Chp 3: Scalar Advection and Linear Hyperbolic Systems. By Prof. Dinshaw S. Balsara

Transcription

1 Chp 3: Scalar Advecton and Lnear Hyperbolc Systems By Prof. Dnshaw S. Balsara

2 3.) Introducton We have seen the need for consstency and stablty n FDAs of PDEs. However, for the advecton equaton, whch s lnear, the approach faled badly for square pulses. We study why n ths chapter. Our analyss s based on a pctoral approach for advecton. Concept of total varaton dmnshng (TVD) schemes that s later formalzed. Need two nsghts for treatng systems: ) How to treat lnear hyperbolc systems w/o oscllatons (ths chapter). 2) How to deal wth non-lneartes n the hyperbolc system (next 3 chapters). Wll also dscuss the Remann problem for lnear hyperbolc systems. These topcs are mportant buldng blocks that wll be used over and over agan later n scheme desgn. Also address boundary condtons. 2

3 3.2) Qualtatve Introducton to Non-Lnear Hybrdzaton for Scalar Advecton The Dlemma: Godunov s theorem : There are no lnear, second order schemes for treatng lnear advecton whch would always reman postvty preservng. Free of oscllatons monotncty preservng. The Way Out: van Leer desgned nherently non-lnear schemes for treatng the lnear advecton problem, thereby escapng the clutches of Godunov s theorem! We follow Godunov s orgnal dea of lterally movng slabs of flud n order to show the evoluton of flud from tme t n to t n+ = t n + t. We focus on what happens to the dscontnuty n the th zone when we do that. Consder x = and t = 0.4,.e. CFL of 0.4. The blue represents the flud (.e. flux) that flows nto zone. The pnk represents the flux flowng out of the same zone. 3

4 a b u n 2 u n u n + 2 u n + a t u n + u n u n + u n + 2 u n + + u n Questons: Can we develop a noton of monotoncty preservng advecton? Can we vsually show that ths advecton s monotoncty preservng? 4

5 a b u n 2 u n u n + 2 u n + a t u n + u n u n + u n + 2 u n + + u n Tme average of left flux : t f = (a t) u n+ /2 n /2 Tme average of rght flux : t f = (a t) u n+ /2 + /2 Conservaton law n ntegral form : x u = x u + t f t f n n+ n n+ /2 n+ /2 /2 + /2 ( ) ( ) n+ n n n n n u = u µ u u = µ u + µ u Queston: What can you say about postvty for the above scheme? 5

6 Tme average of left flux : t f = (a t) u n+ /2 n /2 Tme average of rght flux : t f = (a t) u n+ /2 + /2 n Conservaton law n ntegral form : + x u = x u + t f t f u = u µ ( u u ) = ( µ ) u + µ u n+ n n+ /2 n+ /2 /2 + /2 n n n n n n 6

7 We see by sheer constructon that the frst order upwnd scheme wll produce no new extrema because each mesh pont at t n+ s a convex combnaton of two neghborng mesh ponts at tme t n. A scheme wth such a property of not generatng any new extrema n the soluton that were not present ntally s called a monotoncty preservng scheme. We wsh to explore monotoncty preservng schemes whch are second order accurate extensons of the frst order upwnd scheme. To get to second order accuracy, endow slabs wth slopes. 3 choces: ) Left-based 2) Central dfference 3) Rght-based Formally, ths s called reconstructon. We are carryng out pecewse lnear reconstructon. Reconstructon s an mportant buldng block n scheme desgn. We have: n n n u n u ( x) = u + ( x x) u s called a " slope" or undvded dfference 7 x

8 a b u n 2 u n u n + 2 u n + a t u n + u n u n u n u n + + u n Questons: Can we vsually show that ths second order advecton s not monotoncty preservng? 8

9 a b u n 2 u n u n + 2 u n + a t u n + u n u n u n u n + + u n n+ /2 n Tme average of left flux : t f /2 = (a t) u + ( µ ) u 2 n+ /2 n Tme average of rght flux : t f + /2 = (a t) u + ( µ ) u 2 n+ n n+ /2 n+ /2 Conservaton law n ntegral form : x u = x u + t f /2 t f + /2 n n µ n n ( ) ( ) = 2 Queston: What can you say about postvy t for the above scheme? n+ n n n u u µ u u µ u u 9

10 a b u b a t u n 2 u n u n + 2 u n + u c A = a t u b u 2 ( ) ( + ) c a t u n + u n u n u n u n + + u n

11 n+ /2 n Tme average of left flux : t f /2 = (a t) u + ( µ ) u 2 n+ /2 n Tme average of rght flux : t f + /2 = (a t) u + ( µ ) u 2 n+ n n+ /2 n+ /2 Conservaton law n ntegral form : x u = x u + t f /2 t f + /2 µ = 2 n n ( n n ) n+ n n n u u µ u u µ u u

12 We see that we have generated a new extremum. Rght slopes yeld Lax-Wendroff scheme; central slopes yeld Fromm scheme; left slopes yeld Beam-Warmng scheme. All choces are ll! The only soluton les n restrctng,.e. lmtng, the pecewse lnear profle wthn each zone so that t does not produce any new extrema n the advected profle that were not ntally present n the orgnal slabs of flud. Such lmters are called slope lmters. A smple example the MnMod lmter (.e. choose between the left and rght slopes based on whch one s smaller; 0 f opposte slopes) : n n n n n n n n ( ( + ) ( ) ) ( + ) n u = sgn u u + sgn u u mn u u, u u 2 There are several other/better lmters n the text. All lmters are nonlnear. Thus success on lnear advecton came at the cost of non-lnear hybrdzaton. 2

13 a b u n 2 u n u n + 2 u n + a t u n + u n u n u n u n + + u n Queston: Based on the mnmod lmter, justfy the slopes n the profle to the left. Wth lmter, the spurous extrema are gone! Ths does come at a prce: The order of the method s locally reduced by a lmter n those zones where t s actvated. The lmter acheves ts salutary effect by provdng strong dsspaton (and order reducton) where t s needed to prevent dspersve rpples that would otherwse form n a second order scheme. 3

14 a b Notce that the square pulse s oscllaton free. However, the top of the Gaussan has been clpped! Even the Lax-Wendroff scheme dd better than the non-lnearly hybrdzed schemes. The reason : n n u u left slope Lmter only examnes the rato : θ = = n n u u rght slope + physcal extremum and a spurous oscllaton. When θ, the lmter reduces to central dfference, whch s good and stable. When θ or θ, slopes are truncated, whether that s good or not. Can't tell the dfference between a 4

15 a b All lmters clp extrema. The better ones, lke the MC lmter shown above, fare better. However, even the MC lmter clps the top of the Gaussan. Only recourse les n schemes lke WENO or PPM wth larger stencls. We call the MC lmter a compressve lmter because t allows the scheme to use larger slopes. As a result, t yelds a sharper profle. 5

16 Comparng reconstucton by MnMod lmter to MC lmter. Latter produces steeper slopes sharper profles. Questons: Can you justfy the slopes to the left and rght based on the form of the lmters? Dsspaton α jumps. Whch lmter dsspates less? n n n n n n n n n n n n n u = mnmod ( u+ u, u u ) = sgn ( u + u ) + sgn ( u u ) mn u + u, u u 2 u = MC ( u u, u u ) = ( sgn ( u u ) + sgn ( u u )) mn u u, 2 u u, 2 u u 2 2 n n n n n n n n n n n n n n n

17 3.3) The Total Varaton Dmnshng Property and Understandng the Lmters Monotoncty preservng property s hard to formulate mathematcally; total varaton dmnshng (TVD) s much easer to formulate. A TVD scheme s monotoncty preservng Harten. Helps wth postvty. n n n + = n+ n TV( u ) { + } n n n n n For a mesh functon u..., u, u, u,... at tme t we defne the total varaton as: TV u = u u We say that the scheme s total varaton dmnshng (TVD) f we have: TV u Check out the graph to see that the scheme s not TVD. If t were TVD, could the wggles arse? Queston: Can you evaluate TV for the plot to the rght? 7

18 Harten realzed that many update schemes can be wrtten as: C n+ n n n n n u = u C, /2 u u + +, + /2 u+ u If the scheme s non - lnearly stablzed, C and C may have strongly non-lnear dependence on the mesh functon., /2 +, + /2 Harten ' s Theorem : When C 0, C 0 and C + C for all zones "", the scheme s TVD., /2 +, + /2, /2 +, + /2 Proof: By shftng ndces n the above equaton we get u n+ + n = ( n n) ( n n u ) + C, + /2 u+ u + C+, + 3/2 u+ 2 u+ subtractng one from the other we wrte: u u = C u u + C C u u n+ n+ n n n n +, /2, + /2 +, + /2 + ( n n ) + C u u +, + 3/

19 Now apply the Schwartz nequalty (.e ( C C ) n n n n n +, + /2 +, + /2 + = = n n, /2 ) to get: TV u u u u u + C u u = a+ b a + b Now shftng the summaton ndces n the last two sums gves: ( C C ) + C u u = TV u u u u u n n n n n +, + /2 +, + /2 + = = n n +, + 3/ n n n n, + /2 + C+, + /2 + = + C u u + u u = = u u TV u = n n n + Ths completes our proof! The TVD property s a very mportant buldng block for numercal schemes for hyperbolc systems. It ensures a postvty & monotoncty preservng propertes n the resultng scheme. Postvty for systems s much harder. 9

20 Studyng our schemes n the context of Harten s theorem: The scheme wll be TVD f the fluxes have a specal form. The frst order upwnd scheme s monotone but has dffusve fluxes. The fluxes for the second order schemes have extra terms. We call these extra underlned terms the antdffusve fluxes. (Queston:Why ant?) They reduce dffuson n the 2 nd order scheme. But reduce dsspaton carefully, or else oscllatons appear. The ncluson of the antdffusve terms n red (underlned), turn the donor cell fluxes nto the Lax-Wendroff fluxes Frst order fluxes (Donor cell): f = a u ; n+ /2 n /2 f = a u n+ /2 + /2 Second order fluxes (Lax-Wendroff): f f n = a u + 2 n µ u u + /2 /2 = a u n u n + + µ + u 2 n+ /2 n /2 n ( n ) ( n) 20 ;

21 The prevous second order Lax-Wendroff fluxes, wth ther ant-dffusve parts, are not TVD! Snce the Lmter only depends on the rato θ ( n n µ φ θ ) + u u u n n = n n u + u we wrte the ( n n ) ( n n ) φ θ TVD fluxes wth lmter as (Use u u u u ): n+ /2 n f /2 = a u u n u n + µ φ θ ; 2 n+ /2 f + /2 = a u n + u 2 We seek a lmter φ ) φ θ 2) φ θ ( θ ) wth the followng attrbutes: = 0 for θ < 0,.e. to avod extrema. > 0 for θ > 0,.e. slope has correct sgn. 3) φ θ = for θ =,.e. to be second order accurate for smooth flow. 2

22 The second order accurate scheme becomes: (Questons: Can you demonstrate that the eqn below retreves the Donor Cell and Lax-Wendroff schemes? For what functonal choce of φ θ t retreve the Beam-Warmng scheme?) wll n+ n t ( n+ /2 n+ /2 u u f ) = + /2 f /2 x µ = µ µ φ θ φ θ 2 + n+ n n n n n n n u u u u u u u u 22

23 n+ n t n+ /2 n+ /2 u = u ( f+ /2 f /2 ) x µ = µ µ φ θ φ θ 2 n n ( + ) ( + ) ( ) ( ) n+ n n n n n n n u u u u u u u u n n ( u u ) Use u u = to rewrte t as: θ µ φ θ = µ + ( µ ) φ θ ( ) ( ) n+ n n n u u u u 2 θ µ φ θ C, /2 = µ + ( µ ) φ( θ ) and C+, + /2 = 0 2 θ Harten's Theorem requres 0 C, /2 23

24 Harten's Theorem requres φ θ 2 φ( θ ) 2 θ ( ), /2 Snce θ and θ are ndependent, and φ θ and φ θ 0 C whch for 0 µ mples : are +ve, the only way out s to take: φ θ 0 2 and 0 φ( θ) 2 θ Ths gves us the TVD regon φ (θ) Questons: Where does the Donor 2 Cell scheme ft n the TVD regon? What about Lax-Wendroff scheme? What about Beam-Warmng scheme? What about Fromm scheme? θ

25 4) The scheme should also be second order. Thus φ θ should le wthn the lmts blocked out by : a) the Lax-Wendroff scheme wth ts rght-based slopes ( φ θ = ) b) the Beam-Warmng scheme wth ts left-based slopes ( φ θ = θ) Ths further restrcts the TVD regon ( Sweby regon) 5) The lmter should also symmetrcal: φ ( / θ) φ θ = θ be 2 φ (θ) Beam- Warmng, φ (θ)=θ Lax- Wendroff, φ (θ)= θ 25

26 a φ (θ) 2 Mnmod b 2 φ (θ) van Leer θ θ c φ (θ) 2 MC d 2 φ (θ) Superbee θ Queston: Identfy the least and most compressve lmter. What are the consequences? 26 θ

27 Mathematcally, t means that we have lnearly ndependent egenvectors. 27 Sensble soluton methods wthn reach. 3.4) Lnear Hyperbolc Systems and the Remann Problem (Revew.5) 3.4.) Soluton of Lnear Hyperbolc PDEs for Contnuous Intal Data We wll need to learn how to classfy dfferent hyperbolc PDEs. We take only a frst step here. We have seen that any conservaton law U t + F x = 0 can be lnearzed as U t + A U x = 0. For lnear hyperbolc PDEs we lterally have F = A U. We focus on such PDEs. A s an M M matrx wth constant entres. The M M hyperbolc system U t + A U x = 0 s strctly hyperbolc f we have real dsjont egenvalues : λ < λ 2 < < λ M. Physcally, t means that the waves are well-separated.

28 Several very mportant hyperbolc systems are non-strctly hyperbolc,.e. we stll have real egenvalues but : λ λ 2 λ M. Queston: Gve an example of such a system. Physcally, under favorable crcumstances (.e. when we have lnearly degenerate egenvalues), waves may stll reman well-separated. Mathematcally, we have to examne egenstructure more closely. Need to work on solutons for each specfc case. 28

29 We consder the smple case where A s a constant M M matrx and has well-separated egenvalues. m m m m m m A r = λ r ; l A = λ l m=,..., M We can buld the matrx of rght egenvectors and left egenvectors as before to get: A R = R Λ ; L A = Λ L ; L A R=Λ ; A = R Λ L R L Recall: R l λ M 2 2 r r r l 0 λ 0 ; L ; = = = Λ M M l 0 0 λ m m m m Left-multply U + A U = 0 by l to get : w + λ w = 0 for m=,.., M t x m m where w l U s called the egenweght or characterstc varable. t x m m m th Notce from w + λ w = 0 that the m characterstc varable moves t x m th wth speed λ along the m characterstc of the hyperbolc system. 29

30 The characterstc decomposton descrbed above s one of the standard buldng blocks used n desgnng numercal schemes. Physcally, It tells us that the characterstc varables undergo smple m m m advecton wth a speed that s gven by the egenvalue. w + λ w = 0 It helps us solve the Cauchy problem for hyperbolc systems: Gven dfferentable ntal condtons on non-characterstc surfaces n spacetme, the system can be evolved further n tme at least for a small tme. 0 ( x) = = Thus gven ntal condtons U m m m 0 0 0, we form the characterstc varables w x l U x for m,..., M.e. w x are known functons of x. Then to evolve the system n tme, advect the profles to get : m 0 = 0 λ The soluton at a later tme s gven by : U xt, w x t r M m= t m m m x ( m λ ) w x t 30

31 3.4.2) Soluton of Lnear Hyperbolc PDEs for Dscontnuous Intal Data: Smple waves and the Remann Problem Lnear hyperbolc PDEs also admt dscontnuous solutons (weak solutons). For frst order conservaton laws, one can have dscontnuous solutons even though the dfferental form of the PDE does not admt them. Reason: The ntegral form of the PDE admts such weak solutons. Study weak solutons n two stages: a) Smple waves b) Remann Problem 3

32 0 ( x) t= 0 x= 0 Integral form U + A U dx dt = 0 gves: X U X U X + A U T U T = 0 A U U = U U T X must match up wth an egenvalue of A : A ( U U m R L) = λ ( U R U L) T Thus wth ntal condtons: U = The t= Tx= X t x L R R L R L R L m th m m U for x<0 ; U x = U = U + α r for x 0 L 0 R L smple wave s gven by: ( xt) x t ( x ) m m m m U, = U for < λ ; U, t = U = U + α r for x λ t 32 L R L

33 (0,T) t Dscontnuty (X,T) t= Tx= X Integral form U + A U dx dt = 0 t= 0 x= 0 t x U L U R (0,0) (X,0) x X A ( UR U L) = ( UR U L) match up wth A r =λ r T m m m 33

34 Notce: A smple wave s a very specal wave structure where the jump U R U L s restrcted to le parallel to the m th rght egenvector. Queston: What f we have a very general jump U R U L wth no partcular arrangement between left and rght state? Answer: Then we have to solve the Remann problem! As long as the space of rght egenvectors s complete (whch s guaranteed for a strctly hyperbolc lnear system) we can always make the projecton: M U U = α r where α l U U m m m m R L R L m= α m s an egenweght of r m.but what does t mean physcally? It means that we have a set of M constant states between U L and U R where the m th constant state U (m) les between x=λ m t and x=λ m+ t 34

35 Physcally: We have a smlarty soluton wth m smple waves. The Remann problem s a super-mportant buldng block for numercal schemes. ( xt) U, = U for m M ( m) p p p p m m = U U L + α r = U R α r for λ < < λ +, m=,..., M p= p= m+ t = U L R x t < λ M for λ < x t x 35

36 x t m = λ 0 m 0 t x = λ + t x = λ 2 t x = λ t x = λ 3 t U (2) U () U m 0 U (M ) x = λ M t x = λ M t U L U R U () = x U (2) = U (3) = 36

37 x t m = λ 0 m 0 t x = λ + t x = λ 2 t x = λ t x = λ 3 t U (2) U () U m 0 U (M ) x = λ M t x = λ M t U L U R U ( M ) = x U ( M 2) = U ( M 3) = 37

38 3.4.3) The Remann Problem as a Buldng Block for the Numercal Soluton of Hyperbolc Systems Here we focus on formulatng a frst order (Godunov) scheme for lnear hyperbolc systems that can handle dscontnutes. Hgher order monotoncty preservng schemes wll also produce jumps at zone boundares, albet the jumps wll be smaller. Ether way, we have to learn how to handle such jumps for the case of hyperbolc systems. The Remann problem from the prevous secton s well-suted for dong that and comes to our rescue. The Remann solver s a very mportant buldng block for schemes that solve lnear and non-lnear hyperbolc problems. 38

39 u x u x t t n+ t n x 39

40 n { } Let us start wth a d mesh functon U on a mesh wth zone sze x. For Godunov's method, we wsh to evolve the soluton from n n tme t to t + t : t x ( + /2 /2 ) n+ n n n U = U F F n The goal s to fnd properly upwnded F + /2 at the zone boundares,.e. fluxes that buld n the realzaton that there are dscontnutes n the soluton and ncorporate the fact that the dscontnuty at each boundary wll splt nto a Remann fan of smple waves that move n dfferent drectons.. Shft coordnate system so that the orgn les at the zone boundary of nterest. 40

41 x t m = λ 0 m 0 t x= λ + t x = λ 2 t x = λ t x = λ 3 t U (2) U () U m 0 U (M ) x = λ M t x = λ M t U L U R x 4

42 We seek the resolved state,.e. the soluton of the Remann problem that overles the zone boundary x=0 of nterest. It wll then be easy to average that state n ( m ) RS 0 space and tme to get the numercal orresolved flux. I.e. we want state U U m0 m0 wth the property that λ < 0 λ + See prevous fg. Fnd state that overles tme axs. U ( RS ) ( m ) U 0 = U L m M p p p p r R r p= p= m f 0 < λ m = U + α = U α f λ < 0 λ L M = U f λ 0 R 0 0 m + m0 m0 Notce that the resolved state s bounded by the characterstcs x = λ t and x= λ + t. Obtanng the resolved flx u requres us to dstngush between waves that move to the rght and those that move to the left. F ( RS ) ( RS ) A U Because the resolved flux s of great mportance to us, we obtan compact, computatonally effcent expressons for t. 42

43 Dstngushng between waves that move to the rght and those that move to the left s most easly done by defnng: +, m m max,0 ;, m m λ λ λ mn λ,0 The left and rght fluxes are also best wrtten as: F A U ; F A U L L R R Automatc expressons for evaluatng the resolved flux are then obtaned as: F M RS, m m m = F L + λ α r m= = F R M m= λ α +, m m m = ( F R + F L) 2 2 r M m= λ α r m m m Queston: What makes the eqns. above so desrable for computer mplementaton? m m Set α l U U n the above formulae to obtan even more compact expressons. R L 43

44 ( m ) 0 0 p p 0 0 RS m m U U = U + α r f λ < 0 λ + To fnd the flux, do F L m p= ( RS ) ( RS ) A U M ( RS ), m m m, m ( m ) L λ α r λ λ F = F + wth mn,0 m= 44

45 ( m ) RS 0 p p m0 m0 U U = U α r f λ < 0 λ + To fnd the flux, do F R M p= m + 0 ( RS ) ( RS ) A U M ( RS ), m m m, m ( m ) R λ + α r λ + λ F = F wth max,0 m= 45

46 RS M m L + λ m α m r m=, F = F = F R M m= λ α r +, m m m M F = ( F R + F L) 2 2 RS m m m m= λ α r 46

47 Our fnal computer-frendly expressons for the resolved flux are: ( RS ) F = F + A U U + = F A U U = F F A U U 2 2 wth the defntons: ( + ) ( ) { } { } + +, +,2 +,M,,2,M Λ dag λ, λ,..., λ ; Λ dag λ, λ,..., λ ; + + A Λ ; L R L R R L R L R L { 2 M } Λ dag λ, λ,..., λ ; R L A RΛ L ; A R Λ L + The matrces A, A and A are evaluated once and for all for a gven lnear hyperbolc system! 47

48 M ( RS ), m m m m m F = F L + λ α r wth α = l UR UL m= ( RS ) F = F + A U U wth A = Λ L R L R L 48

49 M ( RS ) +, m m m m m F = F R λ α r wth α = l UR UL m= ( RS ) F = F A U U wth A Λ R R L = R L 49

50 Notce that we can now show consstency of our scheme. ( RS ) F F U s guaranteed when U U and U U. + /2 /2 n+ n n n As a result, the FDA U = U F F wll converge to the PDE U + A U = 0. t x t x The present Remann solver goes through wth small modfcatons for non-lnear hyperbolc systems, hence ts great mportance. L R 50

51 Example: The lnearzed Euler system vx0 ρ0 0 ρ ρ v x + 0 v x0 v x = 0 t ρ 0 x P P 2 0 ρ0c0 v x0 Recall the characterstc analyss λ = v c ; λ = v ; λ = v + c 2 3 x0 0 x0 x0 0 ρ0 ρ0 2 3 r = c 0 ; r = 0 ; r = c0 2 2 ρ0 c 0 0 ρ0 c l = 0 ; l = 0 ; l = c0 2 ρ0 c0 c0 2 c0 2 ρ0 c0 Now consder the Remann egenweghts T ( ρ ) ( ρ ) problem wth U =, v,p and U =, v, P. L L xl L R R xr R Our are gven by: α vxr vxl + P P ; 2 R L 2 c0 2 ρ0 c0 α ρ ρ α + 5 ( P P ) ; ( v v ) ( P P ) 2 3 R L 2 R L xr xl 2 R L c0 2 c0 2 ρ0 c0 T

52 2 3 l = 0 ; = 0 ; = 0 2 l 2 l 2 2 c0 2 ρ0 c0 c0 2 c0 2 ρ0 c0 ρl ρr Now consder the Remann problem wth UL = v xl and UR = v xr. P L P R Our egenweghts are gven by: α vxr vxl + P P ; 2 R L 2 c0 2 ρ0 c0 α ρ ρ α + ( P P ) ; ( v v ) ( P P ) 2 3 R L 2 R L xr xl 2 R L c0 2 c0 2 ρ0 c0 52

53 The soluton to the Remann Problem s then gven by: x U ( xt, ) = U L for < vx0 c0 t () x = U U L + α r = U R α r α r for vx0 c0 < < vx0 t (2) x = U U L + α r + α r = U R α r for vx < < vx + c t x = U R for v x0+c0 < t Takng the transonc case 0 < v < c we get m =. x The Resolved Flux s now gven by: ( RS ) F = A U = A U () L = A U R + λ α r λ α r λ α = A U R A U L + + r ( + ) ( λ α r λ α r λ α r )

54 3.5) Numercal Boundary Condtons for Lnear Hyperbolc Systems Recall that for scalar advecton u t + a u x = 0 wth a>0, we only need to specfy the boundary condtons at the left boundary. Specfyng them at the rght boundary would overspecfy the problem. We want to understand how t goes for the M component lnear hyperbolc system U t + A U x = 0. 54

55 Because we apply lmters to obtan the slopes, we wll need two zones out from each physcal boundary. Called ghost zones. Queston: Why two? 55

56 x t m = λ 0 m 0 t x= λ + t x = λ 2 t x = λ t x = λ 3 t U (2) U () U m 0 U (M ) x = λ M t x = λ M t U L U R x Say we have m 0 left-gong waves and M m 0 rght gong waves: At left boundary: m 0 waves leave and M m 0 enter the doman. At rght boundary: M-m 0 waves leave and m 0 enter the doman. 56

57 We classfy boundary condtons as follows: Boundary condtons that permt a wave to leave the computatonal doman wthout generatng a back-reacton are called radatve or nonreflectve boundary condtons. Boundary condtons that specfy the ampltude of a wave that should flow nto a computatonal doman are called nflow boundary condtons. Our phlosophy n developng the boundary condtons s that we should apply the same numercal algorthm, f ths s at all possble, to all the zones of the mesh. 57

58 m m m The ndvdual waves obey : w + λ w = 0 Consder the left boundary: t x We may want the frst back reacton. I.e. we want : 0 waves to leave the doman wthout generatng any m w 0 for m m we should contrve to have 0 for. t m m = < 0 wx = m< m0 Soluton: Do ths by settng : w = w = w for m< m. m m m 0 0 For m> m, the waves are comng nto the doman from the left. We are free 0 to specfy them as we wsh n accordance wth the physcs of our problem. Soluton: For m> m, you are free to set w and w n accordance wth your needs. m m 0 0 The Result: We have all the characterstc weghts determned for zones M M n m m n m m and 0. So we can set : U 0 = w0 r and U = w r m= m= 58

59 For perodc boundary condtons we set : U = U, U = U, U = U, U = U n n n n n n n n 0 N N N+ N+ 2 2 For outflow boundary condtons at left boundary we set : U = U = U n n n 0 59

60 3.6) Second Order Upwnd Schemes for Lnear Hyperbolc Systems Three Schemes: A) Lax-Wendroff B) Two-stage Runge-Kutta C) Predctor-Corrector Each of the three schemes descrbed here can be extended to non-lnear hyperbolc systems. The latter two are also relatvely easy to extend to hgher orders. Consder the same mesh that we used for descrbng boundary condtons. 60

61 Recap Lax-Wendroff for scalar advecton:- u + a u = 0 t x ) { } 2 I I n n n n n Start wth mesh functon at tme t : u, u,..., u, u, 2) Usng a lmter, construct lmted undvded dfferences n u n each zone. 3) Construct fluxes at zone boundares: n+ /2 n t n f + /2 = a u + a u for a 0 2 x t 2 x n n = a u+ a u + for a < 0 t 4) Update: u u f x = ( f ) n+ n n+ / n+ /2 + /2 /2 6

62 3.6.) Second Order Accurate Extenson of the Lax-Wendroff Scheme wth Lmters ) Envson the computatonal problem n characterstc varables : m m n m m m w = l U for m=,..., M wth w + λ w = 0 t x 2) In words, we should form the characterstc varables of the mesh functon and advect them around wth second order accuracy. To do that, we frst reconstruct the undvded dfferences n each zone: + m m m m m w = Lmter w w, w w for m =,..., M 3) Usng these, we could of course construct the fluxes boundares, f, for the characterstc varables. m + /2 at the zone Small problem: Now some waves can be left-gong whle other waves may be rght-gong. We need an upgrade forour flux formula. 62

63 m +, m m t m m, m m t m m f+ /2 = λ w + λ w+ /2 + λ w+ λ w+ /2 2 x 2 x wth w = w for λ 0 m m m + /2 m = w for λ < 0 (Notce, only one of λ m + +, m or, m λ s non-zero) M n+ /2 m m + /2 = + / 2 r m= The resultng flux for the whole system can be wrtten as : F f The flux can be wrtten n a more elegant form that s more nstructve as: F n+ /2 + / 2 = A + U n M n F /2 F m m m / 2 = λ λ w+ /2 2 m= x + A U + where t r m Queston: Interpret the terms n the above flux formula. 4) Fnally the Update : t U = U F x + + ( + / 2 F / 2 ) n+ n n /2 n /2 63

64 m +, m m t m m, m m t m m f+ /2 = λ w + λ w+ /2 + λ w+ λ w+ /2 2 x 2 x M n+ /2 m m + /2 = + /2 r m= F f F A U + A U + F where F M n+ /2 + n n m m m m + /2 = + + /2 + /2 = λ λ w+ /2 r 2 m= x 64 t

65 3.6.2) Second Order Accurate, Two-Stage Runge-Kutta Scheme wth Lmters The methods descrbed here are amongst the most popular ones. Reason: They have a very appealng plug-and-chug qualty. Each stage of a Runge-Kutta method looks exactly lke the other. Whle each stage only needs to attan the desred spatal accuracy, the whole scheme wll have the desred spatal and temporal accuracy. Thus RK schemes are called method of lnes or sem-dscrete methods. U t = F x It s possble, though, to obtan schemes that are more effcent than Runge-Kutta. Two possble second order accurate Runge Kutta schemes: Modfed Euler Approxmaton: ( F+ /2 ( U ) F /2 ( U )) t U = U 2 x t U = U F U F x n+ /2 n n n n n ( + /2 /2 ( U )) n+ n n+ /2 n+ /2 n+ /2 n+ /2 Improved Euler Approxmaton: ( + /2 ( U ) F /2 ( U )) () n t n n n n U = U F x t U = U + U F U F U 2 2 x 65 n + /2 /2 n+ () () () () ()

66 66

67 We only need to descrbe one stage, snce they all look alke: Step : We have to obtan the undvded dfferences of the conserved varables. Ths can be done usng two dfferent styles of lmtng: Lmtng on the Characterstc Varables: m m a) Obtan characterstc varables : w = l U ( + ) b) Lmt characterstc varables : w = Lmter w w, w w c) Project back to obtan undvded dfferences n conserved varables : U = w M m= Lmtng on the Conserved Varables: m m m m m 2 M a) Wrte conserved varables as vector : U u, u,..., u b) Lmt each of the components of the vector: m m r m ( 2 M ) m m m m u = Lmter u+ u, u u gves U u, u,..., u T T Queston: What are the strengths and weaknesses of each lmtng strategy? 67

68 Step 2: Obtan the left and rght states at the zone boundary: (.e. snce we know the varaton wthn the zone, we can obtan the values of the soluton at any pont n the zone, ncludng the facecenters.) U U + ΔU ; U U ΔU 2 2 L; + /2 R; + /2 + + Step 3: Treat the Remann solver as a machne that accepts two states and spts out a flux. Feed the above left and rght states nto the Remann solver and obtan a properly upwnded flux. F = F U, U + /2 RS L; + /2 R; + /2 Step 4: These fluxes can be used n each of the two stages, as needed, to obtan the full update. 68

69 It s very nterestng and mportant to show that the mproved Euler Runge -Kutta scheme s strong stablty preservng;.e. f each stage s TVD n a certan sense, then the whole scheme s TVD. Queston: But n whch varables s t TVD? To see ths, make the mn, m n w l U do ths for all m=,..., M transformaton: m,() mn, t () n t n n n n U = U F+ /2 ( U ) F /2 ( U ) ( m mn, m mn, w f ) = w + /2 w f /2 w x U = U t x F U F U n+ n (2) U = ( U + U ) 2 n+ n ( + /2 /2 ) (2) () /2 () () Convex combnaton Snce the frst stage s TVD, we have : TV Snce the second stage s TVD, we have: TV x w = w t x w w mn, + mn, m,(2) w = ( w + w ) 2 U f ( f ) + /2 /2 m,(2) m,() m m,() m m,() M n+ mn, + m w r m= m,() mn, ( w ) TV( w ) m,(2) m,() ( w ) TV( w ) So m=,.., M we can wrte: TV TV TV TV TV TV 2 2 ( w ) ( w ) + ( w ) ( w ) + ( w ) ( w ) mn, + m,(2) mn, m,() mn, mn, 69

70 () n t n n n n u = u f u f u x t u = u f u f u x n+ ( n (2) u u u ) = + 2 ( ) + /2 /2 ( n+ n ) + /2 /2 (2) () /2 () () 70

71 3.6.3) Predctor-Corrector Formulaton It s based on the dea that f we have the pecewse lnear slopes wthn a zone, we can also predct the tme-evoluton of the hyperbolc system for at least a small tme nterval wthn that zone. u(x) u n n u x n ( x) = u + ( x x ) /2 u n + R ; /2 /2 u n + L ; + /2 u n + u(x) u n + u /2 u n + R ; /2 n n u x n ( x) = u + ( x x ) /2 u n + L ; + /2 u n u n u n - + x - + The panel to the left shows a mesh functon that s not decreasng wth x. The panel to the rght shows a mesh functon that s not ncreasng wth x. We solve the advecton equaton u t + a u x = 0 wth a>0. In each case the dashed lne shows the pecewse lnear reconstructed profle n x at tme t n whle the dotted lne shows the same profle 7 at tme t n + t / 2. u n x

72 Step : We have to obtan the undvded dfferences of the conserved varables; same as before. Step 2: Realze that we can obtan the tme evoluton wthn a zone n because: U U + A = 0 t x Use ths to drectly obtan tme-centered left and rght states at the zone boundares. Ths tme-centerng makes the scheme second order. n+ /2 n n U U L ; + /2 U + ΔU + t 2 2 t n n t n = U + ΔU A ΔU 2 2 x n+ /2 n n U+ U R ; + /2 U + ΔU + + t 2 2 t n t = U + ΔU A ΔU 2 2 x Ths s the predctor step. n n

73 Step 3: Treat the Remann solver as a machne that accepts two states and spts out a flux. Feed the above left and rght states nto the Remann solver and obtan a properly upwnded flux. n+ /2 n+ /2 F = F U, U + /2 RS L; + /2 R; + /2 Step 4: Use the fluxes to obtan the update step: t U = U F U, U F U, U x Ths s the corrector step. ( ) ; + /2 ; + /2 ; /2 ; /2 n+ n n+ /2 n+ /2 n+ /2 n+ /2 RS L R RS L R Questons: How would the speed of ths method compare to that of the Runge-Kutta method? Whch steps are repeated n the Runge-Kutta method? Whch steps domnate the cost? 73

74 3.6.4) Numercal Results from the Prevous Three Schemes Queston: Compare and contrast the results of the three schemes: 74