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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.
AMS212BPerturbationMethods
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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+
AMS212BPerturbationMethods
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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:
AMS212BPerturbationMethods
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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
⎛⎝⎜
⎞⎠⎟
AMS212BPerturbationMethods
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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
AMS212BPerturbationMethods
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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
AMS212BPerturbationMethods
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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
AMS212BPerturbationMethods
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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:
AMS212BPerturbationMethods
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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
AMS212BPerturbationMethods
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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
AMS212BPerturbationMethods
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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⎛
⎝⎜⎞
⎠⎟
AMS212BPerturbationMethods
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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
AMS212BPerturbationMethods
- 17 -
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.
AMS212BPerturbationMethods
- 18 -
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
AMS212BPerturbationMethods
- 19 -
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