AMS 212B Perturbation Methods - University of California...

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- 1 - AMS 212B Perturbation Methods Lecture 06 Copyright by Hongyun Wang, UCSC Recap: Question: Where is the boundary layer? The case of constant coefficients ε ′′ y + a y + by = 0 y 0 () = α , y 1 () = β , a 0, ε→ 0 We have λ 1 b a (regular root) λ 2 a ε (singular root) When λ 2 a ε → −∞ , a boundary layer is at x = 0. When λ 2 a ε +, a boundary layer is at x = 1. We continue the discussion on the location of boundary layer. A more general case (the case of variable coefficients) ε ′′ y + ax ( ) y + bx ( ) y = 0 y 0 () = α , y 1 () = β , ε→ 0 + Here we focus on the case of ϵ → 0+. The case of ϵ → 0- can be treated similarly. We first freeze the coefficients at x = 0. Near x = 0, the ODE is approximately ε ′′ y + a 0 () y + b 0 () y = 0 If a(0) > 0, then λ 2 a 0 () ε → −∞ . ==> there is a boundary layer at x = 0.

Transcript of AMS 212B Perturbation Methods - University of California...

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AMS212BPerturbationMethodsLecture06

CopyrightbyHongyunWang,UCSC

Recap:

Question: Whereistheboundarylayer?Thecaseofconstantcoefficients

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

⎧⎨⎪

⎩⎪, a≠0 , ε→0

Wehave

λ1 ≈

−ba (regularroot)

λ2 ≈

−aε (singularroot)

Whenλ2 ≈

−aε→−∞ ,aboundarylayerisatx=0.

Whenλ2 ≈

−aε→ +∞ ,aboundarylayerisatx=1.

Wecontinuethediscussiononthelocationofboundarylayer.Amoregeneralcase(thecaseofvariablecoefficients)

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

⎧⎨⎪

⎩⎪, ε→0+

Herewefocusonthecaseofϵ→0+.Thecaseofϵ→0-canbetreatedsimilarly.

Wefirstfreezethecoefficientsatx=0.

Nearx=0,theODEisapproximately

ε ′′y +a 0( ) ′y +b 0( ) y =0

Ifa(0)>0,thenλ2 ≈

−a 0( )ε

→−∞ .

==> thereisaboundarylayeratx=0.

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Ifa(0)<0,thenλ2 ≈

−a 0( )ε

→ +∞ ,

==> thereisnoboundarylayeratx=0.Note: a(0)canonlytellusaboutwhathappensatx=0.Wecannotpredictanythingat

positionx=1bylookingata(0).

Nextwefreezethecoefficientsatx=1.

Nearx=1,theODEisapproximately

ε ′′y +a 1( ) ′y +b 1( ) y =0

Ifa(1)>0,thenλ2 ≈

−a 1( )ε

→−∞ .

==> thereisnoboundarylayeratx=1.

Ifa(1)<0,thenλ2 ≈

−a 1( )ε

→ +∞ ,

==> thereisaboundarylayeratx=1.

Note: a(1)canonlytellusaboutwhathappensatx=1.

Summary(forthecaseofϵ→0+)

Ifa(0)>0anda(1)>0,thenthereisaboundarylayeratx=0butnoboundarylayeratx=1.

Ifa(0)<0anda(1)<0,thenthereisaboundarylayeratx=1butnoboundarylayeratx=0.

Ifa(0)>0anda(1)<0,thenthereareboundarylayersatbothx=0andx=1.

Ifa(0)<0anda(1)>0,thenthereisnoboundarylayeratx=0andthereisnoboundarylayeratx=1.

==> thereisaninternalboundarylayer.

Example: (boundarylayersatbothx=0andx=1)

ε ′′y − x − 12⎛⎝⎜

⎞⎠⎟

′y − y +2 x − 12⎛⎝⎜

⎞⎠⎟=0

y 0( ) =0 , y 1( ) =0

⎨⎪⎪

⎩⎪⎪

, ε→0+

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a x( ) = − x − 12

⎛⎝⎜

⎞⎠⎟

==>a 0( ) = 12 >0 , a 1( ) = −1

2 <0

==> boundarylayersatbothx=0andx=1

Inthecalculationbelow,wefindonlytheleadingtermoftheexpansion.

Thesolutionoftwo-termexpansionisgivenintheAppendix.

• Outerexpansion

Weseekanexpansionoftheform

yout( ) x( ) = a0 x( )+O ε( )

Note: noboundaryconditionisimposedony(out).

Substitutingintoequationyields

− x − 12⎛⎝⎜

⎞⎠⎟

′a0 −a0 +2 x − 12⎛⎝⎜

⎞⎠⎟=0

==>x − 12

⎛⎝⎜

⎞⎠⎟a0 x( )⎡

⎣⎢

⎦⎥′=2 x − 12

⎛⎝⎜

⎞⎠⎟

==>x − 12

⎛⎝⎜

⎞⎠⎟a0 x( ) = x − 12

⎛⎝⎜

⎞⎠⎟

2

+ c

==>

a0 x( ) = x − 12⎛⎝⎜

⎞⎠⎟+ c

x − 12⎛⎝⎜

⎞⎠⎟

a0(x)isfinite

==> c=0

==>a0 x( ) = x − 12

⎛⎝⎜

⎞⎠⎟

==>y out( ) x( ) = x − 12

⎛⎝⎜

⎞⎠⎟

• Innerexpansionatx=0:

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(WeassumethewidthofboundarylayerisO(ε).Wewilldiscusshowtodeterminethewidthofboundarylayerlater.)

Letu= x

εbetheinnervariable. Wehavex=ϵuand

dydx

= dydu

⋅1ε, d2 y

dx2= d

2 ydu2

⋅ 1ε2

==>ε ′′y u( ) 1

ε2− εu− 12⎛⎝⎜

⎞⎠⎟

′y u( )1ε − y u( )+2 εu− 12⎛⎝⎜

⎞⎠⎟=0

==>′′y u( )+ 12 ′y u( )+ ε −u ′y u( )− y u( )−1( ) =0

==>′′y u( )+ 12 ′y u( )+O ε( ) =0

ThisistheODEintermsofinnervariableu=x/ϵafterscaling.Weseekanexpansionoftheform

yinnL( ) u( ) = a0 u( )+O ε( )

Boundaryconditionatx=0: y(x)|x=0=0, x=ϵu

==> a0(0)=0

Note: Onlytheboundaryconditionatx=0isimposedintheinnerexpansionatx=0.Substitutingintoequationyields

′′a0 +12 ′a0 =0

a0 0( ) =0

⎨⎪

⎩⎪⎪

==>a0 u( ) = c0 1−e

−u2⎛

⎝⎜⎞

⎠⎟

==>y innL( ) u( ) = c0 1−e

−u2⎛

⎝⎜⎞

⎠⎟

• Matchingatx=0

Recalltheouterexpansion: y out( ) x( ) = x − 12

⎛⎝⎜

⎞⎠⎟

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Theinnerlimitofouterexpansionatx=0is

y out( ) innL( ) u( ) = x − 12

⎛⎝⎜

⎞⎠⎟= −12

Theouterlimitofinnerexpansion:

yinnL( ) out( ) u( ) = c0

Thematchingcondition:

yout( ) innL( ) u( ) = y innL( ) out( ) u( )

==> c0=-1/2

==>y mL( ) u( ) = −1

2

y innL( ) u( ) = −1

2 1−e−u2⎛

⎝⎜⎞

⎠⎟

y innL( ) u( )− y mL( ) u( ) = 12e

−u2

Note: y(innL)(u)–y(m)(u)isneededwhenwewriteoutthecompositeexpansion.

• Innerexpansionatx=1:

Letv = 1− x

εbetheinnervariable. Wehavex=1−ϵvand

dydx

= dydv

⋅ −1ε

⎛⎝⎜

⎞⎠⎟, d2 y

dx2= d

2 ydv2

⋅ 1ε2

==>ε ′′y v( ) 1

ε2− 12− εv

⎛⎝⎜

⎞⎠⎟

′y v( ) −1ε

⎛⎝⎜

⎞⎠⎟− y v( )+2 1

2− εv⎛⎝⎜

⎞⎠⎟=0

==>′′y v( )+ 12 ′y v( )+ ε −v ′y v( )− y v( )+1( ) =0

==>′′y v( )+ 12 ′y v( )+O ε( ) =0

Weseekanexpansionoftheform

yiR( ) v( ) = a0 v( )+O ε( )

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Boundaryconditionatx=1: y(x)|x=1=0, (1-x)=ϵv.

==> a0(0)=0

Note: Onlytheboundaryconditionatx=1isimposedintheinnerexpansionatx=1.Substitutingintoequationyields

′′a0 +12 ′a0 =0

a0 0( ) =0

⎨⎪

⎩⎪⎪

==>a0 v( ) = d0 1−e

−v2⎛

⎝⎜⎞

⎠⎟

==>y innR( ) v( ) = d0 1−e

−v2⎛

⎝⎜⎞

⎠⎟

• Matchingatx=1

Recalltheouterexpansion: y out( ) x( ) = x − 12

⎛⎝⎜

⎞⎠⎟

Theinnerlimitofouterexpansionatx=1

y out( ) innR( ) v( ) = x − 12

⎛⎝⎜

⎞⎠⎟= 12− 1− x( ) = 12+!

Theouterlimitofinnerexpansion:

yinnR( ) out( ) v( ) = d0

Thematchingcondition

yout( ) innR( ) v( ) = y innR( ) out( ) v( )

==> d0=1/2

==>y mR( ) v( ) = 12

y innR( ) v( ) = 12 1−e

−v2⎛

⎝⎜⎞

⎠⎟

y innR( ) v( )− y mR( ) v( ) = −1

2 e−v2

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Thecompositeexpansionis

y c( ) x( ) = y out( ) x( )+ y innL( ) u( )− y mL( ) u( )( )

u=xε

+ y innR( ) v( )− y mR( ) v( )( )v=1−xε

==>y c( ) x( ) = x − 12

⎛⎝⎜

⎞⎠⎟+ 12e

−x2ε − 12e

− 1−x( )2ε

DerivationofthetwotermasymptoticexpansionisgiveninAppendix.

Thefigurebelowcomparesthe“exactsolution”andthecompositeexpansionsforε=0.05.

Example: (internalboundarylayer)

ε ′′y +2 x − 12⎛⎝⎜

⎞⎠⎟

′y +4 x − 12⎛⎝⎜

⎞⎠⎟

2

y =0

y 0( ) =1 , y 1( ) = −2

⎨⎪⎪

⎩⎪⎪

, ε→0+

a x( ) =2 x − 12

⎛⎝⎜

⎞⎠⎟

==> a(0)=-1<0, a(1)=1>0

==> Thereisaninternalboundarylayer.

LetxBbethelocationoftheinternalboundarylayer.

Question: xB=?

Claim: xBsatisfiesa(xB)=0.

0 0.2 0.4 0.6 0.8 1-0.3

-0.2

-0.1

0

0.1

0.2

0.3

x

y

ε = 0.05

Exact solutionasymptotic with 1 termasymptotic with 2 terms

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Ifa(xB)>0,thenweconsidertheinterval[0,xB].

a(xB)>0 ==> noboundarylayeratx=xB.

Ifa(xB)<0,thenweconsidertheinterval[xB,1].

a(xB)<0 ==> noboundarylayeratx=xB.

Therefore,xBmustsatisfya(xB)=0.

• OuterexpansionWeseekanexpansionoftheform

y x( ) = a0 x( )+O ε( ) Substitutingintoequationyields

2 x − 12⎛⎝⎜

⎞⎠⎟

′a0 +4 x − 12⎛⎝⎜

⎞⎠⎟

2

a0 =0

==>′a0 +2 x − 12

⎛⎝⎜

⎞⎠⎟a0 =0

==> ex−12

⎝⎜

⎠⎟

2

a0

⎛⎝⎜

⎞⎠⎟′

=0

==> a0 x( ) = ce− x−12⎛

⎝⎜

⎠⎟

2

Theleftsegmentofouterexpansion,

Boundaryconditionatx=0: y(x)|x=0=1

==> a0(0)=1

Note: Onlytheboundaryconditionatx=0isimposedontheleftsegment.

==> ce−14 =1

==> c = e14

==> youtL( ) x( ) =e

14− x−12

⎝⎜

⎠⎟

2

Therightsegmentofouterexpansion,

Boundaryconditionatx=1: y(x)|x=1=-2

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==> a0(1)=-2

Note: Onlytheboundaryconditionatx=1isimposedontherightsegment.

==> ce−14 = −2

==> c = −2e14

==> youtR( ) x( ) =−2e

14− x−12

⎝⎜

⎠⎟

2

• InnerexpansionQuestion: Howtofindthewidthofaboundarylayer?

Answer: Principleofleastdegeneracy

Tryu= 1

εαx − 12

⎛⎝⎜

⎞⎠⎟whereα>0.Wehave

x − 12 = ε

αu

dydx

= dydu

⋅ 1εα, d2 y

dx2= d

2 ydu2

⋅ 1ε2α

SubstitutingintotheODEyields

ε ′′y u( ) 1

ε2α+ εα2u ′y u( ) 1

εα+ ε2α4u2 y u( ) =0

==> ′′y u( )+ ε2α−12u ′y u( )+ ε4α−14u2 y u( ) =0 Sinceα>0,wealwayshaveϵ2α-1≫ϵ4α-1.

Thatis,wealwayshavey’(u)term≫y(u)term.Asaresult,weonlyneedtocompare y’’(u)termandy’(u)term.

§ Forα<0.5(thesmallerα,thelargerϵα),wehave

y’(u)term≫y’’(u)term ==> theODEisdegenerate

§ Forα>0.5,wehave

y’’(u)term≫y’(u)term ==> theODEisdegenerate§ Atα=0.5,wehave

y’’(u)term~y’(u)term ==> theODEislessdegenerate

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Theprincipleofleastdegeneracytellsusthatαshouldtakethevaluethatmakestheequationleastdegenerate.

Thus,weconcludeα=1/2.

Thedifferentialequationafterscalingis

′′y u( )+2u ′y u( )+ ε4u2 y u( ) =0 Twomoreexamplesfortheprincipleofleastdegeneracy:

Beforewecontinuewiththeinnerexpansionofinternalboundarylayer,weapplytheprincipleofleastdegeneracytotwomoreexampleswestudiedpreviously.

Example(determinethewidthofboundarylayer)

ε ′′y − y = −2sin x − 12

⎛⎝⎜

⎞⎠⎟, ε→0+

Tryu= x

εα, α>0

==>′′y u( )− ε2α−1 ′y u( ) = −ε2α−12sin εαu− 12

⎛⎝⎜

⎞⎠⎟

==> 2α-1=0

==> α=1/2.Example(determinethewidthofboundarylayer)

ε ′′y − x − 12

⎛⎝⎜

⎞⎠⎟

′y − y +2 x − 12⎛⎝⎜

⎞⎠⎟=0 , ε→0+

Tryu= x

εα, α>0

==>′′y u( )− εα−1 εαu− 12

⎛⎝⎜

⎞⎠⎟

′y u( )+ ε2α−1 y u( )+ ε2α−12 εαu− 12⎛⎝⎜

⎞⎠⎟=0

==> α–1=0

==> α=1

Nowbacktotheexampleofinternalboundarylayer.

• Innerexpansion(continued)RecallthattheODEinuis

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′′y u( )+2u ′y u( )+ ε4u2 y u( ) =0 Weseekanexpansionoftheform

yinn( ) u( ) = a0 u( )+O ε( )

Note: noboundaryconditionisimposedony(inn).

Substitutingintoequationyields

′′a0 u( )+2u ′a0 u( ) =0

==> eu2 ′a0 u( )( )′ =0

==> ′a0 u( ) = c0e−u2

==>a0 u( ) = c1 + c0 e−v

2dv

0

u

∫ = c1 + c0π2 Erf u( )

wheretheerrorfunctionErf(z)isdefinedas

Erf z( ) = 2

πe−v

2dv

0

z

Renamingc0 = c0

π2 ,wehave

a0 u( ) = c1 + c0Erf u( ) Theinnerexpansionis

yinn( ) u( ) = c1 + c0Erf u( )

• Matchingattheleftsideofboundarylayer

First,werecallthattheerrorfunctionsatisfiesErf(0)=0, Erf(∞)=1, Erf(-∞)=-1

Andrecallthetwosegmentsofouterexpansion:

youtL( ) x( ) =e

14− x−12

⎝⎜

⎠⎟

2

, y outR( ) x( ) =−2e14− x−12

⎝⎜

⎠⎟

2

Theinnerlimitoftheleftsegmentofouterexpansion:

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youtL( ) inn( ) = lim

x→0.5y outL( ) x( ) = e

14

Theleftouterlimitofinnerexpansion:

yinn( ) outL( ) = lim

u→−∞y inn( ) u( ) = c1 − c0

Thematchingcondition:

youtL( ) inn( ) = y inn( ) outL( )

==> c1 − c0 = e14

whichprovidesoneequationforc0andc1.

• MatchingattherightsideofboundarylayerTheinnerlimitoftherightsegmentofouterexpansion:

youtR( ) inn( ) = lim

x→0.5y outR( ) x( ) = −2e

14

Therightouterlimitofinnerexpansion:

yinn( ) outR( ) = lim

u→+∞y inn( ) u( ) = c1 + c0

Thematchingcondition

youtR( ) inn( ) = y inn( ) outR( )

==> c0 + c1 = −2e14

Thus,wehavetwoequationsforthetwounknowncoefficients

c1 − c0 = e14

c1 + c0 = −2e14

⎨⎪⎪

⎩⎪⎪

==>

c0 = −32e

14

c1 = −12e

14

⎨⎪⎪

⎩⎪⎪

==>y inn( ) u( )− y mL( ) = −32e

14 1+Erf u( )( )

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y inn( ) u( )− y mR( ) = 32e

14 1−Erf u( )( )

Thecompositeexpansionis

y c( ) x( ) =y outL( ) x( )+ y inn( ) u( )− y mL( )( )

u=x−0.5ε

, x <0.5

y outR( ) x( )+ y inn( ) u( )− y mR( )( )u=x−0.5

ε

, x >0.5

⎪⎪

⎪⎪

=

e14− x−12

⎝⎜

⎠⎟

2

− 32e14 1+Erf x −0.5

ε

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎠⎟ , x <0.5

−2e14− x−12

⎝⎜

⎠⎟

2

+ 32e14 1−Erf x −0.5

ε

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎠⎟ , x >0.5

⎪⎪⎪

⎪⎪⎪

Thefigurebelowcomparestheexactsolutionandthecompositeexpansion(withonlytheleadingterm)forε=0.01.Theanalyticalsolutionoftheproblemisactuallyunknown.Herethe“exactsolution”meansaveryaccuratenumericalsolution.

AppendixDerivationofatwo-termexpansionfortheboundaryvalueproblem

0 0.2 0.4 0.6 0.8 1

-2

-1

0

1

x

y

ε = 0.01

Exact solutionasymptotic

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ε ′′y − x − 12⎛⎝⎜

⎞⎠⎟

′y − y +2 x − 12⎛⎝⎜

⎞⎠⎟=0

y 0( ) =0 , y 1( ) =0

⎨⎪⎪

⎩⎪⎪

, ε→0+

(boundarylayersatbothx=0andx=1)

• OuterexpansionWeseekanexpansionoftheform

yout( ) x( ) = a0 x( )+ εa1 x( )+O ε2( )

Note: noboundaryconditionisimposedony(out).

Substitutingintoequationyields

ε ′′a0( )− x − 12

⎛⎝⎜

⎞⎠⎟

′a0 + ε ′a1( )− a0 + εa1( )+2 x − 12⎛⎝⎜

⎞⎠⎟=0

==> − x − 12⎛⎝⎜

⎞⎠⎟

′a0 −a0 +2 x − 12⎛⎝⎜

⎞⎠⎟

⎣⎢

⎦⎥+ ε ′′a0 − x − 12

⎛⎝⎜

⎞⎠⎟

′a1 −a1⎡

⎣⎢

⎦⎥+!=0

Allcoefficientsmustbezero.

ε0:− x − 12⎛⎝⎜

⎞⎠⎟

′a0 −a0 +2 x − 12⎛⎝⎜

⎞⎠⎟=0

==>a0 x( ) = x − 12

⎛⎝⎜

⎞⎠⎟

ε1:x − 12

⎛⎝⎜

⎞⎠⎟

′a1 +a1 = ′′a0 =0

==>x − 12

⎛⎝⎜

⎞⎠⎟a1

⎣⎢

⎦⎥′=0

==>a1 x( ) = c

x − 12

a1(x)issmooth ==> c=0

==> a1 x( )≡0 Theouterexpansionis

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y out( ) x( ) = x − 12

⎛⎝⎜

⎞⎠⎟+O ε2( )

• Innerexpansionatx=0:

Letu= x

εbetheinnervariable.Wehavex=ϵuand

dydx

= dydu

⋅1ε, d2 y

dx2= d

2 ydu2

⋅ 1ε2

==>ε ′′y u( ) 1

ε2− εu− 12⎛⎝⎜

⎞⎠⎟

′y u( )1ε − y u( )+2 εu− 12⎛⎝⎜

⎞⎠⎟=0

==>′′y u( )+ 12 ′y u( )+ ε −u ′y u( )− y u( )−1( ) =0

Weseekanexpansionoftheform

yinnL( ) u( ) = a0 u( )+ εa1 u( )+O ε2( )

Boundaryconditionatx=0: y(x)x=0=0, x=ϵu

==> a0 0( )+ εa1 0( )+!=0

==> a0(0)=0, a1(0)=0

Note: Onlytheboundaryconditionatx=0isimposedintheinnerexpansionatx=0.

Substitutingintotheequationyields

′′a0 + ε ′′a1( )+ 12 ′a0 + ε ′a1( )+ ε −u ′a0( )− a0( )−1( ) =0

==>

′′a0 +12 ′a0

⎣⎢

⎦⎥+ ε ′′a1 +

12 ′a1 −u ′a0 −a0 −1

⎣⎢

⎦⎥+!=0

Allcoefficientsmustbezero.

ε0:

′′a0 +12 ′a0 =0

a0 0( ) =0

⎨⎪

⎩⎪⎪

==>a0 u( ) = c0 1−e

−u2⎛

⎝⎜⎞

⎠⎟

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ε1:

′′a1 +12 ′a1 =1+a0 +u ′a0 =1+ c0 1−e

−u2⎛

⎝⎜⎞

⎠⎟+c02 ue

−u2

a1 0( ) =0

⎨⎪⎪

⎩⎪⎪

LetA s( ) = L a1 u( )⎡⎣ ⎤⎦ andc1=a1’(0).

TakingLaplacetransformyields

s2A s( )− c1 + 12 sA s( ) = 1s + c0

1s− 1s + 1

2

⎝⎜⎞

⎠⎟+c02

1s + 1

2( )2

==>s2 + 12 s

⎛⎝⎜

⎞⎠⎟A s( ) = c1 + 1s +

c02

1s + 1

2( )1s+ 1s + 1

2

⎝⎜⎞

⎠⎟

==>

A s( ) =2c1 1

s− 1s + 1

2

⎝⎜⎞

⎠⎟+ 2s1s− 1s + 1

2

⎝⎜⎞

⎠⎟+ c0

1s + 1

2( )1s2

− 1s + 1

2( )2⎛

⎝⎜⎜

⎠⎟⎟

==>

A s( ) = 2c1 −4−4c0( ) 1s −

1s + 1

2

⎝⎜⎞

⎠⎟+2+2c0s2

− c01

s + 12( )3

Renamec1 = 2c1 −4−4c0( ) ,wehave

A s( ) = c1 1

s− 1s + 1

2

⎝⎜⎞

⎠⎟+2+2c0s2

− c01

s + 12( )3

==>a1 u( ) = L−1 A s( )⎡⎣ ⎤⎦ = c1 1−e−

u2⎛

⎝⎜⎞

⎠⎟+ 2+2c0( )u− c02 u

2e−u2

Theinnerexpansionatx=0is

y innL( ) u( ) = c0 1−e

−u2⎛

⎝⎜⎞

⎠⎟+ ε c1 1−e−

u2⎛

⎝⎜⎞

⎠⎟+ 2+2c0( )u− c02 u

2e−u2⎡

⎣⎢⎢

⎦⎥⎥+O ε2( )

• Matchingatx=0

Recalltheouterexpansion: y out( ) x( ) = x − 12

⎛⎝⎜

⎞⎠⎟+O ε2( )

Theinnerlimitofouterexpansionatx=0

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y out( ) innL( ) u( ) = −1

2 + εu

Theouterlimitofinnerexpansionatx=0

yinnL( ) out( ) u( ) = c0 + ε c1 + 2+2c0( )u⎡⎣ ⎤⎦

Thematchingcondition

yout( ) innL( ) u( ) = y innL( ) out( ) u( )

==> c0=-1/2, c1=0

==>y mL( ) u( ) = −1

2 + εu

y innL( ) u( ) = −1

2 1−e−u2⎛

⎝⎜⎞

⎠⎟+ ε u+ 14u

2e−u2⎡

⎣⎢

⎦⎥+O ε2( )

y innL( ) u( )− y mL( ) u( ) = 12e

−u2 + ε

4u2e

−u2

• Innerexpansionatx=1:

Letv = 1− x

εbetheinnervariable.Wehave(1−x)=ϵvand

dydx

= dydv

⋅ −1ε

⎛⎝⎜

⎞⎠⎟, d2 y

dx2= d

2 ydv2

⋅ 1ε2

==>ε ′′y v( ) 1

ε2− 12− εv

⎛⎝⎜

⎞⎠⎟

′y v( ) −1ε

⎛⎝⎜

⎞⎠⎟− y v( )+2 1

2− εv⎛⎝⎜

⎞⎠⎟=0

==>′′y v( )+ 12 ′y v( )+ ε −v ′y v( )− y v( )+1( ) =0

Weseekanexpansionoftheform

yinnR( ) v( ) = a0 v( )+ εa1 v( )+O ε2( )

Boundaryconditionatx=1: y(x)|x=1=0, (1−x)=ϵv.

==> a0 0( )+ εa1 0( )+!=0

==> a0(0)=0, a1(0)=0

Note: Onlytheboundaryconditionatx=1isimposedintheinnerexpansionatx=1.

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Substitutingintoequationyields

′′a0 + ε ′′a1( )+ 12 ′a0 + ε ′a1( )+ ε −v ′a0( )− a0( )+1( ) =0

==>

′′a0 +12 ′a0

⎣⎢

⎦⎥+ ε ′′a1 +

12 ′a1 − v ′a0 −a0 +1

⎣⎢

⎦⎥+!=0

Allcoefficientsmustbezero.

ε0:

′′a0 +12 ′a0 =0

a0 0( ) =0

⎨⎪

⎩⎪⎪

==>a0 v( ) = d0 1−e

−v2⎛

⎝⎜⎞

⎠⎟

ε1:

′′a1 +12 ′a1 = −1+a0 + v ′a0 = −1+d0 1−e

−v2⎛

⎝⎜⎞

⎠⎟+d02 ve

−v2

a1 0( ) =0

⎨⎪⎪

⎩⎪⎪

LetA s( ) = L a1 v( )⎡⎣ ⎤⎦ andd1=a1’(0).

TakingLaplacetransformyields

s2A s( )−d1 + 12 sA s( ) = −1

s+d0

1s− 1s + 1

2

⎝⎜⎞

⎠⎟+d02

1s + 1

2( )2

==>s2 + 12 s

⎛⎝⎜

⎞⎠⎟A s( ) = d1 − 1s +

d02

1s + 1

2( )1s+ 1s + 1

2

⎝⎜⎞

⎠⎟

==>

A s( ) =2d1 1

s− 1s + 1

2

⎝⎜⎞

⎠⎟− 2s1s− 1s + 1

2

⎝⎜⎞

⎠⎟+d0

1s + 1

2( )1s2

− 1s + 1

2( )2⎛

⎝⎜⎜

⎠⎟⎟

==>

A s( ) = 2d1 +4−4d0( ) 1s −

1s + 1

2

⎝⎜⎞

⎠⎟+−2+2d0s2

−d01

s + 12( )3

Renamed1 = 2d1 +4−4d0( ) ,wehave

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A s( ) = d1 1

s− 1s + 1

2

⎝⎜⎞

⎠⎟+−2+2d0s2

−d01

s + 12( )3

==>a1 v( ) = L−1 A s( )⎡⎣ ⎤⎦ = d1 1−e−

v2⎛

⎝⎜⎞

⎠⎟+ −2+2d0( )v − d02 v

2e−v2

==>y innR( ) v( ) = d0 1−e

−v2⎛

⎝⎜⎞

⎠⎟+ ε d1 1−e−

v2⎛

⎝⎜⎞

⎠⎟+ −2+2d0( )v − d02 v

2e−v2⎡

⎣⎢⎢

⎦⎥⎥+O ε2( )

• Matchingatx=1:

Recalltheouterexpansion: y out( ) x( ) = x − 12

⎛⎝⎜

⎞⎠⎟+O ε2( )

Theinnerlimitofouterexpansionatx=1:

y out( ) innR( ) v( ) = 12− 1− x( ) = 12− εv

Theouterlimitofinnerexpansionatx=1:

yinnR( ) out( ) v( ) = d0 + ε d1 + −2+2d0( )v⎡⎣ ⎤⎦

Thematchingcondition:

yout( ) innR( ) v( ) = y innR( ) out( ) v( )

==>d0 =

12 , d1 =0

y mR( ) v( ) = 12− εv

y innR( ) v( )− y mR( ) v( ) = −1

2 e−v2 − ε

4 v2e

−v2

• Thecompositeexpansion:

y c( ) x( ) = y out( ) x( )+ y innL( ) u( )− y mL( ) u( )( )

u=xε

+ y innR( ) v( )− y mR( ) v( )( )v=1−xε

= x − 12⎛⎝⎜

⎞⎠⎟+ 12e

−x2ε − 12e

− 1−x( )2ε + ε

4xε

⎛⎝⎜

⎞⎠⎟

2

e−x2ε − ε

41− xε

⎛⎝⎜

⎞⎠⎟

2

e− 1−x( )2ε