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✓ 7 results · all verified · 6 also independently AI-judged

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High-resolution TVD, ENO and WENO reconstruction

1 · Prerequisites

2 · Summary

Objective

This page separates nodal ENO2 from finite-volume cell-average reconstruction, defines minmod MUSCL, WENO5-JS and the WENO-Z comparison, and proves their stated smooth-region face accuracy. The scalar linear monotonicity barrier is proved by a finite-stencil moment argument. The unchanged direct reconstructed-upwind forward-Euler MUSCL/minmod update is TVD under the sharp CFL bound $∣ a ∣ Δ t / h \leq 2 / 3$ , proved in Direct forward-Euler MUSCL/minmod is TVD under the sharp two-thirds CFL bound. The exact CFL-one counterexample remains in The specified direct MUSCL/minmod flux update is not TVD at CFL one; the theorem also shows failure for every larger CFL number through one.

For the cell-average convention, each rectangle below is one half-open cell and its label is the cell average. The dashed vertical line is the target face. The three lower bars show the five-cell WENO5 candidate stencils; the nodal ENO2 rule uses separate point samples and is defined by its own divided-difference formula in Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions. The diagram records stencil geometry only; the proofs use the displayed formulas and moment calculations.

\begin{tikzpicture}[x=1cm,y=0.75cm]
  \draw (0,0) rectangle (1,0.6);
  \draw (1,0) rectangle (2,0.6);
  \draw (2,0) rectangle (3,0.6);
  \draw (3,0) rectangle (4,0.6);
  \draw (4,0) rectangle (5,0.6);
  \node at (0.5,0.3) {$U_{i-2}$};
  \node at (1.5,0.3) {$U_{i-1}$};
  \node at (2.5,0.3) {$U_i$};
  \node at (3.5,0.3) {$U_{i+1}$};
  \node at (4.5,0.3) {$U_{i+2}$};
  \draw[red,dashed] (3,-0.2) -- (3,1.4);
  \node[red,right] at (3,1.25) {$x_{i+1/2}$};
  \draw[thick] (0,-0.35) -- (3,-0.35);
  \node at (1.5,-0.62) {$S_0$};
  \draw[thick] (1,-0.9) -- (4,-0.9);
  \node at (2.5,-1.17) {$S_1$};
  \draw[thick] (2,-1.45) -- (5,-1.45);
  \node at (3.5,-1.72) {$S_2$};
\end{tikzpicture}

3 · Logical flowchart

4 · Definitions, theorems and proofs

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

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions

Definition

Let $h > 0$ , let $I_{i} = [i h, (i + 1) h)$ , let $x_{i} = (i + \frac{1}{2}) h$ be its center, and write $x_{i + 1 / 2} = (i + 1) h$ and $x_{i - 1 / 2} = i h$ for its right and left faces. A cell average of a locally integrable scalar function $u$ is $U_{i} = h^{- 1} \int_{I_{i}} u (x) d x$ . The cells are half-open only to form a partition; the choice of endpoint representatives does not change an integral.

Here MUSCL/minmod means the piecewise-linear reconstruction with the limiter defined next; nodal ENO2 means the two-stencil quadratic selector defined below; and cell-average WENO5-JS/WENO-Z means the three-candidate reconstructions with the normalized nonlinear weights displayed below. These names specify the constructions and do not, by themselves, assert an accuracy or stability property.

For real $a, b$ , define $minmod (a, b) = {\begin{cases} sgn (a) \min (∣ a ∣, ∣ b ∣), & a b > 0, \\ 0, & a b \leq 0. \end{cases}$ The MUSCL/minmod reconstruction assigns the cell slope $s_{i} = h^{- 1} minmod (U_{i} - U_{i - 1}, U_{i + 1} - U_{i})$ and the two face states $U_{i - 1 / 2}^{+, M} = U_{i} - h s_{i} / 2$ and $U_{i + 1 / 2}^{-, M} = U_{i} + h s_{i} / 2$ . Here $+$ denotes a trace from the right cell and $-$ a trace from the left cell.

For separate nodal data $V_{j} = u (y_{j})$ at $y_{j} = j h$ , define the second divided difference by $[y_{a}, y_{b}, y_{c}] V = (((V_{c} - V_{b}) / (y_{c} - y_{b})) - ((V_{b} - V_{a}) / (y_{b} - y_{a}))) / (y_{c} - y_{a})$ . At the face $y_{i + 1 / 2} = (i + \frac{1}{2}) h$ , set $d_{L} = ∣ [y_{i - 1}, y_{i}, y_{i + 1}] V ∣$ and $d_{R} = ∣ [y_{i}, y_{i + 1}, y_{i + 2}] V ∣$ . Choose the left triple when $d_{L} \leq d_{R}$ and otherwise choose the right triple. For the selected triple $(y_{a}, y_{b}, y_{c})$ define its Newton quadratic by $P_{a, b, c} (y) = V_{a} + \frac{V_{b} - V_{a}}{y_{b} - y_{a}} (y - y_{a}) + [y_{a}, y_{b}, y_{c}] V (y - y_{a}) (y - y_{b}) .$ The nodal ENO2 face value is $P_{a, b, c} (y_{i + 1 / 2})$ . This nodal interpolation branch is distinct from the cell-average WENO branch below.

For cell averages $U_{i - 2}, \dots, U_{i + 2}$ , define the left-biased WENO5 candidate face functionals $p_{0} = (2 U_{i - 2} - 7 U_{i - 1} + 11 U_{i}) / 6, p_{1} = (- U_{i - 1} + 5 U_{i} + 2 U_{i + 1}) / 6, p_{2} = (2 U_{i} + 5 U_{i + 1} - U_{i + 2}) / 6.$ Define the Jiang–Shu indicators $β_{0} = \frac{13}{12} (U_{i - 2} - 2 U_{i - 1} + U_{i})^{2} + \frac{1}{4} (U_{i - 2} - 4 U_{i - 1} + 3 U_{i})^{2},$ $β_{1} = \frac{13}{12} (U_{i - 1} - 2 U_{i} + U_{i + 1})^{2} + \frac{1}{4} (U_{i - 1} - U_{i + 1})^{2}, β_{2} = \frac{13}{12} (U_{i} - 2 U_{i + 1} + U_{i + 2})^{2} + \frac{1}{4} (3 U_{i} - 4 U_{i + 1} + U_{i + 2})^{2} .$ Fix a positive parameter $ϵ_{0}$ with the same units as each $β_{k}$ , independently of $h$ , and ideal weights $(d_{0}, d_{1}, d_{2}) = (1 / 10, 6 / 10, 3 / 10)$ . If $u$ is nondimensionalized, $ϵ_{0}$ is a fixed positive number. The WENO-JS weights and left face value are $α_{k}^{J} = d_{k} / (ϵ_{0} + β_{k})^{2}$ , $ω_{k}^{J} = α_{k}^{J} / \sum_{l = 0}^{2} α_{l}^{J}$ , and $U_{i + 1 / 2}^{-, J} = \sum_{k = 0}^{2} ω_{k}^{J} p_{k}$ . For the WENO-Z comparison, use the fifth-order $p = 2$ weights: with $τ_{5} = ∣ β_{0} - β_{2} ∣$ , set $α_{k}^{Z} = d_{k} [1 + (τ_{5} / (ϵ_{0} + β_{k}))^{2}]$ , $ω_{k}^{Z} = α_{k}^{Z} / \sum_{l = 0}^{2} α_{l}^{Z}$ , and $U_{i + 1 / 2}^{-, Z} = \sum_{k = 0}^{2} ω_{k}^{Z} p_{k}$ . All denominators are positive because $ϵ_{0} > 0$ and every $β_{k}$ is a sum of squares. The right face state is obtained by reflection about that face: replace each stencil value $U_{k}$ in the displayed left formula by $U_{2 i + 1 - k}$ .

For scalar linear advection $u_{t} + a u_{x} = 0$ , a direct reconstructed-upwind forward-Euler update uses $ν = a Δ t / h$ , the flux $F_{i + 1 / 2} = a U_{i + 1 / 2}^{-, M}$ when $a \geq 0$ and $F_{i + 1 / 2} = a U_{i + 1 / 2}^{+, M}$ when $a < 0$ , and $U_{i}^{n + 1} = U_{i}^{n} - (Δ t / h) (F_{i + 1 / 2} - F_{i - 1 / 2})$ . The total variation of a sequence is $TV (U) = \sum_{i \in Z} ∣ U_{i + 1} - U_{i} ∣$ when the sum is finite; TVD means that this quantity does not increase in one update.

For a vector system whose flux Jacobian at an interface is diagonalizable as $A = R Λ R^{- 1}$ with real diagonal $Λ$ , a characteristic-wise reconstruction applies a chosen scalar reconstruction to the components of $R^{- 1} U$ and maps the reconstructed vector back by $R$ . This definition alone gives no TVD or stability result for systems.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A monotone linear one-step advection scheme cannot be second order at noninteger CFL

Statement

Let $a \neq 0$ , $h > 0$ and $Δ t > 0$ , and put $ν = a Δ t / h$ . Fix a noninteger $ν$ . Consider a finite-stencil, translation-invariant linear one-step scheme $U_{j}^{n + 1} = \sum_{ℓ = - r}^{r} c_{ℓ} U_{j + ℓ}^{n}$ with real coefficients that may depend on the fixed $ν$ but are independent of $h$ , $j$ and $n$ . Assume the scheme is conservative, is consistent with $u_{t} + a u_{x} = 0$ , and is second-order accurate in the local truncation sense. For these fixed coefficients, second-order consistency requires exact matching of the constant, linear and quadratic Taylor moments for the one-step translation. If the update is monotone in the componentwise order on grid sequences, then these assumptions are inconsistent. Thus at a noninteger Courant number no such linear monotone scheme can be second order.

Facts & Assumptions

Given: A finite radius $r$ , a fixed noninteger Courant number $ν$ , a linear translation-invariant one-step update with coefficients $c_{ℓ}$ , conservation and second-order consistency through quadratic data.

[A1]

Monotone means order preserving: if $V_{j} \geq U_{j}$ for every integer $j$ , then $V_{j}^{n + 1} \geq U_{j}^{n + 1}$ for every $j$ .

[A2]

With coefficients fixed as $h \to 0$ , second-order local accuracy requires the update moments on $1$ , $x$ and $x^{2}$ to equal those of $u (x) \mapsto u (x - a Δ t)$ .

Proof

1.1givenA1choose

Fix an integer offset $ℓ$ . Compare the zero input sequence with the sequence that is $1$ at the single index $j + ℓ$ and $0$ elsewhere. The first is componentwise below the second; by [A1], their outputs at index $j$ are ordered. Linearity makes those outputs $0$ and $c_{ℓ}$ , respectively, so $c_{ℓ} \geq 0$ . This holds for each $ℓ$ .

1.2givenA2algebra

Apply [A2] to the constant polynomial $u (x) = 1$ . Its exact translate is $1$ , so $\sum_{ℓ} c_{ℓ} = 1$ . Applying [A2] to $u (x) = x$ gives $\sum_{ℓ} c_{ℓ} (x + ℓ h) = x - a Δ t = x - ν h$ ; subtracting $x \sum c_{ℓ} = x$ yields $\sum_{ℓ} ℓ c_{ℓ} = - ν$ . Applying [A2] to $u (x) = x^{2}$ gives $\sum_{ℓ} c_{ℓ} (x + ℓ h)^{2} = (x - a Δ t)^{2}$ ; expanding both sides and using the first two identities yields $\sum_{ℓ} ℓ^{2} c_{ℓ} = ν^{2}$ .

2.1A1step 1.1step 1.2algebra∎

The three moment identities imply $\sum_{ℓ} c_{ℓ} (ℓ + ν)^{2} = \sum_{ℓ} ℓ^{2} c_{ℓ} + 2 ν \sum_{ℓ} ℓ c_{ℓ} + ν^{2} \sum_{ℓ} c_{ℓ} = ν^{2} - 2 ν^{2} + ν^{2} = 0$ . Every summand is nonnegative by step 1.1, so any $ℓ$ with $c_{ℓ} > 0$ must satisfy $ℓ = - ν$ . Since $ℓ$ is an integer and $ν$ is not, all coefficients vanish, contradicting $\sum_{ℓ} c_{ℓ} = 1$ from step 1.2.

Remarks

The noninteger restriction is necessary: when $ν$ is an integer, an exact grid shift is monotone and reproduces every polynomial. The theorem is the scalar linear barrier; nonlinear limiters and nonmonotone high-order schemes lie outside its hypotheses.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

The selected nodal ENO2 quadratic reconstructs the face to third order

Statement

Let $h > 0$ , let $y_{j} = j h$ , and let $u \in C^{3} ([y_{i - 1}, y_{i + 2}])$ with $M : = \sup ∣ u^{'''} ∣ < \infty$ . Form the nodal ENO2 value at $y_{i + 1 / 2}$ by the rule in Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions. Then $∣ U_{i + 1 / 2}^{E N O 2} - u (y_{i + 1 / 2}) ∣ \leq M h^{3} / 16.$ The conclusion holds for either selected triple, so it includes the left-tie case.

Facts & Assumptions

Given: Uniform distinct nodes $y_{j} = j h$ , nodal values $U_{j} = u (y_{j})$ from a C3 function, and the ENO2 selector and Newton quadratic.

[F1]

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions: ENO2 compares the absolute divided differences on $(y_{i - 1}, y_{i}, y_{i + 1})$ and $(y_{i}, y_{i + 1}, y_{i + 2})$ , chooses the smaller (ties left), and evaluates the corresponding Newton quadratic at $y_{i + 1 / 2}$ .

[F2]

Existence and uniqueness of the cardinal-polynomial interpolant: for distinct real nodes, the data have a unique interpolating polynomial of degree at most the number of nodes minus one.

[F3]

Pointwise Lagrange interpolation remainder: for three distinct nodes and $u \in C^{3}$ , the quadratic interpolant error at a non-node $x$ is $u^{'''} (ξ) (x - y_{a}) (x - y_{b}) (x - y_{c}) / 3!$ for some $ξ$ between the nodes and $x$ .

Proof

1.1givenF1F2algebra

For either candidate triple $y_{a} < y_{b} < y_{c}$ , let $D_{a b} = (U_{b} - U_{a}) / (y_{b} - y_{a})$ and $D_{b c} = (U_{c} - U_{b}) / (y_{c} - y_{b})$ . The Newton polynomial in [F1] takes values $U_{a}$ and $U_{b}$ at the first two nodes. At $y_{c}$ , its value is $U_{a} + D_{a b} (y_{c} - y_{a}) + (D_{b c} - D_{a b}) (y_{c} - y_{b}) = U_{b} + D_{b c} (y_{c} - y_{b}) = U_{c}$ , using the divided-difference definition. By [F2] it is the unique quadratic interpolant to that triple.

1.2givenF3algebra

Apply [F3] to the left triple. The offsets from $y_{i + 1 / 2}$ are $- 3 h / 2, - h / 2, h / 2$ , whose product has magnitude $3 h^{3} / 8$ ; hence the error is at most $M (3 h^{3} / 8) / 6 = M h^{3} / 16$ . For the right triple the offsets are $- h / 2, h / 2, 3 h / 2$ , with the same product magnitude, so it has the same bound.

2.1F1step 1.2∎

The selector in [F1] returns one of those two quadratics, including the left one at a tie. Since both candidates satisfy the bound from step 1.2, whichever is selected satisfies it.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

The WENO5 ideal linear weights reproduce quartics at the right face

Statement

Let $h > 0$ , let $I_{j} = [j h, (j + 1) h)$ , and let $x_{i + 1 / 2} = (i + 1) h$ . For $u \in C^{5}$ on the union of cells $I_{i - 2}, \dots, I_{i + 2}$ , let $p_{0}, p_{1}, p_{2}$ be the cell-average functionals in Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions. Each candidate reproduces every polynomial of degree at most two exactly at $x_{i + 1 / 2}$ . The ideal combination $p_{o p t} = p_{0} / 10 + 3 p_{1} / 5 + 3 p_{2} / 10$ reproduces every polynomial of degree at most four and satisfies $∣ p_{o p t} - u (x_{i + 1 / 2}) ∣ \leq (23 / 15) (3^{5} / 5!) \sup ∣ u^{(5)} ∣ h^{5}$ . Each candidate separately satisfies $∣ p_{k} - u (x_{i + 1 / 2}) ∣ \leq 15 \sup ∣ u^{'''} ∣ h^{3}$ .

Facts & Assumptions

Given: Uniform cell averages of a C5 function, the three candidate functionals and the ideal weights in Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions.

[F1]

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions: the candidate coefficient vectors, in order $(U_{i - 2}, U_{i - 1}, U_{i}, U_{i + 1}, U_{i + 2})$ , are $(2, - 7, 11, 0, 0) / 6$ , $(0, - 1, 5, 2, 0) / 6$ , and $(0, 0, 2, 5, - 1) / 6$ ; the ideal weights are $(1 / 10, 3 / 5, 3 / 10)$ .

[F2]

A uniform derivative bound gives a uniform Taylor remainder bound: if $∣ u^{(m + 1)} ∣ \leq M$ on the interval between a center $c$ and $x$ , the degree- $m$ Taylor polynomial at $c$ has remainder bounded by $M ∣ x - c ∣^{m + 1} / (m + 1)!$ .

Proof

1.1F1algebra

Combining the three coefficient vectors in [F1] gives $p_{o p t} = (2 U_{i - 2} - 13 U_{i - 1} + 47 U_{i} + 27 U_{i + 1} - 3 U_{i + 2}) / 60$ .

1.2F1F2algebra

For $p_{0}$ , the coefficient vector $(2, - 7, 11)$ has dot products $6$ , $0$ , and $0$ with the average vectors $(1, 1, 1)$ , $(- 5 / 2, - 3 / 2, - 1 / 2)$ , and $(19 / 3, 7 / 3, 1 / 3)$ . For $p_{1}$ , $(- 1, 5, 2)$ has dot products $6$ , $0$ , and $0$ with $(1, 1, 1)$ , $(- 3 / 2, - 1 / 2, 1 / 2)$ , and $(7 / 3, 1 / 3, 1 / 3)$ . For $p_{2}$ , $(2, 5, - 1)$ has dot products $6$ , $0$ , and $0$ with $(1, 1, 1)$ , $(- 1 / 2, 1 / 2, 3 / 2)$ , and $(1 / 3, 1 / 3, 7 / 3)$ . After division by $6$ , each candidate exactly reproduces $1, z, z^{2}$ at the face. The degree-two Taylor polynomial at $x_{i + 1 / 2}$ therefore cancels from $p_{k} - u (x_{i + 1 / 2})$ . By [F2], the pointwise remainder on each stencil is bounded by $M_{3} (3 h)^{3} / 3!$ , where $M_{3} = \sup ∣ u^{'''} ∣$ ; averaging preserves this bound. The largest sum of absolute candidate coefficients is $10 / 3$ , giving $∣ p_{k} - u (x_{i + 1 / 2}) ∣ \leq (10 / 3) M_{3} (3 h)^{3} / 6 = 15 M_{3} h^{3}$ .

2.1step 1.1algebra

Write $z = (x - x_{i + 1 / 2}) / h$ . The vectors of cell averages of $1, z, z^{2}, z^{3}, z^{4}$ over the five cells, ordered from $i - 2$ through $i + 2$ , are respectively $(1, 1, 1, 1, 1)$ , $(- 5 / 2, - 3 / 2, - 1 / 2, 1 / 2, 3 / 2)$ , $(19 / 3, 7 / 3, 1 / 3, 1 / 3, 7 / 3)$ , $(- 65 / 4, - 15 / 4, - 1 / 4, 1 / 4, 15 / 4)$ , and $(211 / 5, 31 / 5, 1 / 5, 1 / 5, 31 / 5)$ . Direct multiplication gives dot products $60 / 60$ , $0 / 60$ , $0 / 180$ , $0 / 240$ , and $0 / 300$ with $(2, - 13, 47, 27, - 3) / 60$ , respectively. Thus the functional in step 1.1 returns the value at $z = 0$ for every polynomial of degree at most four.

3.1F1F2step 2.1algebra∎

Let $T_{4}$ be the degree-four Taylor polynomial of $u$ at $x_{i + 1 / 2}$ and put $M_{5} = \sup ∣ u^{(5)} ∣$ on the stencil. By [F2], the cell-average remainder after subtracting $T_{4}$ has magnitude at most $M_{5} (3 h)^{5} / 5!$ in each cell. Step 2.1 reproduces $T_{4}$ exactly, and the absolute coefficients of $p_{o p t}$ sum to $(2 + 13 + 47 + 27 + 3) / 60 = 23 / 15$ . Therefore $∣ p_{o p t} - u (x_{i + 1 / 2}) ∣ \leq (23 / 15) M_{5} (3 h)^{5} / 5!$ , as claimed.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

MUSCL/minmod face states are second order on smooth monotone data

Statement

Let $u \in C^{2}$ on a neighborhood of a cell and its two neighbors, and suppose $∣ u^{'} (x) ∣ \geq c > 0$ there. Let $U_{j}$ be the averages of $u$ over uniform cells of width $h$ , centered at $x_{j} = x_{0} + j h$ . For sufficiently small $h$ , the MUSCL/minmod face states of Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions satisfy $∣ U_{i} + h s_{i} / 2 - u (x_{i} + h / 2) ∣ \leq C h^{2}, ∣ U_{i} - h s_{i} / 2 - u (x_{i} - h / 2) ∣ \leq C h^{2},$ where $C$ depends only on $c$ and a bound for $∣ u^{''} ∣$ on the stated neighborhood.

Facts & Assumptions

Given: The stated smoothness and nonvanishing derivative, uniform cell averages, and minmod MUSCL slopes.

[F1]

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions: $h s_{i} = minmod (U_{i} - U_{i - 1}, U_{i + 1} - U_{i})$ and the two face states are $U_{i} \pm h s_{i} / 2$ .

[F2]

A uniform derivative bound gives a uniform Taylor remainder bound: with $∣ u^{''} ∣ \leq M_{2}$ , the degree-one Taylor remainder at distance $r$ is bounded by $M_{2} ∣ r ∣^{2} / 2$ .

Proof

1.1givenF2algebra

Let $M_{2} = \sup ∣ u^{''} ∣$ on the cell and its neighbors. Applying [F2] about $x_{i}$ and averaging over the symmetric central cell cancels the linear term, giving $∣ U_{i} - u (x_{i}) ∣ \leq M_{2} h^{2} / 24$ . Applying [F2] at $x_{i}$ to the neighboring centers and using the same cell-average estimate gives $U_{i} - U_{i - 1} = h u^{'} (x_{i}) + O (M_{2} h^{2})$ and $U_{i + 1} - U_{i} = h u^{'} (x_{i}) + O (M_{2} h^{2})$ .

2.1F1step 1.1algebra

Because $∣ u^{'} (x_{i}) ∣ \geq c$ , both increments in step 1.1 have the same nonzero sign for all sufficiently small $h$ . Minmod selects one of them, and either equals $h u^{'} (x_{i}) + O (M_{2} h^{2})$ ; hence $h s_{i} = h u^{'} (x_{i}) + O (M_{2} h^{2})$ .

3.1F2step 1.1step 2.1algebra∎

By step 1.1 and step 2.1, $U_{i} + h s_{i} / 2 = u (x_{i}) + h u^{'} (x_{i}) / 2 + O (M_{2} h^{2})$ . Applying [F2] at distance $h / 2$ gives $u (x_{i} + h / 2) = u (x_{i}) + h u^{'} (x_{i}) / 2 + O (M_{2} h^{2})$ , so subtraction proves the right-face estimate. The same calculation with $- h / 2$ proves the left-face estimate.

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

Direct forward-Euler MUSCL/minmod is TVD under the sharp two-thirds CFL bound

Statement

Let $h > 0$ , $Δ t \geq 0$ , $a \in R$ , and $μ = ∣ a ∣ Δ t / h$ . Use the direct reconstructed-upwind forward-Euler update for scalar linear advection from Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions, with the stated MUSCL/minmod reconstruction and no time predictor; write $U^{+}$ for the sequence after one such update. The data are either a real bi-infinite sequence with finite discrete total variation $TV (U) = \sum_{i \in Z} ∣ U_{i} - U_{i - 1} ∣ < \infty,$ or a real $N$ -periodic sequence, for an integer $N \geq 1$ , with ${TV}_{N} (U) = \sum_{i = 0}^{N - 1} ∣ U_{i} - U_{i - 1} ∣,$ where periodic indices are taken modulo $N$ . If $0 \leq μ \leq 2 / 3$ , one update satisfies $TV (U^{+}) \leq TV (U)$ in the bi-infinite case and ${TV}_{N} (U^{+}) \leq {TV}_{N} (U)$ in the periodic case.

The constant $2 / 3$ is sharp within the original range $0 \leq μ \leq 1$ : for every $2 / 3 < μ \leq 1$ there are finite-support data of variation $2$ whose updated variation is $3 μ > 2$ . For every period $N \geq 7$ , the same witness periodized with period $N$ has the same strict increase.

Facts & Assumptions

Given: $h > 0$ , $Δ t \geq 0$ , $a \in R$ , the direct forward-Euler MUSCL/minmod update for scalar linear advection, and either a bi-infinite real sequence of finite discrete total variation or an $N$ -periodic real sequence with $N \geq 1$ .

[F1]

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions: the MUSCL/minmod slope is $s_{i} = h^{- 1} minmod (U_{i} - U_{i - 1}, U_{i + 1} - U_{i})$ , with face states $U_{i} \pm h s_{i} / 2$ . Minmod vanishes when either argument is zero or their signs differ, and otherwise has their common sign and magnitude no larger than either argument's magnitude.

[F2]

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions: for $a > 0$ , the direct update is $U_{i}^{+} = U_{i} - μ [d_{i} + (m_{i} - m_{i - 1}) / 2]$ , where $d_{i} = U_{i} - U_{i - 1}$ and $m_{i} = h s_{i}$ . For $a < 0$ , the fluxes are $F_{i + 1 / 2} = a (U_{i + 1} - m_{i + 1} / 2)$ and $F_{i - 1 / 2} = a (U_{i} - m_{i} / 2)$ .

Proof

1.1givenF1F2algebra

First suppose $a > 0$ . If $d_{i} \neq 0$ , then $m_{i} / d_{i}$ and $m_{i - 1} / d_{i}$ lie in $[0, 1]$ : in each minmod, $d_{i}$ is one argument, so a nonzero output has the sign of $d_{i}$ and magnitude at most $∣ d_{i} ∣$ . Set $C_{i} = μ [1 + \frac{m_{i} / d_{i} - m_{i - 1} / d_{i}}{2}]$ in this case. If $d_{i} = 0$ , then $m_{i} = m_{i - 1} = 0$ ; set $C_{i} = μ$ . In either case $μ / 2 \leq C_{i} \leq 3 μ / 2$ , and the exact update satisfies $U_{i}^{+} = U_{i} - C_{i} d_{i} = (1 - C_{i}) U_{i} + C_{i} U_{i - 1}$ . Thus $0 \leq C_{i} \leq 1$ when $μ \leq 2 / 3$ , including equality.

2.1step 1.1algebra

Subtract the identity for $U_{i - 1}^{+}$ in step 1.1 from that for $U_{i}^{+}$ to obtain $d_{i}^{+} = (1 - C_{i}) d_{i} + C_{i - 1} d_{i - 1}$ . The coefficients are nonnegative, so $∣ d_{i}^{+} ∣ \leq (1 - C_{i}) ∣ d_{i} ∣ + C_{i - 1} ∣ d_{i - 1} ∣$ . For periodic data, $C_{i}$ is $N$ -periodic because it is defined from shifts of $U$ ; summing over one period and cyclically reindexing gives ${TV}_{N} (U^{+}) \leq \sum_{i = 0}^{N - 1} [(1 - C_{i}) + C_{i}] ∣ d_{i} ∣ = {TV}_{N} (U)$ . This also covers $N = 1$ , when every $d_{i} = 0$ . For bi-infinite data, both nonnegative series on the right are bounded by $\sum_{i} ∣ d_{i} ∣ < \infty$ . Reindexing the absolutely convergent second series gives $TV (U^{+}) \leq \sum_{i} [(1 - C_{i}) + C_{i}] ∣ d_{i} ∣ = TV (U)$ , so the updated sequence still has finite variation.

3.1F1F2step 2.1algebra

Now suppose $a < 0$ . At face $i + 1 / 2$ , [F2] gives $F_{i + 1 / 2} = a (U_{i + 1} - m_{i + 1} / 2)$ and $F_{i - 1 / 2} = a (U_{i} - m_{i} / 2)$ , hence $U_{i}^{+} = U_{i} + μ [(U_{i + 1} - U_{i}) - (m_{i + 1} - m_{i}) / 2]$ . Reflect the data by $V_{j} = U_{- j}$ . Its minmod increments obey $m_{j}^{V} = - m_{- j}$ , because reflection reverses each adjacent difference and minmod is odd under simultaneous sign reversal. With $i = - j$ , the positive-speed expression for $V_{j}$ is $V_{j} - μ [(V_{j} - V_{j - 1}) + (m_{j}^{V} - m_{j - 1}^{V}) / 2] = U_{i} + μ [(U_{i + 1} - U_{i}) - (m_{i + 1} - m_{i}) / 2]$ , exactly the negative-speed update just computed. Also $V_{j} - V_{j - 1} = - (U_{- j + 1} - U_{- j})$ , so reflection only reindexes the absolute increments and preserves both the bi-infinite and periodic variation sums. Step 2.1 therefore proves the claim for negative velocity. If $a = 0$ , both fluxes vanish; if $Δ t = 0$ , their difference is multiplied by zero in the update. In either case the update is the identity, covering $μ = 0$ .

4.1F1F2step 2.1step 3.1algebra∎

Fix $2 / 3 < μ \leq 1$ , any $h > 0$ and any $a \neq 0$ , and take $Δ t = μ h / ∣ a ∣$ . For positive velocity choose $U_{- 1} = U_{1} = 1 / 2$ , $U_{0} = 1$ , and $U_{i} = 0$ otherwise; its initial variation is $∣ 1 / 2 ∣ + ∣ 1 / 2 ∣ + ∣ - 1 / 2 ∣ + ∣ - 1 / 2 ∣ = 2$ . The only nonzero differences are $d_{- 1} = d_{0} = 1 / 2$ and $d_{1} = d_{2} = - 1 / 2$ , so the only nonzero minmod increments are $m_{- 1} = 1 / 2$ and $m_{1} = - 1 / 2$ . Substituting these in [F2], the four affected updates are $U_{- 1}^{+} = 1 / 2 - μ (1 / 2 + 1 / 4) = A$ , $U_{0}^{+} = 1 - μ (1 / 2 - 1 / 4) = B$ , $U_{1}^{+} = 1 / 2 - μ (- 1 / 2 - 1 / 4) = C$ , and $U_{2}^{+} = 0 - μ (- 1 / 2 + 1 / 4) = D$ , where $A = 1 / 2 - 3 μ / 4$ , $B = 1 - μ / 4$ , $C = 1 / 2 + 3 μ / 4$ , and $D = μ / 4$ ; every other update is zero because its state and all adjacent minmod increments are zero. Here $A < 0$ , $B \geq 3 / 4 > 0$ , $C - B = μ - 1 / 2 > 0$ , $D > 0$ , and $C - D = 1 / 2 + μ / 2 > 0$ , so $A < 0 < B < C$ and $0 < D < C$ . Hence $TV (U^{+}) = - A + (B - A) + (C - B) + (C - D) + D = 2 (C - A) = 3 μ > 2$ . For negative velocity reflect this witness; step 3.1 preserves its updated variation. Periodizing the data with any $N \geq 7$ leaves at least four zero initial cells between successive three-cell supports and at least three zero updated entries between the four affected entries; since a positive-velocity update uses only cells $i - 2, \dots, i + 1$ and a negative-velocity update uses only cells $i - 1, \dots, i + 2$ , no stencil meets two distinct support copies, and the periodic variation is the same $3 μ$ . At $μ = 2 / 3$ , the coefficient argument in steps 1.1–2.1 still proves TVD, and the witness updates to $(0, 5 / 6, 1, 1 / 6)$ with variation $2$ ; every larger CFL number through one fails.

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The specified direct MUSCL/minmod flux update is not TVD at CFL one

Statement refuted

The direct reconstructed-upwind forward-Euler update in Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions is TVD for every scalar grid sequence whenever $∣ a ∣ Δ t / h \leq 1$ .

Facts & Assumptions

Given: $a = h = Δ t = 1$ , so the CFL number is $1$ ; the minmod slope and direct forward-Euler flux update of Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions. For a finite-support sequence, $TV (U) = \sum_{i} ∣ U_{i + 1} - U_{i} ∣$ .

[F1]

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions: the slope increments are $σ_{i} = h s_{i} = minmod (U_{i} - U_{i - 1}, U_{i + 1} - U_{i})$ and the positive-velocity face flux is $F_{i + 1 / 2} = U_{i} + σ_{i} / 2$ .

Counterexample

1.1givenF1algebra

Set $U_{- 3} = U_{3} = 0$ , $U_{- 2} = U_{2} = 0$ , $U_{- 1} = U_{1} = 1 / 2$ , $U_{0} = 1$ , and $U_{i} = 0$ for all other indices. The initial variation is $1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 2$ . By [F1], the nonzero slope increments are $σ_{- 1} = 1 / 2$ and $σ_{1} = - 1 / 2$ ; the central slope is $σ_{0} = minmod (1 / 2, - 1 / 2) = 0$ , and the adjacent outer slopes are zero.

2.1F1step 1.1algebra

The relevant right-going face fluxes from [F1] are $F_{- 3 / 2} = 0$ , $F_{- 1 / 2} = 3 / 4$ , $F_{1 / 2} = 1$ , $F_{3 / 2} = 1 / 4$ , and $F_{5 / 2} = 0$ . Substituting in the stated update gives $U_{- 1}^{n + 1} = - 1 / 4$ , $U_{0}^{n + 1} = 3 / 4$ , $U_{1}^{n + 1} = 5 / 4$ , and $U_{2}^{n + 1} = 1 / 4$ , with neighboring states zero.

3.1step 1.1step 2.1algebra∎

The new variation is $1 / 4 + 1 + 1 / 2 + 1 + 1 / 4 = 3$ , which exceeds the initial value $2$ . The data satisfy the claimed CFL bound with equality, so the proposed all-data TVD claim is false.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Fixed-epsilon WENO-JS and WENO-Z face reconstructions are fifth order

Statement

Let $u \in C^{5}$ on the five-cell stencil, let $\sup ∣ u^{'''} ∣ < \infty$ and $\sup ∣ u^{(5)} ∣ < \infty$ , and let $ϵ_{0} > 0$ be fixed independently of $h$ . For the cell-average WENO-JS and WENO-Z states in Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions, there is a constant $C$ independent of sufficiently small $h$ such that $∣ U_{i + 1 / 2}^{-, J} - u (x_{i + 1 / 2}) ∣ \leq C h^{5}, ∣ U_{i + 1 / 2}^{-, Z} - u (x_{i + 1 / 2}) ∣ \leq C h^{5} .$ The same result holds for the reflected right states.

Facts & Assumptions

Given: The fixed positive $ϵ_{0}$ , uniform cell averages of a C5 scalar function, the WENO-JS/WENO-Z formulas and ideal weights, and the finite-volume candidate estimates.

[F1]

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions: each $β_{k}$ is a sum of squares, $τ_{5} = ∣ β_{0} - β_{2} ∣$ , $ϵ_{0} > 0$ , and both normalized weight triples are positive and sum to one.

[F2]

The WENO5 ideal linear weights reproduce quartics at the right face: $\sum d_{k} p_{k} = u (x_{i + 1 / 2}) + O (h^{5})$ and each $p_{k} = u (x_{i + 1 / 2}) + O (h^{3})$ for the stated smooth data.

Proof

1.1givenF1algebra

Each first-difference combination in [F1] is $O (h)$ on C1 data, and each second-difference combination is $O (h^{2})$ on C2 data. Squaring shows $β_{k} = O (h^{2})$ for all three stencils; consequently $τ_{5} = O (h^{2})$ . For the constant data case all these differences vanish and both schemes return the ideal weights exactly.

2.1F1step 1.1algebra

For WENO-JS, $α_{k}^{J} = d_{k} ϵ_{0}^{- 2} (1 + β_{k} / ϵ_{0})^{- 2} = d_{k} ϵ_{0}^{- 2} (1 + O (h^{2}))$ . Since $\sum d_{k} = 1$ , normalization gives $ω_{k}^{J} = d_{k} + O (h^{2})$ . For WENO-Z, $ϵ_{0} + β_{k} \geq ϵ_{0}$ , so $(τ_{5} / (ϵ_{0} + β_{k}))^{2} = O (h^{4})$ ; hence $α_{k}^{Z} = d_{k} (1 + O (h^{4}))$ and normalization gives $ω_{k}^{Z} = d_{k} + O (h^{4})$ . The positive fixed $ϵ_{0}$ makes all constants uniform for sufficiently small $h$ .

3.1F2step 2.1algebra∎

For either normalized weight vector $ω$ , $\sum_{k} (ω_{k} - d_{k}) = 0$ . Therefore $\sum_{k} ω_{k} p_{k} - \sum_{k} d_{k} p_{k} = \sum_{k} (ω_{k} - d_{k}) (p_{k} - u (x_{i + 1 / 2}))$ . By [F2] the right side is $O (h^{5})$ for WENO-JS and $O (h^{7})$ for WENO-Z. The ideal combination has error $O (h^{5})$ by [F2], so both left states have error $O (h^{5})$ . Reflection changes only the stencil orientation and preserves the same estimates.

Remarks

This is the fixed- $ϵ_{0}$ asymptotic branch. The estimate does not claim fifth order for arbitrary choices $ϵ = ϵ (h)$ ; the critical-point example on the companion page computes a scale-dependent branch separately.

5 · Examples, counterexamples and false statements

None yet.

High-resolution TVD, ENO and WENO reconstruction

1 · Prerequisites

2 · Summary

Objective

3 · Logical flowchart

4 · Definitions, theorems and proofs

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions

Definition

A monotone linear one-step advection scheme cannot be second order at noninteger CFL

Statement

Facts & Assumptions

Proof

Remarks

The selected nodal ENO2 quadratic reconstructs the face to third order

Statement

Facts & Assumptions

Proof

The WENO5 ideal linear weights reproduce quartics at the right face

Statement

Facts & Assumptions

Proof

MUSCL/minmod face states are second order on smooth monotone data

Statement

Facts & Assumptions

Proof

Direct forward-Euler MUSCL/minmod is TVD under the sharp two-thirds CFL bound

Statement

Facts & Assumptions

Proof

The specified direct MUSCL/minmod flux update is not TVD at CFL one

Statement refuted

Facts & Assumptions

Counterexample

Fixed-epsilon WENO-JS and WENO-Z face reconstructions are fifth order

Statement

Facts & Assumptions

Proof

Remarks

5 · Examples, counterexamples and false statements

Sources