i+1 by A and imposes Ax

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS 09.9 NUMERICAL FLUID MECHANICS FALL 009 Mody, October 9, 009 QUIZ : SOLUTIONS Notes: ) Multple solutos re provded for some problems. ) Aswers provded re ofte muc more complete t requred for full credt. Problem I (0 pots) ) No. It s te reltve, ot bsolute roud-off error tt s costt. I flotg pot represetto, umber x s represeted wt fte set of mce dgts (bts) wc defes te mtss m: x = m b e, b - m < b 0. Ts qutzto of rel umbers leds to roud-off (copoff) error o te mtss tt s costt d bsolute error o x tt s fucto of te expoet. ( pts) b) Yes, t s symmetrc. If t ws ot, exterl torque (e.g. due to mgeto-ydrodymcs) would be ecessry to blce te rte of cge of gulr mometum of flud prcel. Wtout suc exterl torque, f te stress tesor ws ot symmetrc, te gulr ccelerto would go to fty s te volume of te flud prcels goes to zero (ts s becuse, s te volume goes to zero, te momet of ert decys fster t te torque exerted by te vscous stresses). (.5 pts) c) Precodtog A x = b cosst of pre-multplyg by te verse of o-sgulr mtrx M d solvg sted: M - A x = M - b or A M - (M x) = b. It c ccelerte te covergece of solver for A x = b becuse te covergece propertes of te precodtoed system re bsed o M - A (or A M - ) sted of A. (.5 pts) d) Usully, oe left multples x + = x +α + v + by A d mposes Ax + = b. Oe te obts b= Ax + α + Av +, wc s projected log te serc drecto, for exmple v +, to gve α + : T T T v + (b Ax ) = v + r = α + v + Av +. (.5 pts) e) Te et coducto equto oe dmeso spce. My oters equtos too. ( pts) f) If te order of ccurcy of FD sceme s p, te FD sceme coverges wt tructo error τ = L ( ) Lˆ ( ) O x p Δx φ Δx φ (Δ ) for Δx 0. If te order of ccurcy s p for oe FD sceme d p for te oter, ts mes tt we te grd Δx s refed, te dscretzto error of te st sceme wll decy s O(Δ x p ) wle tt of te d wll decy s O(Δ x p ). Note tt te order of covergece does ot defe te sze of te totl (dscretzto) errors; t oly dctes te rte t wc tese errors decy s te grd s refed. (.5 pts) Problem II: Totl Numercl Error Forwrd Euler Sceme. (0 pots) I prts )-c) we re pproxmtg te vlue of te dervtve wt fucto evlutos. I prt d) we re tryg to pproxmte te vlue of te fucto by evlutg d ddg ts dervtves. ) df t () f ( t +Δt) f () t Δt d f ( ξ ) = dt Δt dt ' f + f '' f = f ξ () b) Substtute f ( t ) = f ( t ) + E to () - -

2 f f ' + E '' f = + f ξ f f f ' + = ' E ' '' ε = f f = fξ E '' E ε f '' f '' = f + E + ξ ξ ξ ε c) To fd te vlue of tt mmzes te error, fd suc tt = 0 '' ε f ξ ω s ω t ω = E + = E + E + = 0 E = ω Te optml vlue of creses lerly wt ω - (te frequecy of te sgl) d wt te squre-root of te roud-off error. d) Wt o pproxmtos, te fucto wll evolve s follows: f + f '' ' f ξ = f '' ' + = + ξ f f f + f Wt pproxmto, te fucto evolves s: f f + = f ' f = f + f ', were f ' = f ' + E ' + Tg te dfferece betwee te two, d supposg tt t =0, f 0 = f 0 : '' ' f ε f ξ E f = = f f = = + f E ξ ξ = '' ' '' ' ε ε = f E '' ' ξ = ε = f E ε f '' E ' ω E ' ' ξ + E ω s ω t + + = = = ε ω E ' + Note, ere we ve bouded te error o E by E. - -

3 Ts sows tt bot te tructo error d roud-off error from oe step compouds, so tt te error t steps s -tmes te frst error. For > E, or lrge gulr frequecy, te errors due to tructo wll domte. Te errors for tructo d roud-off wll be of te sme order we: ' E ' ω = E = ω Here te optml vlue of creses qudrtclly wt ω - d lerly wt E. Te m pot ere s tt ofte tructo error domtes, d oly for very well-resolved problem wll te roud-off error ve effect. Notes:. We could ve bouded te error E by ωe, rgug tt te fucto evluto o cos(ωt) s E ' resposble for te roud-off error. I tt cse = ω. We could ve cluded roud-off error due to te summto of te terms te umercl sceme. Tt s: f = f f ' E, wc cse our error would be bouded by: ω ε E ' + + E. Wle ts wll ot ffect te optml vlue of, (sce te dervtve wt respect to of E s zero), ts term my domte te E term, for smll vlues of. I fct, oe my cosder te reltve szes of te terms te error. To do so, dvde ' ω E E ec term by ω :,,. Te cses re: ) ω>e ½, te te tructo error ω ω ω domtes (commo sce ω s ofte close or bt smller t ); ) ω<e ½, te oe of te roud-off errors wll domte, eter te term wt E (roud-off due to ddtos) we s smller or te oe wt E (roud-off due to te dervtve evluto) we s lrger. Problem III: Guss Elmto & LU decomposto for postve symmetrc mtrces (5 pots) ) Soluto : Cosder ( + ) ( ) ( ) = m j j j = ( ) ( ) ( ) ( ) j j ( ) j For symmetrc mtrces: j = j ( ) ( ) ( + ) ( ) ( ) ( ) j ( ) ( ) ( ) ( +) ( ) ( ) j = j j = j = j m j = j Hece te sub-mtrces re symmetrc. Soluto : Proof by ducto. By specto, te sttemet s true for systems of sze x d x. Assume te sttemet s true for system of sze x. - -

4 ,, +,, + # # # # [Performstep of Guss Elmto],,, +, +, + +, +,, + 0,,, +, + # # # #,,, 0,,,, +, +, +, +, + 0, +, +, + + Sce te bottom x mtrx s symmetrc, d t ws ssumed te sttemet ws true for x mtrces, te result s prove for + x + mtrces. Tus te result true geerl for y mtrx, d te requred s prove. b) If A s ot postve defte, te result obted ) s ot lwys true, becuse tt cse pvotg my be requred for stblty, d pvotg cuses te symmetry to be lost. Problem IV: Ler Solver for Bloc Trdgol Mtrces (5 pots) ) Soluto To verfy te soluto, strt by performg step of Guss elmto: ' (row) : F =F [ucged] () ' () () () ' () ' ' (row ) : m = β = E (F ), = m F =F G (row ) : E E (F ) F = E F = 0 [o], F =F G, G =G (row ) :for =, o elmto eeded, lredy zero -β 0 [ucged] ' ' ' ' (row ) : E E (F ) F = E F = 0 [o], F =F G, G =G -β 0 [ucged] ' ' ' ' (row ): E E (F ) F = 0, β = E (F ), F =F G, G =G -β 0 Now provg te bc d forwrd substtutos: Iy = d β y + Iy = d y = d y β y + Iy = d y = d y - 4 -

5 F' x = y x = F' N N N N N y N F' x + G x = y x = F ' ( y G x ) N N N N N N N N N N F' x + G x = y x = F ' (y G x ) Soluto : We c prove te sttemet by multplyg L d U togeter, d sowg tt te orgl mtrx s recovered. Followg ts, we ve to sow tt te forwrd d bc-substtutos re correct. I F ' G β I F ' G LU = % % % % β I F ' G N N N β N I F ' N F ' G β F ' β G + F ' G = % % % β F ' β G + F ' G N N N N N N β F ' β G + F ' N N N N N F G E ( F ' ) F ' β G + F ' G = % % % E ( F ' ) F ' β G + F ' G N N N N N N N E ( F ' ) F ' β G + F ' N N N N N N F G E β G + F G G = % % % E β G + F G G N N N N N N N F β G N N N N N N E β G + F G E F G = % % % E F G N N N E N F N Tere, te bloc LU decomposto s, fct, correct

6 Now cecg te forwrd d bc substtutos: I y d β I y d % % # = # β I y d N N- N β I y d N N N Iy = d β y + Iy = d y = d y β y + Iy = d y = d y F ' G x y F ' G x y # = # F ' G x y N N N- N- F ' x y N N N F' x = y x = F ' y N N N N N N F ' x + G x = y x = F ' (y G x N N N N N N N N N N F ' x + G x = y x = F ' (y G x ) Ad te requred s prove. ) b) Domt order of operto cout. Soluto : Soluto : β = E F ( ' ) : M (Iverse) + M (Mtrx Mtrx mult.) F ' = F G : M (Mtrx Mtrx mult.) BlocTotl:6 M Totl : 6M N ( ) : LU fctorzto ) β = E F' M ( F ' = F G F ' = F E (F ' ) G : M (bc fwd substtutos ) + M (Mtrx Mtrx mult.) 4 4 BlocTotl : M Totl : M N - 6 -

7 I eter cse, ts s muc smller t te Guss elmto o te etre mtrx, wc would be of order (MN) c) If we could solve M systems usg te Toms lgortm, te domt operto cout would be 8MN, wc s cosderbly smller t full Guss elmto d te lgortm bove. C. Numercl code Evluto Problems ) Te equto beg solved s prbolc. It s te D ustedy et dffuso equto wt source term: u u ν = f ( xt, ) t x Ad te dscretzto sceme used s bcwrd dfferece tme, d cetrl dfferece spce: u u u + u + u ν = f ( xδt), Δt Δx b) Drclet boudry codtos re mplemeted. Tese re mplemeted troug te rgt-dsde vector, d s ot cluded s prt of te mtrx. c) Te Successve Over-Relxto (SOR) Guss-Sedel lgortm s used

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