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AM213BNumericalMethodsfortheSolutionofDifferentialEquations

Lecture05CopyrightbyHongyunWang,UCSC

Listoftopicsinthislecture

• ConsistencyofLMM,characteristicpolynomials,consistencycondition

• Stabilityandzero-stabilityofLMM,rootcondition

• Stabilitytheorem:relationbetweenstabilityandzero-stability

• Dahlquistequivalencetheorem:consistency+stability=convergence

• Stiffproblems,verydifferenttimescalesinaproblem

ConsistencyofLMM(linearmulti-stepmethods)LocaltruncationerrorofanLMMisdefinedthesamewayasbefore:

Localtruncationerror

=residualtermwhensubstitutinganexactsolutionintonumericalmethod.

en(h)= α j u(tn + jh)

j=0

r

∑ −h β j f u(tn + jh),tn + jh( )j=0

r

∑

ConsistencyofLMMisdefinedthesamewayasbefore:

AnLMMisconsistentifen(h)=O(hp+1)withp>0.

Consistencycondition

Weinvestigatetheconditiononcoefficients{αj,βj}forconsistency.

WeintroducecharacteristicpolynomialsofanLMM(therearetwoofthem).RecallthegeneralformofLMM

α j un+ j

j=0

r

∑ = h β j f (un+ j ,tn+ j )j=0

r

∑ , αr =1

Definition:ThetwocharacteristicpolynomialsofanLMMaredefinedas


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ρ(ξ)= α j ξj

j=0

r

∑ ← coefficientsontheleftsideofLMM

σ(ξ)= β j ξj

j=0

r

∑ ← coefficientsontherightsideofLMM

Theorem(consistencycondition):AnLMMisconsistentifandonlyif

ρ(1)=0ρ′(1)= σ(1)

⎧⎨⎪

⎩⎪

Proof:

RecallthatconsistencymeansO(1)andO(h)termsdisappearinthelocaltruncationerror.Intheexpressionoflocaltruncationerror,wedoTaylorexpansionaroundtnandcalculatecoefficientsoftermsuptoO(h).

en(h)= α j u(tn + jh)

j=0

r

∑ −h β j f u(tn + jh),tn + jh( )j=0

r

∑

= α j u(tn)+u′(tn) jh( )

j=0

r

∑ −h β j f u(tn),tn( )j=1

r

∑ +O h2( )

=u(tn) α j

j=0

r

∑ +h u′(tn) α j jj=0

r

∑ − f u(tn),tn( ) β jj=1

r

∑⎛

⎝⎜⎞

⎠⎟+O h2( )

Usef u(tn),tn( ) =u′(tn) , α j

j=0

r

∑ = ρ(1) , α j jj=0

r

∑ = ′ρ (1) , β jj=1

r

∑ = σ(1)

=u(tn)ρ(1)+h ′u (tn) ρ′(1)−σ(1)( )+O(h2) Thus,en(h)=O(h

2) ifandonlyifthecharacteristicpolynomialssatisfy

ρ(1)=0ρ′(1)= σ(1)

⎧⎨⎪

⎩⎪

StabilityofLMM

Recallthestabilityweintroducedearlierforsingle-stepmethodsun+1 = Lnum(un) :

Lnum(un)−Lnum(vn) ≤(1+C ⋅h)un − vn forsmallh (E01)


- 3 -

whereconstantCisindependentofh,unandvn.

Applyingthenumericalmethodovernstepsleadsto

Lnum( )n(u0)− Lnum( )n(v0) ≤CT u0 − v0 forsmallhandnh≤T (E01B)

whereCTisindependentofh,n,u0andv0(aslongasnh≤T).

(E01B)ismoregeneralinthesensethat(E01)implies(E01B).

Foranr-stepLMM,wewriteitinthevector-operatorform:

!un+1 = LLMM(

!un)

where

!un ≡

unun+1"

un+r−1

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

, !u0 ≡

u0u1"ur−1

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

Applyingthenumericalmethodovernstepsleadsto

!un = LLMM( )n(!u0)

While(E01)isconvenienttouseinanalysis,itistoonarrowandisinadequateforLMM.Wedemonstratethisinthesimpleexamplebelow.

Example:WecasttheEulermethodintotheformofanr–stepLMM

un+r =un+r−1 + β j f (un+ j ,tn+ j )

j=0

r

∑ , β j =1 , j = r −10 , otherwise

⎧⎨⎪

⎩⎪

WeapplyittoODEu′=0.Itbecomesun+r=un+r–1 forn≥0.

Invector-operatorform,thissimplenumericalmethod(forODEu′=0)is

!un+1 = LLMM(!un)=

!un(2)!un(3)"!un(r)!un(r)

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟

whereu ⃗n(i)denotesthei-thcomponentofvectoru ⃗n.

Toexaminethevalidityof(E01)and(E01B),weapplyLLMMtotwospecialvectors:


- 4 -

!u0 = ( 0 " 0 1 )T , !v0 = ( 0 " 0 )T .

Wehave

LLMM(!u0)= ( 0 " 0 1 1 )T

LLMM( )n(!u0)= ( 1 " 1 1 )T forn≥(r −1)

LLMM( )n(!v0)=0

Itfollowsthat

•

LLMM( )n(!u0)− LLMM( )n(!v0)( 1 " 1 1 )T−0

# $%%%%% &%%%%%≤ CT

!u0 −!v0

( 0 " 0 1 )T−0#$% &%

istrue.

•

LLMM(!u0)−LLMM(

!v0)( 0 " 0 1 1 )T−0

# $%%% &%%%≤ 1+C ⋅h( ) !u0 −

!v0( 0 " 0 1 )T−0#$% &%

isNOTtrue.

Itisclearthatthissimplemethod(forODEu′=0)shouldbeclassifiedas“stable”.

Therefore,weselect(E01B)todefinethestabilityforLMM.

DefinitionofstabilityforLMM

AnLMM !un+1 = LLMM(

!un) isstableif

LLMM( )n(!u0)− LLMM( )n(!v0) ≤ CT

!u0 −!v0 forsmallhandnh≤T


Remark:

Thisstabilityisdifficulttocheckdirectly.Weneedtoworkwithalternativeconditionsthataremoreconvenienttocheck.

Zero-stabilityofLMMConsideranLMMappliedtothemodelODE: u’=0

Exactsolution:u(t)=const

Numericalsolution:

α j un+ jj=0

r

∑ = 0


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Motivation:

Wewantnumericalsolutionuntoremainboundedasn→∞.

Definitionofzero-stability:

Ifallsolutionsof α j un+ jj=0

r

∑ = 0 remainboundedasn→∞,

thenwesaythecorrespondingLMMiszero-stable.

Remark:

Thezero-stabilityisnotaffectedbycoefficients{βj}.

Nextweconnectthezero-stabilityofanLMMtoitscharacteristicpolynomialρ(ξ).

Inequation α j un+ jj=0

r

∑ = 0 ,weconsidersolutionsoftheform{uk=ξk,k=0,1,2,…}.

Substitutinguk=ξkintotheequation,weget

α j ξ

n+ j

j=0

r

∑ =0 ⎯→⎯ α j ξj

j=0

r

∑ =0

==> ρ ξ( ) = 0

Forpolynomialρ(ξ),let

{ξj}denotesimplerootsand

{qi}denoterootswithmultiplicity>1.

Thegeneralnumericalsolutionhastheform:

un = c1ξ1n + c2ξ2

n +!( )+ b1(0) +b1

(1)n+!( )q1n + b2(0) +b2

(1)n+!( )q2n +!

unremainsboundedasn→∞ifandonlyif

|ξj|≤1and|qi|<1

Thisiscalledtherootcondition.

Definition(rootcondition)

Forapolynomial,if

• allrootssatisfy|ξj|≤1and

• allrootswithmultiplicityabove1 satisfy|qi|<1,

thenwesaythepolynomialsatisfiestherootcondition.


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Wesummarizetheresultweobtainaboveintoatheorem.

Theorem(conditionforzero-stability):

AnLMMiszero-stableifandonlyifitscharacteristicpolynomialρ(ξ)satisfiestherootcondition.

Nextwelookattheconnectionbetweenthezero-stabilityandthestability.Theorem(thestabilitytheorem):

Supposeƒ(u,t)inu′=ƒ(u,t)isLipschitzcontinuous.IfanLMMiszero-stable,thenitisstable.

Thatis,foranyT>0,thereexistsCTsuchthat




Proof:Wewillnotgothroughtheproofindetails.AnoutlineofkeystepsintheproofispresentedinAppendixB.

Thistheoremconnectsthezero-stabilitytothestability.Withthistheoremasthestepping-stone,wenowintroducetheDahlquistequivalencetheorem(whichissimilartotheequivalencetheoremweprovedforsingle-stepmethods:

consistency+stability=convergence

Theorem(Dahlquistequivalencetheorem)

AnLMMmethodisconvergentifandonlyifitiszero-stableandisconsistent.Proof:Wewillnotgothroughtheproofindetails.ThekeystepsforprovingtheDahlquist

equivalencetheoremaresimilartothestepsusedforprovingthestabilitytheorem(AppendixB).TheadaptationofthesestepsforprovingtheDahlquistequivalencetheoremisdiscussedinAppendixC.

Example:Themidpointmethod:

un+2 −un =2h f (un+1 ,tn+1) Itisa2-stepLMM.Thetwocharacteristicpolynomialsare

ρ(ξ)= ξ2 −1 , σ(ξ)=2ξ

Checkingtheconsistencycondition:


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ρ(1)=0, ρ′(1)=2=σ(1)

==> Itisconsistent.Checkingtherootcondition:

ρ(ξ)hastwosimpleroots:

ξ1=1, ξ2=–1

==> Itsatisfiestherootcondition.==> Itiszero-stable.

BytheDahlquistequivalencetheorem,themidpointmethodisconvergent.

Remark: Innumericalsimulations,wefindthatthemidpointmethodhasanexponentiallygrowingerrormodeevenforthesimpleODE:u’=–u.Theexponentiallygrowingerrormodequicklyruinsthenumericalsolutionastimeincreases.ThisbehaviordoesnotcontradicttheDahlquistequivalencetheorem.Atafinitetime,ifweuseaverysmalltimestep,themidpointmethodwillbewellbehaved.

h→0whileT isfixed

⎛⎝⎜

⎞⎠⎟

vs T→∞whilehisfixedalthoughsmall

⎛⎝⎜

⎞⎠⎟

Example:

AllAdamsmethods(bothAdams-BashforthandAdams-Moulton)arezero-stable.Thegeneralformofr-stepAdamsmethods:

un+r =un+r−1 +h β j f (un+ j ,tn+ j )

j=0

r

∑

Characteristicpolynomialρ(ξ):

ρ(ξ)= ξr −ξ(r−1) = ξ(r−1)(ξ−1)

ρ(ξ)hasonesamplerootandonerootofmultiplicity(r–1).

ξ1=1, asimpleroot,

q1=0, arootofmultiplicity(r–1).

==> Itsatisfiestherootcondition.

==> Itiszero-stable.

Question: Iszero-stabilityenoughformakingamethodwellbehaved?

ConsidernumericalsolutionsofODE

u′ = −λsinh u− cos(t)( ) , λ = large>0 (E02)


- 8 -

InAssignment#1,youobservedtheresultsbelow.

ThenumericalsolutionofEulermethodisnotevenboundedunlessthetimestepistiny.TheimplicittrapezoidalmethodperformsmuchbetterthanEulermethod.TheimplicitbackwardEulerisevenbetter.

Thebottomline:

Zero-stabilityimpliesthatwhenthetimestepissmallenough,thenumericalsolutioniswellbehaved.Butforacertaincategoryofproblems,wecannotaffordverytinytimesteps.Weneedothertypesofstabilitytoensurethatthenumericalsolutioniswellbehavedevenwhenthetimestepisnotverysmall.

Stiffproblems

Weconsideralinearversionof(E02)asamodelproblem.

u′ = −λ u− cos(t)( ) , λ = large>0u(0)=0

⎧⎨⎪

⎩⎪ (E03A)

Exactsolution:

u(t)= λ2

1+λ2 cos(t)+λ

1+λ2 sin(t)−λ2

1+λ2 e−λt

(SeeAppendixAforderivation)

Therearetwoverydifferenttimescalesinthisproblem:

• Slowevolutionofcos(t),and

• Veryfastdecayofexp(−λt)

Definition(stiffproblem):Aproblemiscalledstiffifithas(atleast)twoverydifferenttimescales.

InODE(E03A),thetwotimescalesaretangledtogether.Tosimplifythediscussion,weconsideranevensimplerprobleminwhichthetwotimescalesareseparated.

u′ = −λu , λ = large>0′v = −v

⎧⎨⎪

⎩⎪ (E03B)

Wefocusontheu-componentof(E03B)anduseitasamodelproblemforexaminingtheperformanceofvariousnumericalmethods.

Asimplifiedmodel:


- 9 -

u′ = −λu , λ = large>0u(0)=1

⎧⎨⎪

⎩⎪

Exactsolution:

u(t)= exp(−λt) Twopropertiesoftheexactsolution:

1. u(t)decreasestozeroast→∞.

Forlargeλ,u(t)decreasesveryfast.

2. Forafixedvalueoft>0(nomatterhowsmallitis),u(t)→0asλ→∞.Thesearethepropertieswewanttopreserveinnumericalsolutions.

Behaviorsofnumericalsolutionsofu′=–λu

Eulermethod:

un+1 =un +h(−λun)

==> un+1 =un(1−λh)

==> un =u0(1−λh)n

Asn→∞,undecreasesinabsolutevalueifandonlyif 1− λh <1 ,

ifandonlyif0 < λh < 2 ,

ifandonlyifh < 2λ.

Forlargeλ,thisconditionisveryrestrictive.

λ=108 ==> h<2×10-8

Foruntodecreasewithoutoscillatinginsign,weneed h ≤1λ.

ConclusionforEulermethod:

• Forlargeλ,Eulermethodhastouseatinytimestepjusttoensurethatthenumericalsolutiondecreasesinabsolutevalueasn→∞.Inotherwords,thenumericalsolutioniswellbehavedONLYwhenthefastevolutioncomponentisresolved,whichrequiresatinytimestep.

• Inastiffproblem,therequirementofatinytimestepiscoupledwiththeneedofsimulatingtheslowevolutioncomponentoveralongtimeperiod.

==> Extremelylargenumberoftimestepsisneeded.


- 10 -

Forexample,toaccommodateλ=108andtosimulatetoT=10,weneed

h<2/λ=2×10–8

N=T/h>5×108=500milliontimesteps

BackwardEulermethod:

un+1 =un +h(−λun+1) , λ = large>0

==> un+1(1+λh)=un

==> un+1 = un1

1+ λh( )

==> un = u01

1+ λh( )n

Property1:

Asn→∞,undecreasesto0withoutoscillatinginsign,foranyvalueofh>0.

Property2:

Whenhisfixed,wehaveu1 = u01

1+ λh( )→ 0 asλ→∞.

ConclusionforBackwardEulermethod:

• Evenforverylargeλ,thenumericalsolutionofbackwardEulermethoddecreasesto0withoutoscillatinginsign,asn→∞,foranytimesteph>0.ThebackwardEulermethodpreservesbothproperties1and2.

• Inotherwords,thenumericalsolutionisalwayswellbehavedevenwhenthefastevolutioncomponentisnotresolved.Wecanselectthetimestepbasedontheneedofresolvingtheslowevolutioncomponent.

Trapezoidalmethod:

un+1 = un +h2

−λun − λun+1( )

==> un+1 1+λh2

⎛⎝⎜

⎞⎠⎟ = un 1−

λh2

⎛⎝⎜

⎞⎠⎟

==> un+1 = un1− λh

21+ λh

2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟


- 11 -

==> un = u01− λh

21+ λh

2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

n

Themultiplicationfactorsatisfies

1− λh2

1+ λh2

<1 for all values of h > 0

Property1:

Asn→∞,undecreasesinabsolutevalueto0,foranyvalueofh>0.

However,forλh>2,themultiplicationfactorisnegative.

==> Forλh>2,unoscillatesinsignwhiledecreasingto0asn→∞.

Property2:

Whenhisfixed,wehave

u1 =1− λh

21+ λh

2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟u0→ (−u0) asλ→∞.

Conclusionfortrapezoidalmethod:

• Thenumericalsolutionoftrapezoidalmethoddecreasesto0asn→∞,foranytimesteph>0.Sothenumericalsolutionwillalwaysremainboundedforanytimestep.Wedon’tneedtorestricttheselectionoftimesteptomakethenumericalsolutionbounded.

• However,forlargeλ,asn→∞,thenumericalsolutionoftrapezoidalmethodoscillatesinsignwithaveryslowdecayinamplitude.Thetrapezoidalmethodpreservesproperty1,butnotproperty2.

BasedontheperformancesofEuler,backwardEulerandtrapezoidalmethodsanalyzedabove,weintroducetheA-stabilityandtheL-stabilitytomeasuretheperformanceofanumericalmethodforsolvingstiffproblems.

Intuitively,wedefineA-stabilityandL-stabilityasfollows:A-stabilitydescribeswhetherornotproperty1ispreserved.

L-stabilitydescribedwhetherornotbothproperties1&2arepreserved.


- 12 -

AppendixA: ExactsolutionoftheIVP

u′ = −λ u− cos(t)( )u(0)=0

⎧⎨⎪

⎩⎪

WewritetheODEas

u′+λu= λcos(t) Weusetheintegratingfactormethod.Multiplyingbyeλt,leadsto

eλtu′+λeλtu= λeλt cos(t)

==> eλtu( )′ = λeλt cos(t)

Integratingfrom0tot,weget

eλtu(t)= λ eλ s cos(s)ds

0

t

∫

Weusetheintegrationformula

eλ s cos(s)ds

0

t

∫ =Real e(λ+i )s ds0

t

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥=Real 1

λ+ ie(λ+i )t −1( )⎡

⎣⎢

⎤

⎦⎥

=Real λ− i

λ2 +1 eλt cos(t)−1+ ieλt sin(t)( )⎡

⎣⎢

⎤

⎦⎥ =

λλ2 +1 eλt cos(t)−1( )+ 1

λ2 +1eλt sin(t)

ThesolutionoftheIVPis

u(t)= e−λtλ eλ s cos(s)ds

0

t

∫

= λ2

1+λ2 cos(t)+λ

1+λ2 sin(t)−λ2

1+λ2 e−λt

AppendixB: Proofofthestabilitytheorem

Theorem(thestabilitytheorem):

Supposef(u,t)inu’=ƒ(u,t)isLipschitzcontinuous.

IfanLMMiszero-stable,thenforanyT>0,thereexistsCTsuchthat


- 13 -




Outlineofkeystepsintheproof:

1. WefirstlookattheLMMappliedtosolvingu’=0.Ithastheform

α jun+ jj=0

r

∑ = 0 , αr = 1

Inthematrix-vectorform,wecanwriteitas

!un+1 = A

!un

wherevectoruandmatrixAare

!un ≡

unun+1"

un+r−1

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟, A ≡

0 1 0" # #0 $ 0 1

−α0 −α1 $ −αr−1

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

MatrixAisaconstantmatrix,independentofh.ThecharacteristicpolynomialofmatrixAisρ(ξ).Polynomialρ(ξ)determinesthezero-stabilityoftheLMM.WhentheLMMiszero-stable,theeigenvaluesofmatrixAsatisfytherootcondition.BywritingmatrixAintheJordancanonicalform,wecanshowthatthereexistsCA≥1suchthat

An ≤CA for all n ≥ 0

2. Next,welookattheLMMappliedtosolvingu’=ƒ(u,t).Ithastheform

α jun+ jj=0

r

∑ = h β j f un+ j , tn+ j( )j=0

r

∑ , αr = 1

Forsimplicity,wefocusonexplicitmethods.Wewriteitinthematrix-vectorform

!un+1 = A

!un + h!φ !un , tn( ) (E05)

wherefunctionφ(u,t)satisfiesLipschitzcontinuitywithrespecttou:

!φ !u, t( )− !φ !v, t( ) ≤CL

!u − !v

NotethatwhilemappingAuislinear,functionφ(u,t)isnon-linear.Let

Δ!un ≡!un −!vn and Δ

!φn ≡!φ !un , tn( )− !φ !vn , tn( )

TheLipschitzcontinuitygivesus


- 14 -

Δ!φn ≤CL Δ!un

Takingthedifferenceof(E05)betweenuandvyields

Δ!un+1 = AΔ

!un + hΔ!φn (E06)

Substituting(E06)atn=0into(E06)atn=1,andtheninto(E06)atn=2,…weobtain

Δ!u2 = A

2Δ!u0 +h AΔ!φ0 +Δ

!φ1( )

Δ!un+1 = A

n+1Δ!u0 +h An− jΔ!φ j

j=0

n

∑ (E07)

Takingnormofbothsides,usingAn ≤CA andtheLipschitzcontinuity

Δ!φn ≤CL Δ!un ,

wearriveatarecursiveinequalityfor Δ!un

Δ!un+1 ≤CA Δ!u0 + hCACL Δ!uj

j=0

n

∑ (E08)

3. Nowwesolvetherecursiveinequality(E08).

Weintroduce{gn}recursivelyas

g0 = CA Δ!u0

gn+1 = g0 + hCACL gjj=0

n

∑ (E09)

Itisstraightforwardtoshowthat

Δ!un ≤ gn for all n ≥ 0

Tocalculategn,were-writetherecursiveequation(E09)as

gn+1 = gn +hCACL gn = (1+hCACL)gn Applyingthisrelationsuccessivelyfromindex0toindex(n–1),yields

gn = (1+hCACL)n g0 ≤ exp CACLT( ) Δ!u0 fornh≤T

Using Δ!un ≤ gn andg0 =CA Δ!u0 ,wehave

Δ!un ≤CAexp CACLT( ) Δ!u0 fornh≤T

Therefore,wefinallyarriveat

LLMM( )n !u0 − LLMM( )n !v0 ≤CAexp CACLT( ) !u0 − !v0 fornh≤T


- 15 -

whichimpliestheLMMisstable,bydefinition.

AppendixC: KeystepsforprovingDahlquistequivalencetheorem

BelowwediscussbrieflythekeystepsforprovingtheDahlquistequivalencetheorem.ThesestepsareadaptedfromthestepsinAppendixBforprovingthestabilitytheorem.

Again,wewritetheLMMinthematrix-vectorform

!un+1 = A

!un + h!φ !un , tn( )

Weintroducethevectorversionsofexactsolutionandlocaltruncationerror.

!vn ≡!u tn( ) =

u tn( )u tn+1( )"

u tn+r−1( )

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

, !en h( ) ≡00"

en h( )

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

=O hp+1( )

Numericalsolutionuandexactsolutionvsatisfy,respectively

!un+1 = A!un + h

!φ !un , tn( )

!vn+1 = A!vn + h

!φ !vn , tn( ) + !en h( )

(E25)

Let Δ!un ≡!un −!vn andΔ

!φn ≡!φ !un ,tn( )− !φ !vn ,tn( ) .Takingthedifferencein(E25)givesus

Δ!un+1 = AΔ

!un +hΔ!φn −!en(h) (E26)

Substituting(E26)atn=0into(E26)atn=1,andtheninto(E26)atn=2,…weobtain

Δ!u2 = A

2Δ!u0 +h AΔ!φ0 +Δ

!φ1( )− A!e0(h)+

!e1(h)( )

Δ!un+1 = A

n+1Δ!u0 + h An− jΔ!φ j

j=0

n

∑ − An− j !en h( )j=0

n

∑ (E27)

Takingnormofbothsides,usingAn ≤CA andtheLipschitzcontinuity

Δ!φn ≤CL Δ!un ,

wearriveatarecursiveinequalityfor Δ!un

Δ!un+1 ≤CA Δ!u0 +hCACL Δ!uj

j=0

n

∑ + n+1( )CACehp+1 (E28)

Tosolverecursiveinequality(E28),weintroduce{gn}recursivelyas

g0 = CA Δ!u0


- 16 -

gn+1 = g0 +hCACL gj

j=0

n

∑ + n+1( )CACehp+1 (E29)

whichensuresthat

Δ!un ≤ gn for all n ≥ 0

Tocalculategn,were-writerecursiveequation(E29)as

gn+1 = 1+hCACL( )gn +CACehp+1 ==> 1+hCACL( )−(n+1) gn+1 ≤ 1+hCACL( )−n gn + 1+hCACL( )−(n+1)CACehp+1

==>1+hCACL( )−n gn ≤ g0 +CACehp+1 1+hCACL( )−(n+1)

j=0

n−1

∑

≤ g0 +CACeh

p+1 1+hCACL( )−1 1− 1+hCACL( )−n1− 1+hCACL( )−1

==>gn ≤ 1+hCACL( )n g0 +CACehp+1

1+hCACL( )n −1CACLh

≤ exp(nhCACL)g0 +

exp(nhCACL)−1CL

Cehp

Using Δ!un ≤ gn andg0 =CA Δ!u0 ,weobtain

Δ!un ≤CAexp CACLT( ) Δ!u0 +

exp CACLT( )−1CL

Cehp fornh≤T

Whentheinitialvalueu0isatleastp-thorderaccurate: Δ!u0 = !u0 −

!v0 ≤ cinthp ,the

differencebetweenthenumericalsolutionuandexactsolutionvisboundedby

!un −!vn ≤ CAexp CACLT( )cint +

exp CACLT( )−1CL

⎛

⎝⎜⎜

⎞

⎠⎟⎟hp fornh≤T

whichimpliestheLMMisconvergent,bydefinition.