Numerical Solution of Equations

Transcription

1 School of Mechaical Aerospace ad Civil Egieerig Numerical Solutio of Equatios T J Craft George Begg Buildig, C4 TPFE MSc CFD- Readig: J Ferziger, M Peric, Computatioal Methods for Fluid Dyamics HK Versteeg, W Malalasekara, A Itroductio to Computatioal Fluid Dyamics: The Fiite Volume Method SV Patakar, Numerical Heat Trasfer ad Fluid Flow Notes: - People - T Craft - Olie Teachig Material Itroductio The Navier-Stokes equatios are a set of coupled, o-liear, partial differetial equatios Solvig these umerically cosists of two steps: Approimatio of the differetial equatios by algebraic oes Solutio of the system of algebraic equatios Before cosiderig how to approimate ad solve such systems, it is useful to review briefly how we use umerical methods for approimate solutios of sigle o-liear algebraic equatios These are relevat, sice the differece equatios we obtai from discretizig the Navier-Stokes equatios are o-liear Sice most solutio methods for o-liear equatios are iterative, this itroduces a umber of cocepts ad geeric treatmets that will also be met later whe dealig with iterative solutio methods for large sets of coupled equatios Numerical Solutio of Equatios 00/ / 8 The geeral problem to be cosidered is that of solvig the equatio f() = 0 () where f is some arbitrary (o-liear) fuctio For some methods we eed to reformulate this as = F() () May methods will probably have bee met i earlier courses, but it is useful to review these, to uderstad their behaviour ad eamie some of their advatages ad weakesses Direct Iteratio Method This is a fairly simple method, which requires the problem to be writte i the form = f() for some fuctio f() We start with a iitial guess to the solutio,, ad the calculate a ew estimate as = f() (3) This process is cotiued, at each step geeratig a ew approimatio f() + = f() (4) Iteratios are stopped whe the differece betwee successive estimates becomes less tha some prescribed covergece criterio ε ie whe + < ε If the process is coverget, takig a smaller value for ε results i a more accurate solutio, although more iteratios will be eeded Numerical Solutio of Equatios 00/ 3 / 8 Numerical Solutio of Equatios 00/ 4 / 8

2 As a eample, cosider solvig the equatio (which has the eact solutio = 4) = 0 (5) We first eed to write the equatio i the form = f(), ad there is more tha oe way of doig this Oe way is to write the equatio as = ( ) /3 so f() = ( ) /3 (6) Startig with a iitial guess of = 5, the above iterative method gives: Iteratio f() Iteratio f() The scheme coverges, although ot particularly rapidly If, however, we rewrite equatio (5) i the form = ( 3 3 4)/3 so f() = ( 3 3 4)/3 (7) ad agai take = 5, the the iteratio process leads to: Iteratio f() I this case the process is clearly divergig Eve if we had started from a iitial guess much closer to the real solutio, with this formulatio the iterative process would have diverged As ca be see from this eample, the direct method as described is ot guarateed to coverge The particular formulatio chose for f() ca be highly ifluetial i determiig the covergece behaviour To uderstad why this should be, we ote that the differece betwee successive iteratios will be + = f() f( ) (8) Numerical Solutio of Equatios 00/ 5 / 8 Numerical Solutio of Equatios 00/ 6 / 8 Assumig that we are close to the true solutio, r, we ca approimate f() ad f( ) by Taylor series epasios about r : f() f(r)+( r)f (r)+ (9) f( ) f(r)+( r)f (r)+ (0) where f deotes the derivative df/d Substitutig the epressios ito equatio (8) gives + f (r) () Thus, if the gradiet of f is such that f (r) <, the differece betwee successive approimatios decreases, ad the scheme will coverge If f (r) > the differece betwee successive approimatios icreases ad the method will ot coverge f() f() The graphs of the two particular forms of f() eamied i the eample earlier are show below + = ( )/3 coverges + = ( 3 3 4)/3 diverges This behaviour ca also be iferred graphically Covergece Divergece This highlights a importat aspect of umerical solutios: oe eeds a good uderstadig of the problem to be solved ad solutio methods i order to select the most appropriate scheme Numerical Solutio of Equatios 00/ 7 / 8 Numerical Solutio of Equatios 00/ 8 / 8

3 The Bisectio Method This is desiged to solve a problem formulated as f() = 0 We start off with two poits ad, chose to lie o opposite sides of the solutio Hece f() ad f() have opposite sigs We the bisect this iterval, so take 3 = 05( + ), ad evaluate f(3) For the et iteratio we retai 3 ad whichever of or gave the opposite sig of f to f(3) The solutio thus lies betwee the two poits we retai We cotiue bisectig the iterval as above util it becomes sufficietly small, ie < ε for some small covergece criteria ε Clearly, as we reduce the covergece criteria ε we get a more accurate approimatio to the solutio, but have to perform more iteratios f() Numerical Solutio of Equatios 00/ 9 / 8 Solvig the same eample as earlier we write f() = = 0 () Applyig the bisectio method iteratio, with iitial poits = 5 ad = 5 ow gives: Iteratio ( ) (,) (3,) (3,4) (3,5) f() Iteratio ( ) (6,5) (7,5) (7,8) (7,9) (0,9) f() The method will always coverge, sice the iterval size always decreases The method ca be rather slow, sice the iterval size is oly halved i each iteratio Numerical Solutio of Equatios 00/ 0 / 8 The Secat Method This method solves the system f() = 0 It also requires two startig poits, ad, but they eed ot be o opposite sides of the eact solutio f() With the earlier eample, solvig f() = = 0, ad with the same startig poits, we would get: Iteratio f() The process shows a fairly rapid covergece We ow fit a straight lie through the two poits (,f()) ad (,f()), ad take the et estimate as the poit at which this lie cuts the ais This iteratio method ca be formulated mathematically as ( ) + = f() f() f( ) (3) The process is agai repeated util the covergece criteria is reached Numerical Solutio of Equatios 00/ / 8 Numerical Solutio of Equatios 00/ / 8

4 However, if we istead start with the two poits = 0 ad = 5, we get the sequece: Iteratio f() Iteratio f() I this case the process does ot appear to be covergig By plottig the sequece, we ca see that the iterative process oscillates across the local maimum of f aroud = 04 Uder-Relaatio Uder-relaatio is commoly used i umerical methods to aid i obtaiig stable solutios Essetially it slows dow the rate of advace of the solutio process by liearly iterpolatig betwee the curret iteratio value, ad the value that would otherwise be take at the et iteratio level This prevets the scheme from makig such large chages i solutio estimate from iteratio to iteratio I the preset eample of usig the Secat method, equatio (3) could be modified to read ( ) + = ωf() (4) f() f( ) where the uder-relaatio factor ω is take betwee 0 ad Numerical Solutio of Equatios 00/ 3 / 8 Numerical Solutio of Equatios 00/ 4 / 8 If, for eample, we take ω = 05, the Secat method applied to the previous eample gives: Iteratio f() Iteratio f() By reducig the jump betwee successive values of i the early iteratios the solutio estimates stay to the right had side of the local miimum i f, ad the process ow coverges Note that as the solutio gets closer to covergig, the uder-relaatio factor could be icreased to speed up covergece Covergece Criteria I the eamples preseted so far we tested for covergece by checkig o the differece betwee successive solutio estimates We thus took the solutio as coverged whe + < ε for some predefied covergece criteria ε However, this coditio should ot be applied i a blid fashio For eample, if the gradiet f is very steep, the secat method might oly give small chages i betwee iteratios, eve if still far from the true solutio, particularly if usig heavy uder-relaatio f() 3 A small value of + could simply idicate very slow covergece A safer way of checkig for covergece is to also test whether the actual fuctio value f() is sufficietly small Numerical Solutio of Equatios 00/ 5 / 8 Numerical Solutio of Equatios 00/ 6 / 8

5 The Newto-Raphso Method This also solves the equatio f() = 0 I this method we start with a iitial guess We draw the taget to the curve of f at, ad take our et estimate to be the poit where this taget cuts the ais f() Mathematically, we ca write this iteratio process as + = f() f () where f deotes the derivative df/d We agai cotiue the process util we satisfy some suitable covergece criteria 4 3 (5) Usig the same eample as before, = 0, with a startig poit of = 5, we obtai the followig sequece: Iteratio f() f () Iteratio f() f () 8 07 Note the large jump i the first step, which occurs sice f () is small After this the covergece is quite rapid Numerical Solutio of Equatios 00/ 7 / 8 Numerical Solutio of Equatios 00/ 8 / 8 If, however, we had started the iteratio process from a iitial value of = 0, we would have got: Iteratio f() f () Iteratio f() f () Because of the shape of f, the method speds a umber of iteratios oscillatig aroud the local maimum at 04 Numerical Solutio of Equatios 00/ 9 / 8 I this case covergece ca agai be improved by itroducig some uder-relaatio With a uder-relaatio factor of 09, takig we get the iteratio sequece: + = 09f()/f () (6) It f() f () The solutio rapidly reaches a poit ear the maimum of f where f is small eough ad positive so that the et value of lies to the right of the miimum at 4 Covergece could be further speeded up by usig less uder-relaatio after iteratio 4 Numerical Solutio of Equatios 00/ 0 / 8

6 Systems of Algebraic Equatios Most of the methods outlied for solvig a sigle equatio ca be eteded to apply to a system of equatios of the form: f(,,,) = 0 f(,,,) = 0 f(,,,) = 0 The fuctios ff could be liear or o-liear fuctios of As i the case of sigle equatios, to solve the system efficietly (ad sometimes to solve it at all) requires a uderstadig of the method adopted, ad of the ature of the equatios to be solved The Direct Iteratio Method For the equivalet of the direct iteratio method, we first have to rewrite the system of equatios ito the form = f(,,,) = f(,,,) = f(,,,) This ca be writte more compactly i vector form: = f() Oe the proceeds, with a startig guess 0, costructig the sequece + = f( ) (7) stoppig the iteratio whe + < ε for a suitably small covergece criterio ε The orm + is simply a scalar measure of the size of the vector + For eample, it could be the L orm L = () / = ( i ) / (8) Numerical Solutio of Equatios 00/ / 8 Numerical Solutio of Equatios 00/ / 8 As before the method is ot always guarateed to coverge I additio to covergece behaviour beig depedet o the precise formulatio used for f i i the equatio i = f i(,,,) (9) the choice of which equatio is used for updatig which variable is ow also importat It is possible to devise epressios for the coditios ecessary to esure covergece, i the same way as was doe i the sigle equatio case However, the result is rather comple ad will ot be cosidered here I geeral, a rule of thumb is to write the equatios i a form where the right had sides cotai the variables raised to powers less tha oe Uder-relaatio is ofte also used, to aid covergece by avoidig large chages i the estimates for from iteratio to iteratio With this icluded, the iteratio sequece ca be writte as + = ( ω) + ωf( ) (0) where ω is the uder-relaatio factor, with 0 < ω < Numerical Solutio of Equatios 00/ 3 / 8 The Newto-Raphso Method Other schemes ca also be adapted to obtai solutios to a system of equatios To see how the Newto-Raphso method ca be applied to a system of equatios, we first otice that its form for a sigle equatio ca be devised by retaiig the first term i a Taylor series epasio of f about the poit If r is the actual root, the oe ca write 0 = f(r) f()+(r )f ()+ () Neglectig the higher order terms ad rearragig leads to the approimatio r f()/f () () Numerical Solutio of Equatios 00/ 4 / 8

7 I the case of a system of equatios a Taylor series epasio about the root, retaiig oly first order terms, gives 0 = f (r) f (i) +((r) (i) 0 = f (r) f (i) +((r) (i) 0 = f (r) f (i) +( (r) (i) +( (r) (i) +( (r) (i) +( (r) (i) + +( (r) (i) + +( (r) (i) + +( (r) (i) ) f(i) ) f(i) ) f(i) where f (i) deotes the value f( i ) ad superscripts r deote the actual root Rearragig, ad takig (r) as the estimate to the solutio at iteratio level i +, leads to a system of liear equatios for the elemets of the vector (i+) (i) : f (i) (i+) (i) f (i) (i+) (i) f (i) = f (i) (i+) (i) f (i) which ca be solved usig a variety of methods, some of which will be discussed later Uder-relaatio ca also be applied i this method (3) Whe the scheme coverges it geerally does so more rapidly tha the simpler direct iteratio method Numerical Solutio of Equatios 00/ 5 / 8 Numerical Solutio of Equatios 00/ 6 / 8 However, more work has to be doe per iteratio, sice the derivatives f i/ j have to be computed, ad the matri system of equatio (3) solved I comple problems the gradiets f i/ j might also have to be computed umerically I practical problems, the Jacobia matri [ f i/ j] might ot be recomputed every iteratio, but perhaps oly every few iteratios Methods for solvig matri systems such as equatio (3) will be eamied later i the course Summary We have eamied a umber of iterative umerical methods for solvig sigle o-liear equatios of the form f() = 0 ad systems of the form f() = 0 Some schemes, such as the bisectio method, will always coverge, but may do so rather slowly Others coverge more rapidly for some cases, but are ot guarateed to always coverge Uder-relaatio ca be applied i some cases to aid covergece by reducig the chage i solutio estimates betwee iteratios The uder-relaatio factor ca ofte be icreased as the true solutio is approached To select ad apply the most appropriate method for a particular problem requires a good uderstadig of the characteristics of the method ad of the problem beig solved Numerical Solutio of Equatios 00/ 7 / 8 Numerical Solutio of Equatios 00/ 8 / 8