AMS 212B Perturbation Methods Lecture 05 Copyright by Hongyun Wang, UCSC

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AMS B Pertrbation Methods Lectre 5 Copright b Hongn Wang, UCSC Recap: we discssed bondar laer of ODE Oter epansion Inner epansion Matching: ) Prandtl s matching ) Matching b an intermediate variable (Skip it) Consider the range = η ϵ α where η > and / < α < are fied as +. Epand ( ot )( ) =η α Epand ( inn ) =η α Match (ot) and (inn) Note: This matching method does not prodce a composite epansion. We appl matching b an intermediate variable to or eample + + = =, ( ) = ( ot) inn = e + ( ), + ( ) = c ( e )+ c ( e ) c ( + e ) Epand ( ot )( ) and ( inn) =η α in and keep terms p to O(). =η α ot ( ) = =η α e e ηα + η α +! + = e η α +! - -

AMS B Pertrbation Methods innt = e( η α + )+! = c ( e ηα )+ c ( e ηα ) c ( + e ηα )η α +! =η α = c + c c η α +! = c c η α + c +! Matching condition: ( ot ) = ( inn ) =η α =η α ==> e( η α + )+!= c c η α + c +! ==> c = e, c = e Again, this matching method does not prodce a composite epansion. 3) Van Dke s matching We etend Prandtl s matching b considering more sophisticated limits, respectivel for small and for large. Inner limit of the oter epansion (ot)(inn) (): To calclate (ot)(inn) (), we do a) epand ( ot )( ) in the Talor series arond =, and then b) sbstitte = to write it in terms of. Oter limit of the inner epansion (inn)(ot) (): To calclate (inn)(ot) (), we treat e - as zero and keep other terms intact. The accrac of this epression improves eponentiall as increases. Note: Both (ot)(inn) () and (inn)(ot) () are fnctions, not scalars. Van Dke s matching condition: (ot)(inn) () = (inn)(ot) () Composite epansion: To constrct the composite epansion, we define - -

AMS B Pertrbation Methods ( ) ( inn )( ot) m = ot ( inn) Note: (m) () is a fnction, not a scalar. Using (m) () we write ot the composite epansion: ( ) = ( ot )( )+ ( inn )( ) ( m ) c ( ) = Remark: ) In the composite epansion, it is more convenient if we first calclate ( inn )( ) ( m )( ) and then evalate it at =. ) The composite epansion is one epression valid for the whole interval [, ]. We appl Van Dke s matching to or eample. + + = =, ( ) = ( ot) inn = e + ( ), + ( ) = c ( e )+ c ( e ) c ( + e ) The oter limit of inner epansion: ( inn )( ot) ( ) = c + c c The inner limit of oter epansion: We do Talor epansion arond =, and keep terms p to O(). Note that = ϵ = O()). ( ot )( ) = e e + e ( ) = e + +O ==> ( ot )( inn) Matching condition: ( inn )( ot) ( ) = ( ot )( inn) ==> c + c c ( ) = e + = e + - 3 -

AMS B Pertrbation Methods ==> c = e, c = e ==> ( m ) = e + e e Using c = e and c = e, we write ot (inn) () ( inn )( ) = e( e )+ e( e ) e( + e ) We calclate (inn) () - (m) () in preparation for writing ot the composite epansion ( inn) ( ) ( m) The composite epansion is c = e e + e = e + + ( ) = ( ) = ( ot )( )+ ( inn )( ) ( m ) = e + e + + The figre below compares the eact soltion and composite epansions for =...5.5 =. Eact soltion asmptotic with term asmptotic with terms..4.6.8 Qestion: Where is the bondar laer? We first consider the simple case of constant coefficients with a. We hope to find some gidelines from this simple case. - 4 -

AMS B Pertrbation Methods +a + b = = α, ( ) = β Characteristic eqation: λ +aλ + b =, a, At =, it becomes a linear eqation, aλ + b =. ==> λ ( ) = b a For, the characteristic eqation has two roots: λ λ ( ) b a ( )λ = b ==> λ ( ) a A general soltion of the ODE is = c e λ + c e λ (the reglar root) (the singlar root) The bondar laer is cased b the term e λ. To determine the location of bondar laer, we std Behavior of e λ when λ is large: For λ, inepect the behavior of mae λ eλ = e λ. << ==> For λ, the bondar laer is at =. For λ +, inepect the behavior of Conclsion:. mae λ eλ = e λ << ==> For λ +, the bondar laer is at =. When λ ( ) a, the bondar laer is at =. - 5 -

AMS B Pertrbation Methods When λ ( ) a +, the bondar laer is at =. Eample: + + = As +, λ ( ) a. ==> the bondar laer is at =. As -, λ ( ) a +, ==> the bondar laer is at =. Remark: The sign of a is determined b and coefficient a. It is not affected b fnction term (b ) in ODE ϵ + a + b =. We can tr to appl the reslt to an ODE with non-linear fnction term. Eample (nonlinear ODE with bondar laer at = ) + + = =, ( ) =, As -, a = + Highlights of the soltion: ==> the bondar laer is at = ) The ODE is non-linear. There is no eact soltion! ) Since the bondar laer is at =, the inner variable is selected as = 3) After scaling, ODE ( )+ ( )+ f ( ) = becomes - 6 -

( ) ( )+ f AMS B Pertrbation Methods ( ) = This confirms that the fnction term ƒ(()) in ODE does not affect the leading term behavior of bondar laer (inclding the location and width). 4) For (, ), as ϵ - we have (instead of + ). In Van Dke matching, when calclating ( inn )( ot) ( ), we treat e as zero. 5) The composite epansion is ( ) = c ( +) e + e + log ( +)+ log ( + ) (See Appendi A for the derivation) e 8 e e ( ) Qestion: What happens if a =? Answer: Two possibilities: *) bondar laers; *) no soltion. Eample (Two bondar laers) = sin ( ) =, ( ) = Claim: It has two bondar laers., + We can arrive at this conclsion in two was. View #: (a simple view) At =, the eqation becomes = sin ==> it does not need or accommodate an bondar condition. Ths, we need two bondar laers to accommodate the two bondar conditions. View #: (a more detailed view) - 7 -

AMS B Pertrbation Methods Characteristic eqation of the ODE is λ = ==> λ =, λ = Recall the behavior of e λ. λ = + ==> a bondar laer is at =. λ = ==> a bondar laer is at =. Therefore, we have two bondar laers. The epression of λ and λ also sggests that both bondar laers have the width ϵ. Now we derive the epansions from the differential eqation assming that both bondar laers have the width ϵ. Later on, we will discss how to find the width of a bondar laer. Oter epansion: We seek an epansion of the form ( ot )( ) = a ( )+O( ) Note: no bondar condition is imposed on (ot). Sbstitting into eqation ields ( ot) = sin Inner epansion at = : Let = be the inner variable. We have d d = d d, d d = d d =, sin sin - 8 -

AMS B Pertrbation Methods In terms of inner variable, the eqation becomes ( ) = sin = sin cos +O( ) We seek an epansion of the form ( innl )( ) = a ( )+ a ( )+O Bondar condition at = : () = =, = ϵ. ==> a ( )+ a ( )+!= ==> a() =, a() = Note: Onl the bondar condition at = is imposed on the inner epansion at =. Sbstitting into eqation ields ( a ( )+ a ( ) ) a + a ( ) = sin cos +O( ) ==> a ( ) a ( ) sin + Both coefficients mst be zero. a a ( )+cos +O = : a a = sin a ( ) = ==> a = sin ( e )+ c e e (See Appendi B for the derivation) / : a a = cos a ( ) = ==> a = cos + c e e (See Appendi C for the derivation) The inner epansion at = is - 9 -

AMS B Pertrbation Methods ( ) = sin ( e )+ c ( e e )+ cos + c ( e e ) innl Matching at = : Recall the oter epansion: ( ) = sin ot The inner limit of oter epansion: We do Talor epansion arond =, and keep terms p to O( ϵ).. Note that = = O ( ot )( innl) ( ) = sin = sin +cos +! = sin + cos +! The oter limit of inner epansion: As +, = + and we treat e- as zero. ( innl )( ot) ( ) = sin + cos c e c e +! Van Dke matching condition: ( innl )( ot) ( ) = ( ot )( innl) ( ) ==> c =, c = ==> ( ml ) ( innl) = sin = sin ==> ( innl) ( ) ( ml) + cos ( e )+ cos = sin e - -

AMS B Pertrbation Methods Inner epansion at = : Note: We select the inner variable sch that ) v = at = and ) v + as moves ot of the bondar laer and into the oter epansion region. Let v = ( v) v be the inner variable. The ODE becomes = sin v + The inner epansion at = is innr = sin +cos v +O( ) ( v) = sin ( e v )+ d ( e v e v )+ cos v + d e v e v Matching at = : Recall the oter epansion: ( ) = sin ot The inner limit of oter epansion: We do Talor epansion arond =. ( ot )( innr) ( v) = sin = sin ( ) = sin cos ( )+! = sin cos v +! ( innr) ( v) ( mr) v = sin e v The composite epansion (inclding two bondar laers) is ( ) = ( ot )( )+ ( innl )( ) ( ml ) c ( ) = ( ) v= + ( innr )( v) ( mr ) v - -

AMS B Pertrbation Methods ( ) = sin +sin e sin e c Remark: In the composite epansion above, the ϵ term disappears. This is attribted to two factors: The roots of the characteristic eqation have the form of λ = c + c +! with c =. There is no ϵ term in the oter epansion If we change the problem to + = =, ( ) =, + there will be ϵ term in the composite epansion. The figre below compares the eact soltion and the composite epansion for =.5..3.. =.5 -. -. -.3 Eact soltion asmptotic with term..4.6.8 Eample (Asmptotic epansion does not eist) - -

AMS B Pertrbation Methods = sin ( ) =, ( ) =, We epand both () and sin in Forier sine series. (The se of sine series is dictated b the bondar conditions () = () =.) sin = k= a sin kπ k = α k sin( kπ) k= Sbstitting into the ODE ields + α k kπ sin kπ = a k sin kπ k= a ==> α k = k kπ + k= When =, we have ϵ(k π) + = and the BVP has no soltion. ( kπ) As ϵ -, it will enconter -/(k π) infinitel man times. Therefore, this problem does not have an asmptotic epansion. Appendi A Derivation of a two-term epansion for the bondar vale problem + + = =, ( ) =, As -, we have a = +. Oter epansion: ==> The bondar laer is at =. - 3 -

AMS B Pertrbation Methods We seek an epansion of the form ( ot )( ) = a ( )+ a ( )+! Bondar condition at = : () = = ==> a ( )+ a ( )+!= ==> a() =, a() = Note: When the bondar laer is at =, onl the bondar condition at = is imposed on the oter epansion. Sbstitting into eqation ( a +! )+( a + a +! )+ a + a +! = ==> a + a + a +a a + a +!= : a + a = a = ==> a a = ==> = a ==> = + c ==> a a ( ) = + c Using the bondar condition a() =, we obtain a ( ) = + : a +a a = a = a ( ) = ( +) 3 ==> a + + a = + 3 ==> + a +( +)a = ( +) - 4 -

AMS B Pertrbation Methods a ==> + ==> a = ( +) = + log + + ==> ( +) a = log( +)+ c c ( +) Using the bondar condition a() =, we obtain a ( ) = log + ( +) The oter epansion is ( ) = ot ( +) log + + Inner epansion: Let = be the inner variable. We have d d = d d, d d = d d Sbstitting into eqation, we have d d d d + = ==> ( ) ( )+ ( ) = This is the ODE in terms of the inner variable after scaling. We seek an epansion of the form ( inn )( ) = a ( )+ a ( )+! Bondar condition at = : () = =, ( ) = ϵ. ==> a ( )+ a ( )+!= ==> a() =, a() = Note: Onl the bondar condition at = is imposed on the inner epansion at = Sbstitting into eqation - 5 -

AMS B Pertrbation Methods ( a + a +! ) ( a + a +! )+ a +! ==> a a + a a + a +!= = : : a a = a = ==> a ( ) = c ( e ) a a = a = c ( e ) = c e + e a ( ) = Let A s = L a ( ) and c = a (). Taking Laplace transform of both sides, we have s A( s) sa ( ) a ( ) sa( s) a ==> ( s s)a s ==> A s = c = c c = c s s + s s s + s s c s s s s s + s = c 5 c s + c s s + s c s s Renaming c = c 5 c and taking inverse transform, we get a ( ) = L A( s) = c ( e )+ c ( +e ) c e e In the above, we have sed L e a = s + a and Ths, the inner epansion is L e a = s + a - 6 -

AMS B Pertrbation Methods ( ) = c ( e )+ c ( e )+ c ( +e ) c ( e e ) inn where coefficients c and c are to be determined in matching. Matching: Recall the oter epansion: ( ) = ot The oter limit of inner epansion: ( +) log + + For (, ), as ϵ - we have = (-)/ϵ. Ths, we treat e as zero. ( inn )( ot) ( ) = c + c + c The inner limit of oter epansion: We do Talor epansion arond = and keep terms p to O(). Note that (-) = ϵ = O()). ( ot )( inn) ( ) = ( ) = + 4 log ( ) Van Dke s Matching condition: ( inn )( ot) ( ) = ( ot )( inn) ( ) ==> c + c + c = + 4 log ==> c =, c = log ==> ( m ) inn = + 4 log ( ) ( ) = ( e )+ log ( ) = ( + ) log ( e )+ ( 4 +e ) ( 8 e e ) 4 log ( ) - 7 -

AMS B Pertrbation Methods ( ) ( m ) inn The composite epansion is c = e + log ( ) ( ) = ( ot )( )+ ( inn )( ) ( m ) = ( +) e ( ) = + e + log ( +)+ log ( + ) e + e 8 e e e 8 e e ( ) The figre below compares the eact soltion and composite epansions for =...8 =..6.4. Eact soltion asmptotic with term asmptotic with terms..4.6.8 Appendi B Soltion of Let A s a a = sin a ( ) = = L a ( ) and c = a (). Taking Laplace transform of both sides, we have - 8 -

AMS B Pertrbation Methods s A( s) sa ( ) a ( ) A s ==> ( s )A s ==> A( s) = c ==> A( s) = c ==> A s = c = c +sin = sin s s +sin s s s s + s + sin s + + s s + sin s + s +sin s + s Renaming c = c + sin and taking inverse transform, we obtain a ( ) = L A( s) = c ( e e )+sin e In the above, we have sed L e a = s + a s Appendi C Soltion of Let A s a a = cos a ( ) = = L a ( ) and c = a (). Taking Laplace transform of both sides, we have s A( s) sa ( ) a ( ) A s ==> ( s )A s = c cos = cos s s - 9 -

AMS B Pertrbation Methods ==> A( s) = c ==> A s = cos s cos s s c s + s +cos s Renaming c = cos c and taking inverse transform, we arrive at a ( ) = L A( s) = c ( e e )+cos - -