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10.539Lecture9ApproximateSolutionofLinearDifferential

EquationsI

Prof.DeanWangWhentheredoesnotexistanexactclosed-formsolutionofadifferentialequationorwhentheexactsolutionistoocomplicatedtobeuseful,thenoneshouldtrytheapproximatesolution.Localanalysisisaveryimportanttechniquetoobtainanapproximatesolution,i.e.,theapproximatesolutionaroundalocalpoint.Thebehaviorofthesolutionoveranentireintervalmaybefoundbyputtingtogetherneighborhoodsinwhichthelocalbehaviorisknown.Example1: !!!

!!!= 𝑥! + sin𝑥 𝑦 (1)

1. ClassificationofSingularPointsofHomogeneousLinearEquationsInthissectionweclassifyapoint𝑥!asanordinarypoint,aregularsingularpoint,oranirregularsingularpointofahomogeneouslinearequation.Thebehaviorofthesolutionnear𝑥!maysuggesttheappropriateapproximation.Annth-orderhomogeneouslinearequation 𝐿𝑦 𝑥 = 𝑦 ! 𝑥 + 𝑝!!! 𝑥 𝑦 !!! 𝑥 +⋯+ 𝑝! 𝑥 𝑦 𝑥 = 0, (2)Where 𝑦 ! 𝑥 = !!

!!!𝑦 𝑥 (3)

OrdinaryPointsThepoint𝑥! ≠ ∞ iscalledanordinarypointof(2)ifthecoefficientfunctions𝑝! 𝑥 ,… ,𝑝!!! 𝑥 areallanalyticalinaneighborhoodof𝑥!.Note:Inmathematics,ananalyticfunctionisafunctionthatislocallygivenbyaconvergentpowerseries(i.e.,infinitelydifferentiable).Example2Ordinarypoints

a. 𝑦!! = 𝑒!𝑦.Everypoint𝑥! ≠ ∞ isanordinarypoint.b. 𝑥!𝑦!!! = 𝑦.Everypoint𝑥!except𝑥! = 0 and ∞isanordinarypoint.

RegularSingularPointsThepoint𝑥! ≠ ∞ iscalledaregularsingularpointof(2)ifNOTallofthecoefficientfunctions𝑝! 𝑥 ,… ,𝑝!!! 𝑥 areallanalyticalbutifallof

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𝑥 − 𝑥! !𝑝! 𝑥 , 𝑥 − 𝑥! !!!𝑝! 𝑥 ,… , 𝑥 − 𝑥! 𝑝!!! 𝑥 areanalyticalinaneighborhoodof𝑥!.Example3Regularsingularpoints

a. 𝑥 − 1 𝑦!!! = 𝑦hasaregularsingularpointat1.b. 𝑥!𝑦!! + 𝑥𝑦! = 𝑦hasaregularsingularpointat0.c. 𝑥!𝑦! = 𝑥 + 1 𝑦doesnothavearegularsingularpointat0.

Asolutionof(2)maybeanalyticalataregularsingularpoint.Ifitisnotanalytical,itssingularitymustbeeitherapoleoranalgebraicorlogarithmicbranchpoint.Thereisalwaysatleastonesolutionoftheform 𝑦 𝑥 = 𝑥 − 𝑥! !𝐴 𝑥 , (4)where𝛼isanumbercalledtheindicialexponentand𝐴 𝑥 isafunctionwhichisanalyticalat𝑥!andwhichhasaTaylorserieswhoseradiusofconvergenceisatleastaslargeasthedistancetothenearestsingularityofthecoefficientfunctionsinthecomplexplane.If(2)isoforder𝑛 ≥ 2,thenthereisasecondlinearlyindependentsolutionhavingoneofthepossibleforms: 𝑦 𝑥 = 𝑥 − 𝑥! !𝐴 𝑥 , (6) 𝑦 𝑥 = 𝑥 − 𝑥! !𝐴 𝑥 ln 𝑥 − 𝑥! + 𝐶 𝑥 𝑥 − 𝑥! ! , (7)IrregularSingularPointsThepoint𝑥! ≠ ∞ iscalledaregularsingularpointof(2)ifitisneitheranordinarypointnoraregularsingularpoint.Typically,atanirregularsingularpoint,allsolutionsexhibitanessentialsingularity,oftenincombinationwithapoleoranalgebraicorlogarithmicbranchpoint.Butthisisnotalwaysthecase.ClassificationofthePoint𝒙𝟎 = ∞Wecanmapthepointatinfinityintotheoriginusingtheinversiontransformation 𝑥 = !

! (8)

!!"= −𝑡! !

!" (9)

!!

!!!= 𝑡! !!

!!!+ 2𝑡! !

!" (10)

andsoon,andthenclassifyingthepoint𝑡 = 0.Thepoint𝑥! = ∞iscalledanordinary,aregularsingular,oranirregularsingularpointifthepointat𝑡 = 0arecorrespondinglyclassified.Example4

a. !"!"− !

!𝑦 = 0 !"

!"+ !

!!!𝑦 = 0,ordinarypointsexceptfor∞,whichisa

irregularsingularpointb. !"

!"− !

!!𝑦 = 0

!!!/! !"!"+ !

!!𝑦 = 0, ordinarypointsexceptfor0and∞,which

areregularsingularpoints

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c. !"!"− !

!!!𝑦 = 0 !"

!"+ !

!𝑦 = 0, ordinarypointsexceptfor0whichisa

irregularsingularpointExample5Taylorseriesaboutanordinarypoint 𝑦! + !

!!!𝑦 = 0

whichhasregularsingularpointsat1and∞.Thesolution𝑦 𝑥 = !!!!

hasapoleat𝑥 = 1andanalyticalat∞.Taylorseriesofthesolutionabout𝑥 = 0is 𝑦 𝑥 = 𝑐 𝑥!!

!!! ,hasradiusofconvergence1,whichisthedistancetotheregularpointat1.Example6Analyticalsolutionsnearsingularpoints 𝑦!! + !!!

!𝑦! − !

!!= 0

Theaboveequationhasaregularsingularpointat0andanirregularsingularpointat∞.Onesolution𝑦! 𝑥 = !!!!!!

!,isanalyticalat𝑥 = 0buthasanessential

singularityat𝑥 = ∞.Theothersolution𝑦! 𝑥 = !!!!hasapoleat𝑥 = 0butis

analyticalat𝑥 = ∞.

2. LocalBehaviornearOrdinaryPointsofHomogeneousLinearEquationsIngeneral,itisveryusefultoknowhowtoobtainanswerstohardproblemsintermsofinfiniteseries.Usually,thefirstfewtermsoftheseriesaresufficienttogiveaveryaccurateapproximationtothelocalbehaviorofthesolutiontoadifferentialequation.ToobtainaseriessolutionaboutanordinarypointwesubstitutetheTaylorseries 𝑦 𝑥 = 𝑎! 𝑥 − 𝑥! !!

!!! (11)intothedifferentialequationanddeterminethecoefficient𝑎!bysolvingarecursionrelation.Example7.Taylorseriessolutionofa1st-orderdifferentialequation.

𝑦! = 2𝑥𝑦𝑦 0 = 1 (12)

Since𝑥 = 0isanordinarypoint,wemayseekasolutionintheformoftheTaylorseries 𝑦 𝑥 = 𝑎! 𝑥 − 0 ! = 𝑎!𝑥!!

!!!!!!! (13)

Substituting(13)into(12)anddifferentiatingtermbytermgives

𝑛𝑎!𝑥!!! = 2𝑥 𝑎!𝑥!!!!! = 2 𝑎!!!𝑥!!!!

!!!!!!!𝑎! = 1 (14)

SincethecoefficientsofaTaylorexpansionofafunctionisunique,thecoefficientsof(14)mustagree.Sowehave

𝑛𝑎! = 0 𝑛 = 0,1𝑛𝑎! = 2𝑎!!! 𝑛 = 2,3… (15)

Togetherwith𝑎! = 1wehave

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𝑎! = 1 𝑎! = 0 𝑎!! =

!!!

𝑎!!!! = 0So 𝑦 𝑥 = 𝑎!𝑥!!

!!! = !!!

!!!!!! = 𝑒!! (16)

3. LocalSeriesExpansionaboutRegularSingularPointsofHomogeneousLinearEquations

Wehaveseeninsection2thatTaylorseriesaregoodwaytorepresentthelocalbehaviorofasolutiontoadifferentialequationnearanordinarypoint.WhathappensifwetrytorepresentthelocalbehaviorofasolutionneataregularsingularpointbyTaylorseries?Example8.BreakdownofaTaylorseriesrepresentationataregularsingularpoint.ObtainthesolutiontothefollowingequationusingtheTaylorseries. 𝑦!! + !

!!!𝑦 = 0 (17)

SubstitutingtheTaylorseriesabout𝑥 = 0intheformof𝑦 𝑥 = 𝑎!𝑥!!!!! into(17)

gives 𝑛 𝑛 − 1 𝑎!𝑥!!! +!

!!!!!!!

𝑎!𝑥! = 0!!!! (18)

⇓𝑛 𝑛 − 1 + !

!𝑎!𝑥! = 0!

!!! ⇓

𝑛 − !!

!𝑎!𝑥! = 0!

!!! ⇓

𝑛 − !!

!𝑎! = 0

⇓ 𝑎! = 0 for all 𝑛Thisisatrivialsolution,noprogress!ThereasonforthefailureTaylorseriesforthisequationisthattheTaylorseriesarenotgeneralenoughtocapturethelocalbehaviorofthesolutionneartheregularsingularpoints.If𝑥 = 𝑥!isaregularsingularpoint,onesolutionmusthavetheform(4): 𝑦 𝑥 = 𝑥 − 𝑥! !𝐴 𝑥 (4)Since𝐴 𝑥 isanalytical,itcanbeexpandedinaTaylorseries: 𝑦 𝑥 = 𝑥 − 𝑥! !𝐴 𝑥 = 𝑥 − 𝑥! ! 𝑎! 𝑥 − 𝑥! !!

!!! (19)Example9Localanalysisataregularsingularpoint.Solvingexample8.Let𝑥! = 0, anddifferentiate(19)=,wehave 𝑦! 𝑥 = 𝛼𝑥!!! 𝑎!𝑥!!

!!! + 𝑥! 𝑛𝑎!𝑥!!!!!!!

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𝑦!! 𝑥 = 𝛼 𝛼 − 1 𝑥!!! 𝑎!𝑥!!!!! + 2𝛼𝑥!!! 𝑛𝑎!𝑥!!!!

!!! + 𝑥! 𝑛 𝑛 −!!!!

1 𝑎!𝑥!!! (20)Substituting(20)into(17)gives 𝛼 𝛼 − 1 𝑎! + 2𝛼𝑛𝑎! + 𝑛 𝑛 − 1 𝑎! +

!!𝑎! = 0 (21)

Assuming𝑎! ≠ 0gives 𝛼 𝛼 − 1 + !

!= 0

⇓𝛼 = !

!

𝑎! = 𝑎! = 𝑎! = ⋯ = 0Sowehavethesolutionto(17)as

𝑦 𝑥 = 𝑎! 𝑥 (22)Whichistheexactsolution.