A third-order unconditionally positivity-preserving scheme for. production-destruction equations with applications to non-equilibrium flows.

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1 A thrd-order ucodtoally postvty-preservg scheme for producto-destructo equatos wth applcatos to o-equlbrum flows Jutao Huag 1, Wefeg Zhao 2 ad Ch-Wag Shu 3 Abstract I ths paper, we exted our prevous work [10] ad develop a thrd-order ucodtoally postvty-preservg modfed Patakar Ruge-Kutta MPRK method for productodestructo equatos. The ecessary ad suffcet codtos for the method to be of thrd-order accuracy are derved. Wth the same approach as [10], ths tme tegrato method s the geeralzed to solve a class of ODEs arsg from sem-dscrete schemes for PDEs ad coupled wth the postvty-preservg fte dfferece weghted essetally ooscllatory WENO schemes for o-equlbrum flows. Numercal expermets are provded to demostrate the performace of our proposed scheme. Keywords: Compressble Euler equatos; postvty-preservg; chemcal reactos; producto-destructo equatos; fte dfferece; thrd-order accuracy 1 Zhou Pe-Yua Ceter for Appled Mathematcs, Tsghua Uversty, Bejg , Cha. E-mal: huagjt13@mals.tsghua.edu.c 2 Departmet of Appled Mathematcs, Uversty of Scece ad Techology Bejg, Bejg , Cha. E-mal: wfzhao@ustb.edu.c 3 Dvso of Appled Mathematcs, Brow Uversty, Provdece, RI 02912, USA. E-mal: shu@dam.brow.edu. Research supported by ARO grat W911NF ad NSF grat DMS

2 1 Itroducto Ths paper s cocered wth postvty-preservg umercal methods for o-equlbrum flow models wthout coducto or radato [17], a system of hyperbolc balace laws U t + FU x = SU. 1.1 Here U, FU ad SU are vectors wth m = s + 2 compoets where s s the umber of speces: U = ρ 1,, ρ s, ρu, ρe 0 T, FU = ρ 1 u,, ρ s u, ρu 2 + p, ρue 0 + up T, SU = s 1,, s s, 0, 0 T, where ρ s the desty of the -th speces, = 1, 2,, s, u s the velocty ad e 0 s the total eergy per ut mass of mxture. The total desty s defed as ρ = s =1 ρ ad the pressure p s gve by p = RT s where R s the uversal gas costat ad M s the molar mass of the -th speces. The total eergy has the expresso: =1 ρ M, s s ρe 0 = ρ e, T + ρ h ρu2, =1 where e, T = C T s the teral eergy of the -th speces wth C = 3R/2M ad =1 5R/2M for mooatomc speces ad datomc speces, respectvely, ad the ethalpes h 0 are costats. The source term SU descrbes the chemcal reactos occurrg gas flows whch result chages the amout of mass of each chemcal speces. It s assumed that there are J reactos of the form ν 1,j X 1 + ν 2,j X ν s,j X s ν 1,j X 1 + ν 2,j X ν s,j X s, j = 1,, J, 2

3 where ν,j ad ν,j are respectvely the stochometrc coeffcets of the reactats ad products of the -th speces the j-th reacto. For o-equlbrum chemstry, the rate of producto of speces due to chemcal reacto, may be wrtte as J s = M ν,j ν,j s k f,j ρ s s ν s,j kb,j ρ s ν s,j, = 1,, s. M s M s j=1 s=1 For each reacto j, the forward ad backward reacto rates, k f,j ad k b,j are gve fuctos of the temperature. Notce that the reactve Euler equatos, whch are ofte used to model detoato waves [16], ca also be wrtte ths form 1.1. For more detals, please refer to our prevous work [10]. Physcally, the desty ad the pressure should both be postve. I practcal computato, the falure of preservg postvty of such quattes may cause blow-ups of the computato because the learzed system may become ll-posed. I recet years, there have bee sgfcat progresses developg hgh-order postvty-preservg schemes for hyperbolc systems sce the poeerg works by Zhag ad Shu [19, 20, 21, 22]. Readers are referred to the revew papers [18, 14] for detals. We remark that almost all these works rely o the strog-stablty-preservg SSP tme tegrato [4], whch could be take as a covex combato of Euler forward method ad thus aturally preserve the postvty of umercal solutos. However, for equatos wth stff chemcal reacto sources 1.1, the explct tme tegrato may suffer from very severe restrcto o the tme step ad thus causes large computatoal cost. Recetly, there are a few works o postvty-preservg tme dscretzatos for PDEs wth stff source terms. A famly of sem-mplct schemes for a partcular class of ODEs wth stff terms were proposed [2] ad further adapted to shallow water equatos wth frcto terms [3]. We developed a class of mplct-explct IMEX RK methods for the systems of ODEs arsg from the sem-dscretzato of the Kerr-Debye model [8]. I [9], a class of expoetal SSP hgh order tme tegrato wth boud-preservg property was costructed ad appled to scalar hyperbolc equatos wth stff dsspatve source terms. Most recetly, Hu et al. proposed IMEX RK schemes [7] ad expoetal RK schemes [6] 3 s=1

4 for ketc equatos, whch share both the asymptotc-preservg ad postvty-preservg propertes. Ths paper s devoted to the dervato of postvty-preservg schemes for ths class of hyperbolc equatos 1.1. Followg our prevous work [10], we frst gore the covecto terms 1.1 ad move to a larger class of ODEs called producto-destructo equatos whch have the form: wth ad dc dt = P c D c, = 1, 2,, N, 1.2 P c = N p j c, D c = j=1 N d j c, j=1 p j c = d j c 0. Here c = c 1, c 2,, c N T ad c = c t 0 deotes the cocetrato of the -th compoet. The producto fucto p j c deotes the rate at whch the j-th compoet trasforms to the -th compoet, whle the destructo fucto d j c deotes the rate at whch the -th compoet trasforms to the j-th compoet. The exact solutos to 1.2 share the coservato property,.e., N =1 c t remas uchaged wth respect to tme t. Also, the postvty of the soluto s guarateed as log as the tal codto s postve ad d j c = 0 for c = 0 [1]. There have bee may lteratures o developg umercal methods for 1.2 whch ucodtoally preserve the coservato ad the postvty of the solutos. Here we oly commet o the most recet works by Kopecz ad Mester [13, 12] where they formulated a class of modfed Patakar-Ruge-Kutta MPRK methods. The methods adapt explct Ruge-Kutta schemes a way to esure the coservato ad the postvty of the soluto depedet of the tme step sze. Ispred by ther work, we modfy the explct RK scheme Shu-Osher form [15] stead of the classcal form ad develop aother class of secodorder MPRK schemes whch are the successfully appled to sem-dscrete schemes arsg from PDEs [10]. Combg wth the postvty-preservg fte-dfferece WENO schemes, 4

5 we obta a postvty-preservg WENO scheme for o-equlbrum flows 1.1. For more detaled revews, please refer to our prevous work [10]. I ths paper, we cotue our work [10] ad costruct a thrd-order MPRK scheme for producto-destructo equatos 1.2 ad a class of sem-dscrete schemes for PDEs. Followg the les of [13, 12, 10], the MPRK scheme s appled to a specal class of ODEs ad thus the ecessary codto for thrd order accuracy s derved. By drect Taylor expaso, we fd that the ecessary codto s ot a suffcet codto. For elmatg the redudat terms, a addtoal costrat o the coeffcets of the method s proposed. The MPRK scheme s the exteded to solve sem-dscrete schemes arsg from PDEs after dscretzato space ad the order of accuracy s valdated. The paper s orgazed as follows. I secto 2, we formulate a class of MPRK scheme the Shu-Osher form ad the derve the ecessary ad suffcet codtos for the scheme to be thrd-order accurate. I secto 3, the explct form of udetermed fucto the scheme s costructed to satsfy the suffcet codto. The scheme s the exteded to sem-dscrete schemes arsg from PDEs after dscretzato space. Numercal results for the ODEs ad PDEs are preseted secto 4. Some cocludg remarks are gve secto 5. 2 The ODE solver I ths part, we frst formulate a class of modfed Patakar RK scheme the Shu-Osher form. Followg the procedures [13], we apply t to a specal class of ODEs ad derve the ecessary codto for ths solver to be thrd-order accurate. However, we further fd that ths ecessary codto s ot a suffcet codto. By elmatg the redudat terms Taylor expasos, the suffcet codto s obtaed. The three-stage explct RK scheme the Shu-Osher form s c 0 = c, 2.1a 5

6 c 1 = α 10 c 0 + t j c 2 = α 20 c 0 + α 21 c 1 + t j β 10 p 0 j β 10 d 0 j, 2.1b β 20 p 0 j + β 21 p 1 j β 20d 0 j + β 21 d 1 j, 2.1c c +1 = α 30 c 0 + α 31 c 1 + α 32 c 2 + t β 30 p 0 j + β 31 p 1 j + β 32 p 2 j β 30d 0 j + β 31 d 1 j + β 32 d 2 j. 2.1d j Here ad afterwards, the summato s from 1 to N, uless otherwse stated. The superscrpt deotes the stage whch the fucto evaluates, e.g., p k j = p j c k for k = 0, 1, 2. Followg the modfcato [10], the explct RK scheme 2.1 becomes c 0 = c, c 1 = α 10 c 0 + t j β 10 p 0 j c 2 = α 20 c 0 + α 21 c 1 + t j 2.2a β 10 d 0 c 1 j π j π, 2.2b c 1 j β 20 p 0 j + β 21 p 1 c2 j j β 20 d 0 j + β 21 d 1 ρ j c2 j ρ c +1 = α 30 c 0 + α 31 c 1 + α 32 c 2 c+1 β 30 p 0 j + β 31 p 1 j + β 32 p 2 j β 30 d jσ 0 j + β 31 d 1 j + β 32 d 2 j + t j c+1 j σ, 2.2c. 2.2d Here π, ρ ad σ are fuctos of c k, k = 0, 1, 2 ad wll be determed later. The coeffcets are assumed to be postve ad satsfy the order codtos for the explct RK scheme 2.1: α 10 = 1, α 20 + α 21 = 1, α 30 + α 31 + α 32 = 1, α 31 β 10 + α 32 α 21 β 10 + β 20 + β 21 + β 30 + β 31 + β 32 = 1, β 10 α 32 β 21 + β 31 + β 32 α 21 β 10 + β 20 + β 21 = 1 2, β10 2 α 32β 21 + β 31 + β 32 α 21 β 10 + β 20 + β 21 2 = 1 3, β 10 β 21 β 32 =

7 Sce 2.2 s oly a modfcato of 2.1, t s atural to assume that 2.1 Necessary codtos π = c 1 ρ = c 2 σ = c +1 + O t, + O t, + O t. To derve the ecessary codtos for 2.2 to be thrd-order accurate, we cosder a specfc class of ODEs wth the source to be of lear ad quadratc forms [12]: For I, J {1, 2,, N} ad I J, ad wth costat µ > 0 ad κ = 1, 2. µc κ I, = J, j = I, p j c = 0, otherwse, µc κ I, = I, j = J, d j c = 0, otherwse, We splt our procedure to three steps: 1. Frst, by Taylor expasos, we express the exact soluto c t +1 terms of c ad other quattes at tme level up to O t 4. Smplfy the fal stage 2.2d for the specfc class of ODEs; 2. Secod, derve the expressos of c 1 ad c 2 ad 2.2c. by the frst ad the secod stage 2.2b 3. Fally, obta the ecessary codtos terms of c1 π, c2 ρ ad c+1 σ. We start wth the frst step. Step 1: By takg the dervatve wth respect to t o the ODEs 1.2, we have d 2 c dt 2 = d dt P c D c = k P D P k D k = P D P D, c k c 7

8 wth the vector P = P 1, P 2,, P N T ad D = D 1, D 2,, D N T, ad furthermore, d 3 c dt 3 = k,l 2 P D c k c l P l D l P k D k + P D c k P k D k c l P l D l = P D T H P D P D + k P D P k D k P D, c k c where H φ deotes the Hessa matrx of some fucto φ = φc. Therefore, by Taylor expasos, we obta c t +1 = c + tdc dt + t2 d 2 c 2 dt + t3 d 3 c 2 6 dt + 3 O t4 = c + tp D + t2 P D P D 2 c + t3 6 P D T H P D P D + k Substtutg the above expresso to the fal stage 2.2d yelds α 30 c + α 31c 1 + α 32 c 2 + t j c tp D t2 P D P D 2 c t3 6 P D T H P D P D + k P D Pk D k P D + O t 4. c k c 2.5 β 30 p j + β 31p 1 j + β 32 p 2 j c+1 jσ j β 30 d j + β 31d 1 j + β 32 d 2 P D c+1 j Pk D k P D = O t 4. c k c By takg = I, rememberg that p Ij c = 0 for ay j, ad rewrtg I as, we have α 31 c 1 c + α 32 c 2 c tβ 30 D + β 31 D 1 + td t2 2 + β 32 D 2 c+1 σ D D c + t3 2 D D 6 c t3 6 D 2 D = O t 4, 2.6 c where we used the relato α 30 + α 31 + α 32 = 1. Note that the above equato holds for ay, sce I could be ay umber from 1 to N. Expad D k = D c k = µc k κ for k = 1, 2 at c : σ D k = D + D c k c D c k c 2 c 2 c 2, 2.7 8

9 where the hgher order terms vash due to the fact that D k s a lear or quadratc fucto of c k. Thus, the lear combato β 30 D + β 31D 1 + β 32 D 2 s smplfed as follows: β 30 D + β 31 D 1 + β 32 D 2 = β 30 D + β 31D + D c 1 c c D 2 c 2 c 1 c 2 + β 32 D + D c 2 c c D c 2 2 c 2 c 2 = β 30 + β 31 + β 32 D + D β 31 c 1 c + β 32 c 2 c D β c 2 c 2 31 c 1 c 2 + β 32 c 2 c 2 By substtutg the above relato to 2.6 ad rearragg the terms, we obta: α 31 c 1 c + α 32c 2 t 2 D c t 2 D 2 c 2 c + td 2 D + tβ 31c 1 c + β 32c 2 β 31 c 1 c 2 + β 32 c 2 1 β 30 + β 31 + β 32 c+1 c 2 c+1 σ c+1 c σ σ + t3 6 D 2 D c = O t t2 3 D 2 Step 2: I the ext, we expad c 1 ad c 2 the frst stage 2.2b ad the secod stage 2.2c. By the frst stage 2.2b, we mmedately get c 1 = α 10 c tβ 10D c 1 π = c tβ 10D c 1 π. 2.9 By the secod stage 2.2c, we obta c 2 = α 20 c + α 21 c 1 Substtute 2.7 for k = 1 ad 2.9 to 2.10: c 2 tβ 20 D + β 21 D 1 = α 20 c + α 21c tβ 10D c 1 π t β 20 D + β 21 D + D c 1 c D c 1 c 2 c 2 c 2 = α 20 + α 21 c tα 21 β 10 D c 1 π t β 20 + β 21 D + β 21 D c tβ 10 D c ρ c 2 ρ c π 2 β 2 D 21 tβ c 2 10 D c 1 2 π c 2 ρ 9

10 = c tα 21β 10 D t c2 ρ c 1 π β 20 + β 21 D + β 21 D c tβ 10 D = c td α c 1 21β 10 + β 20 + β 21 c2 + t 2 β 21 β 10 D π ρ t 31 2 β 21β 2 10 D 2 2 D c 2 c1 c π 2 β 2 D 21 tβ c 2 10 D D c 1 c 2 c π ρ c 1 2 π 2c π ρ Step 3: I ths step, we wll substtute the expressos of c ad c to 2.8, ad derve the ecessary codtos. We frst focus o the case κ = 1,.e., the ODEs wth lear source terms. I ths case, the followg equalty holds: 2 D c 2 = 0, ad the 2.8 s smplfed as follows: α 31 tβ 10 D c 1 + α 32 td c 1 α 21 β 10 π + td D c tβ 32 1 β 30 + β 31 + β 32 c+1 σ + β 20 + β 21 c2 + t 2 β 21 β 10 D π ρ D c t 2 2 D D c tβ 31 tβ 10 D td c 1 α 21 β 10 + β 20 + β 21 c2 + t 2 β 21 β 10 D π ρ D c 1 c 2 c π D c 1 c+1 π σ c 1 c 2 c +1 c π ρ σ + t3 6 D c 2 D = O t O left had sde of 2.12, the O t term s D c 1 c 1 α 31 β 10 + α 32 α 21 β 10 + β 20 + β 21 c2 + β 30 + β 31 + β 32 c+1 1 t π π ρ σ = µc c 1 α 31 + α 32 α 21 β 10 + α 32 β 20 + β 21 c2 + β 30 + β 31 + β 32 c+1 1 t. π ρ σ The O t 2 term s α 32 β 21 β 10 D D c 1 c 2 1 c π ρ 2 + D c +1 β 31 β 10 D c σ c 1 D c D + β 32 D π α c 1 21β 10 + β 20 + β 21 c2 t 2 π ρ 10 ρ

11 = D D c = µ 2 c c 1 c 2 α 32 β 21 β 10 + β 31 β 10 + β 32 α 21 β 10 c1 c +1 π ρ π c 1 c 2 α 32 β 21 β 10 + β 31 β 10 + β 32 α 21 β 10 c1 c +1 π ρ π The O t 3 term s σ σ + β 32 β 20 + β 21 c2 + β 32 β 20 + β 21 c2 D c +1 β 32 β 21 β 10 D D c 1 c c σ c π ρ 6 D 2 D t 3 c = D D 2 c 1 c 2 c +1 β 32 β 21 β t 3 c π ρ σ 6 = µ 3 c c 1 c 2 c +1 β 32 β 21 β t 3. π ρ σ 6 Sce the costat µ s arbtrary, we have c 1 α 31 + α 32 α 21 β 10 + α 32 β 20 + β 21 c2 + β 30 + β 31 + β 32 c+1 π ρ c 1 c 2 c 1 c +1 α 32 β 21 β 10 + β 31 + β 32 α 21 β 10 π ρ π c 1 c 2 c +1 β 32 β 21 β 10 = 1 π ρ σ 6 σ σ + β 32 β 20 + β 21 c2 c +1 1 t 2 ρ σ 2 c +1 1 t 2. ρ σ 2 = 1 + O t 3, 2.13a c +1 = 1 ρ σ 2 + O t2, 2.13b + O t. 2.13c I the ext, we set κ = 2. Note that the terms 2.8, whch do ot volve 2 D c 2 are balaced as log as the codto 2.13 holds. We oly focus o the terms whch are relevat to 2 D c 2. O the left had sde of 2.8, the term c 1 The term c 2 c has t3 1β 2 21β10 2 D 2 2 D c 2 sde of 2.8, we have β 31 c 1 c 2 + β 32 c 2 c 2 = β 31 tβ 10 D = t 2 D 2 c 1 c1 π 2 c2 c does ot volve 2 D c 2 ρ. For the last term o the left had 2 + β 32 td π α c 1 21β 10 + β 20 + β 21 c2 2 π ρ β 31 β10 2 c1 2 c 1 + β 32 α 21 β 10 + β 20 + β 21 c2 2 π π ρ.,. Therefore, the O t 3 term volvg 2 D c 2 α β 21β10 2 D D 2 2 c1 2c2 c 2 π ρ s 11

12 1 2 D 2 c 2 = 1 2 D 2 2 D c 2 D 2 β 31 β10 2 c1 2 c 1 + β 32 α 21 β 10 + β 20 + β 21 c2 2 c+1 π π ρ σ 1 3 D 2 β 31 β10 2 c1 2 c 1 + β 32 α 21 β 10 + β 20 + β 21 c2 2 c+1 π π ρ ad thus the codto s derved: σ + α 32 β 21 β10 2 c1 2c2 1 π ρ 3 β 31 β10 2 c1 2 c 1 + β 32 α 21 β 10 + β 20 + β 21 c2 2 c+1 π π ρ σ c1 + α 32 β 21 β10 2 2c2 π ρ = O t By the order codtos for the coeffcets, 2.13c ad 2.14 are automatcally satsfed. Therefore, we oly cosder the frst two codtos 2.13a ad 2.13b afterwards. We summarze the above results the followg theorem: Theorem 2.1 ecessary codto. The ecessary codto for 2.2 to be thrd-order accurate s c 1 α 31 + α 32 α 21 β 10 + α 32 β 20 + β 21 c2 + β 30 + β 31 + β 32 c+1 π ρ c 1 c 2 c 1 c +1 α 32 β 21 β 10 + β 31 + β 32 α 21 β 10 π ρ π σ σ + β 32 β 20 + β 21 c2 = 1 + O t 3, 2.15a c +1 = 1 ρ σ 2 + O t b Remark 2.1. If restrctg to the case of classcal RK form,.e., α 21 = α 31 = α 32 = 0, the ecessary codtos 2.15 reduce to the codtos gve Theorem 2.6 [12]. 2.2 Suffcet codto I ths part, we try to derve the suffcet codto for 2.2 to be thrd-order accurate. Dfferet from the works [13, 12, 10], the ecessary codto s o loger a suffcet codto. For smplcty, we take π = c. The procedure s splt to three steps: 1. Expad c 1 2. Expad c 2 the frst stage 2.2b; the secod stage 2.2c; 12

13 3. Expad c +1 the fal stage 2.2d. c 1 Step 1: By the frst stage 2.2b, we have the expresso of c 1 = c + tβ 10P D + t2 β t 3 β10 3 p j p jk j,k P k D k c k c j j P p j Dj j c j d Pj D j jk c j To smplfy the dervato, we troduce some otatos: d P D j c d j p Pk D k k c k c by teratos: d P D k c + O t R := j,k p j p jk Q := j P k D k c k S := P Q, P p j Dj j c j c j d Pj D j jk c j d P D j c d j p Pk D k k c k, c d P D k c, wth the vector S = S 1, S 2,, S N T, Q = Q 1, Q 2,, Q N T ad R = R 1, R 2,, R N T. The 2.16 s rewrtte as c 1 = c + tβ 10S + t 2 β 2 10 Q + t 3 β 3 10 R + O t Now we trasfer the ecessary codto 2.15 to a equvalet form. Assume c 2 /ρ ad c +1 /σ have the expaso: c 2 ρ = 1 + X t + T t 2 + O t 3, c +1 σ = 1 + Y t + N t 2 + O t 3. Substtutg ths to the ecessary codto 2.15 gves α 31 + α 32 α 21 β10 2 S + α c 32 β 20 + β 21 X + β 30 + β 31 + β 32 Y = a α 32 β 21 β 10 β 10 S c + X + β 31 + β 32 α 21 β 10 β 10 S c + Y + β 32 β 20 + β 21 Y + X = 0, 2.18b α 31 + α 32 α 21 β10 3 Q + α c 32 β 20 + β 21 T + β 30 + β 31 + β 32 N = c 13

14 From 2.18a ad 2.18b, we solve out wth x, y beg costats satsfyg X = x S c, Y = y S, c α 31 + α 32 α 21 β α 32β 20 + β 21 x + β 30 + β 31 + β 32 y = a α 32 β 21 β 10 β 10 + x + β 31 + β 32 α 21 β 10 β 10 + y + β 32 β 20 + β 21 x + y = b Note that the codtos 2.15 are equvalet to 2.18c-2.19a-2.19b. Gve the coeffcets 2.2, the costats x ad y are uquely determed ad ca be solved out by I the followg, we wll focus o the equvalet codtos 2.18c-2.19a-2.19b. Step 2: We proceed to derve the expaso of c 2 φ k = φc k at c wth k = 1, 2 ad φ = p j, d j : by the secod stage 2.2c. Expad φ k = φc k = φ + φ c ck c ck c T H φ ck c + O t Takg k = ad substtutg 2.17 to 2.20, we have wth φ = p j, d j. φ 1 = φ + tβ 10 φ c S + t2 β 2 10 φ c Q β2 10 t2 S T H φ S + O t3,

15 Now by the secod stage 2.2c, the expaso of c 2 s smplfed as follows: c 2 = α 20 c + α 21c 1 + t j β 20 p j + β 21p 1 c2 j j ρ j t j β 20 d j + β 21d 1 c2 j ρ = c + α 21 β 10 ts + α 21 β 2 10 t 2 Q + α 21 β 3 10 t 3 R + β 20 t j p j1 + X j t + T j t 2 β 20 t j d j1 + X t + T t 2 + β 21 t j p j + β 21 β 10 t 2 j p j c S + β 21β 2 10 t 3 j p j c Q β 21β 2 10 t 3 j S T H p j S + β 21 t 2 j p jx j + β 21 β 10 t 3 j p j c SX j + β 21 t 3 j p jt j β 21 t j d j β 21 β 10 t 2 j d j c S β 21β 2 10 t 3 j d j c Q 1 2 β 21β 2 10 t 3 j S T H d j S β 21 t 2 j d jx β 21 β 10 t 3 j d j c SX β 21 t 3 j d jt + O t 4 = c + α 21 β 10 + β 20 + β 21 ts + β 21 β 10 t 2 S c S + α 21β β 20 + β 21 x t 2 Q + α 21 β 3 10 t 3 R + β 20 + β 21 t 3 j β 21β 2 10 t 3 j Step 3: We expad c +1 p jt j j S T H p j H d j S + β 21 β 10 x t 3 j the fal stage 2.2d ths step. d jt + β 21 β 2 10 t 3 S c Q p j c SS j c j Takg k = ad substtutg 2.22 to 2.20, we have φ 2 = φ + tα 21 β 10 + β 20 + β 21 φ c S + β 21β 10 t 2 k φ c k S k c S j d j c SS + O t 4. c α 21 β β 20 + β 21 x t 2 φ c Q α 21β 10 + β 20 + β 21 2 t 2 S T H φs + O t 3 wth φ = p j, d j. It mmedately follows that β 30 φ + β 31 φ 1 + β 32 φ 2 = β 30 + β 31 + β 32 φ + t[β 31 β 10 + β 32 α 21 β 10 + β 20 + β 21 ] φ c S + β 32β 21 β 10 t 2 k φ c k S k c S + t 2 β 31 β β 32 α 21 β β 32 β 20 + β 21 x φ c Q β 31β β 32 α 21 β 10 + β 20 + β 21 2 t 2 S T H φs + O t 3. 15

16 The the lear combato terms 2.2d are smplfed: α 30 c + α 31c 1 + α 32 c 2 = c + tα 31 β 10 + α 32 α 21 β 10 + β 20 + β 21 S + α 32 β 21 β 10 t 2 S c S ad + α 31 β α 32 α 21 β α 32 β 20 + β 21 x t 2 Q + α 31 + α 32 α 21 β10 t 3 3 R + t 3 α 32 β 20 + β 21 p jt j d jt + α 32 β 21 β10 t 2 3 S c Q j α 32β 21 β10 t 2 3 S T H p j H d j S + α 32 β 21 β 10 x t p 3 j c SS j d j c j j j c SS + O t 4 c t j β 30 p j + β 31p 1 j + β 32 p 2 c+1 j j σ j t j β 30 d j + β 31d 1 j + β 32 d 2 c+1 j σ = β 30 + β 31 + β 32 ts + t 2 β 31 β 10 + β 32 α 21 β 10 + β 20 + β 21 S c S + β 32 β 21 β 10 t 3 S S k c k c S + t3 β 31 β β 32 α 21 β β 32 β 20 + β 21 x S c Q k β 31β β 32 α 21 β 10 + β 20 + β 21 2 t 3 j S T H p j H d j S + β 30 + β 31 + β 32 y t 2 Q + t 3 β 31 β 10 + β 32 α 21 β 10 + β 20 + β 21 y j p j c SS j c j d j c SS c + t 3 β 30 + β 31 + β 32 j p jn j d jn + O t 4. 16

17 Now we have the expaso of c +1 : c +1 = α 30 c + α 31c 1 + α 32 c 2 + t j β 30 p j + β 31p 1 j + β 32 p 2 c+1 j j σ j t j β 30 d j + β 31d 1 j + β 32 d 2 c+1 j σ = c + t[α 31 β 10 + α 32 α 21 β 10 + β 20 + β 21 + β 30 + β 31 + β 32 ]S + α 32 β 21 β 10 + β 31 β 10 + β 32 α 21 β 10 + β 20 + β 21 t 2 S c S + β 32β 21 β 10 t 3 k S c k S k c S + α 31 β α 32α 21 β α 32β 20 + β 21 x + β 30 + β 31 + β 32 y t 2 Q + t 3 α 31 + α 32 α 21 β 3 10 R + t 3 α 32 β 20 + β 21 j p j T j d j T + t 3 β 30 + β 31 + β 32 j p j N j d j N t3 β 31 β β 32α 21 β 10 + β 20 + β α 32 β 21 β 2 10 j S T H p j H d j S + t 3 β 31 β β 32α 21 β β 32β 20 + β 21 x + α 32 β 21 β10 2 S c Q + t 3 α 32 β 21 β 10 x + β 31 β 10 y + β 32 α 21 β 10 + β 20 + β 21 y p j c SS j c j j d j c SS + O t 4. c By the order codtos for the thrd-order explct RK scheme ad the codto 2.18c, the above expaso ca be smplfed as follows: c +1 = c + ts t2 S c S t3 k S c k S k c S t3 j S T H p j H d j S + t 3 β 31 β β 32α 21 β β 32β 20 + β 21 x + α 32 β 21 β10 2 S c Q + t 3 α 32 β 21 β 10 x + β 31 β 10 y + β 32 α 21 β 10 + β 20 + β 21 y p j c SS j c j j d j c SS + O t 4 c = c + ts t2 S c S t3 k S c k S k c S t3 j + t 3 β 31 β β 32α 21 β β 32β 20 + β 21 x + α 32 β 21 β10 2 [ S c Q p ] j c SS j d j c j j c SS + O t 4. c S T H p j H d j S 2.23 I the last equalty, we used the relato α 32 β 21 β 10 x + β 31 β 10 y + β 32 α 21 β 10 + β 20 + β 21 y = [α 32 β 21 β β 31 + β 32 α 21 β β 32β 20 + β 21 x], 17

18 whch s derved from 2.19b. It s observed 2.23 that, the expaso of c +1 cocde wth that of the exact soluto 2.5, besdes the last term. The last addtoal term does ot vash for the geeral case. However, t vashes for ODEs wth oly oe compoet, amely, fc, = J, j = I, fc, = I, j = J, p j c = d j c = 0, otherwse, 0, otherwse, where fc s some fucto of c. It s easly valdated by the followg computato: S c Q p j c SS j d j c j j c SS c = j,k,l p j d j c k p Pl Dl kl c l d Pk D k kl c k j p j c SP j D j c j j 2.24 d j c SP D c = p JI d IJ d IJ c I c I p JI c I d IJ d IJ c I = 0. I the geeral case, to acheve thrd order accuracy, t s requred that 2.25 β 31 β β 32α 21 β β 32β 20 + β 21 x + α 32 β 21 β10 2 = Ths relato s automatcally satsfed by the RK scheme classcal form α 21 = α 31 = α 32 = 0 sce y = 0 ad β 31 β β 32β 20 + β 21 x = 0. For the SSP RK3 scheme wth the optmal coeffcets [4], t s deduced from 2.19 that y = 0 ad x = 1 ad thereby β 31 β β 32α 21 β β 32β 20 + β 21 x + α 32 β 21 β10 2 = 1 6. Thus the scheme wth the coeffcets the optmal SSP RK3 scheme s ot thrd order accurate. Moreover, we pot out that, the valdato 2.25 dcates that the addtoal relato 2.26 caot be obtaed by testg the specal class of ODEs wth polyomals of hgh degree 2.3 ad 2.4. By expressg x terms of α ad β from 2.19 ad substtutg t to 2.26, we fally arrve at 0 =β 32 β 20 + β 21 α 31 + α 32 α 21 α 32 β 21 β [α 32 β 20 + β 21 α 32 β 21 β β 30 + β 31 + β 32 α 32 β 21 β 10 ]β 31 + β 32 α 21 + α 32 β

19 Ths s the addtoal codto for the scheme to be thrd order accurate. The above result s summarzed the followg theorem: Theorem 2.2 suffcet codto. A suffcet codto for the scheme 2.2 to be thrdorder accurate s c 1 α 31 + α 32 α 21 β 10 + α 32 β 20 + β 21 c2 + β 30 + β 31 + β 32 c+1 π ρ c 1 c 2 c 1 c +1 α 32 β 21 β 10 + β 31 + β 32 α 21 β 10 π ρ π σ σ + β 32 β 20 + β 21 c2 = 1 + O t 3, 2.27a c +1 = 1 ρ σ 2 + O t2, β 32 β 20 + β 21 α 31 + α 32 α 21 α 32 β 21 β α 32β 20 + β 21 α 32 β 21 β b + β 30 + β 31 + β 32 α 32 β 21 β 10 β 31 + β 32 α 21 + α 32 β 21 = c 3 MPRK schemes I ths secto, we frst focus o the MPRK scheme wth the coeffcets the SSP RK3 method, whch s thrd-order accurate for ODEs wth oly oe compoet, see Next, we move to the suffcet codtos the geeral case. We wll fd the optmal coeffcets such that the thrd codto 2.27c s fulflled. The, we try to costruct the explct form of ρ ad σ such that the frst two suffcet codtos 2.27a ad 2.27b are satsfed. Lastly, we exted the MPRK scheme to sem-dscrete schemes arsg from PDEs after dscretzato space ad valdate the order of accuracy. 3.1 Optmal RK3 method For the optmal SSP RK3 method wth coeffcets α 10 = β 10 = 1, α 20 = 3 4, α 21 = β 21 = 1 4, β 20 = 0, α 30 = 1 3, α 32 = β 32 = 2 3, α 31 = β 30 = β 31 = 0, 19

20 the frst two codtos 2.27 reduce to 1 6 c 1 c 1 π c 2 ρ c 2 + c1 c +1 π ρ π σ c +1 σ = 1 + O t 3, + c2 c +1 = 3 + O t 2. ρ σ To smplfy the above codto, we gve the followg lemma: Lemma 3.1. Cosder three fuctos y = y t for = 1, 2, 3. Assume that y = 1 + O t, = 1, 2, 3, ad y 1 + y 2 + 4y 3 = 6 + O t 3. The y 1 y 2 + y 2 y 3 + y 3 y 1 = 3 + O t 2 s equvalet to y 1 + y 2 = 2 + O t 2. Proof. y 1 y 2 + y 2 y 3 + y 3 y 1 = 3 + O t 2, y 1 y 2 + y 1 + y 2 y 3 = 3 + O t 2, y 1 y 2 + y 1 + y y 1 y 2 + O t 3 = 3 + O t 2, 6y 1 + y 2 y 1 y = O t 2, 6y 1 + y 2 12 = O t 2. By the lemma, we have the equvalet order codtos c 1 + c2 + 4 c+1 = 6 + O t 3, 3.1a π ρ σ 20

21 c 1 Take π = c ad recall from 2.16 ad 2.22 that c 1 π c 2 c + c2 = 2 + O t b π ρ = c1 c = 1 + t S c = ts c Wth these relatos, we have c c 2 c 2 c Accordg to 3.1b, c2 ρ ad the = ts c c 1 c t2 S c ca be expaded as c 2 ρ = 1 t S c + t 2 Q + O t 3, t2 S c S c t2 S c c 2 ρ = 1 + t 2 Q + T t 2 It follows from above expasos of c1 π σ c = 1 + t S c ad c2 ρ Combg ths wth 3.2 ad 3.4, we take σ c = 4 c2 c + O t 3. c 2 4 c S c + O t T t 2 + O t S c ad 3.1a that t2 Q + T t2 S c c2 c 2 + O t S c c c 4 + O t 3. c 1 c c 2 ρ. 3.5 Its valdty does ot deped o the specfc expresso of ρ. O the other had, from 3.3 we have Thus ρ ca be take as ρ = 1 2 c ρ c c 1 c = ts c + c1 2 c + O t 2. or c c c Wth the coeffcets for the optmal RK3 method ad the expressos of σ ad ρ 3.5 ad 3.6, the scheme 2.2 s thrd-order accurate for ODEs wth oly oe compoet. 21

22 3.2 Optmal coeffcets Next we focus o the thrd codto 2.27c ad try to fd the optmal coeffcets. To ths ed, oe eeds to solve the optmzato problem maxmze m{ α 10 β 10, α 20 β 20, α 21 β 21, α 30 β 30, α 31 β 31, α 32 β 32 }, s.t. α 10, β 10, α 20, β 20, α 21, β 21, α 30, β 30, α 31, β 31, α 32, β 32 0, α 10 = 1, α 20 + α 21 = 1, α 30 + α 31 + α 32 = 1, α 31 β 10 + α 32 α 21 β 10 + β 20 + β 21 + β 30 + β 31 + β 32 = 1, β 10 α 32 β 21 + β 31 + β 32 α 21 β 10 + β 20 + β 21 = 1 2, β 2 10 α 32β 21 + β 31 + β 32 α 21 β 10 + β 20 + β 21 2 = 1 3, β 10β 21 β 32 = 1 6, β 32 β 20 + β 21 α 31 + α 32 α 21 α 32 β 21 β α 32β 20 + β 21 α 32 β 21 β β 30 + β 31 + β 32 α 32 β 21 β 10 β 31 + β 32 α 21 + α 32 β 21 = 0. We use the MATLAB fucto fmco to solve the above optmzato problem. It s easy to verfy that α 10 = 1, β 10 = 2 3, α 20 = 35 36, α 21 = 1 36, β 20 = , β 21 = 3 8, α 30 = , α 31 = , α 32 = 1 9, β 30 = , β 31 = 1 24, β 32 = 2 3, satsfy all the codtos the optmzato problem. We take these as the tal data for terato fmco. The fal result s α 10 = 1, α 20 = E-01, α 21 = E-02, α 31 = E-10, β 10 = E-01, β 21 = E-01, β 31 = E-10, α 30 = E-01, α 32 = E-01, β 20 = E-02, β 30 = E-01, β 32 = E wth m{ α 10 β 10, α 20 β 20, α 21 β 21, α 30 β 30, α 31 β 31, α 32 β 32 } =

23 The above results may be locally optmal sce t s dffcult to solve the global optmzato problem. I the followg, we fx the coeffcets 3.7 ad try to fd the explct form of ρ ad σ such that the suffcet codtos 2.27a ad 2.27b are satsfed. 3.3 Costructo of ρ ad σ Note that the suffcet codtos 2.27a ad 2.27b are equvalet to Also, gve α ad β, the values of x ad y are determed by We splt the procedure to several steps. Frst, we calculate the expaso of ρ ad try to approxmate ρ by approprate combatos of c 1 ad c. Step 1: Note that ρ c = c2 /c = c 2 /ρ Wth the coeffcets 3.7, we have 1 + α 21 β 10 + β 20 + β 21 t S c 1 + x t S + O t c 2 = 1 + α 21 β 10 + β 20 + β 21 x t S c + O t 2. + O t 2 β 10 < α 21 β 10 + β 20 + β 21 x < 2β Thus, we try to take ρ c the form wth 1 ad 2 are gve by ρ c c 1 = 1 c + 2 c c 1 = α 21β 10 β 20 β 21 + x + 2β 10 β 10 > 0, 2 = 1 1 = α 21β 10 + β 20 + β 21 x β 10 β 10 > 0. Remark 3.1. Note that we ca also take other forms of combato, such as ρ c m 2 c1 3 as log as the coeffcet m c 1 ad m 2 are o-egatve c = m 1 1 c + Gve the explct expresso of ρ 3.9, we proceed to hgher order expaso of ρ. Recall from 2.17 that c 1 c P D = 1 + β 10 t + β c 10 2 Q t 2 + O t 3. 23

24 Wth ths, 3.9 ca be expaded to hgher order as ρ c c 1 = 1 c + 2 c1 2 c = 1 + α 21 β 10 + β 20 + β 21 x t S c + α 21 β 10 + β 20 + β 21 xβ 10 Q t 2 Thus, wth the expaso + α 21 β 10 + β 20 + β 21 x β 10 β 10 S c 2 t 2 + O t 3. c 2 ρ = c2 /c ρ /c = 1 + x S t + T c t 2 + O t 3 we have T = β 20 β 10 β 21 β 10 + β 10 + β 20 + β 21 xq + β 21 β 10 S c α 21 β 10 + β 20 + β 21 x β 10 β 10 + xα 21 β 10 + β 20 + β 21 x S c S c O the other had, t follows from 2.18c that N = α 31 + α 32 α 21 β 3 10 β 30 + β 31 + β 32 Q α 32β 20 + β 21 β 30 + β 31 + β 32 T

25 3.12 Step 2: The we have the expaso of σ σ c = c+1 /c c +1 = 1 + /σ S t + 1 S c 2 c S t 2 + O t 3 c 1 + y S t + N c t 2 + O t 3 = y S t N c t 2 y1 y S c 2 t S 2 c S t 2 + O t 3 = y S t + α 31 + α 32 α 21 β10 3 Q c t 2 + α 32β 20 + β 21 T t 2 β 30 + β 31 + β 32 β 30 + β 31 + β 32 2 S y1 y t S S t 2 + O t 3 2 c 3.11 c = y S t c c + α 31 + α 32 α 21 β α 32β 20 + β 21 β 20 β 10 β 21 β 10 + β 10 x + β 20 x + β 21 x Q t 2 β 30 + β 31 + β 32 y1 y + α 32β 20 + β 21 α 21 β 10 + β 20 + β 21 x β 10 β 10 + xα 21 β 10 + β 20 + β 21 x β 30 + β 31 + β 32 2 S t 2 c α 32β 21 β 10 β 20 + β 21 S S t 2 + O t 3. β 30 + β 31 + β 32 c c 3.13 Note that there exsts S c 2 the above equalty. It seems mpossble to geerate approxmatos to ths term. Here, we use c2 ρ to elmate S c c Deote z := y1 y + α 32β 20+β 21 β 30 +β 31 +β 32 α 21 β 10 + β 20 + β 21 x β 10 β 10 + xα 21 β 10 + β 20 + β 21 x α 21 β 10 + β 20 + β 21 x β 10 β 10 + xα 21 β 10 + β 20 + β 21 x y1 y = α 21 β 10 + β 20 + β 21 x β 10 β 10 + xα 21 β 10 + β 20 + β 21 x + α 32β 20 + β 21 > 0 β 30 + β 31 + β ad take a c = σ c z c ρ Sce z s postve, σ determed by 3.15 s postve f so s a. Such a expresso of a 25

26 wll be costructed as follows. Before that, we use 3.15 to gve the expaso of a : where a c = 1 z + 1 y zx S t c + M β 20 β 10 β 21 β 10 + β 10 + β 20 + β 21 xz Q t α 32β 21 β 10 β 20 + β 21 S S β 21 β 10 z t 2 + O t 3, β 30 + β 31 + β 32 c M = α 31 + α 32 α 21 β α 32 β 20 + β 21 β 20 β 10 β 21 β 10 + β 10 x + β 20 x + β 21 x β 30 + β 31 + β 32. For the sake of coveece, we rewrte the above expaso as a c S = r 1 + r 2 t + r c 3 Q t 2 S + r 4 c where the costats r, = 1, 2, 3, 4 are gve by r 1 = 1 z, r 2 = 1 y xz, r 3 = M [ β 20 β 10 β 21 β 10 + β 10 + β 20 + β 21 x] z, r 4 = α 32β 21 β 10 β 20 + β 21 β 30 + β 31 + β 32 β 21 β 10 z. c S t 2 + O t Step 3: Now we follow the dea [12] ad sert a addtoal stage to costruct a. c Set ad costruct aother stage: a = η 1 c + η 2 c 1 + t j µ = c c1 s. c η 3 p j + η 4 p 1 j a j η 3 d j + η 4 d 1 j µ a j µ 3.17 wth coeffcets η k > 0, k = 1, 2, 3, 4 to be determed. It follows from the above equato that ad thus a c a a µ = η 1 + η 2 + O t = η 1 + η 2 + [η 2 β 10 + η 1 + η 2 η 3 + η 4 ] S t + O t 2, c = η 1 + η 2 + [η 2 β 10 + η 1 + η 2 η 3 + η 4 sβ 10 ] S t + O t 2. µ c 26

27 Wth ths ad 2.21, we proceed to hgher order expaso: a c = η 1 + η 2 + [η 2 β 10 + η 1 + η 2 η 3 + η 4 ] S t c + {η 2 β [η 2β 10 + η 1 + η 2 η 3 + η 4 sβ 10 ]η 3 + η 4 }Q t 2 + η 4 β 10 η 1 + η 2 S c S t 2 + O t 3. c 3.18 Comparg 3.18 ad 3.16 gves r 1 = η 1 + η 2, r 2 = η 2 β 10 + η 1 + η 2 η 3 + η 4 = η 2 β 10 + r 1 η 3 + η 4, r 3 = η 2 β [η 2β 10 + η 1 + η 2 η 3 + η 4 sβ 10 ]η 3 + η 4, r 4 = η 4 β 10 η 1 + η 2 = η 4 β 10 r 1, from whch η 1, η 3, η 4 ad s ca be expressed terms of η 2 : η 1 = r 1 η 2, η 3 = r 2 η 2 β 10 r 1 r 4 β 10 r 1, η 4 = r 4 β 10 r 1, s = r 2 β 10 r 1 r 3 η 2 β 2 10 β 10 r 2 β 10 η To esure the postvty of η k, t s requred that 0 < η 2 < m{r 1, r 2 r 4 }, 3.20 β 10 β10 2 I summary, the fal scheme for the geeral ODEs 1.2 s c 0 = c c 1 = α 10 c 0 + t j ρ = 1 c 1 c 2 = α 20 c 0 + t j c1 + 2 c 2, c + α 21 c 1 β 10 p 0 c 1 j j c 0 j β 20 p 0 j + β 21 p 1 d 0 j c2 j j c 1 c a, 3.21b β 20 d 0 j + β 21 d 1 ρ j c2 j ρ, 3.21c 27

28 c µ = c c1 s, a = η 1 c + η 2 c 1 + t j η 3 p j + η 4 p 1 j a j η 3 d j + η 4 d 1 j µ a, 3.21d j µ σ = a + Zc c 2 ρ, c +1 = α 30 c 0 + α 31 c 1 + α 32 c 2 c+1 β 30 p 0 j + β 31 p 1 j + β 32 p 2 j β 30 d jσ 0 j + β 31 d 1 j + β 32 d 2 j + t j c+1 j σ, 3.21e where the coeffcets α j ad β j are gve by 3.7, 1, 2 are gve by 3.10, η 1, η 3, η 4, s are gve by 3.19, ad z s gve by The parameter η 2 s a free parameter satsfyg the costrat I partcular, we take η 2 = 1/3 satsfyg the codto 3.20 ad the other parameters are gve by 1 = E-01, 2 = E-01, Z = E-01, η 1 = E-02, η 2 = 1/3, η 3 = E-01, η 4 = , s = Ths scheme s of thrd-order accuracy ad ucodtoally postvty-preservg for Exteso to sem-dscrete scheme Followg our prevous work [10], we formulate a system of ODEs the followg form, to cover the sem-dscrete scheme for the PDEs: dc k, dt = F k, c + P k, c D k, c, k = 1,, M, = 1,, N Here c k, = c k, t deotes the cocetrato of the -th speces at the k-th grd pot, N ad M deote the umber of speces ad odes, respectvely. The vector c s the collecto of all ukow varables wth the form c := c 11, c 12,, c 1N,, c M1, c M2,, c MN T ad s of legth M N. F k, = F k, c deotes the cotrbutos of the covecto terms after spatal 28

29 dscretzatos the PDEs. The producto ad destructo terms are P k, = P k, c = N j=1 p k,,jc ad D k, = D k, c = N j=1 d k,,jc whch satsfy p k,,j c = d k,j, c,, j, k ad c 0. We make the followg assumpto o 3.23: Assumpto 3.1. The Euler forward method for the covecto term satsfes the postvtypreservg property: f c k, 0 for all k,, the c k, + tf k,c 0, for all k, ad t t 0. Remark 3.2. Ths famly of ODEs 3.23 covers the sem-dscrete fte dfferece WENO scheme for chemcal reactg flows 1.1 by settg the producto ad destructo terms to be zero the mometum ad eergy equatos. The varable c k, does ot ecessarly deote the cocetrato or desty ths case. We geeralze the MPRK scheme 3.21 by corporatg the covecto term F k, usg the Euler forward each stage ad arrve at c 0 k, = c k, c 1 k, = α 10c 0 k, + tβ 10F k, c 0 + t j β 10 p 0 c 1 k,j k,j c k,j d 0 k,j 3.24a c 1 k,, 3.24b c k, c1 ρ k, = 1 c 1 k, + 2c k, k, 2, 3.24c c k, c 2 k, = α 20c 0 k, + α 21c 1 k, + tβ 20F k, c 0 + tβ 21 F k, c 1 + t j β 20 p 0 k,j + β 21p 1 c2 k,j k,j β 20 d 0 k,j ρ + β 21d 1 k,j c2 k, k,j ρ k,, 3.24d µ k, = c k, c1 k, s, c k, a k, = η 1 c k, + η 2c 1 k, + tη 3η 1 + η 2 F k, c 0 + tη 4 η 1 + η 2 F k, c e 29

30 + t j η 3 p k,j + η 4p 1 k,j a k,j µ k,j t j η 3 d k,j + η 4d 1 k,j a k, µ k,, 3.24f σ k, = a k, + Zc 0 c 2 k, k,, ρ k, c +1 k, = α 30 c 0 k, + α 31c 1 k, + α 32c 2 k, + tβ 30F k, c 0 + tβ 31 F k, c 1 + tβ 32 F k, c 2 + t j β 30 p 0 k,j + β 31p 1 k,j + β 32p 2 c+1 k,j k,j β 30 d 0 k,j σ + β 31d 1 k,j + β 32d 2 k,j c+1 k, k,j σ k, 3.24g. 3.24h Except for the stage 3.24f, the other steps are oly drect extesos of those The thrd-order accuracy of ths scheme s show the appedx. We summarze the result the followg theorem. Theorem 3.1. The MPRK scheme 3.24 s thrd-order accurate for the system of ODEs It s coservatve: k, k, k, c 1 k, = k, c 2 k, = α 20 c +1 k, c k, + tβ 10 F k, c, k, c k, + α 21 k, k, = α 30 c k, + α 31 k, k, c 1 k, + t β 20 F k, c + β 21 F k, c 1, k, c 1 k, + α 32 k, c 2 k, + t β 30 F k, c + β 31 F k, c 1 + β 32 F k, c 2, k, where the summato of the covecto terms wll vash wth perodc boudary codtos. It s postvty-preservg: f c k, 0 for all k,, the c1 k, k, ad t m{ α 10 β 10, α 20 β 20, α 21 β 21, α 30 β 30, α 31 β 31, α 32 β 32 } t 0. 0, c2 k, 0 ad c+1 k, 0 for all 4 Numercal tests I ths part, the umercal results wll be preseted. We frst dscuss the covergece order for the o-stff problems ad the performace o the stff problems solvg the ODEs. The we move to reactve Euler equatos ad Euler equatos wth three speces reactos ad geeral equato of state. The ffth-order fte-dfferece WENO spatal dscretzato [11] s appled here. For the mplemetato detals of our tme tegrato ad the postvtypreservg WENO schemes ad the lmter, please refer to our prevous work [10]. 30

31 4.1 ODEs Example 4.1 lear case. The lear test case s wth costat a > 0, ad tal value dc 1 dt = c 2 ac 1, dc 2 dt = ac 1 c 2, c 1 0 = c 0 1, c 20 = c 0 2. The exact soluto s c 1 t = 1 + b exp a + 1tc 1, c 2t = c c0 2 c 1t, wth the parameters c 1 ad b determed by c 1 = c0 1 + c0 2 a + 1, b = c0 1 c 1 1. I the umercal expermet, we take c 0 1 = 4.5, c 0 2 = 3.2, a = 2.7 ad the fal tme t = 1. The errors betwee umercal solutos ad exact solutos at the fal tme are lsted Table 4.1. The thrd-order accuracy s clearly observed. Table 4.1: Example 4.1: Error table for lear ODEs. t error order 1/ e-04-1/ e / e / e / e Example 4.2 olear case. To test the accuracy of our solver 3.21, we make up a o-stff olear problem: dc 1 dt = F 1c c 1c 2 c 1 + 1, 4.1a 31

32 dc 2 dt = F 2c + c 1c 2 c ac 2, dc 3 dt = F 3c + ac b 4.1c where F 1 c, F 2 c, F 3 c deotes covecto terms. To express ths system of ODEs the form of producto-destructo equatos, we set p 21 c = d 21 c = c 1c 2 c 1 + 1, p 32c = d 32 c = ac 2, ad p j = d j = 0 for other sets of, j. I the umercal examples, the tal codtos are set as c 0 1 = 9.98, c 2 = 0.01 ad c 3 = The covecto terms are F 1 c, F 2 c, F 3 c = c 1 c 2 c 3, c 3 c 2, c 1 c 2 c 2 3. The fal tme s t = 1 ad the parameter a = 1. The errors are lsted Table 4.2, whch shows thrd-order accuracy. Table 4.2: Example 4.2: Error table for olear ODEs 4.1. t error order 1/ e-04-1/ e / e / e / e To test our solver 2.2 wth optmal RK3 coeffcets whch s thrd-order accuracy for ODEs wth oly oe compoet, we take the ODEs to be dc 1 dt = c 1c 2 c 1 + 1, dc 2 dt = c 1c 2 c 1 + 1, dc 3 dt = a 4.2b 4.2c The results are lsted Table 4.3, where the thrd-order accuracy s observed. Moreover, the errors wth optmal SSP RK3 coeffcets are much smaller tha those wth optmal coeffcets

33 Example 4.3 stff case. Oe of the most promet examples of the stff ODEs s the Robertso test case, whch descrbes the chemcal reactos [5]: wth tal values c 0 = c 0 for = 1, 2, 3. dc 1 dt = 104 c 2 c c 1, dc 2 dt = c c 2 c c 2 2, dc 3 dt = c 2 2, I ths problem, the producto ad destructo terms are p 12 c = d 21 c = 10 4 c 2 c 3, p 21 c = d 21 c = c 1, p 32 c = d 23 c = c 2 2, ad p j = d j = 0 for other sets of, j. I the umercal smulatos, we take c 0 1 = 1 ad c 0 2 = c 0 3 = 0. Followg [13], the tme step sze the k-th tme step s chose as t k = 2 k 1 t 1 wth the tal tme step sze t 1 = The small tal tme step was set to obta a adequate resoluto of the compoet c 2 the startg tme terval. To vsualze the evoluto of c 2, t s multpled by 10 4 Fg From Fg. 4.1, we observe the excellet accuracy of our scheme the case of a hghly stff problem. 4.2 Reactve Euler equatos We cosder the reactve Euler equatos whch are ofte used to model the detoato waves [16] 2D case: U t + FU x + GU y = SU, 4.3 Table 4.3: Example 4.2: Error table for olear ODEs wth oly oe compoet 4.2. The SSP RK3 coeffcets vs. coeffcets 3.7. t error SSP RK3 order error optmal order 1/ e e-08-1/ e e / e e / e e / e e

34 c 1 exa c 1 um 10 4 c 2 exa 10 4 c 2 um c 3 exa c 3 um Fgure 4.1: Example 4.3: Tme evoluto of c, = 1, 2, 3. wth U = ρ, m,, E, ρy, FU = m, ρu 2 + p, ρuv, E + pu, ρuy, GU =, ρuv, ρv 2 + p, E + pu, ρvy, SU = 0, 0, 0, 0, ω, ad m = ρu, = ρv, E = 1 2 ρu2 + v 2 + p + ρqy. 4.4 γ 1 Here q > 0 s the heat release of reacto, γ s the specfc heat rato ad 0 Y 1 deotes the reactat mass fracto. The source term s assumed to be a Arrheus form ω = KρY e T/T, 4.5 where T = p/ρ s the temperature, T > 0 s the actvato costat temperature ad K > 0 s a costat. To ft to our framework, we rewrte 4.3 a equvalet form the same approach [10]. Here, we omt the detals for savg space ad preset the umercal results the followg part. 34

35 y 1.00 ρ x x y p x x a b Fgure 4.2: Example 4.4: Numercal covergece study. Left: colored cotour map of desty ad pressure of our scheme; rght: cut alog y = 0. Up: desty; dow: pressure. The symbols ad the sold les represet the results of our scheme ad the SSP RK3 method, respectvely. Example 4.4 Numercal covergece study. We study the umercal covergece of our scheme ths test. The same parameters ad tal ad boudary codtos wth Example 4.4 [16] are appled here. The tal codto s, f x 2 + y , the ρ, u, v, p, Y = 1, 0, 0, 80, 0; otherwse, ρ, u, v, p, Y = 1, 0, 0, 10 9, 1. The boudary codtos for the bottom ad the left are reflectve. The termal tme s t = 0.2. The mesh s uformly rectagular. The umercal results wth our tme tegrato ad the SSP RK3 method are show Fg The mesh szes are x = y = 1/120. We observe the good agreemet. Example 4.5 Detoato dffracto problems. We test the detoato dffracto ths example. The same parameters ad tal codtos wth [16] are appled here. 35

36 y 6 y x x y 25 y x x a b Fgure 4.3: Example 4.5: Detoato dffracto problems. Top: colored cotour map ad cotour le of desty; bottom: colored cotour map ad cotour le of pressure. The tal codtos are, f x < 0.5, the ρ, u, v, E, Y = 11, 6.18, 0, 970, 1; otherwse, ρ, u, v, E, Y = 1, 0, 0, 55, 1. The boudary codtos are reflectve except that at x = 0, ρ, u, v, E, Y = 11, 6.18, 0, 970, 1. The termal tme s t = 0.6. The parameters are γ = 1.2, q = 50, T = 50 ad K = The umercal results wth x = y = 1/48 are show Fgure 4.3, whch are comparable to the results [16]. 4.3 Euler equatos wth three speces reactos ad geeral equato of state We cosder the three speces model wth a more geeral equato of state [17] U = ρ 1, ρ 2, ρ 3, ρu, E T, 36

37 FU = ρ 1 u, ρ 2 u, ρ 3 u, ρu 2 + p, E + pu T, SU = 2M 1 ω, M 2 ω, 0, 0, 0. ad ρ = 3 ρ s, s=1 p = RT 3 s=1 ρ s M s, E = 3 ρ s e s T + ρ 1 h ρu the teral eergy e s T = 3RT/2M s ad 5RT/2M s for mooatomc ad datomc speces s=1 respectvely. The rate of chemcal reacto s gve by ω = k f T ρ 2 k b T ρ ρ s 4.7 M 2 M 1 M s k f = CT 2 e E/T, k b = k f / expb 1 + b 2 log z + b 3 z + b 4 z 2 + b 5 z 3, z = 10000/T 4.8 where b, C ad E are costats whch ca be foud [21]. s=1 Example 4.6. We take the same parameters as [17, 22]. The parameters are M 1 = 0.016, M 2 = 0.032, M 3 = 0.028, h 0 1 = , R = , C 0 = m 3, E 0 = K, ad b 1 = 2.855, b 2 = 0.988, b 3 = 6.181, b 4 = 0.023, b 5 = The tal codtos are: the destes ρ 1, ρ 2 ad ρ 3 are , , o the left, ad , , o the rght. The veloctes are zero. The pressures are 1000 o the left ad 1 o the rght. The fal tme s t = ad the mesh sze s x = 1/4000. The profles of desty, velocty ad pressure are preseted Fgure. 4.4, where good agreemet of our scheme ad the SSP RK3 method s observed. 5 Cocludg remarks I ths paper, a thrd-order tme tegrato method for the producto-destructo equatos, whch s coservatve ad ucodtoally postvty-preservg, s costructed. The ecessary ad suffcet codtos for the methods to be thrd-order accurate are derved. Ths ODE solver s the exteded to cover a class of sem-dscrete schemes for PDEs ad 37

38 ρ1 um ρ1 exa ρ2 um ρ2 exa x ρ3 um ρ3 exa x x u um u exa x p um p exa x Fgure 4.4: Example 4.6: Three speces reacto problem at t = Fte dfferece. The sold les are the referece solutos geerated wth SSP-RK3 whle the symbols are the umercal solutos wth our scheme. 38

39 successfully appled to fte dfferece WENO schemes for o-equlbrum flows. A varety of umercal examples are coducted to valdate the performace of the tme tegrato method coupled wth the ffth-order WENO scheme ad the postvty-preservg lmter. Ackowledgemets We would lke to thak Xagxog Zhag from Purdue Uversty for may frutful dscussos. A Proof of accuracy for the MPRK scheme 3.24 I ths appedx, we show that the MPRK scheme 3.24 s thrd-order accurate for the system of ODEs Accordg to 3.24a ad 3.24b, c 1 k, ca be expaded as c 1 k, = c k, + tβ 10S k, + t 2 β 2 10 Q k, + t 3 β 3 10 R k, + O t 4, A.1 wth S k, := F k, + P k, D k,, Q k, := j R k, := j,r F p k,j + P k,j D k,j k,j c k,j j d k,j p k,j p Fk,r +P k,r D k,r k,jr d c k,jr k,r j,r c k,j Fk, + P k, D k,, c k, Fk,j +P k,j D k,j c k,j d k,j p Fk,r +P k,r D k,r k,r d c k,r k,r c k, F k, +P k, D k, c k,. Thus ρ k, c k, = 1 c 1 k, c k, + 2 c1 k, 2 c k, = 1 + α 21 β 10 + β 20 + β 21 x t S k, + α c 21 β 10 + β 20 + β 21 xβ 10 Q k, t 2 k, 2 S k, + α 21 β 10 + β 20 + β 21 x β 10 β 10 t 2 + O t 3. c k, A.2 39

40 Recall from 2.20 that φ l wth l = 1, 2 ad φ = p k,j, d k,j, F k, ca be expaded as φ l = φc l = φ + φ c cl c cl c T H φc l c + O t 3. A.3 Takg l = 1 ad substtutg A.1 to the above equato, we have φ 1 = φ + tβ 10 φ c S + t2 β 2 10 φ c Q β2 10 t2 S T H φ S + O t3 A.4 wth φ = p k,j, d k,j, F k,. It follows from ths, 3.24d ad A.2 that ad c 2 k, =α 20c 0 k, + α 21c 1 k, + tβ 20F k, c 0 + tβ 21 F k, c 1 + t j β 20 p 0 k,j + β 21p 1 c2 k,j k,j =c k, + α 21 β 10 + β 20 + β 21 S k, t + O t 2 The above two equatos further yeld c 2 k, ρ k, = c k, + xs k, t + O t 2. β 20 d 0 k,j ρ + β 21d 1 k,j c2 k, k,j ρ k, c 2 k, = c k,+α 21 β 10 +β 20 +β 21 ts k, +β 21 β 10 t 2 S k, c S+ t2 [α 21 β 2 10+β 20 +β 21 x]q k, +O t 3 ad wth c 2 k, ρ k, = c k, + xs k, t + T k, t 2 + O t 3 T k, = [ β 20 β 10 β 21 β 10 + β 10 + β 20 + β 21 x] Q k, + β 21 β 10 S k, c [α 21 β 10 + β 20 + β 21 x β 10 β 10 + xα 21 β 10 + β 20 + β 21 x] S c k, S k, c k, 2. A.5 Furthermore, we deote X k, := x S k, c k, 40

41 ad have the expaso of c 2 k, : c 2 k, =α 20c 0 k, + α 21c 1 k, + tβ 20F k, c 0 + tβ 21 F k, c 1 + t j β 20 p 0 k,j + β 21p 1 c2 k,j k,j β 20 d 0 k,j ρ + β 21d 1 k,j =c k, + tα 21β 10 S k, + t 2 α 21 β 2 10 Q k, + t 3 α 21 β 3 10 R k, c2 k, k,j ρ k, + tβ 20 + β 21 Fk, + F t2 k, β 10 β 21 c S + t3 β10 2 β Fk, 21 c Q β2 10 β 21 t 3 S T H F k, S + β 20 t j p k,j 1 + X k,j t + T k,j t 2 β 20 t j d k,j 1 + X k, t + T k, t 2 + β 21 t j p k,j + β 21 β 10 t 2 j p k,j c S + β 21β 2 10 t 3 j p k,j c Q β 21β 2 10 t 3 j S T H p k,j S + β 21 t 2 j p k,jx k,j + β 21 β 10 t 3 j p k,j c SX k,j + β 21 t 3 j p k,jt k,j β 21 t j d k,j + β 21β 10 t 2 j d k,j c S β 21β 2 10 t3 j d k,j c Q 1 2 β 21β 2 10 t3 j S T H d k,j S β 21 t 2 j d k,j X k, + β 21 β 10 t 3 j d k,j c SX k, + β 21 t 3 j d k,j T k, + O t 4 =c k, + α 21β 10 + β 20 + β 21 ts k, + β 21 β 10 t 2 S k, c S + [α 21β β 20 + β 21 x] t 2 Q k, + t 3 α 21 β 3 10 R k, + t 3 β 20 + β 21 j p k,j T k,j j d k,j T k, + β 21 β 2 10 t3 S k, c Q β 21β10 2 t3 S T H Fk, + H p k,j H d k,j S j j + β 21 β 10 x t p 3 k,j c SS k,j d k,j c j k,j c SS k, + O t 4. c k, A.6 Next we expad a k, accordg to 3.24e ad 3.24f. Note that µ k, = c k, c1 k, s = c c k, + sβ 10 S k, t + O t 2 k, 41

42 ad thereby a k, =η 1 c k, + η 2 c 1 k, + tη 3η 1 + η 2 F k, c 0 + tη 4 η 1 + η 2 F k, c 1 + t j η 3 p k,j + η 4p 1 k,j a k,j µ k,j t j η 3 d k,j + η 4d 1 k,j a k, µ k,, a k, =η 1 + η 2 c k, + tη 2 β 10 S k, + tη 1 + η 2 η 3 + η 4 F k, + η 1 + η 2 η 3 + η 4 P k, D k, t + O t 2 =η 1 + η 2 c k, + [η 2β 10 + η 1 + η 2 η 3 + η 4 ]S k, t + O t 2, =η 1 + η 2 + [η 2 β 10 + η 1 + η 2 η 3 + η 4 sβ 10 ] S k, t + O t 2. µ k, Wth these, we further expad a k, c k, as c,k a k, c k, =η 1 + η 2 + [η 2 β 10 + η 1 + η 2 η 3 + η 4 ] S k, t c,k + {η 2 β [η 2β 10 + η 1 + η 2 η 3 + η 4 sβ 10 ]η 3 + η 4 }Q k, t 2 + η 4 β 10 η 1 + η 2 S k, c S c k, t 2 + O t 3 S k, =r 1 + r 2 t + r c 3 Q k, t 2 S k, + r 4,k c S c k, t 2 + O t 3, A.7 where r, = 1, 2, 3, 4 are defed below It follows from ths ad 3.13 that σ k, =a k, + zc 0 c 2 k, k, ρ k, =1 + 1 ys k, t + α 31 + α 32 α 21 β10 3 Q k, t 2 + α 32β 20 + β 21 T k, t 2 β 30 + β 31 + β 32 β 30 + β 31 + β 32 2 y1 yc S k, k, t S k, c k, 2 c S t2 + O t 3. A.8 Moreover, we substtute expaso A.6 to A.3 wth l = 2 ad obta φ 2 =φ + tα 21 β 10 + β 20 + β 21 φ c S + β 21β 10 t 2 m,r φ c m,r S m,r c S + t 2 [α 21 β β 20 + β 21 x] φ c Q α 21β 10 + β 20 + β 21 2 t 2 S T H φ S + O t3 A.9 42