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

- 1 -

AMS212BPerturbationMethodsLecture04

CopyrightbyHongyunWang,UCSC

Recap

• Non-dimensionalization,identifyingtimescale,lengthscaleidentifyingsmallparameter,

• Asymptoticseriesofƒ(x,ϵ)IVPofODEoscillationofpendulumstraightforwardexpansionisinvalidforlarget.periodofoscillation

AbrieflookatLaplacetransformConsiderfunctionƒ(t).

Laplacetransformofƒ(t)isdefinedas

L f t( )⎡⎣ ⎤⎦ s( )Fullnotation! "# $##

≡ e−st f t( )dt0

∞

∫

Othernotations: F s( ) , L f t( )⎡⎣ ⎤⎦

WestartwiththeL-transformofeat.

Example:

L eat⎡⎣ ⎤⎦ = e− s−a( )t dt

0

∞

∫ = 1s −a

L-transformofderivativeswithrespecttoaparameter

L dda

f t ,a( )⎡

⎣⎢

⎤

⎦⎥ =

ddaL f t ,a( )⎡⎣ ⎤⎦

Example:

1= eat

a=0

AMS212BPerturbationMethods

- 2 -

==>L 1⎡⎣ ⎤⎦ =

1s −a

⎛⎝⎜

⎞⎠⎟a=0

= 1s

Similarly,wecanderive

L tn⎡⎣ ⎤⎦ =

n!sn+1

Example:

L sinh at( )⎡⎣ ⎤⎦ = L

eat −e−at

2⎡

⎣⎢

⎤

⎦⎥ =

12

1s −a

− 1s +a

⎛⎝⎜

⎞⎠⎟= as2 −a2

L eiat⎡⎣ ⎤⎦ =

1s −ai

L sin at( )⎡⎣ ⎤⎦ = L

eiat −e− iat

2i⎡

⎣⎢

⎤

⎦⎥ =

12i

1s −ai

− 1s +ai

⎛⎝⎜

⎞⎠⎟= as2 +a2

L-transformofderivativeswithrespecttoindependentvariable

L ′f t( )⎡⎣ ⎤⎦ = sL f t( )⎡⎣ ⎤⎦− f 0( )

L ′′f t( )⎡⎣ ⎤⎦ = s2L f t( )⎡⎣ ⎤⎦− s f 0( )− ′f 0( )

FindinginverseL-transformusingderivativewithrespecttoaparameter

L−1 d

dag s ,a( )⎡

⎣⎢

⎤

⎦⎥ =

ddaL−1 g s ,a( )⎡⎣ ⎤⎦

Example:

L−1 s

s2 +a2( )2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= t2asin at( )

L−1 1

s2 +a2( )2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= ? Weusedifferentiationtofindthisone.


- 3 -

1s2 +a2( )2

= 1−2a( )

dda

1s2 +a2

⎛⎝⎜

⎞⎠⎟

L−1 1

s2 +a2( )2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 1

−2a( )L−1 dda

1s2 +a2

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ =

1−2a( )

ddaL−1 1

s2 +a2⎡

⎣⎢

⎤

⎦⎥

= 1

−2a( )dda

1asin at( )⎛

⎝⎜⎞⎠⎟=sin at( )2a3 −

t cos at( )2a2

L−1 1

s2 +a2( )2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=sin at( )2a3 −

t cos at( )2a2

BoundaryValueProblem(BVP)ofODERegularperturbation

Example:

′′y − ε ′y −9y =0y 0( ) =0 , y 1( ) =1

⎧⎨⎪

⎩⎪

Weseekanexpansionoftheform

y x ,ε( ) = a0 x( )+ εa1 x( )+!

Boundarycondition:

y(0)=0 ==> a0 0( )+ εa1 0( )+!=0

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

==> a0 1( ) =1 , a1 1( ) =0 Substitutingtheexpansionintotheequation,wehave

′′a0 + ε ′′a1⎡⎣ ⎤⎦− ε ′a0⎡⎣ ⎤⎦−9 a0 + εa1⎡⎣ ⎤⎦ =0

==> ′′a0 −9a0⎡⎣ ⎤⎦+ ε ′′a1 −9a1 − ′a0⎡⎣ ⎤⎦+!=0


- 4 -

ε0:

′′a0 −9a0 =0a0 0( ) =0 , a0 1( ) =1

⎧⎨⎪

⎩⎪

WeuseLaplacetransformtosolveit.

LetA s( ) = L a0 x( )⎡⎣ ⎤⎦

Recall

L ′f x( )⎡⎣ ⎤⎦ = sL f x( )⎡⎣ ⎤⎦− f 0( )

L ′′f x( )⎡⎣ ⎤⎦ = s2L f x( )⎡⎣ ⎤⎦− s f 0( )− ′f 0( )

TakingLaplacetransformofbothsides,wehave

s2A s( )− sa0 0( )− ′a0 0( )−9A s( ) =0

Letα=a0’(0),unknownforthetimebeing.Wehave

s2 −32( )A s( ) =α

==>A s( ) =α 1

s −3( ) s +3( )

==>A s( ) = α

61s −3−

1s +3

⎛⎝⎜

⎞⎠⎟

==>a0 x( ) = L−1 A s( )⎡⎣ ⎤⎦ =

α6 e3x −e−3x( )

Enforcingtheboundaryconditiona0(1)=1todetermineα

==>a0 x( ) = L−1 A s( )⎡⎣ ⎤⎦ =

α6 e3x −e−3x( )

ε1:

′′a1 −9a1 = ′a0 =3

e3 −e−3e3x +e−3x( )

a1 0( ) =0 , a1 1( ) =0

⎧

⎨⎪⎪

⎩⎪⎪

(Skipthederivationinlecture)WeuseLaplacetransformtosolveit.

LetA s( ) = L a1 x( )⎡⎣ ⎤⎦



- 5 -

s2A s( )− sa1 0( )− ′a1 0( )−9A s( ) = q 1

s −3+1s +3

⎛⎝⎜

⎞⎠⎟, q= 3

e3 −e−3

Letα=a1’(0).Wehave

==>s2 −32( )A s( ) =α+q 1

s −3+1s +3

⎛⎝⎜

⎞⎠⎟

==>A s( ) = 16

1s −3−

1s +3

⎛⎝⎜

⎞⎠⎟

α+q 1s −3+

1s +3

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

==>

A s( ) = α

61s −3−

1s +3

⎛⎝⎜

⎞⎠⎟+ q6

1s −3( )2

− 1s +3( )2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

==>

a1 x( ) = L−1 A s( )⎡⎣ ⎤⎦ =

α6 e3x −e−3x( )+ q6L

−1 1s −3( )2

− 1s +3( )2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Recall

L eat⎡⎣ ⎤⎦ =

1s −a

Differentiatingwithrespecttoa

==>

L t eat⎡⎣ ⎤⎦ =

1s −a( )2

L−1 1

s −a( )2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= t eat

Usingthisresulttocalculatea1(x),weobtain

a1 x( ) = α

6 e3x −e−3x( )+ q6 x e3x −e−3x( )

Enforcingtheboundaryconditiona1(1)=0todetermineα

==> α=−q

==>a1 x( ) = q6 x −1( ) e3x −e−3x( ) Recall

q= 3

e3 −e−3

==>a1 x( ) = 1

2 e3 −e−3( ) x −1( ) e3x −e−3x( )

Thus,thesolutionofBVPhastheexpansion


- 6 -

y x ,ε( )~ e

3x −e−3x

e3 −e−31+ 12ε x −1( )⎡

⎣⎢

⎤

⎦⎥

Incomparison,theexactsolutionofBVPis

yext x( ) = e

λ1x −eλ2x

eλ1 −eλ2

where

λ1 =

ε+ ε2 +362 ≈3 , λ1 =

ε− ε2 +362 ≈ −3

Thefigureabovecomparestheexactsolutionandasymptoticexpansionsforε=1.

Singularperturbation

(Boundarylayersandmatchedasymptoticexpansions)

WefirststudytheexactsolutionoftheBVPbelow.Thegoalofexaminingtheexactsolutionistoseethebehaviorofboundarylayerandtoseewhatweneedtodotocapturethesingularboundarylayer.

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

⎧⎨⎪

⎩⎪, ε→0+

Thecharacteristicequationis

ϵλ2+λ+1=0

0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

x

y ε = 1

Exact solutionasymptotic with 1 termasymptotic with 2 terms


- 7 -

Ithastwodistinctroots:

λ1 =

−1+ 1−4 ε2ε , λ2 =

−1− 1−4 ε2ε

Ageneralsolutionis

yext x( ) = c1eλ1x + c2eλ2x Enforcingtheboundaryconditionyields

yext x( ) = 1

eλ1 −eλ2eλ1x −eλ2x( )

Nowwelookatthebehaviorsofvarioustermsintheexactsolutionasε→0+.

Asε→0+,wehave

λ1 =−1+ 1−4 ε

2ε =−1+ 1− 12 ⋅4ε+O ε2( )⎛

⎝⎜⎞⎠⎟

2ε= −1+O ε( )→−1

λ2 =−1− 1−4 ε

2ε =−1− 1− 12 ⋅4ε+O ε2( )⎛

⎝⎜⎞⎠⎟

2ε

= −1ε+O 1( )→−∞

eλ1 = e−1+O ε( ) → e−1

eλ1x = e−1+O ε( )⎛

⎝⎜⎞⎠⎟x → e−x

eλ2x = e−1εx+O 1( )x ≈

0 , x >>O ε( )e−1ε x , x =O ε( )

⎧

⎨⎪

⎩⎪

==>

yext x( )≈e1−x , x >>O ε( )e1−x − e

1−1εx

boundarylayer!"# $# , 0≤ x ≤O ε( )

⎧

⎨⎪⎪

⎩⎪⎪

Thefigurebelowshowstheboundarylayerintheexactsolutionasε→0+


- 8 -

Theboundarylayersuggeststhatweneedtwoasymptoticexpansions:

• Outerexpansion:outsidetheboundarylayer: x >>O ε( )

• Innerexpansion:insidetheboundarylayer: x =O ε( )

Belowwefirstusetheexactsolutiontoillustratetheouterexpansionandinnerexpansion.Later,wewillderivetheouterexpansionandinnerexpansiondirectlyfromdifferentialequationwithoutusingtheexactsolution.Outerexpansion:

Atafixedx>0(outsidetheboundarylayer),

limε→0+ yext x( ) = e1−x (Hereyext(x)istheexactsolution)

==> yout( ) x( ) = e1−x +!

Observation:

y(out)(x)satisfiesonlytheboundaryconditionatx=1

(attheendawayfromtheboundarylayer)Theboundaryconditionsare y(0)=0andy(1)=1.

y(out)(x)satisfiesonlyy out( ) x( )

x=1= e1−x

x=1=1 .

Atx=0,y out( ) x( )

x=0= e1−x

x=0= e ≠0 .

Innerexpansion:

0 0.2 0.4 0.6 0.8 1

0

0.5

1

1.5

2

2.5

x

y

ε = 0.05ε = 0.01


- 9 -

Fromtheexactsolution,weknowthewidthofboundarylayer=O(ε)asε→0+.

Inotherwords,asε→0+,thewidthofboundarylayer→0.

Togetridofthissingularityandtocapturetheboundarylayerasε→0+,weusescalingtointroduceaninnervariable

u= x

ε

Intermsofu,thewidthoftheboundarylayerisO(1).

Notation(Becareful!Thisnotationmaybeabitconfusing):

yext u( )≡ yext x( ) x=εu

Atafixedu(insidetheboundarylayer)

limε→0+ yext u( ) = limε→0+ e e

−εu −e−u( ) = e 1−e−u( ) ==> y

inn( ) u( ) = e 1−e−u( )+!

Observation:

y(inn)(u)satisfiesonlytheboundaryconditionatx=0

(attheboundarylayer)Theboundaryconditionsare y(0)=0andy(1)=1.

y(inn)(u)satisfiesonlyy inn( ) x

ε⎛⎝⎜

⎞⎠⎟x=0

= e 1−e−xε

⎛

⎝⎜⎞

⎠⎟x=0

=0 .

Atx=1,y inn( ) x

ε⎛⎝⎜

⎞⎠⎟x=1

= e 1−e−xε

⎛

⎝⎜⎞

⎠⎟x=1

≈ e ≠1 .

Letussummarizewhatwelearnedfromexaminingtheexactsolution.Summary:

Weneedtwoexpansions:

y(out)(x): outsidetheboundarylayerand

y(inn)(u): insidetheboundarylayerwithinnervariableu= x

ε.

Onlyoneboundaryconditionisimposedoneachofy(out)(x)andy(inn)(u).

y(out)(x)satisfiesonlytheboundaryconditionawayfromtheboundarylayer.

y(inn)(u)satisfiesonlytheboundaryconditionattheboundarylayer.


- 10 -

Question: In

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

⎧⎨⎪

⎩⎪,whathappensifε→0-?

Weagainusetheexactsolutiontoanswerthisquestion.

yext x( ) = e

λ2x −eλ1x

eλ2 −eλ1

Asε→0-,wehave

λ1 =−1+ 1−4 ε

2ε =−1+ 1− 12 ⋅4ε+O ε2( )⎛

⎝⎜⎞⎠⎟

2ε= −1+O ε( )→−1

λ2 =−1− 1−4 ε

2ε =−1− 1− 12 ⋅4ε+O ε2( )⎛

⎝⎜⎞⎠⎟

2ε

= −1ε+O 1( )→ +∞

Were-writetheexactsolutionas

yext x( ) = e

λ2x −eλ1x

eλ2 −eλ1= e

λ2 x−1( ) −eλ1x−λ21−eλ1−λ2

eλ1−λ2 →0 eλ1x−λ2 →0

eλ2 x−1( ) = e

−1ε+O 1( )⎛

⎝⎜

⎞

⎠⎟ x−1( )

≈0 , 1− x( ) >>O ε( )

e−1ε x−1( ) , 1− x( ) =O ε( )

⎧

⎨⎪

⎩⎪

==>

yext x( )≈0 , 1− x( ) >>O ε( )e−1ε x−1( )

boundarylayer! "# $# , 1− x( ) =O ε( )

⎧

⎨⎪⎪

⎩⎪⎪

Thefigurebelowshowstheboundarylayerintheexactsolutionasε→0-.Remark:


- 11 -

ε→0+andε→0-leadtoverydifferentsolutions.Inparticular,theboundarylayerisatdifferentlocations.

Nowwederiveasymptoticexpansionsdirectlyfromthedifferentialequationwithoutusingtheexactsolution.Weproceedwithtwoassumptions:

1) weknowthattheboundarylayerisatx=0and

2) weknowthatthewidthoftheboundarylayerisO(ε).

Lateron,wewilldiscusshowtofindthelocationandthewidthofaboundarylayer.

Example

(Derivingtheouterexpansion,theinnerexpansion,andthecompositeexpansion):

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

⎧⎨⎪

⎩⎪, ε→0+

Outerexpansion(outsidetheboundarylayer)

Weseekanexpansionoftheform

yout( ) x( ) = a0 x( )+ εa1 x( )+!

Ony(out)(x),weenforcetheboundaryconditionawayfromtheboundarylayer.

Boundaryconditionatx=1: y(1)=1

0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

x

y

ε = -0.05ε = -0.01


- 12 -

==> a0 1( )+ εa1 1( )+!=1

==> a0(1)=1, a1(1)=0

Note: Whentheboundarylayerisatx=0,onlytheboundaryconditionatx=1isimposedontheouterexpansion.

Substitutingintoequation

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

ε0:

′a0 +a0 =0a0 1( ) =1

⎧⎨⎪

⎩⎪

==> a0 x( ) = e1−x

ε1:

′a1 +a1 = − ′′a0 = −e1−x

a1 1( ) =0⎧⎨⎪

⎩⎪

(Skipthederivationinlecture.)WeusetheLaplacetransformtosolveit.

LetA s( ) = L a1 x( )⎡⎣ ⎤⎦

TakingLaplacetransformofbothsidesyields

s A s( )−a1 0( )+ A s( ) = −e

s +1

Letα=a1(0).Wehave

s +1( )A s( ) =α− e

s +1

==>

A s( ) = α

s +1 −e

s +1( )2

==> a1 x( ) = L−1 A s( )⎡⎣ ⎤⎦ =αe−x − xe1−x

Imposingconditiona1(1)=0

==> a1 x( ) = 1− x( )e1−x


- 13 -

Theouterexpansionis

yout( ) x( )~e1−x 1+ ε 1− x( )⎡⎣ ⎤⎦

Innerexpansion(insidetheboundarylayer)Rescaling: considertheinnervariable

u= x

ε

Note: Therelationbetweentheinnervariableuandoriginalvariablexdependson1)thelocationand2)thewidthoftheboundarylayer.

Notation(maybeabitconfusing):

y u( )≡ y x( ) x=εu

Letusderivethedifferentialequationfory(u).

dydx

= dydu

⋅1ε

d2 ydx2

= d2 ydu2

⋅ 1ε2

Substitutingintoequation,wehave

ε⋅ 1ε2d2 ydu2

+ 1εdydu

+ y =0

==> ′′y u( )+ ′y u( )+ ε y u( ) =0 Remark: Intermsofinnervariableu,itisaregularperturbationproblem.Weseekanexpansionoftheform

yinn( ) u( ) = a0 u( )+ εa1 u( )+!

Ony(inn)(u),weenforcetheboundaryconditionattheboundarylayer.Boundaryconditionatu=0: y(0)=0

==> a0 0( )+ εa1 0( )+!=0

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

Note: Onlytheboundaryconditionattheboundarylayerisimposedontheinnerexpansion.

Substitutingtheexpansionintoequation


- 14 -

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

ε0:

′′a0 + ′a0 =0a0 0( ) =0

⎧⎨⎪

⎩⎪

Result#1

Thesolutionof

′′a0 + ′a0 =0a0 0( ) =0

⎧⎨⎪

⎩⎪ is

a0 u( ) = c0 1−e−u( ) (SeeAppendixAforthederivation).

ε1:

′′a1 + ′a = −a0 = −c0 1−e−u( )a1 0( ) =0

⎧⎨⎪

⎩⎪

Result#2

Thesolutionof

′′a1 + ′a = −a0 = −c0 1−e−u( )a1 0( ) =0

⎧⎨⎪

⎩⎪ is

a1 u( ) = c1 1−e−u( )− c0u 1+e−u( ) (SeeAppendixBforthederivation).

Thus,theinnerexpansionis

yinn( ) u( ) = c0 1−e−u( )+ ε c1 1−e−u( )− c0u 1+e−u( )⎡

⎣⎤⎦

wherecoefficientsc0andc1aretobedeterminedinmatching.

Matching

y(out)(x)andy(inn)(u)needtobematchedinthetransitionzonebetweentheboundarylayerandtheouterregion.


- 15 -

Tofacilitatethediscussion,wewritetheouterexpansionandinnerexpansioninthegeneralabstractform.Outerexpansion:

yout( ) x( ) = y0out( ) x( )+ ε y1out( ) x( )+!

Innerexpansion:

yinn( ) u( ) = y0inn( ) u( )+ ε y1inn( ) u( )+!

Wearegoingtolookatthreematchingmethods.

• Prandtle’smatching:Itworksonlyfortheleadingterm.

• Matchingbyanintermediatevariable:Itdoesnotproduceacompositeexpansion.

• VanDyke’smatchingItworksformultipletermsandproducesacompositeexpansion.

1) Prandtl’smatching(Itworksonlyfortheleadingterm!)

Werequire

limx→0y0out( ) x( ) = lim

u→∞y0inn( ) u( )

Weliketohaveaunifiedexpansionthatisvalidbothinsidetheboundarylayerandoutsidetheboundarylayer.Let

ym( ) ≡ lim

x→0y0out( ) x( ) = lim

u→∞y0inn( ) u( )

Note: y(m)isascalar,notafunction.

Thecompositeexpansion:

y c( ) x( ) = y0out( ) x( )+ y0

inn( ) u( )− y m( )( )u=xε

Remarks:1) Inthecompositeexpansion,itismoreconvenientifwefirstcalculate

y0inn( ) u( )− y m( )( ) andthenevaluateitat

u= x

ε.

2) Thecompositeexpansionisoneexpressionvalidforthewholeinterval[0,1].

Outsidetheboundarylayer,wehaveu= x

ε>>1 and


- 16 -

y0inn( ) u( )≈ y m( )

y c( ) x( ) = y0out( ) x( )+ y0

inn( ) u( )− y m( )( )u=xε

≈ y0out( ) x( )

Insidetheboundarylayer,wehavex =O ε( )→0 and

y0out( ) x( )≈ y m( )

y c( ) x( ) = y0out( ) x( )+ y0

inn( ) u( )− y m( )( )u=xε

≈ y0inn( ) x

ε⎛⎝⎜

⎞⎠⎟

NowletusapplyPrandtl’smatchingtotheexampleweareworkingon.

Recallthatintheexampleweobtained

y0out( ) x( ) = e1−x

y0inn( ) u( ) = c0 1−e−u( )

Calculatingthelimits,wehave

limx→0y0out( ) x( ) = e

limu→∞y0inn( ) u( ) = c0

Prandtl’smatchingcondition:

limx→0y0out( ) x( ) = lim

u→∞y0inn( ) u( )

==> c0=e

Thecompositeexpansion:

y(m)=e

==> y0inn( ) u( )− y m( ) = −e1−u

==>y c( ) x( ) = y0out( ) x( )+ y0

inn( ) u( )− y m( )( )u=xε

= e1−x −e1−xε

Finally,wearriveattheleadingtermcompositeexpansion:

yc( ) x( ) = e1−x −e1−

xε


- 17 -

Appendix A DerivationofResult#1:

Thesolutionof

′′a0 + ′a0 =0a0 0( ) =0

⎧⎨⎪

⎩⎪is

a0 u( ) = c0 1−e−u( ) Solution:

LetA s( ) = L a0 u( )⎡⎣ ⎤⎦


s2A s( )− sa0 0( )− ′a0 0( )+ sA s( )−a0 0( ) =0

Letc0=a0’(0).WewriteA(s)as

==> s2 + s( )A s( ) = c0

==>A s( ) = c0 1

s s +1( )

==>A s( ) = c0 1

s− 1s +1

⎡

⎣⎢

⎤

⎦⎥

==> a0 u( ) = L−1 A s( )⎡⎣ ⎤⎦ = c0 1−e−u( )

Appendix B DerivationofResult#2:

Thesolutionof

′′a1 + ′a = −c0 1−e−u( )a1 0( ) =0

⎧⎨⎪

⎩⎪is

a1 u( ) = c1 1−e−u( )− c0u 1+e−u( ) Solution:

LetA s( ) = L a1 u( )⎡⎣ ⎤⎦



- 18 -

s2A s( )− sa1 0( )− ′a1 0( )+ sA s( )−a1 0( ) = −c0 1

s− 1s +1

⎡

⎣⎢

⎤

⎦⎥

Letc1=a1’(0).WewriteA(s)as

==>s2 + s( )A s( ) = c1 − c0 1

s− 1s +1

⎡

⎣⎢

⎤

⎦⎥

==>

A s( ) = c1 1

s− 1s +1

⎡

⎣⎢

⎤

⎦⎥− c0

1s2

− 2s s +1( ) +

1s +1( )2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= 2c0 + c1( ) 1s −

1s +1

⎡

⎣⎢

⎤

⎦⎥− c0

1s2

+ 1s +1( )2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

(Renaming2c0+c1asc1)

= c1

1s− 1s +1

⎡

⎣⎢

⎤

⎦⎥− c0

1s2

+ 1s +1( )2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

==> a1 u( ) = L−1 A s( )⎡⎣ ⎤⎦ = c1 1−e−u( )− c0u 1+e−u( )

Intheabove,wehaveused

L e−au⎡⎣ ⎤⎦ =

1s +a

==>L−1 1

s +a⎡

⎣⎢

⎤

⎦⎥ = e

−au

Differentiatingwithrespecttoa,weobtain

L ue−au⎡⎣ ⎤⎦ =

1s +a( )2

==>

L−1 1

s +a( )2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=ue−au

Note: DifferentiatingwithrespecttoaparameterisagoodwayofderivingmoreformulasfortheforwardLaplacetransformandfortheinverseLaplacetransform.

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

Documents

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