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AMS212BPerturbationMethodsLecture07

CopyrightbyHongyunWang,UCSC

Revisittheissueofboundarylayerlocation

Welookattheissuefromadifferentangle.Specifically,weanalyzetheproblemusingthecharacteristicsoffirstorderPDEs.

Considertheboundaryvalueproblem(BVP)

ε ′′y +a x( ) ′y +b x( ) y − c x( )( ) =0y 0( ) =α , y 1( ) =β

⎧⎨⎪

⎩⎪, ε→0+

Keystrategy:

WeviewthesolutionoftheBVPasthesteadstatesolutionofatimeevolutionPDE.

∂ y∂t

= ε ∂2 y∂x2

+a x( )∂ y∂x +b x( ) y − c x( )( ) Physicalmeaningoftheterms:

ε ∂

2 y∂x2

: diffusion

a x( )∂ y∂x : transportation

b x( ) y − c x( )( ) : growth/decay

Note: ε≥0isrequiredforthewell-posednessofthePDE.

Inthecaseofε→0-,weredefine

εnew = −εold

anew x( ) = −aold x( )

bnew x( ) = −bold x( )

Weconsider

∂ y∂t

= εnew∂2 y∂x2

+anew x( )∂ y∂x +bnew x( ) y − c x( )( ) , εnew →0+

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TheunperturbedPDE(ε=0)isafirstorderlinearPDE.

∂ y∂t

= a x( )∂ y∂x +b x( ) y − c x( )( ) Characteristics

dxdt

= −a x( ) Alongacharacteristic,wehave

dy x t( ) ,t( )

dt= ∂ y∂x

⋅dxdt

+ ∂ y∂t

= ∂ y∂x

⋅ −a x( )( )+a x( )∂ y∂x +b x( ) y − c x( )( )

= b x t( )( ) y − c x t( )( )( )

Itisclearthatthesolutionpropagatesalongcharacteristics.

Example:

∂ y∂t

= ∂ y∂x

+ y

Characteristics:

dxdt

= −1 ==> x(t)=-t+x(0)

Alongacharacteristic,thesolutionsatisfies

dydt

= y ==> y x t( ) ,t( ) = y x 0( ) ,0( )et Thefigurebelowshowsthecharacteristicsinthex-tplane.

x

t

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Fromthefigure,weseethatinthecaseofε=0,§ aboundaryconditionshouldbeimposedatx=1;

§ noboundaryconditionshouldbeimposedatx=0;ifaboundaryconditionisimposedatx=0,thenthereisadiscontinuityatx=0.

Inthecaseofε>0,thePDEisdiffusive

∂ y∂t

= ε ∂2 y∂x2

+ ∂ y∂x

+ y

§ Atx=0,thereisamismatch(discontinuity)betweentheboundaryconditionandthepropagationalongcharacteristics.

§ Themismatchissmoothedbythesmalldiffusionterm;butthetransportation(propagationalongcharacteristics)preventstheeffectoftheboundaryconditionfromreachingsignificantlyinside(0,1).

==> Themismatchbecomesanarrowtransitionnearx=0.==> Thereisaboundarylayeratx=0.

Summary:Aboundarylayeriscausedbyamismatch

§ eitherbetweenboundaryconditionandpropagationalongcharacteristics

§ orbetweenpropagationsalongtwodifferentgroupsofcharacteristics

NowweexamineseveralcasesofthegeneralBVP

ε ′′y +a x( ) ′y +b x( ) y − c x( )( ) =0 ThecorrespondingfirstorderPDEforϵ=0:

∂ y∂t

= a x( )∂ y∂x +b x( ) y − c x( )( ) Characteristics:

dxdt

= −a x( )

Case1: a x( ) >0forx∈ 0,1⎡⎣ ⎤⎦

Arepresentativediagramofcharacteristicsisshownbelow.

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==> Thereisamismatchatx=0only.

==> Thereisaboundarylayeratx=0only.

Case2: a x( ) <0forx∈ 0,1⎡⎣ ⎤⎦

Arepresentativediagramofcharacteristicsisshownbelow.

==> Thereisamismatchatx=1only.==> Thereisaboundarylayeratx=1only.

Case3:

a x( ) = <0 , forx∈ 0, x0⎡⎣ )

>0 , forx∈ x0 ,1( ⎤⎦

⎧⎨⎪

⎩⎪

Arepresentativediagramofcharacteristicsisshownbelow.

x

t

x

t

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==> Thereisamismatchatx=x0only.

==> Thereisaninternalboundarylayeratx=x0only.

Exampleofcase3:

ε ′′y +2 x − 12

⎛⎝⎜

⎞⎠⎟

′y +4 x − 12⎛⎝⎜

⎞⎠⎟

2

y =0 (wesolveditinLecture6).

Case4:

a x( ) = >0 , forx∈ 0, x0⎡⎣ )

<0 , forx∈ x0 ,1( ⎤⎦

⎧⎨⎪

⎩⎪

Arepresentativediagramofcharacteristicsisshownbelow.

==> Therearemismatchesatx=0andatx=1.

==> Thereareboundarylayersatx=0andatx=1.Exampleofcase4:

ε ′′y − x − 12

⎛⎝⎜

⎞⎠⎟

′y − y +2 x − 12⎛⎝⎜

⎞⎠⎟=0 (wesolveditinLecture6).

xx0

t

xx0

t

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Remark: Thesolutioninthemiddleisdetachedfrombothboundaryconditions.Asaresult,itmaybevirtuallyundetermined.Intheexampleabove,settingx=½,theleadingtermsoftheODEgiveus

y(1/2)=0,whichservesasaanchoringcondition.WewillcomebacktotheanchoringproblemshortlyafterthediscussionofCases5A&B.

Case5A: a(x)≡0, b(x)<0

Characteristics:dx/dt=-a(x)=0,whichareverticallinesinthex-tplane.

Alongeachcharacteristic,wehave

dydt

= b x( ) y − c x( )( ) b(x)<0

==> y(x,t)convergestoc(x)ast→∞.==> y(x,∞)isdeterminedindependentoftwoboundaryconditions.

==> Therearemismatchesatx=0andatx=1.

==> Thereareboundarylayersatx=0andatx=1.Exampleofcase5A:

ε ′′y − y = −2sin x − 12

⎛⎝⎜

⎞⎠⎟, ε→0+ (wesolveditinLecture5)

Case5B: a(x)≡0, b(x)>0b(x)>0

==> y(x,t)divergesto±∞ast→∞.==> TheBVPisill-posed.

Exampleofcase5B:

ε ′′y − y = −2sin x − 12

⎛⎝⎜

⎞⎠⎟, ε→0− (wesolveditinLecture5)

Well-posednessoftheBVP(anchoringproblemincase4)

Whentherearetwoboundarylayers,anouterexpansionisdetachedfromimposedboundaryconditions.Asaresult,theouterexpansionmaybeundetermined.Thisisnotthe

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defectoftheasymptoticanalysis/method.Rather,thisiscausedbythefactthattheunderlyingBVPisalmostill-posed.Theexamplebelowillustratesthisill-posedness.Example:

ε ′′y −2 x − 12( ) ′y =0

y 0( ) =α , y 1( ) =β⎧⎨⎪

⎩⎪, ε→0+

Coefficientofy’:

a x( ) = −2 x − 1

2( ) = >0 , forx∈ 0, 12⎡⎣ )<0 , forx∈ 1

2 ,1( ⎤⎦

⎧⎨⎪

⎩⎪

==> Itiscase4.Outerexpansion:

Weseekanexpansionoftheform

y x( ) = a0 x( )+!

SubstitutingintotheODEyields

−2 x − 12

⎛⎝⎜

⎞⎠⎟a0′ =0

==> a0(x)=c

Question: Whyisconstantcundetermined?Answer: TheBVPisalmostsingular(ill-posed).

Anecessaryconditionforwell-posednessisthatthehomogeneousBVP

ε ′′y −2 x − 12( ) ′y =0

y 0( ) =0 , y 1( ) =0⎧⎨⎪

⎩⎪, ε→0+

hasonlythetrivialsolution,y(x)≡0.

ToshowthattheBVPisalmostsingular,weconstructafunctionu(x)suchthat§ u(x)almostsatisfiestheODE,

§ u(x)satisfiestheboundaryconditions,and§ u(0.5)≈1,

Theconstructionoffunctionu(x)isdescribedinAppendixA.

Methodofmulti-scaleexpansion

Wegobacktoinitialvalueproblems(IVP).

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Example:

′′y + ε ′y + y =0y 0( ) =1 , ′y 0( ) =0

⎧⎨⎪

⎩⎪, ε→0+

Physicalbackgroundoftheproblem:Oscillationofamass-springsystemwithsmallfriction.

(Drawamass-springsystem;introduceacoordinatesystemwithzpointingdownward)Newton’s2ndlaw:

m ′′z = − c ′zDragforce!"# −k z −LR( )

Elasticforce! "$ #$

+ gmGravity%

whereLRistherestlengthofthespring.Wewritetheequationas

m ′′z + c ′z +k z −LR −

gmk

⎛⎝⎜

⎞⎠⎟=0

Lety = z −LR −

gmk.Notethaty=0istherestposition.

PositionyisgovernedbytheIVP

′′y + cm

′y + kmy =0

y 0( ) = y0 , ′y 0( ) =0

⎧

⎨⎪

⎩⎪⎪

Note: Herethegravitydoesnotaffecttheequationaslongasthegravityisaconstant,independentofy.Later,wewillstudyacasewherethegravitychangeswithy.

Whenc=0,wehaveaharmonicoscillation.

′′y + k

my =0

==> ′′y +ω2 y =0

whereω = k

misthefrequencyoftheharmonicoscillation.

Ageneralsolutionis

y t( ) = c1 cos ωt( )+ c2sin ωt( ) Question: Whatdoes“smallfriction”mean?

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Answer: Weneedtolookatadimensionlessquantity.

Non-dimensionalization:

Wefirstidentifyatimescaleandalengthscalefornon-dimensionalization.

1time = ω⎡⎣ ⎤⎦ =

km

⎡

⎣⎢⎢

⎤

⎦⎥⎥

length = y0⎡⎣ ⎤⎦

Let

tnew = told

km, ynew =

yoldy0

Wehave

dydtold

= dydtnew

km

dydtold

2 = d2 ydtnew

2km

Substitutingintoequationyields

km

′′y tnew( )+ cmkm

′y tnew( )+ km y =0

==>′′y + c

mk′y + y =0

Weidentifythesmallparameter:

ε ≡ c

mk

“Smallfriction”meansε ≡ c

mk<<1 .

Thenon-dimensionalIVPis

′′y + ε ′y + y =0y 0( ) =1 , ′y 0( ) =0

⎧⎨⎪

⎩⎪

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Twotimescalesintheexactsolution:

Whenε=0,wehaveaperfectharmonicoscillation.

y(t)=cos(t)

Whenε<<1,wehaveanearlyharmonicoscillationandtheamplitudedecaysslowly.

Therearetwotimescalesintheproblem:a) theperiodofthenearlyharmonicoscillation: O(1)

b) thetimescaleoftheamplitudedecaying: O(1/ϵ)Thetwotimescalescanbeseenmathematicallyintheexactsolution.

Exactsolution:

Characteristicequation:

λ2 + ελ+1=0

==>λ1,2 =

−ε± ε2 −42 = −ε

2 ± i 1− ε2

4

Ageneralsolutionis

y t( ) = e

−ε2 t c1 cos 1− ε2

4 t + c2sin 1− ε2

4 t⎛

⎝⎜⎜

⎞

⎠⎟⎟

Applyingtheinitialconditionsy(0)=1andy’(0)=0,wehave

yext t( ) = e−ε2 t cos 1− ε2

4 t +ε21− ε2

4

sin 1− ε2

4 t

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

Theexactsolutionshowstwotimescales:

a) theperiodofthenearlyharmonicoscillation: 2π

b) thehalf-lifeoftheamplitudedecaying: 2εlog 2( )

Straightforwardexpansion:

AderivationofthestraightforwardexpansionfromthedifferentialequationisgivenintheAppendixB.Hereweobtainthestraightforwardexpansionviaashortcut,byTaylorexpandingtheexactsolution.WekeepO(1)termsandO(ϵ)termsintheTaylorexpansion.

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yext t( ) = e−ε2 t cos 1− ε2

4 t +ε21− ε2

4

sin 1− ε2

4 t

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

= 1− ε

2t +!⎛⎝⎜

⎞⎠⎟cost + ε

2sint +!⎛⎝⎜

⎞⎠⎟

= cost + ε 1

2sint −t2cost

⎛⎝⎜

⎞⎠⎟+!

Atwo-termstraightforwardexpansionis

yasy t( ) = cost + ε 1

2sint −t2cost

⎛⎝⎜

⎞⎠⎟

Letuslookattheperformanceoftheleadingterm,usingthesecondtermasanerrorindicator.

Error t( ) = yext t( )− cos t( )Leadingtermofyasy

!"# $#≈ ε2sint −

εt2 cost

Secondtermofyasy

! "### $###

Fort∈[0,O(1)],wehave

Error t( ) =O ε( )

Fort=O(1/ϵ),wehave

Error t( ) =O 1( ) Conclusion: Thestraightforwardexpansionisonlyvalidfort∈[0,O(1)].

NewGoal: Tofindaleadingtermexpansionthatisvaliduniformlyforlongtime.Thelargeerrorinthestraightforwardexpansionatlongtimeiscausedbytheterm(ϵtcost),whichiscalledasecularterm(aseculartermisanoscillatingtermwithagrowingamplitude).

Coefficient(ϵt)insecularterm(ϵtcost)comesfromexpandingexp(-ϵt/2).

Toaccommodatet∈[0,O(1/ϵ)],weshouldavoidexpandingexp(-ϵt/2).Letusseewhathappensifwedonotexpandexp(-ϵt/2).

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yext t( ) = e−ε2 t cos 1− ε2

4 t +ε21− ε2

4

sin 1− ε2

4 t

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

= e

−ε2 t cos t +O ε2t( )( )+O ε( )⎡

⎣⎤⎦

= e−ε2 t cos t( )+O ε2t( )+O ε( )⎡

⎣⎤⎦

==> yext t( ) = e−ε2 t cost +O ε( ) isvaliduniformlyfort∈[0,O(1/ϵ)].

Note: Fort≫O(1/ϵ),thesolutionisvirtuallyzerobecauseofthefactorexp(-ϵt/2).

Summary:

Toaccommodatet∈[0,O(1/ϵ)],weshouldavoidexpandingexp(-ϵt/2).

Strategyofmulti-scaleexpansion:

WetreatT0=tandT1=ϵtastwoseparatevariables.

(andthus,exp(-ϵt/2)iskeptasexp(-T1/2),insteadofbeingexpanded).

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AppendixAWeconstructafunctionu(x)suchthat

§ u(x)almostsatisfiestheODE: ε ′′y −2 x − 1

2( ) ′y = o εn( ) foranyn,§ u(x)satisfiestheboundaryconditions:

u(0)=0andu(1)=0, and

§ u(0.5)≈1,

Toconstructthefunction,westartbyexaminingtheexactsolutionofε ′′y −2 x − 12( ) ′y =0 .

Exactsolution:

ε ′′y −2 x − 12

⎛⎝⎜

⎞⎠⎟

′y =0

==>exp

− x − 12( )2

ε

⎛

⎝⎜

⎞

⎠⎟ ′y

⎛

⎝⎜

⎞

⎠⎟′

=0

==>exp

− x − 12( )2 + 1

4

ε

⎛

⎝⎜

⎞

⎠⎟ ′y = c0

==>′y x( ) = c0 exp

− 14 + x − 1

2( )2ε

⎛

⎝⎜

⎞

⎠⎟

==>y x( ) = c0 exp

− 14 + s − 1

2( )2ε

⎛

⎝⎜

⎞

⎠⎟ ds

0

x

∫ + c1

Weset y(0)=0andy(1)=2

(Notethaty(1)=2isnotthehomogeneousboundarycondition!)Weobtain

c1=0

c0 =2

exp − 14 + s − 1

2( )2ε

⎛

⎝⎜

⎞

⎠⎟ ds

0

1

∫= 1ε1+O ε( )( )

(Wewillstudyasymptoticexpansionsofintegralslater)

y 12

⎛⎝⎜

⎞⎠⎟=1 (symmetry)

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′y 12

⎛⎝⎜

⎞⎠⎟= c0 exp

− 14 + x − 1

2( )2ε

⎛

⎝⎜

⎞

⎠⎟x=12

= c0 exp−14ε

⎛⎝⎜

⎞⎠⎟ ≡ η ε( )

whereη ε( ) = 1

εexp −1

4ε⎛⎝⎜

⎞⎠⎟ 1+O ε( )( ) = o εn( ) foranyn

′′y 12

⎛⎝⎜

⎞⎠⎟= c0

2 x − 12( )

εexp

− 14 + x − 1

2( )2ε

⎛

⎝⎜

⎞

⎠⎟x=12

=0

Nextweconsider

y1 x( ) = y x( )− η ε( )x Functiony1(x)satisfies

y1(0)=0

y1′

12

⎛⎝⎜

⎞⎠⎟=0 , y1′′

12

⎛⎝⎜

⎞⎠⎟=0

ε ′′y1 −2 x − 12

⎛⎝⎜

⎞⎠⎟

′y1 = −2 x − 12⎛⎝⎜

⎞⎠⎟−η ε( )x( )′ =2η ε( ) x − 12

⎛⎝⎜

⎞⎠⎟

(Recallthaty(x)satisfiestheODEexactly.)

y1

12

⎛⎝⎜

⎞⎠⎟=1− η ε( )12 ≈1

Finally,weconsider

u x( ) = y1 x( ) , x ≤ 1

2

y1 1− x( ) , x > 12

⎧⎨⎪

⎩⎪

Functionu(x)satisfies

§ ε ′′y −2 x − 12( ) ′y = o εn( ) foranyn,

§ u(0)=0andu(1)=0, and§ u(0.5)≈1

Therefore,weconcludethattheBVP

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ε ′′y −2 x − 12( ) ′y =0

y 0( ) =α , y 1( ) =β⎧⎨⎪

⎩⎪, ε→0+

isveryclosetobeingsingular(althoughitisnotexactlysingular).

AppendixBWederiveastraightforwardexpansionoftheIVP

′′y + ε ′y + y =0y 0( ) =1 , ′y 0( ) =0

⎧⎨⎪

⎩⎪

Weseekanexpansionoftheform

y t( ) = a0 t( )+ εa1 t( )+!

Initialcondition:

a0(0)+ϵa1(0)+…=1 0 0( ) + 1 0( ) +! =1==> a0(0)=1, a1(0)=0

a0’(0)+ϵa1’(0)+…=0

==> a0’(0)=0, a1’(0)=0

Substitutingintotheequationyields

′′a0 + ε ′′a1( )+ ε ′a01( )+ a0 + εa1( ) =0 ==> ′′a0 +a0⎡⎣ ⎤⎦+ ε ′′a1 +a1 + ′a0⎡⎣ ⎤⎦ =0

Allcoefficientsmustbezero.

ε0:

′′a0 +a0 =0a0 0( ) =1 , ′a0 0( ) =0

⎧⎨⎪

⎩⎪

==> a0 t( ) = cos t( )

ε1:

′′a1 +a1 = − ′a0 = sinta1 0( ) =0 , ′a1 0( ) =0

⎧⎨⎪

⎩⎪

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LetA s( ) = L a1 t( )⎡⎣ ⎤⎦ .

TakingLaplacetransformyields

s2A s( )+ A s( ) = 1

s2 +1

==>

A s( ) = 1

s2 +1( )2

Recall

L−1 1

s2 +ω2( )⎡

⎣⎢⎢

⎤

⎦⎥⎥= sinωt

ω

Differentiatingwithrespecttoωyields

L−1 −2ω

s2 +ω2( )2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= −sinωt

ω2 + t cosωtω

==>

L−1 1

s2 +ω2( )2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 12ω2

sinωtω

−t cosωt⎛⎝⎜

⎞⎠⎟

==>

a1 t( ) = L−1 1

s2 +1( )2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 12 sint −t cost( )

==>yasy t( ) = cos t( )+ ε12 sint −t cost⎡⎣ ⎤⎦