅uh∥22, so Bh((uh,ph),(uh,ph))=∥(uh,ph)∥h2.

2.1A1step 1.1algebra

If this norm is zero, ∇uh=0 on each cell; continuity and zero boundary values force uh=0. Also ∇ph=0 on each cell, so ph is constant on each cell. Continuity across shared facets and connected cell adjacency force one global constant; its zero mean forces ph=0. Thus the norm vanishes only at zero.

3.1step 1.1step 2.1givenalgebra∎

If zh is in the kernel of the bilinear map, then step 1.1 gives ∥zh∥h2=Bh(zh,zh)=0, and step 2.1 forces zh=0. Thus the linear map from Vh×Qh to its dual is injective and, by equal finite dimensions, bijective; the discrete problem has a unique solution. Testing that solution equation with itself gives ∥zh∥h2=Fh(zh)≤∥Fh∥h,∗∥zh∥h. If zh=0 the estimate is immediate; otherwise division proves ∥zh∥h≤∥Fh∥h,∗.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Finite-dimensional VMS residual closure has an explicit Schur defect bound

Statement

Use the block operators and residual closure of Finite-dimensional resolved–unresolved VMS split and residual closure. Equip Xc,Xf and their duals with norms, and use the corresponding induced operator norms. Let G:=Aff−1, assume the exact Schur system has a solution uc, and put rf:=Ff−Afcuc. Suppose the closure map T has S~:=Acc−AcfTAfc invertible, and let u~c solve the approximate Schur system. Then ∥u~c−uc∥≤∥S~−1∥ ∥Acf∥ ∥T−G∥ ∥rf∥, and for the modeled fine component u~f=T(Ff−Afcu~c), ∥u~f−uf∥≤∥T−G∥ ∥rf∥+∥T∥ ∥Afc∥ ∥u~c−uc∥. The exact fine scale is uf=Grf. These bounds concern finite-dimensional algebraic blocks; they do not certify a universal PDE subgrid model.

Facts & Assumptions

Given: The direct-sum block system and residual closure in Finite-dimensional resolved–unresolved VMS split and residual closure, an invertible exact fine block Aff, and an invertible approximate Schur operator S~.

[F1]

Finite-dimensional resolved–unresolved VMS split and residual closure: exact elimination uses G=Aff−1 and residual rf=Ff−Afcuc; the closure replaces G by T in both the Schur operator and load.

Proof

1.1F1givenalgebra

Write the exact Schur operator and load as S=Acc−AcfGAfc and g=Fc−AcfGFf, so Suc=g. Put Δ:=T−G. The approximate quantities satisfy S~=S−AcfΔAfc and g~=g−AcfΔFf.

2.1step 1.1givenalgebra

Since S~u~c=g~, step 1.1 gives S~(u~c−uc)=g~−S~uc=−AcfΔ(Ff−Afcuc)=−AcfΔrf. Apply the induced operator norms and multiply by ∥S~−1∥ to prove the coarse bound.

3.1F1step 2.1algebra∎

The exact fine component is uf=Grf. The modeled component is u~f=T(rf−Afc(u~c−uc)). Subtracting gives u~f−uf=(T−G)rf−TAfc(u~c−uc). The triangle inequality and operator norms give the stated fine-scale bound.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

The one-dimensional SUPG parameter is positive and bounded

Statement

Let s>0 and define m(s):=coth⁡s−1/s and c∗:=min⁡{(3sinh⁡1)−1,coth⁡1−1}>0. Then c∗min⁡{s,1}≤m(s)<1,m(s)≤s/2. Consequently, for the parameter in One-dimensional residual SUPG form with the exact cell parameter, 0<c∗2min⁡{h22ϵ,h∣b∣}≤τ<h2∣b∣,τ≤h28ϵ.

Facts & Assumptions

Given: s>0, ϵ>0, b≠0, h>0, and Pe=∣b∣h/(2ϵ).

[F1]

One-dimensional stationary convection–diffusion model and uniform P1 space: Pe>0 is the dimensionless parameter ∣b∣h/(2ϵ).

[F2]

One-dimensional residual SUPG form with the exact cell parameter: τ=h(2∣b∣)−1m(Pe).

Proof

1.1givenalgebra

Put q(s)=scosh⁡s−sinh⁡s. Then q(0)=0 and q′(s)=ssinh⁡s>0 for s>0, so q(s)>0 and m(s)=q(s)/(ssinh⁡s)>0.

2.1step 1.1algebra

The identity coth⁡s−1=2/(e2s−1) and e2s−1=∫02set dt>2s give m(s)<1.

2.2step 1.1algebra

For 0<s≤1, q(s)=∫0stsinh⁡t dt≥s3/3. The function sinh⁡s/s is increasing because its derivative is q(s)/s2>0, hence sinh⁡s≤ssinh⁡1 on this range. Therefore m(s)≥s/(3sinh⁡1). For every s>0, monotonicity of sinh⁡ also gives q(s)≤sinh⁡(s)∫0st dt=s2sinh⁡(s)/2, whence m(s)≤s/2.

3.1F1F2step 2.2algebra∎

Since sinh⁡s>s for s>0, m′(s)=s−2−(sinh⁡s)−2>0. Thus for s≥1, m(s)≥m(1)=coth⁡1−1. Together with step 2.2 this proves the two-sided bound. Substituting s=Pe into [F2] gives the stated lower and first upper bounds on τ, using h(2∣b∣)−1min⁡{Pe,1}=12min⁡{h2/(2ϵ),h/∣b∣}. The last upper bound follows from m(Pe)≤Pe/2.

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Multidimensional P1 SUPG consistency, coercivity, and error bound

Statement

Use the model of Stationary multidimensional convection–diffusion–reaction P1 model and form/parameter of Multidimensional P1 SUPG residual form and parameter scaling. Assume Countable Choice, and let u∈C2(Ω‾) solve the stated strong problem. For every continuous, zero-boundary function w whose restriction to each cell has square-integrable values and square-integrable weak first derivatives, define ∥w∥SD,h2:=ϵ∥∇w∥L2(Ω)2+ω∥w∥L2(Ω)2+∑KδK∥b⋅∇w∥L2(K)2. M2:=∣Ω∣1/2∥D2u∥L∞(Ω),B:=∥b∥L∞(Ω),C:=∥c∥L∞(Ω),δ∗:=max⁡KδK. Then the discrete form is coercive: ah(vh,vh)≥34∥vh∥SD,h2(vh∈Vh), the discrete problem has a unique solution uh, and the method is consistent, ah(u,vh)=Fh(vh). Moreover, with C∗ depending only on dimension, shape regularity, and the interpolation constant, ∥u−uh∥SD,h≤C∗M2[(ϵ+Bω)h+(ω+Cω)h2+Bδ∗ h+δ∗(d ϵ+Bh+Ch2)]. This is a P1 estimate for the stated strong regularity and parameter branch; it does not claim stability for unrestricted multidimensional parameters or weaker regularity.

Facts & Assumptions

Given: Countable Choice; a connected polyhedral domain with a finite shape-regular affine simplicial mesh; ϵ>0, b∈C1, c∈C, c−12∇⋅b≥ω>0, the parameter bound in Multidimensional P1 SUPG residual form and parameter scaling, and an exact C2 solution with zero boundary values.

[F1]

Stationary multidimensional convection–diffusion–reaction P1 model: the mesh is matching and shape regular, Vh is continuous P1 with zero boundary values, and the coefficient and strong-solution hypotheses above hold.

[F2]

Multidimensional P1 SUPG residual form and parameter scaling: ah is the Galerkin form plus the cell residual pairing with b⋅∇vh, and δK∥c∥∞,K2≤ω/4.

[F3]

Shape-regular affine P1 interpolation estimates for C2 functions: for η=u−Ihu, ∥η∥2≤CIh2M2 and ∥∇η∥2≤CIhM2.

[F4]

Gauss–Green formula on an affine simplex under The Axiom of Countable Choice (ACω): on each affine simplex, a C1 vector field has volume divergence integral equal to its outward facet flux sum.

[A1]

Countable Choice is used only through the imported simplex Gauss–Green identity [F4] (The Axiom of Countable Choice (ACω)).

Proof

1.1A1F1F4given

For vh∈Vh, apply [F4] on each cell to G=bvh2/2. Interior fluxes cancel because b and vh are continuous and the outward normals oppose; boundary fluxes vanish because vh=0. Hence (b⋅∇vh,vh)Ω=−12(∇⋅b,vh2)Ω.

1.2A1F1F2F4given

For the exact u, apply [F4] cellwise to G=ϵvh∇u. The interior fluxes cancel and the boundary flux vanishes, so ϵ(∇u,∇vh)+(b⋅∇u+cu,vh)=(f,vh). Since RK(u)=f, adding ∑KδK(f,b⋅∇vh)K gives ah(u,vh)=Fh(vh).

1.3F1F2F3algebra

Let η=u−Ihu. Enlarge the interpolation constant if necessary so CI≥1; this preserves [F3]. By [F3], Ihu∈Vh, ∥∇η∥2≤CIhM2, and ∥η∥2≤CIh2M2. Cellwise, ΔKIhu=0, so RK(η)=−ϵΔu+b⋅∇η+cη. The Euclidean matrix norm and Cauchy–Schwarz on the diagonal give ∣Δu(x)∣=∣tr⁡D2u(x)∣≤d ∣D2u(x)∣; integrating yields ∥Δu∥2≤dM2. The triangle inequality in the finite product of cellwise L2 spaces then gives (∑KδK∥RK(η)∥0,K2)1/2≤δ∗(d ϵM2+BCIhM2+CCIh2M2). Also ∑KδK∥b⋅∇η∥0,K2≤δ∗B2∥∇η∥22, so the product norm satisfies ∥η∥SD,h≤CIM2(ϵh+ωh2+Bδ∗h).

2.1F1F2step 1.1algebra

Since vh is affine on each cell, ΔKvh=0. Substituting step 1.1 into [F2] gives ah(vh,vh)=ϵ∥∇vh∥22+(c−12∇⋅b,vh2)Ω+∑KδK∥b⋅∇vh∥0,K2+∑KδK(cvh,b⋅∇vh)K. On each cell, Cauchy–Schwarz and [F2] give ∣δK(cvh,b⋅∇vh)K∣≤14(ω∥vh∥0,K2+δK∥b⋅∇vh∥0,K2): write the left bound as (δK∥c∥∞,K/ω)(ω∥vh∥)(δK∥b⋅∇vh∥), use the first factor ≤1/2, then ab/2≤(a2+b2)/4. Since c−12∇⋅b≥ω, this proves ah(vh,vh)≥34∥vh∥SD,h2.

3.1F1step 2.1givenalgebra

The linear map Vh→Vh∗ induced by ah has zero kernel: if ah(wh,vh)=0 for all vh, test with vh=wh and use step 2.1 to get wh=0. The spaces have equal finite dimension, so this map is bijective and the discrete problem has a unique solution uh.

4.1F2step 3.1step 1.2step 1.3algebra

Subtract the discrete equation from step 1.2 to obtain ah(u−uh,vh)=0 for every vh∈Vh. Put eh:=uh−Ihu∈Vh. Taking vh=eh and using u−uh=η−eh gives ah(eh,eh)=ah(η,eh).

5.1F2step 2.1step 4.1algebra

Cauchy–Schwarz bounds the diffusion, convection, reaction, and stabilization terms in ah(η,eh) by ϵ∥∇η∥2∥eh∥SD,h, B∥∇η∥2∥eh∥SD,h/ω, C∥η∥2∥eh∥SD,h/ω, and (∑KδK∥RK(η)∥0,K2)1/2∥eh∥SD,h, respectively. If eh=0 the bound below is immediate; otherwise divide step 4.1 by its norm and use coercivity in step 2.1 to get ∥eh∥SD,h≤43(ϵ∥∇η∥2+Bω∥∇η∥2+Cω∥η∥2+(∑KδK∥RK(η)∥0,K2)1/2).

6.1F1F2F3step 1.3step 5.1algebra∎

The norm in the statement is the product L2 norm of (ϵ∇w,ωw,(δKb⋅∇w)K), so ∥u−uh∥SD,h≤∥η∥SD,h+∥eh∥SD,h. Using step 1.3 in step 5.1 bounds the sum by CIM2(ϵh+ωh2+Bδ∗h)+43CIM2((ϵ+B/ω)h+(C/ω)h2+δ∗(d ϵ+Bh+Ch2)). Since CI+43CI=73CI, this is at most the displayed right-hand side with C∗:=73CI, which has the claimed dependence.

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One-dimensional P1 SUPG is consistent and coercive

Statement

Assume the one-dimensional model of One-dimensional stationary convection–diffusion model and uniform P1 space with homogeneous data g0=g1=0, and the form of One-dimensional residual SUPG form with the exact cell parameter. Let u∈C2([0,1]) solve −ϵu′′+bu′=f with u(0)=u(1)=0. Then ah(u,vh)=Fh(vh) for every vh∈Vh. For every vh∈Vh, ah(vh,vh)=ϵ∥vh′∥L2(0,1)2+τb2∑K∈Th∥vh′∥L2(K)2. The right-hand side is a squared norm on Vh. Consequently the finite-dimensional SUPG system has a unique solution.

Facts & Assumptions

Given: A uniform interval partition, ϵ>0, b≠0, f∈C([0,1]), the homogeneous problem above, and the SUPG form and parameter.

[F1]

One-dimensional stationary convection–diffusion model and uniform P1 space: Vh consists of continuous piecewise-affine functions vanishing at both endpoints; the exact solution satisfies −ϵu′′+bu′=f.

[F2]

One-dimensional residual SUPG form with the exact cell parameter: ah(w,vh) pairs −ϵw′′+bw′ against τbvh′ cellwise and Fh pairs f against the same test perturbation; for P1 wh, wh′′=0.

[F3]

The one-dimensional SUPG parameter is positive and bounded: τ>0 for ϵ>0, b≠0, and h>0.

Proof

1.1F1givenalgebra

For vh∈Vh, continuity and piecewise absolute continuity give ∫01bvh′vh dx=b2∑K[vh2]∂K=b2(vh(1)2−vh(0)2)=0. The endpoint values vanish by [F1].

1.2F1F2givenalgebra

Integration by parts, with vh(0)=vh(1)=0, gives ϵ(u′,vh′)+(bu′,vh)=(−ϵu′′+bu′,vh)=(f,vh). The stabilization contribution to ah(u,vh) is ∑Kτ(−ϵu′′+bu′,bvh′)K, which equals ∑Kτ(f,bvh′)K by the strong equation. This is exactly the stabilization contribution to Fh, so ah(u,vh)=Fh(vh).

2.1F2F3step 1.1algebra

Substituting step 1.1 into [F2] with wh=vh yields the displayed identity. Its right side is nonnegative by [F3] and ϵ>0. If it is zero, then vh′=0 on every cell; continuity makes vh constant on [0,1], and its zero endpoints force vh=0. Hence this quadratic form is a norm squared.

3.1step 2.1givenalgebra

In a basis of the finite-dimensional space Vh, the discrete form defines a square linear map Vh→Vh∗. If a vector is in its kernel, testing with that vector and using step 2.1 gives that it is zero. Thus the map is injective, hence bijective in finite equal dimensions, so each discrete load has exactly one solution.

4.1step 1.2step 3.1algebra∎

Subtracting the discrete equation from the consistency identity in step 1.2 gives ah(u−uh,vh)=0 for every vh∈Vh; the solution exists by step 3.1.

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One-dimensional SUPG energy error bound with explicit tau dependence

Statement

Assume the homogeneous one-dimensional problem and uniform P1 SUPG form of One-dimensional stationary convection–diffusion model and uniform P1 space and One-dimensional residual SUPG form with the exact cell parameter. Assume Countable Choice, and let u∈C2([0,1]) be the exact solution and uh∈Vh the unique discrete solution from One-dimensional P1 SUPG is consistent and coercive. For every continuous, piecewise-H1 function w with w(0)=w(1)=0, define ∥w∥SD2:=ϵ∥w′∥L2(0,1)2+τb2∥w′∥L2(0,1)2,M:=∥u′′∥L∞(0,1). This is the product L2 norm of (ϵw′,τ∣b∣w′), so it has the triangle inequality. Then ∥u−uh∥SD≤[hϵ+τb2+hϵ+h∣b∣τ+ϵτ+h2τ]M. With c∗ from The one-dimensional SUPG parameter is positive and bounded, there is a universal C∗>0 such that also ∥u−uh∥SD≤C∗(hϵ+h3/2∣b∣)M.

Facts & Assumptions

Given: Countable Choice; ϵ>0, b≠0, a uniform mesh of size h, a homogeneous exact solution u∈C2([0,1]), and its discrete SUPG solution uh.

[F1]

One-dimensional residual SUPG form with the exact cell parameter: for P1 functions ah(vh,vh)=ϵ∥vh′∥22+τb2∥vh′∥22, while for piecewise smooth w its stabilization uses −ϵw′′+bw′; τ=h(2∣b∣)−1(coth⁡(Pe)−1/Pe).

[F2]

One-dimensional P1 SUPG is consistent and coercive: ah(u−uh,vh)=0 for every vh∈Vh, and ah(vh,vh)=∥vh∥SD2.

[F3]

One-dimensional P1 interpolation has a first-derivative error bound under Countable Choice: for the nodal P1 interpolant Ihu, ∥(u−Ihu)′∥2≤h∥u′′∥∞.

[F4]

One-dimensional P1 interpolation has an L2 error bound: ∥u−Ihu∥2≤h2∥u′′∥∞.

[F5]

The one-dimensional SUPG parameter is positive and bounded: τ<h/(2∣b∣), τ≥(c∗/2)min⁡{h2/(2ϵ),h/∣b∣}, and τ≤h2/(8ϵ).

[A1]

Countable Choice is assumed only for the imported interpolation result [F3] (The Axiom of Countable Choice (ACω)); no arbitrary choice is used elsewhere in this finite-mesh proof.

Proof

1.1A1F3F4givenconstruct

Let Ihu be the nodal interpolant, η:=u−Ihu, and eh:=uh−Ihu. The homogeneous endpoint data imply Ihu∈Vh, so eh∈Vh. By [F3] and [F4], ∥η′∥2≤hM and ∥η∥2≤h2M.

2.1F1F2step 1.1algebra

By [F2], ah(eh,eh)=ah(η,eh). The convection term satisfies (bη′,eh)=−(bη,eh′): cell boundary terms cancel at interior nodes because η and eh are continuous, and the two outer terms vanish. The diffusion and P1 stabilization terms are bounded by ϵ∥η′∥2∥eh∥SD and ∣b∣τ∥η′∥2∥eh∥SD. The additional consistency-extension term involving −ϵη′′ is bounded by ϵτ∥η′′∥2∥eh∥SD, because τ∣b∣∥eh′∥2≤∥eh∥SD. The integrated convection term is bounded by ∥η∥2∥eh∥SD/τ, since τ∣b∣∥eh′∥2≤∥eh∥SD. Therefore ∣ah(η,eh)∣≤(ϵ∥η′∥2+∣b∣τ∥η′∥2+ϵτ∥η′′∥2+∥η∥2τ)∥eh∥SD.

3.1F2step 2.1algebra

If eh=0, its norm estimate is immediate. Otherwise, divide step 2.1 by ∥eh∥SD and use [F2] to get ∥eh∥SD≤(ϵ+∣b∣τ)∥η′∥2+ϵτ∥η′′∥2+∥η∥2/τ.

4.1F1F3F4step 1.1step 3.1algebra

Since Ihu is affine on each cell, η′′=u′′ cellwise and ∥η′′∥2≤M. With u−uh=η−eh, the triangle inequality, [F1], [F3], [F4], and step 3.1 give ∥u−uh∥SD≤[hϵ+τb2+hϵ+h∣b∣τ+ϵτ+h2τ]M.

5.1F1F5step 4.1algebra∎

From [F5], h∣b∣τ≤h3/2∣b∣/2. If Pe≤1, then τ≥c∗h2/(4ϵ), so h2/τ≤2c∗−1/2hϵ, and ϵτ≤hϵ/8. If Pe≥1, then τ≥c∗h/(2∣b∣), so h2/τ≤(2/c∗)1/2h3/2∣b∣; also ϵ≤∣b∣h/2 and τ<h/(2∣b∣) imply ϵτ≤h3/2∣b∣/(22). Finally hϵ+τb2≤hϵ+h∣b∣τ. These inequalities prove the second estimate with a universal constant depending only on c∗.

5 · Examples, counterexamples and false statements

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