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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.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Randall J. LeVeque, Numerical Methods for Conservation Laws, full textbook (standard reference, not scraped)