Boundary Closures for Sixth-Order Energy-Stable Weighted Essentially Non-Oscillatory Finite-Difference Schemes

Abstract

A general strategy was presented in 2009 by Yamaleev and Carpenter (J. Comput. Phys. 228(11):4248–4272, 2009; J. Comput. Phys. 228(8):3025–3047, 2009), for constructing Energy Stable Weighted Essentially Non-Oscillatory (ESWENO) finite-difference schemes on periodic domains. Fisher et al. (J. Comput. Phys. 230(10):3727–3752, 2011) provided boundary closures for the fourth-order ESWENO scheme that maintain, the WENO stencil biasing properties and satisfy the summation-by-parts (SBP) operator convention, thereby ensuring stability in an L ₂ norm. Herein, the general capability of finite-domain schemes is extended by providing closures for the sixth-order case. Third-order and fifth-order boundary closures are developed that are stable in diagonal and block norms, respectively, and achieve fourth- and sixth-order global accuracy for hyperbolic systems. A novel set of nonuniform flux interpolation points is necessary near the boundaries to simultaneously achieve (1) accuracy, (2) the SBP convention, and (3) WENO stencil biasing mechanics. Complete implementation details for the diagonal-norm sixth-order operator are provided as well as examples that demonstrate shock capturing and multi-domain capabilities.

Appendix

1.1 8.1 Implementation of WENO

The nonlinear WENO scheme is defined as

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The fluxes are constructed by using the formula

$$ \bar{f}_{j} = + {\bar{w}}^{L2}_{j} {\bar{f}}^{L2}_{j} + {\bar{w}}^{L1}_{j} {\bar{f}}^{L1}_{j} + {\bar{w}}^{R1}_{j} {\bar{f}}^{R1}_{j} + {\bar{w}}^{R2}_{j} {\bar{f}}^{R2}_{j}, \quad 7 \le j \le N-6. $$

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The expressions for $\bar{f}_{j}$ at the six boundary closure points are slightly more complicated but conform to the same matrix conventions,

$$ \bar{f}_{j} = \sum_{i=1}^{n_b} {\bar{w}}^{i}_{j} {\bar{f}}^{i}_{j}, \quad 1 \le j \le6 $$

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where n _b is the number of flux terms in each boundary flux.

The nonlinear weights $\bar{w}^{(r)}$ are defined by

$$ \everymath{\displaystyle} \begin{array} {l} \bar{w}_{j}^{(r)}=\frac{\bar{\alpha}_{j}^{(r)}}{\sum_r \bar{\alpha}_{j}^{(r)}} ; \qquad \bar{ \alpha}_{j}^{(r)}=\bar{d}_{j}^{(r)} \biggl(1+ \frac{\bar{\tau }_{j}}{\varepsilon+ \bar{\beta}_{j}^{(r)}} \biggr) , \quad \forall r, \ 1 \le j \le N-1, \\\noalign{\vspace{8pt}} \bar{w}^{(r)} = \mathrm{Diag}\bigl[\bar{w}^{(r)}_0, \ldots,\bar{w}^{(r)}_N\bigr] , \quad \forall r. \end{array} $$

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The coefficients $\bar{d}_{j}^{(r)}$ are the target weights of the candidate stencils, and $\bar{\beta}_{j}^{(r)}$ and $\bar{\tau}_{j}$ are the smoothness indicators.

The smoothness indicators $\bar{\beta}_{j}^{(r)}$ have the same form at all flux points, although they may not all be defined near the boundaries. Experimentation has determined that the choice of f affects the solution smoothness and dissipation properties, and that the best practice is to base the smoothness indicator on the characteristic variables rather than the characteristic fluxes. The interior smoothness indicators

$$ \begin{array} {rcl} \bar{\beta}_{j}^{L2}&=& (f_{j-3}-3 f_{j-2} + 2 f_{j-1} )^2 + (f_{j-3}-2 f_{j-2} + f_{j-1} )^2 ; \\\noalign{\vspace{8pt}} \bar{\beta}_{j}^{L1}&=& (-f_{j-1} + f_{j-0} )^2 + (f_{j-2}-2 f_{j-1} + f_{j-0} )^2 ; \\\noalign{\vspace{8pt}} \bar{\beta}_{j}^{R1}&=& (-f_{j-1} + f_{j-0} )^2 + (f_{j-1}-2 f_{j-0} + f_{j+1} )^2 ; \\\noalign{\vspace{8pt}} \bar{\beta}_{j}^{R2}&=& (-2 f_{j-0} + 3 f_{j+1} - f_{j+2} )^2 + (f_{j-0}-2 f_{j+1} + f_{j+2} )^2; \end{array} \quad 7 \le j \le N-6. $$

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The indicators $\bar{\beta}^{L3}$ and $\bar{\beta}^{R3}$ (needed only near boundaries) as well as the $\bar{\beta}^{r}$ stencil coefficients used near the boundaries are included in Appendix 8.3.

To guarantee that the downwind stencil weight does not exceed that of the central or upwind weights, the downwind smoothness indicator is modified by using the expression

$$ \bar{\beta}_{j}^d= \biggl( \frac{1}{n_s}\sum_r {\bigl[\bar{\beta }_{j}^{(r)}\bigr]}^k \biggr)^{1/k} $$

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where k is an even integer, and n _s is the number of distinct interpolants contributing to the flux f _j .

An additional stencil biasing parameter, $\bar{\tau}_{j}$, is needed for the ESWENO scheme. Here, $\bar{\tau}_{j}$ is a quadratic function of the sum of fourth-order undivided differences that are available on the stencil for $\bar{f}_{j}$ (see Fig. 1),

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At the points $\bar{x}_{1}$ and $\bar{x}_{2}$, $\bar{\tau}_{j}$ is constructed from a single fourth-order undivided difference, biased toward the interior of the domain as

$$ \bar{\tau}_{j}= ( - f_{1} + 4 f_{2} -6 f_{3 } +4 f_{4} - f_{5})^2. $$

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The $\bar{\tau}_{j}$ is biased in a mirrored fashion at $\bar {x}_{N-2}$ and $\bar{x}_{N-1}$.

The parameter ε is a function of the number of points in the discretization:

$$ \varepsilon=\max\bigl(\|f_0\|,\bigl\|f_0'\bigr\| \bigr)_{x \neq x_d} ({\delta x} )^4, \quad{\delta x}= \frac{1}{N} $$

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where ∥f ₀∥ and ∥f ₀∥′ represent a norm of the flux and the gradient of the flux, respectively, as determined using initial conditions but excluding points near discontinuities.⁵

1.2 8.2 Recipe

Consider the Euler equations U _t +F(U) _x =0. A recipe for calculating the gradient term F(U) _x using a sixth-order ESWENO scheme is summarized below.

• 1. Construct the convective flux F at solution points (x) according to the governing differential equations.

• 2. Construct the flux Jacobian matrices: $A = \frac{\partial F}{\partial U}$ at the flux points ($\bar{\mathbf{x}}$) by using the Roe-averaged variables that are formed from nearest neighbor solution point data. Form the eigenvector decomposition A=SϒS ⁻¹, where S is the matrix of right eigenvectors and ϒ is the diagonal matrix of eigenvalues.

• 3. For the flux point $\bar{x}_{j}$, use the eigenvector matrix $\bar{S}^{-1}_{j}$ to transform the solution and the flux at all points x into characteristic form $U_{c} = \bar{S}^{-1} U ; F_{c}(U) = \bar{S}^{-1} F(U) = \varUpsilon U_{c}$.

  Remark Forming the characteristic variables and fluxes is only necessary for nodal points that are located in the stencil of a given flux point. The presentation herein assumes that the transformed fluxes and variables are available at all points but can be reduced to improve performance.

• 4. Form the Lax-Friedrichs characteristic fluxes $\mathbf{f}_{\mathbf{c}}^{\pm}=\frac{1}{2} (F_{c} \pm\varUpsilon_{max} U_{c} )$, where ϒ _max is a diagonal matrix of the maximum local eigenvalues contained within each stencil.

• 5. Perform interpolations on each candidate stencil, .

• 6. Calculate the stencil biasing parameters:

  • a. Calculate $\bar{\beta}^{r}$ for each flux point according to (71), except at the end points.

  • b. Calculate $\bar{\tau}_{j}$ according to (73) and (74).

    Remark The smoothness indicators are calculated by using the characteristic variables U _c in the stencil of the flux point.

• 7. Calculate and normalize the weights using the stencil biasing parameters and the target weights in (70) (see (15) in text).

• 8. Calculate Λ _i from the weights by using (88) through (91) for the diagonal-norm scheme.

• 9. Modify the diagonal of Λ _i to be smoothly positive according to (32).

• 10. Calculate the WENO flux from the weights and candidate interpolations by using (68) (see (20) in the text).

• 11. Calculate the energy stable flux $\bar{\boldsymbol{\psi}}$ from (33).

• 12. Reconstruct the fluxes in characteristic space, $\bar{f}_{j}=\bar{f}^{+}_{j}+\bar{f}_{j}^{-}$ and $\psi_{j}=\psi_{j}^{+}+\psi_{j}^{-}$, and then transform the fluxes back to physical space by using $\hat{f}_{j} = S (\bar{f}_{j} + \psi_{j} )$.

• 13. After calculating the ESWENO interpolation at all flux points, calculate the gradient using the inverse of the norm, as in (34).

Remark

Note that all of the presented equations are for the forward-propagating waves, $\bar{f}_{j}^{+}$. The equations for interpolation of backward-propagating waves ($\bar{f}_{j}^{-}$) are found in exactly the same manner, except that the downwind stencil in (72) is the far left instead of the far right stencil. The Λ _i terms are to be negative instead of positive; therefore, we use

$$ {[\hat{\lambda}_j]}_i= - \frac{1}{2} \Bigl(\sqrt{ \bigl({[\lambda_j]}_i \bigr)^2 + \delta_i^2}+{[ \lambda_j]}_i \Bigr) , \quad i={0,5}, \ j={0,N}. $$

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1.3 8.3 Sixth-Order Diagonal-Norm Operator:

1.3.1 8.3.1 Differentiation Matrix

The upper left matrix quadrant of the target SBP differentiation operator can be written in the form

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The matrix elements in the lower right quadrant are structurally equivalent relative to the outflow boundary, but with the opposite sign. The diagonal mass matrix is

The skew-symmetric matrix follows immediately from .

The distribution the flux points $\bar{\mathbf{x}}$ as described in Lemma 1 follows immediately from the norm and can be written as

$$ \bar{\mathbf{x}} = \biggl[0,\bar{x}_{1}, \ldots, \bar{x}_{5}, \frac{11 {\delta x}}{2}, \ldots, \biggl(1- \frac{11 {\delta x}}{2} \biggr), (1-\bar{x}_{5} ), \ldots, (1-\bar{x}_{1} ), 1 \biggr]^T $$

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where

$$ \everymath{\displaystyle} \begin{array}{l} \bar{x}_{1} = \frac{13649{\delta x}}{43200} ;\qquad \bar{x}_{2} = \frac{36857{\delta x}}{21600} ;\qquad \bar{x}_{3} = \frac{4201{\delta x}}{1800} ; \\\noalign{\vspace{7pt}} \bar{x}_{4} = \frac{77207{\delta x}}{21600} ;\qquad \bar{x}_{5} = \frac{193799{\delta x}}{43200} . \end{array} $$

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Then, the vector ${\varDelta }\bar{\mathbf{x}}$ is the reciprocal of the diagonal elements of . The product is trivially the identity vector 1 because is a diagonal matrix. Note that all integer coefficients are exact. Case should be taken to ensure that the rational coefficients retain the full working precision of the coding language (e.g. 64 bit arithmetic).

1.3.2 8.3.2 Interpolation Operators

The tri-diagonal interpolation matrices , r={L4,L3,L2,L1,R1,R2,R3,R4} for the ESWENO_3-6-3 scheme, presented in compressed format using the arrays I _r , r={L4,L3,L2,L1,R1,R2,R3,R4}, are given by

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The interpolants I _r , r={R1,R2,R3,R4} are mirror images of I _r , r={L1,L2,L3,L4}. For example, the interpolants I _r , r={R1,R2} are given by

$$ I^{R1}=\left ( \begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & 0 \\\noalign{\vspace{2pt}} \frac{15151}{28800} & \frac{13649}{21600} & -\frac{13649}{86400} \\\noalign{\vspace{5pt}} \frac{269}{2160} & \frac{2507}{2400} & -\frac{3653}{21600} \\\noalign{\vspace{5pt}} \frac{763}{1600} & \frac{5129}{7200} & -\frac{109}{576} \\\noalign{\vspace{5pt}} \frac{127}{450} & \frac{18601}{21600} & -\frac{3097}{21600} \\\noalign{\vspace{5pt}} \frac{29401}{86400} & \frac{5}{6} & -\frac{15001}{86400} \\\noalign{\vspace{5pt}} \frac{1}{3} & \frac{5}{6} & -\frac{1}{6} \\ \vdots& \vdots& \vdots\\ \frac{1}{3} & \frac{5}{6} & -\frac{1}{6} \\\noalign{\vspace{5pt}} \frac{8999}{28800} & \frac{18601}{21600} & -\frac{15001}{86400} \\\noalign{\vspace{5pt}} \frac{931}{2160} & \frac{5129}{7200} & -\frac{3097}{21600} \\\noalign{\vspace{5pt}} \frac{2083}{14400} & \frac{2507}{2400} & -\frac{109}{576} \\\noalign{\vspace{5pt}} \frac{967}{1800} & \frac{13649}{21600} & -\frac{3653}{21600} \\\noalign{\vspace{5pt}} 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right ), \qquad I^{R2}=\left ( \begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & 0 \\\noalign{\vspace{2pt}} \frac{38191}{17280} & -\frac{18751}{10800} & \frac{15151}{28800} \\\noalign{\vspace{5pt}} \frac{10211}{7200} & -\frac{11723}{21600} & \frac{269}{2160} \\\noalign{\vspace{5pt}} \frac{30859}{14400} & -\frac{11663}{7200} & \frac{763}{1600} \\\noalign{\vspace{5pt}} \frac{36889}{21600} & -\frac{4277}{4320} & \frac{127}{450} \\\noalign{\vspace{5pt}} \frac{53401}{28800} & -\frac{25801}{21600} & \frac{29401}{86400} \\\noalign{\vspace{5pt}} \frac{11}{6} & -\frac{7}{6} & \frac{1}{3} \\ \vdots& \vdots& \vdots\\ \frac{11}{6} & -\frac{7}{6} & \frac{1}{3} \\\noalign{\vspace{5pt}} \frac{31079}{17280} & -\frac{11999}{10800} & \frac{8999}{28800} \\\noalign{\vspace{5pt}} \frac{4813}{2400} & -\frac{31027}{21600} & \frac{931}{2160} \\\noalign{\vspace{5pt}} \frac{7097}{4800} & -\frac{4487}{7200} & \frac{2083}{14400} \\\noalign{\vspace{5pt}} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right ) . $$

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Reconstructing from the data I _r is accomplished by populating the appropriate diagonals in the interpolation matrices. The middle column of I ^r coincides with the diagonal of , r={L4,L3,L2,L1,R1,R2,R3,R4} that is shifted {−4,−3,−2,−1,0,1,2,3} relative to the main diagonal.

1.3.3 8.3.3 Target Weights

The target weights for the five interpolation stencils are

$$ \hbox{\small$\displaystyle \mathbf{d}_{\text{3-6-3}}=\left ( \begin{array}{c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c} 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & \frac{14400}{15151} & \frac{3475559}{5786318410} \\\noalign{\vspace{7pt}} 0 & 0 & 0 & \frac{17609}{29224} & \frac{34899909}{19653140} & -\frac {720681767}{494416620} \\\noalign{\vspace{7pt}} 0 & 0 & -\frac{113713}{74988} & \frac{364030091}{204342300} & \frac {6384599}{6180300} & -\frac{73697}{247212} \\\noalign{\vspace{7pt}} 0 & \frac{86153}{1039608} & -\frac{140213393}{604921905} & \frac {903169}{758765} & -\frac{948239}{9439656} & \frac{15}{254} \\\noalign{\vspace{7pt}} -\frac{15025}{2221158} & \frac{6395825773}{172578423705} & -\frac {367011523}{8390397630} & \frac{72016320}{134993999} & \frac {190085760}{441044401} & \frac{1440}{29401} \\\noalign{\vspace{7pt}} 0 & 0 & \frac{1}{20} & \frac{9}{20} & \frac{9}{20} & \frac{1}{20} \\ \vdots& \vdots& \vdots& \vdots& \vdots& \vdots\\ 0 & 0 & \frac{1}{20} & \frac{9}{20} & \frac{9}{20} & \frac{1}{20} \\\noalign{\vspace{7pt}} 0 & 0 & \frac{1440}{29401} & \frac{190085760}{441044401} & \frac {72016320}{134993999} & -\frac{367011523}{8390397630} \\\noalign{\vspace{7pt}} 0 & 0 & \frac{15}{254} & -\frac{948239}{9439656} & \frac {903169}{758765} & -\frac{140213393}{604921905} \\\noalign{\vspace{7pt}} 0 & 0 & -\frac{73697}{247212} & \frac{6384599}{6180300} & \frac {364030091}{204342300} & -\frac{113713}{74988} \\\noalign{\vspace{7pt}} 0 & \frac{58297}{735192} & -\frac{720681767}{494416620} & \frac {34899909}{19653140} & \frac{17609}{29224} & 0 \\\noalign{\vspace{7pt}} -\frac{15025}{2537142} & \frac{13296322573}{242239975305} & \frac {3475559}{5786318410} & \frac{14400}{15151} & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ \end{array} \right )$.} $$

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For brevity, the last two columns are excluded from the table. Recognizing the relationship between columns 4 and 7 (and 5 and 6), columns 6 and 7 can be recovered from the mirror images of columns 1 and 2, respectively. Note that the WENO/ESWENO_3-6-3 scheme requires the inclusion of two extra stencils near each boundary.

1.3.4 8.3.4 Smoothness Indicators Coefficients

The smoothness indicators $\bar{\beta}^{r}$ measure the smoothness of the data used by each interpolant. This is accomplished by comparing the L ₂ of the undivided first and second derivatives, evaluated at the flux point ${\bar{x}}_{j}$.

$$ \bar{\beta}^{r} = {\bigl[{\delta x}f_{x}^{r}({\bar{x}}_{j})\bigr]}^2 + {\bigl[{\delta x}^2 f_{2x}^{r}({ \bar{x}}_{j} )\bigr]}^2 , \quad1 \le j \le N-1. $$

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The leading order truncation term is identical in all cases. For example, the interior terms

$$ \everymath{\displaystyle} \begin{array}{rcl} \bar{\beta}^{L2} &=& {f_{x}}^2 {\delta x}^2 + \biggl({f_{2x}}^2 - \frac{23}{12} f_{x}f_{3x} \biggr){\delta x}^4 + (+ 3 f_{2x}f_{3x} - 2 f_{x}f_{4x} ) {\delta x}^5 + O\bigl({\delta x}^6\bigr), \\\noalign{\vspace{9pt}} \bar{\beta}^{L1} &=& {f_{x}}^2 {\delta x}^2 + \biggl({f_{2x}}^2 + \frac{ 1}{12} f_{x}f_{3x} \biggr){\delta x}^4 + (+ 1 f_{2x}f_{3x} ) {\delta x}^5 + O\bigl({\delta x}^6\bigr), \\\noalign{\vspace{9pt}} \bar{\beta}^{R1} &=& {f_{x}}^2 {\delta x}^2 + \biggl({f_{2x}}^2 + \frac{ 1}{12} f_{x}f_{3x} \biggr){\delta x}^4 + (- 1 f_{2x}f_{3x} ) {\delta x}^5 + O\bigl({\delta x}^6\bigr), \\\noalign{\vspace{9pt}} \bar{\beta}^{R2} &=& {f_{x}}^2 {\delta x}^2 + \biggl({f_{2x}}^2 - \frac{23}{12} f_{x}f_{3x} \biggr){\delta x}^4 + (- 3 f_{2x}f_{3x} + 2 f_{x}f_{4x} ) {\delta x}^5 + O\bigl({\delta x}^6\bigr) . \end{array} $$

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The coefficients needed for the first derivative term ${\delta x} \frac{\partial}{\partial x}$ in the smoothness indicators throughout the spatial domain can be expressed as follows.

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The coefficients needed for the first derivative term ${\delta x} \frac{\partial}{\partial x}$ in the rightward biased stencils (r=R1,R2,R3,R4) are the mirror images of those used for leftward biased stencils (r=L1,L2,L3,L4).

The second derivative stencil coefficients are always

$${\delta x}^2 f_{2x}^{r}({\bar{x}}_{j}) = f_{r+j-1}^{r} - 2 f_{r+j}^{r} + f_{r+j+1}^{r} ; \quad1 \le j \le N-1 , \forall r . $$

1.3.5 8.3.5 Energy-Stable Terms

The stabilization matrix Diag(Λ ₀) is identically zero for the ESWENO_3-6-3 scheme (as it is with every consistent scheme). The nonzero terms of the diagonal matrices Λ _i , i=1,5 are

(88)

(89)

(90)

(91)

$$ \hbox{\small$\displaystyle \mathrm{Diag} (\varLambda_5 )= \left ( \begin{array}{c} 0 \\ 0 \\ 0 \\ (370193 \bar{w}^{L4}_{5}-422857 \bar{w}^{R4}_{1}) / 172800 \\\noalign{\vspace{6pt}} \begin{array}{l} (5850 \bar{w}^{L4}_{6}-5011 \bar{w}^{R4}_{2}) / 2700 \\\noalign{\vspace{3pt}} \quad{} -(69251 \bar{w}^{R4}_{3}) / 28800 \\ \end{array}\\\noalign{\vspace{4pt}} 0 \\[-4pt] \vdots\\ 0 \\ (69251 \bar{w}^{L4}_{N-3}) / 28800 \\\noalign{\vspace{6pt}} (5011 \bar{w}^{L4}_{N-2}-5850 \bar{w}^{R4}_{N-6}) / 2700 \\\noalign{\vspace{6pt}} (422857 \bar{w}^{L4}_{N-1}-370193 \bar{w}^{R4}_{N-5}) / 172800 \\\noalign{\vspace{6pt}} 0 \\ 0 \\ 0 \end{array} \right )$.} $$

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