AMS 212B Perturbation Methodshongwang/AMS212B/Notes/... · 2018-02-16 · AMS 212B Perturbation Methods - 3 - =εb 1 (t)+!→0as ε→0 ==> It is justified if we stop at e 1 ε b−1(t)+b0(t)

- 1 -

AMS212BPerturbationMethodsLecture12

CopyrightbyHongyunWang,UCSC

Question: Atwhichtermoftheexpansioncanwestop?

RecalltheexampleweworkedoutExample:

d2 ydx2

− 1xy =0 asx→ +∞

==> y x( ) = e f0 x( )+ f1 x( )+ f2 x( )+!

Solution#1:f0 x( ) =2 x , f1 x( ) = 14 log x , f2 x( ) = −3

16 x

Solution#2:f0 x( ) = −2 x , f1 x( ) = 14 log x , f2 x( ) = 3

16 x

Question: Canwestopaty x( )~e f0 x( ) insteadofexpandingtoy x( )~e f0 x( )+ f1 x( )+ f2 x( ) ?

Answer: Itisdeterminedbytherelativeerror.

Toillustratetheanswer,letusexamine4simpleexamples.Example1A:

Supposeu(t)hastheexpansion

u t( ) = a0 t( )+ εa1 t( )+!

Relativeerroroftheleadingterm:

u t( )−a0 t( )u t( ) =

εa1 t( )+!a0 t( )+! = ε

a1 t( )a0 t( ) +!→0 asε→0

==> Itisjustifiedifwestopattheleadingterma0(t).

Example1B:

Considertheexpansion

eu t( ) = ea0 t( )+εa1 t( )+! Relativeerroroftheleadingterm:

AMS212BPerturbationMethods

- 2 -

eu t( ) −ea0 t( )

eu t( ) =

ea0 t( ) eεa1 t( )+!−1⎛

⎝⎞⎠

ea0 t( ) ⋅eεa1 t( )+! =

εa1 t( )+!1+!

= εa1 t( )+!→0 asε→0

==> Itisjustifiedifwestopattheleadingtermea0 t( ) .Example2A:

Supposeu(t)hastheexpansion

u t( ) = 1εb−1 t( )+b0 t( )+ εb1 t( )+!


u t( )− 1εb−1 t( )u t( ) =

b0 t( )+!1εb−1 t( )+!

= εb0 t( )b−1 t( ) +!→0 asε→0

==> Itisjustifiedifwestopattheleadingtermb-1(t).

Example2B:

Considertheexpansion

eu t( ) = e1εb−1 t( )+b0 t( )+εb1 t( )+!


eu t( ) −e

1εb−1 t( )

eu t( ) =

e1εb−1 t( ) eb0 t( )+!−1⎛

⎝⎞⎠

e1εb−1 t( ) ⋅eb0 t( )+!

=eb0 t( )+!−1⎛

⎝⎞⎠

eb0 t( )+!

=O 1( )O 1( ) =O 1( ) doesnotconvergetozeroasϵ→0

==> Wecannotstopattheleadingterme1εb−1 t( ) .

Relativeerrorofthefirsttwoterms=

eu t( ) −e

1εb−1 t( )+b0 t( )

eu t( ) =

e1εb−1 t( )+b0 t( ) eεb1 t( )+!−1⎛

⎝⎞⎠

e1εb−1 t( )+b0 t( ) ⋅eεb1 t( )+!

=εb1 t( )+!1+!


- 3 -

= εb1 t( )+!→0 asε→0

==> Itisjustifiedifwestopate1εb−1 t( )+b0 t( ) .

Fromthe4examplesabove,weseethat

Whenexpandingtheexponent,itisOKtostopataterm

ifandonlyifthenexttermiso(1).

Nowbacktotheexampleweworkedout

y x( ) = e2 x+14log x( )− 316 x

+!

Weseethat

1) Wecannotstopattheleadingterme2 x .

2) Itisjustifiedifwestopate2 x+14log x( ) butonlyafterweknow

f2 x( ) = −3

16 x= o 1( ) .

Question: Howaboutx→−∞?Solution#1:

f0 x( ) =2 x = i2 x , f1 x( ) = 14 log x , f2 x( ) = −3

16 x= i 316 x

Solution#2:

f0 x( ) = −2 x = −i2 x , f1 x( ) = 14 log x , f2 x( ) = 3

16 x= −i 3

16 x

Ageneralsolution:

y1 x( ) = Aexp i2 x + 14 log x + i

316 x

+!⎛

⎝⎜⎜

⎞

⎠⎟⎟+B exp −i2 x + 14 log x − i

316 x

+!⎛

⎝⎜⎜

⎞

⎠⎟⎟

Revisittheexampleweworkedout:

′′y x( )− 1x y =0 asx→ +∞


- 4 -

Ithasthespecialform(i.e,itdoesnothavethefirstderivativeterm):

′′y x( )+q x( ) y x( ) =0 Claim:WecanreduceageneralODE

′′y x( )+ p x( ) ′y x( )+q x( ) y x( ) =0 tothespecialform ′′Y x( )+ !q x( )Y x( ) =0 .

Proof:

Leta x( ) = e12 p x( )dx∫ .Thederivativesofa(x)havetheexpressions

′a x( ) = d

dxe12 p x( )dx∫⎡

⎣⎢

⎤

⎦⎥ = a x( )p x( )

2

′′a x( ) = d

dxa x( )p x( )

2⎡

⎣⎢⎢

⎤

⎦⎥⎥= a x( ) p

2 x( )4 +

′p x( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Weconsidertheunknownfunction

Y x( ) = y x( )a x( )

wherey(x)satisfiestheoriginalODE

′′y x( )+ p x( ) ′y x( )+q x( ) y x( ) =0 Usingthesepropertiesofa(x)andY(x),wewriteY”(x)as

′′Y x( ) = ′′y x( )a x( )+2 ′y x( ) ′a x( )+ y x( ) ′′a x( )

= ′′y x( )+ p x( ) ′y x( )+ p2 x( )

4 +′p x( )2

⎛

⎝⎜⎜

⎞

⎠⎟⎟y x( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥a x( )

= ′′y x( )+ p x( ) ′y x( )+q x( ) y x( )

=0! "##### $#####

+p2 x( )4 +

′p x( )2 −q x( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟y x( )

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥a x( )

= − q x( )− p2 x( )4 −

′p x( )2

≡ !q x( )" #$$$ %$$$

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

Y x( )

ItfollowsthatY(x)satisfiesanODEinthespecialform


- 5 -

′′Y x( )+ !q x( )Y x( ) =0 where !q x( ) = q x( )− p

2 x( )4 +

′p x( )2

NowwedemonstratetheadvantageofreducinganODEtothespecialform.

Example:

′′y +2x ′y + y =0 asx→ +∞

Wewritey(x)asy x( ) = e f x( ) andderivetheequationforƒ(x).

′′f + ′f( )2 +2x ′f +1=0 Usingthemethodofdominantbalance,weobtain

f x( )~−12log x( )+ 3

16 ⋅1x2

+!

y x( ) = e f x( )~exp −12log x( )+ 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟

(ThederivationisgivenintheAppendix)

Noticethatweobtainedonlyonesolution.

Tofindtwoindependentsolutions,weneedtoreducetheODEtothespecialform(i.e.,gettingridofthefirstderivativeterm).

Weidentifyp(x)=2xandq(x)=1.

a x( ) = 12 2xdx∫ = 12 x

2

!q x( ) = q x( )− p

2 x( )4 +

′p x( )2 = −x2

TheODEforY x( ) = y x( )a x( ) = y x( )e12x

2isinthespecialform

′′Y x( )− x2Y x( ) =0 Usingthemethodofdominantbalance,weobtaintwosolutions:

Y1 x( )~exp 12 x

2 − 12log x( )+ 316 ⋅

1x2

+!⎛⎝⎜

⎞⎠⎟

Y2 x( )~exp −12 x2 − 12log x( )− 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟


- 6 -

==>

y1 x( ) = exp −12 x2

⎛⎝⎜

⎞⎠⎟Y1 x( )~exp −12log x( )+ 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟

y2 x( ) = exp −12 x2

⎛⎝⎜

⎞⎠⎟Y2 x( )~exp −x2 − 12log x( )− 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟

(ThederivationisgivenintheAppendix)

Nowwelookatanotherapplicationofthemethodofdominantbalance.TheWKBapproximation(Wentzel,KramersandBrillouin)

ConsiderthesecondorderODEwithlargeparameterλ.

′′y x( )+ p x ,λ( ) ′y x( )+q x ,λ( ) y x( ) =0 , λ→∞

(IthasconnectionstoeigenvalueproblemsofODE).Recallthetransformation

ynew x( ) = y x( )exp 1

2 p x( )dx∫⎛⎝⎜

⎞⎠⎟

Weonlyneedtoconsiderthespecialcase(i.e.,nofirstderivativeterm)

′′y x( )+q x;λ( ) y x( ) =0 , λ→∞

Goal: Tofindthebehaviorofsolutionsasλ→∞.LetusfirstlookataphysicalproblemthatisrelatedtothisODE.

Example: Considerthe1-Dwaveequationwithvariablecoefficient

∂2u∂t2

= c2 x( ) ∂2u

∂x2

Solutionbyseparationofvariables: u x ,t( ) =w t( ) y x( )

′′w t( ) y x( ) =w t( )c2 x( ) ′′y x( )

==>

′′w t( )w t( ) =

c2 x( ) ′′y x( )y x( ) = −λ2

whereλisaconstantindependentofxandt.Inthetimedimension,forfunctionw(t),wehave


- 7 -

′′w t( )+λ2w t( ) =0 Areal-valuedgeneralsolutionisgivenby

w t( ) =Cexp iλt( )+C exp −iλt( ) form#1= c1 cos λt( )+ c2sin λt( ) form#2

Alternatively,wemayusetheamplitudeandthephaseangleasthetwofreeparametersinthegeneralsolution.Aswewillsee,usingtheamplitudeandthephaseangleasparametersmaybemoreconvenientinsomesituations.

w t( ) = AAmplitude! cos λt + α

Phaseangle!⎛

⎝⎜

⎞

⎠⎟

form#3

Physicalmeaningofλ: f=λ/(2π)isthefrequency

λistheangularfrequency

Inthespacedimension,forfunctiony(x),wehave

c2 x( ) ′′y x( )y x( ) = −λ2 ==>

′′y x( )+λ2 1

c2 x( ) y x( ) =0

Forlargeλ,(highfrequency,smallwavelength),therearetwodifferentlengthscalesintheproblem:

1) theshortwavelength:O(1/λ)2) thelengthscaleinthevariationofc(x):O(1).

Thiscorrespondstohighfrequencywavespropagatingininhomogeneousmedia.

NowweusethemethodofdominantbalancetofindthebehaviorofsolutionsoftheODE

′′y x( )+q x;λ( ) y x( ) =0 , λ→∞

Weseekanexpansionoftheform

y x( ) = exp δn λ( )φn x( )

n=0

∞

∑⎛⎝⎜

⎞

⎠⎟

where

λ: thelargeparameter

x: independentvariable

δn λ( ){ } : asymptoticsequence(unknownbeforetheexpansion).Itsatisfies


- 8 -

δn+1 λ( ) = o δn λ( )( ) asλ→∞

φn x( ){ } : coefficientfunctions

Note: Theasymptoticsequence{δn(λ)}isunknownbeforetheexpansion.Theyaretobedeterminedalongwiththecoefficientfunctions{ϕn(x)}intheexpansion.

Takingderivativesofy(x),wehave

′y x( ) = δn λ( )φn′ x( )

n=0

∞

∑⎡⎣⎢

⎤

⎦⎥ y x( )

′′y x( ) = δn λ( )φn′′ x( )

n=0

∞

∑ + δn λ( )φn′ x( )n=0

∞

∑⎛⎝⎜

⎞

⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥y x( )

SubstitutingintoODEyieldstheequationfor{δn(λ),ϕn(x)}:

δn λ( )φn′′ x( )

n=0

∞

∑ + δn λ( )φn′ x( )n=0

∞

∑⎛⎝⎜

⎞

⎠⎟

2

= −q x;λ( )

Weusethemethodofdominantbalancetodetermine{δn(λ)}and{ϕn(x)}.

Wefirstworkouttheexamplebelow.

Example:

Highfrequencywavespropagatingininhomogeneousmediawithc(x)=1/(1+x2).

′′y x( )+ λ2 1+ x2( )2q x ;λ( )

! "## $##y x( ) =0 , λ→∞

Substitutingy x( ) = exp δn λ( )φn x( )

n=0

∞

∑⎛⎝⎜

⎞

⎠⎟intoODEyieldstheequationfor{δn(λ),ϕn(x)}:

δn λ( )φn′′ x( )

n=0

∞

∑ + δn λ( )φn′ x( )n=0

∞

∑⎛⎝⎜

⎞

⎠⎟

2

= −λ2 1+ x2( )2 Step1: findδ0(λ)andϕ0(x)

Writingouttheleadingtermofeachgroupontheleftside,wehave

δ0 λ( ) ′′φ0 x( )+ δ0 λ( )⎡⎣ ⎤⎦2

′φ0 x( )⎡⎣ ⎤⎦2+!= −λ2 1+ x2( )2

RHS=O(λ2)


- 9 -

LHS=δ0(λ)or(δ0(λ))2.

Strategy: methodofdominantbalance

WeusethelargesttermontheLHStomatchtheleadingtermontheRHS.

Wetryeachtermindividually.

• δ0 λ( ) =O λ2( ) ==> δ0 λ( )⎡⎣ ⎤⎦

2=O λ4( ) notgood

• δ0 λ( )⎡⎣ ⎤⎦2=O λ2( )

==> δ0 λ( ) =O λ( ) good

Calculatingϕ0(x)

Setting δ0 λ( )⎡⎣ ⎤⎦2

′φ0 x( )⎡⎣ ⎤⎦2= −λ2 1+ x2( )2 ,wehaveδ0(λ)=λand

′φ0 x( )⎡⎣ ⎤⎦2= − 1+ x2( )2

==> ′φ0 x( ) = ±i 1+ x2( )

==>φ0 x( ) = ±i x + x

3

3⎛

⎝⎜⎞

⎠⎟

Step2: findδ1(λ)andϕ1(x)

Updatedequation:

OntheLHS,writeoutmoretermsandmoveknowntermstotheRHS

δ0 λ( ) ′′φ0 x( )+δ1 λ( ) ′′φ0 x( )+ δ0 λ( )⎡⎣ ⎤⎦

2′φ0 x( )⎡⎣ ⎤⎦

2+2δ0 λ( )δ1 λ( ) ′φ0 x( ) ′φ1 x( )+!= −λ2 1+ x2( )2

==> δ1 λ( ) ′′φ1 x( )+2λδ1 λ( ) ′φ0 x( ) ′φ1 x( )+!= −λ ′′φ0 x( ) RHS=O(λ)

LHS=O(λδ1(λ))

==> O(λδ1(λ))=O(λ) ==> δ1(λ)=O(1)

(Thatmeanswecannotstoptheexpansionatδ0(λ)andϕ0(x).)

Calculatingϕ1(x)


- 10 -

Setting2λδ1 λ( ) ′φ0 x( ) ′φ1 x( ) = −λ ′′φ0 x( ) ,wehaveδ1(λ)=1and

′φ1 x( ) = − ′′φ0 x( )

2 ′φ0 x( ) = −12 log ′φ0 x( )( )′

==>φ1 x( ) = −12log ′φ0 x( ) = −12log 1+ x

2( ) Step3: findδ2(λ)

Updatedequation:OntheLHS,writeoutmoretermsandmoveknowntermstotheRHS

δ1 λ( ) ′′φ1 x( )+δ2 λ( ) ′′φ2 x( )+2λδ1 λ( ) ′φ0 x( ) ′φ1 x( )+ δ1 λ( )( )2 ′φ1 x( )( )2 +2λδ2 λ( ) ′φ0 x( ) ′φ2 x( )+!= −λ ′′φ0 x( )

==> δ2 λ( ) ′′φ2 x( )+2λδ2 λ( ) ′φ0 x( ) ′φ2 x( )+!= − ′′φ1 x( )− ′φ1 x( )⎡⎣ ⎤⎦2

RHS=O(1)

LHS=O(λδ2(λ))

==> O(λδ2(λ))=O(1) ==> δ2(λ)=O(1/λ)

(Thatmeanswecanstoptheexpansionatδ1(λ)andϕ1(x))

Ageneralsolutionis

y x( ) =C exp iλ x + x3

3⎛

⎝⎜⎞

⎠⎟− 12log 1+ x

2( )+O 1λ

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜

⎞

⎠⎟

+C exp −iλ x + x3

3⎛

⎝⎜⎞

⎠⎟− 12log 1+ x

2( )+O 1λ

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜

⎞

⎠⎟

form#1

ORy x( ) = 1

1+ x2a⋅cos λ x + x

3

3⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥+b ⋅sin λ x + x

3

3⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎛

⎝⎜

⎞

⎠⎟ +O

1λ

⎛⎝⎜

⎞⎠⎟ form#2

ORy x( ) = A

1+ x2cos λ x + x

3

3⎛

⎝⎜⎞

⎠⎟+α

⎡

⎣⎢⎢

⎤

⎦⎥⎥+O 1

λ⎛⎝⎜

⎞⎠⎟ form#3

ThisiscalledWKBapproximation.

Solutionofthegeneralcase


- 11 -

′′y x( )+λ2 f x( ) y x( ) =0 , λ→∞

whereƒ(x)maybepositiveornegative.

Substitutingy x( ) = exp δn λ( )φn x( )

n=0

∞

∑⎛⎝⎜

⎞

⎠⎟intoODEyieldstheequationfor{δn(λ),ϕn(x)}:

δn λ( )φn′′ x( )

n=0

∞

∑ + δn λ( )φn′ x( )n=0

∞

∑⎛⎝⎜

⎞

⎠⎟

2

= −λ2 f x( )


Writingouttheleadingtermofeachgroupontheleftside,wehave

δ0 λ( ) ′′φ0 x( )+ δ0 λ( )⎡⎣ ⎤⎦2

′φ0 x( )⎡⎣ ⎤⎦2+!= −λ2 f x( )

Fromthepreviousexample,weknowtheRHSisbalancedby[δ0(λ)]2.

Calculatingϕ0(x)

Setting δ0 λ( )⎡⎣ ⎤⎦2

′φ0 x( )⎡⎣ ⎤⎦2= −λ2 1+ x2( )2 ,wehaveδ0(λ)=λand

′φ0 x( )⎡⎣ ⎤⎦2= − f x( )

==> ′φ0 x( ) = ±i f x( )

==>φ0 x( ) = ±i f s( )ds

0

x

∫ +C⎛

⎝⎜

⎞

⎠⎟

Note: *)i f x( ) isrealifƒ(x)<0.

*) WekeepconstantCtomakefunctionϕ0(x)moreflexible.


Updatedequation:

OntheLHS,writeoutmoretermsandmoveknowntermstotheRHS

δ1 λ( ) ′′φ1 x( )+2λδ1 λ( ) ′φ0 x( ) ′φ1 x( )+!= −λ ′′φ0 x( ) Fromthepreviousexample,weknowthattheRHSisbalancedbyλδ1(λ).

Calculatingϕ1(x)

Setting2λδ1 λ( ) ′φ0 x( ) ′φ1 x( ) = −λ ′′φ0 x( ) ,wehaveδ1(λ)=1and


- 12 -

′φ1 x( ) = − ′′φ0 x( )

2 ′φ0 x( ) = −12 log ′φ0 x( )( )′

==>φ1 x( ) = −12log ′φ0 x( ) = −12log f x( )

==>φ1 x( ) = −14 log f x( )

Insummary,obtaintwosolutionsintheformof

y x( ) = exp λφ0 x( )+φ1 x( )+!( )

φ0 x( ) = ±i f s( )ds

0

x

∫ +C⎛

⎝⎜

⎞

⎠⎟ , φ1 x( ) = −14 log f x( )

AppendixPartA: Applythemethodofdominantbalancedirectlytosolve

′′y +2x ′y + y =0 asx→ +∞

(withouttransformingittothespecialform)

Wewritey(x)asy x( ) = e f x( ) andderivetheequationforƒ(x).

′′f + ′f( )2 +2x ′f +1=0

Considertheexpansionforexponentƒ(x)oftheform

f x( ) = f0 x( )+ f1 x( )+ f2 x( )+!

Step1: Findingf0(x)

Theleadingtermofƒ(x)shouldsatisfiesthedifferentialequationapproximately

′′f0 + ′f0( )2 +2x ′f0 +1= o 1( ) Weusethelargesttermtobalance1.Therestisleftastheerror.Wetryeachtermindividually.

§ ′′f0 =O 1( ) ==> ′f0 =O x( )


- 13 -

==> ′′f0 + ′f0( )2 +2x ′f0 +1=O x2( ) notgood

§ ′f0( )2 =O 1( ) ==> ′f0 =O 1( ) , f0′′ =0

==> ′′f0 + ′f0( )2 +2x ′f0 +1=O x( ) notgood

§ x ′f0 =O 1( ) ==> ′f0 =O x −1( ) , f0′′ =O x −2( ) ==> ′′f0 + ′f0( )2 +2x ′f0 +1=O x −2( ) good

Calculatingƒ0(x):

2x ′f0 = −1 ==>′f0 = −

12x

==>f0 = −

12log x( )

Step2: Findingƒ1(x)

Consider g(x)≡ƒ(x)−ƒ0(x)=ƒ1(x)+⋯

Substitutingƒ(x)=g(x)+ƒ0(x)intotheequationforƒ(x)yields

′′g + ′′f0 + ′g + ′f0( )2 +2x ′g + ′f0( )+1=0

==>′′g + ′g( )2 − 1x ′g +2x ′g + 3

4x2 =0

Theleadingtermofg(x)shouldsatisfiesthisdifferentialequationapproximately

′′f1 + ′f1( )2 + −1

x+2x⎛

⎝⎜⎞⎠⎟

′f1 +34x2 = o

1x2

⎛⎝⎜

⎞⎠⎟

Weusethelargesttermtobalance3/(4x2).Therestisleftastheerror.


• ′′f1 =O x −2( ) notgood

• ′f1( )2 =O x −2( ) notgood

• x ′f1 =O x −2( ) good

Calculatingƒ1(x):


- 14 -

2x ′f1 = −

34x2 ==>

′f1 = −

38x3

==>f1 =

316 ⋅

1x2

Insummary,wefoundonlyonesolution.

f0 = −

12log x( ) , f1 =

316 ⋅

1x2

y x( ) = e f0 x( )+ f1 x( )+!~exp −12log x( )+ 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟

PartB: reducetothespecialformandfindtwosolutionsTofindtwoindependentsolutions,wefirstreducetheODEtothespecialform(i.e.,gettingridofthefirstderivativeterm).Weidentifyp(x)=2xandq(x)=1.

a x( ) = exp 1

2 2xdx∫⎛⎝⎜

⎞⎠⎟= exp 1

2 x2⎛

⎝⎜⎞⎠⎟

!q x( ) = q x( )− p

2 x( )4 +

′p x( )2 = −x2

TheODEforY x( ) = y x( )a x( ) = y x( )e12x

2isinthespecialform

′′Y x( )− x2Y x( ) =0 WewriteY(x)as

Y x( ) = e f x( ) andderivetheequationforexponentƒ(x)

′′f + ′f( )2 − x2 =0 Considertheexpansionforƒ(x)oftheform

f x( ) = f0 x( )+ f1 x( )+ f2 x( )+!

Step1: Findingf0(x)

Theleadingtermofƒ(x)shouldsatisfiesthedifferentialequationapproximately

′′f0 + ′f0( )2 − x2 = o x2( ) Weusethelargesttermtobalancex2.Therestisleftastheerror.


- 15 -


§ ′′f0 =O x2( ) ==> ′f0 =O x3( ) ==> ′′f0 + ′f0( )2 − x2 =O x6( ) notgood

§ ′f0( )2 =O x2( ) ==> ′f0 =O x( ) , f0′′ =O 1( ) ==> ′′f0 + ′f0( )2 − x2 =O 1( ) good

Calculatingƒ0(x):

′f0( )2 = x2 ==> ′f0 = ±x

==> Case1: f0 =12 x

2

Case2: f0 = −12 x

2


Consider g(x)≡ƒ(x)−ƒ0(x)=ƒ1(x)+⋯

Substitutingƒ(x)=g(x)+ƒ0(x)intotheequationforƒ(x)yields

′′g + ′′f0 + ′g + ′f0( )2 − x2 =0 ==> Case1: ′′g + ′g( )2 +2x ′g +1=0

Case2: ′′g + ′g( )2 −2x ′g −1=0

Theleadingtermofg(x)shouldsatisfiesthisdifferentialequationapproximately

Case1: ′′f1 + ′f1( )2 +2x ′f1 +1= o 1( ) Case2: ′′f1 + ′f1( )2 −2x ′f1 −1= o 1( )

Weusethelargesttermtobalance1.Therestisleftastheerror.


• ′′f1 =O 1( ) notgood

• ′f1( )2 =O 1( ) notgood

• x ′f1 =O 1( ) good


- 16 -

Calculatingƒ1(x):

ForbothCase1andCase2,wehave

2x ′f1 = −1 ==>f1 = −

12log x( )


Consider h(x)≡g(x)−ƒ1(x)=ƒ2(x)+⋯

Substitutingg(x)=h(x)+ƒ1(x)intotheequationforg(x)yields

′′g + ′′f0 + ′g + ′f0( )2 − x2 =0

Case1:′′h + f1′′ + ′h + f1′( )2 +2x ′h + f1′( )+1=0

Case2:′′h + f1′′ + ′h + f1′( )2 −2x ′h + f1′( )−1=0

==> Case1:′′h + ′h( )2 − 1x ′h +2x ′h + 3

4x2 =0

Case2:′′h + ′h( )2 − 1x ′h −2x ′h + 3

4x2 =0

Theleadingtermofh(x)shouldsatisfiesthisdifferentialequationapproximately

Case1:′′f2 + ′f2( )2 + −1

x+2x⎛

⎝⎜⎞⎠⎟

′f2 +34x2 = o

1x2

⎛⎝⎜

⎞⎠⎟

Case2:′′f2 + ′f2( )2 + −1

x−2x⎛

⎝⎜⎞⎠⎟

′f2 +34x2 = o

1x2

⎛⎝⎜

⎞⎠⎟

Weusethelargesttermtobalance3/(4x2).Therestisleftastheerror.


• ′′f2 =O x −2( ) notgood

• ′f2( )2 =O x −2( ) notgood

• x ′f2 =O x −2( ) good

Calculatingƒ2(x):

Case1:2x ′f2 = −

34x2 ==>

f2 =

316 ⋅

1x2


- 17 -

Case2:2x ′f2 =

34x2 ==>

f2 = −

316 ⋅

1x2

Insummary,wefoundtwosolutions.

Case1:f0 =

12 x

2 , f1 = −12log x( ) , f2 =

316 ⋅

1x2

Case2:f0 = −

12 x

2 , f1 = −12log x( ) , f2 = −

316 ⋅

1x2

==>

Y1 x( ) = e f x( )~exp 12 x

2 − 12log x( )+ 316 ⋅

1x2

+!⎛⎝⎜

⎞⎠⎟

Y2 x( ) = e f x( )~exp −12 x2 − 12log x( )− 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟

==>

y1 x( ) = exp −12 x2

⎛⎝⎜

⎞⎠⎟Y1 x( )~exp −12log x( )+ 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟

y2 x( ) = exp −12 x2

⎛⎝⎜

⎞⎠⎟Y2 x( )~exp −x2 − 12log x( )− 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟

Ageneralsolutionis

y x( ) = c1 exp −12log x( )+ 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟+ c2exp −x2 − 12log x( )− 3

16 ⋅1x2

+!⎛⎝⎜

⎞⎠⎟

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