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The WENO5 ideal linear weights reproduce quartics at the right face

Statement

Let h>0, let Ij=[jh,(j+1)h), and let xi+1/2=(i+1)h. For u∈C5 on the union of cells Ii−2,…,Ii+2, let p0,p1,p2 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 xi+1/2. The ideal combination popt=p0/10+3p1/5+3p2/10 reproduces every polynomial of degree at most four and satisfies ∣popt−u(xi+1/2)∣≤(23/15)(35/5!) sup⁡∣u(5)∣ h5. Each candidate separately satisfies ∣pk−u(xi+1/2)∣≤15sup⁡∣u′′′∣h3.

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 (Ui−2,Ui−1,Ui,Ui+1,Ui+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)∣≤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 popt=(2Ui−2−13Ui−1+47Ui+27Ui+1−3Ui+2)/60.

1.2F1F2algebra

For p0, 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 p1, (−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 p2, (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,z2 at the face. The degree-two Taylor polynomial at xi+1/2 therefore cancels from pk−u(xi+1/2). By [F2], the pointwise remainder on each stencil is bounded by M3(3h)3/3!, where M3=sup⁡∣u′′′∣; averaging preserves this bound. The largest sum of absolute candidate coefficients is 10/3, giving ∣pk−u(xi+1/2)∣≤(10/3)M3(3h)3/6=15M3h3.

2.1step 1.1algebra

Write z=(x−xi+1/2)/h. The vectors of cell averages of 1,z,z2,z3,z4 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 T4 be the degree-four Taylor polynomial of u at xi+1/2 and put M5=sup⁡∣u(5)∣ on the stencil. By [F2], the cell-average remainder after subtracting T4 has magnitude at most M5(3h)5/5! in each cell. Step 2.1 reproduces T4 exactly, and the absolute coefficients of popt sum to (2+13+47+27+3)/60=23/15. Therefore ∣popt−u(xi+1/2)∣≤(23/15)M5(3h)5/5!, as claimed.

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