Poularikas, A. D., Seely, S. Laplace Transforms. The Transforms and Applications Handbook: Second Edition. Ed. Alexander D. Poularikas Boca Raton:


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1 Poularika, A D, Seely, S Laplace Tranform The Tranform and Applicaion Handbook: Second Ediion Ed Alexander D Poularika Boca Raon: CRC Pre LLC,
2 5 Laplace Tranform Alexander D Poularika Univeriy of Alabama in Hunville Samuel Seely (Deceaed) 5 Inroducion 5 Laplace Tranform of Some Typical Funcion 53 Properie of he Laplace Tranform 54 The Invere Laplace Tranform 55 Soluion of Ordinary Linear Equaion wih Conan Coefficien 56 The Inverion Inegral 57 Applicaion o Parial Differenial Equaion 58 The Bilaeral or TwoSided Laplace Tranform Appendix 5 Inroducion * The Laplace ranform ha been inroduced ino he mahemaical lieraure by a variey of procedure Among hee are: (a) in i relaion o he Heaviide operaional calculu, (b) a an exenion of he Fourier inegral, (c) by he elecion of a paricular form for he kernel in he general Inegral ranform, (d) by a direc definiion of he Laplace ranform, and (e) a a mahemaical procedure ha involve muliplying he funcion f() by e d and inegraing over he limi o We will adop hi laer procedure No all funcion f(), where i any variable, are Laplace ranformable For a funcion f() o be Laplace ranformable, i mu aify he Dirichle condiion a e of ufficien bu no neceary condiion Thee are f() mu be piecewie coninuou; ha i, i mu be ingle valued bu can have a finie number of finie iolaed diconinuiie for > f() mu be of exponenial order; ha i, f() mu remain le han Me a o a approache, where M i a poiive conan and a o i a real poiive number For example, uch funcion a: an β, co β, e are no Laplace ranformable Given a funcion f() ha aifie he Dirichle condiion, hen { } ()= () () F f e d wrien f () i called he Laplace ranformaion of f() Here can be eiher a real variable or a complex quaniy Oberve he horhand noaion {f()} o denoe he Laplace ranformaion of f() Oberve alo ha only ordinary inegraion i involved in hi inegral * All he conour inegraion in he complex plane are counerclockwie by CRC Pre LLC
3 To amplify he meaning of condiion (), we conider piecewie coninuou funcion, defined for all poiive value of he variable, for which Funcion of hi ype are known a funcion of exponenial order Funcion occurring in he oluion for he ime repone of able linear yem are of exponenial order zero Now we can recall ha he inegral () f e d converge if c lim f() e =, c = real conan () < = + f e d, σ jω If our funcion i of exponenial order, we can wrie hi inegral a () c ( σ c ) f e e d Thi how ha for σ in he range σ > (σ i he abcia of convergence) he inegral converge; ha i () < ()> f e d, Re c The rericion in hi equaion, namely, Re() = c, indicae ha we mu chooe he pah of inegraion in he complex plane a hown in Figure 5 FIGURE 5 Pah of inegraion for exponenial order funcion by CRC Pre LLC
4 5 Laplace Tranform of Some Typical Funcion We illurae he procedure in finding he Laplace ranform of a given funcion f() In all cae i i aumed ha he funcion f() aifie he condiion of Laplace ranformabiliy Example 5 Find he Laplace ranform of he uni ep funcion f() = u(), where u() =, >, u() =, < Soluion By () we wrie e { ()}= () = = = u u e d e d () The region of convergence i found from he expreion e d = e σ d <, which i he enire righ halfplane, σ > Example 5 Find he Laplace ranform of he funcion f() = π F ()= e d π () To carry ou he inegraion, define he quaniy x =, hen dx = d, from which d = dx = x dx Then Bu he inegral 4 x F ()= x e dx π Thu, finally, x e dx = x π 3 4 F ()= 3 (3) by CRC Pre LLC
5 Example 53 k Find he Laplace ranform of f() = erfc error funcion, erfc, are defined by, where he error funcion, erf, and he complemenary u u erf = e du erfc = e, du, π o π Soluion Conider he inegral u I = e e du d λ π k where λ = (4) Change he order of inegraion, noing ha u = λ, = λ u u I = e e λ ddu= exp u du λ π u u π The value of hi inegral i known = π π e λ, which lead o { } k erfc = exp k (5) Example 54 Find he Laplace ranform of he funcion f() = inh a Soluion Expre he funcion inh a in i exponenial form a a e e inh a = The Laplace ranform become by CRC Pre LLC
6 inha ( a) ( + a) { }= e e d = a a (6) A moderae liing of funcion f() and heir Laplace ranform F() = {f ()} are given in Table 5, in he Appendix 53 Properie of he Laplace Tranform We now develop a number of ueful properie of he Laplace ranform; hee follow direcly from () Imporan in developing cerain properie i he definiion of f() a =, a quaniy wrien f(+) o denoe he limi of f() a approache zero, aumed from he poiive direcion Thi deignaion i conien wih he choice of funcion repone for > Thi mean ha f(+) denoe he iniial condiion Correpondingly, f (n) (+) denoe he value of he nh derivaive a ime = +, and f ( n) (+) denoe he nh ime inegral a ime = + Thi mean ha he direc Laplace ranform can be wrien F lim f e d, R, a R a a + R ()= () > > (3) We proceed wih a number of heorem Theorem 53 Lineariy The Laplace ranform of he linear um of wo Laplace ranformable funcion f() + g() wih repecive abcia of convergence σ f and σ g, wih σ g > σ f, i Proof From (3) we wrie Thu, { ()+ ()}= ()+ () { f ( ) + g ( )} = F + G (3) f g f g e d f e d g e d, Re ()> σ g { f ( ) + g ( )} = F + G A a direc exenion of hi reul, for K and K conan, [ ] = () + () { K f ( ) + K g ( )} = K F + K G (33) Theorem 53 Differeniaion Le he funcion f() be piecewie coninuou wih ecionally coninuou derivaive df()/d in every inerval T Alo le f() be of exponenial order e c a Then when Re() > c, he ranform of df()/d exi and by CRC Pre LLC
7 () df d { f } f( + )= F() f( + ) = () (34) Proof Begin wih (3) and wrie () df d T df = lim e d T d Wrie he inegral a he um of inegral in each inerval in which he inegrand i coninuou Thu, we wrie () T T e f d n L () () = [ ]+ [ ]+ + [ ] Each of hee inegral i inegraed by par by wriing u = e du = e d df dν = d d ν = f wih he reul T () + () + + () + () n T e f e f L e f e f d Bu f() i coninuou o ha f( ) = f( + ), and o forh, hence However, wih lim f()e = (oherwie he ranform would no exi), hen he heorem i eablihed Theorem 533 Differeniaion Le he funcion f() be piecewie coninuou, have a coninuou derivaive f (n ) () of order n and a ecionally coninuou derivaive f (n) () in every finie inerval T Alo, le f() and all i derivaive hrough f (n ) () be of exponenial order e c a Then he ranform of f (n) () exi when Re() > c and i ha he following form: { f (n) ( )} = n F n f (+) n f () (+) L n n f (n ) (+) (35) Proof The proof follow a a direc exenion of he proof of Theorem 53 Example 53 Find { m } where m i any poiive ineger e f () d = f + e f T e f d + + () T T () T by CRC Pre LLC
8 Soluion The funcion f() = m aifie all he condiion of Theorem 533 for any poiive c Thu, By (35) wih n = m + we have I follow, herefore, ha f (+) = f () (+) = L = f ( m ) (+) = f (m ) () = m!, f (m +) () = { f (m +) ( )} = = m + { m } m! m m! { }= m+ Theorem 534 Inegraion If f() i ecionally coninuou and ha a Laplace ranform, hen he funcion ranform given by o f ( ξ) dξ ha he Laplace f F () d = f ( ) ξ ξ + ( + ) (36) Proof Becaue f() i Laplace ranformable, i inegral i wrien Thi i inegraed by par by wriing Then f ξ dξ f ξ dξ e d = u f ξ dξ du f ξ dξ f d d e d e ν = ν = = = = () from which e f ( ξ) dξ = f ( ξ) dξ = f() e d+ f( ξ) + () dξ f e d by CRC Pre LLC
9 where [f ( ) (+)/] i he iniial value of he inegral of f() a = + The negaive number in he brackeed exponen indicae inegraion Example 53 Deduce he value of {in a} from {co a} by employing Theorem 534 Soluion By ordinary inegraion f ( ξ) dξ = ()+ ( + ) F f ( ) coax dx = ina a From Theorem 534 we can wrie, knowing ha {co a} = + a o ha ina = a + a a { in a}= + a Theorem 535 Diviion of he ranform of a funcion by correpond o inegraion of he funcion beween he limi and () F () F = f ( ξ ) dξ ξ = ( λ ) f dλdξ (37) and o forh, for diviion by n, provided ha f() i Laplace ranformable Proof The proof of hi heorem follow from Theorem 534 Theorem 536 Muliplicaion by If f() i piecewie coninuou and of exponenial order, hen each of he Laplace ranform: { f(), {f(), { f(), i uniformly convergen wih repec o when = c, where σ > c, and () n d F n n { f() }=( ) n d (38) by CRC Pre LLC
10 Furher Proof I follow from (3) when hi inegral i uniformly convergen and he inegral converge, ha Furher, i follow ha () = () n d F n lim,,,,, n { f }= n= 3K d F () F () Similar procedure follow for derivaive of higher order { } = () = e f d f { } = () = e f d f Theorem 537 Differeniaion of a Tranform Differeniaion of he ranform of a funcion f() correpond o he muliplicaion of he funcion by ; hu n d F d () = ()= ( ) n ( n) n F f, n= 3,,, (39) Proof Thi i a reaemen of Theorem 536 Thi heorem i ofen ueful for evaluaing ome ype of inegral, and can be ued o exend he able of ranform Example 533 Employ Theorem 537 o evaluae F()/ for he funcion f() = inh a Soluion Iniially we eablih inh a By Theorem 537 from which a a e e a { inh a}= e d = = F a F () = a = inhae d a = a ( a ) = { }= a e inha d inh a a by CRC Pre LLC
11 We can, of coure, differeniae F() wih repec o a In hi cae, Theorem 537 doe no apply However, he reul i ignifican and i F () = e = { }= a d a a a = + a coh coh a a a Theorem 538 Complex Inegraion f() If f() i Laplace ranformable and provided ha lim + exi, he inegral of he funcion F() d correpond o he Laplace ranform of he diviion of he funcion f() by, () f Fd = () (3) Proof Le F() be piecewie coninuou in each finie inerval and of exponenial order Then ()= () F e f d i uniformly convergen wih repec o Conequenly, we can wrie for Re() > c and any a > c Expre hi in he form a () a = () F d e f d d = f() e dd = Now if f()/ ha a limi a, hen he laer funcion i piecewie coninuou and of exponenial order Therefore, he la inegral i uniformly convergen wih repec o a Thu, a a end o infiniy Theorem 539 Time Delay; Real Tranlaion The ubiuion of λ for he variable in he ranform { f ( )} correpond o he muliplicaion of he funcion F() by e λ ; ha i, { f ( λ )} = e λ F (3) Proof Refer o Figure 5, which how a funcion f ()u() and he ame funcion delayed by he ime = λ, where λ i a poiive conan () f ( e e ) d a a F d () = () f by CRC Pre LLC
12 FIGURE 5 A funcion f() a he ime = and delayed ime = λ We wrie direcly Now inroduce a new variable τ = λ Thi conver hi equaion o he form becaue u(τ ) = for λ We would imilarly find ha { f( λ) u( λ) }= f( λ) u( λ) e d { f ( + λ ) u ( + λ )} = e λ F (3) Example 534 Find he Laplace ranform of he pule funcion hown in Figure 53 { }= = = () λ τ λ τ λ f τ u τ e f τ u τ e dτ e f τ e dτ e F λ FIGURE 53 Pule funcion and i equivalen repreenaion Soluion Becaue he pule funcion can be decompoed ino ep funcion, a hown in Figure 53, i Laplace ranform i given by by CRC Pre LLC
13 where he ranlaion propery ha been ued Theorem 53 Complex Tranlaion The ubiuion of + a for, where a i a real or complex, in he funcion F( + a), correpond o he Laplace ranform of he produc e a f() Proof We wrie { [ ]} = u u 5 e () = e 5 5 which i e f() e d = f() e d for Re()> c Re ( a), a + a F ( + a) = {e a f ( )} (33) In a imilar way we find F ( a) = { e a f ( )} (34) Theorem 53 Convoluion The muliplicaion of he ranform of wo ecionally coninuou funcion f () (= F ()) and f () (= F ()) correpond o he Laplace ranform of he convoluion of f () and f () where he aerik i he horhand deignaion for convoluion Proof By definiion, he convoluion of wo funcion f () and f () i F F () = { f ( ) f ()} (35) Thu, f f f τ f τ dτ f τ f τ dτ () ()= = (36) { f() f () }= f( τ) f( τ) dτ = f( τ) f( τ) dτ e d = ( ) f τ dτ f τ e d Now effec a change of variable, wriing τ = ξ and herefore d = dξ, hen = τ ( ξ+ τ) f τ dτ f ξ e dξ by CRC Pre LLC
14 Bu for poiive ime funcion f (ξ ) = for ξ <, which permi changing he lower limi of he econd inegral o zero, and o which i { f ( ) f ( )} = F F Example 535 Given f () = and f () = e a, deduce he Laplace ranform of he convoluion e a by he ue of Theorem 53 Soluion Begin wih he convoluion τ = ξ f e d f e d τ τ ξ ξ, Then a a aτ e e = τ e dτ a τ aτ aτ τ e a = e a a e a = τ a By Theorem 53 we have and a { e }= a a = a ()= { ()}= {}= ()= () { }= { }= F f a, F f e a a { e }= a Theorem 53 The muliplicaion of he ranform of hree ecionally coninuou funcion f (), f (), and f 3 () correpond o he Laplace ranform of he convoluion of he hree funcion { f ( ) f ( ) f 3 ( )} = F F F 3 (37) Proof Thi i an exenion of Theorem 53 The reul i obviou if we wrie F F F 3 = { f ( ) { F F 3 }} Example 536 Deduce he value of he convoluion produc: f(); f() by CRC Pre LLC
15 Soluion By equaion (34) and (36) we wrie direcly F (a) For f () =, f () = f (), { f()} = = f ξ dξ by equaion (37) F ξ (b) For f () =, f () =, f 3 () = f(), { f()} = = f λ dλ dξ () Theorem 533 Frequency Convoluion plane The Laplace ranform of he produc of wo piecewie and ecionally coninuou funcion f () and f () correpond o he convoluion of heir ranform, wih () [ ] { f() f() }= F() F() π j (38) Proof Begin by conidering he following line inegral in he zplane: f ()= F ( z ) z e dz, σ π j = axi of convergence C Thi mean ha he conour inerec he xaxi a x > σ (ee Figure 54) Then we have FIGURE 54 The conour C and he allowed range of () () = () () f f e d π j C ( z ) f d F z e dz Aume ha he inegral of F (z) i convergen over he pah of inegraion Thi equaion i now wrien in he form by CRC Pre LLC
16 () () = () () σ + j f f e d π jσ j σ + j = F() z F z dz f f π j σ j ( z) F z dz f e d { } = () () (39) The Laplace ranform of f (), he inegral on he righ, converge in he range Re( z) > σ, where σ i he abcia of convergence of f () In addiion, Re(z) = σ for he zplane inegraion involved in (38) Thu, he abcia of convergence of f () f () i pecified by Re() > σ + σ (3) Thi iuaion i porrayed graphically in Figure 54 for he cae when boh σ and σ are poiive A far a he inegraion in he complex plane i concerned, he emicircle can be cloed eiher o he lef or o he righ ju o long a F () and F () go o zero a Baed on he foregoing, we oberve he following: Pole of F ( z) are conained in he region Re( z) < σ Pole of F (z) are conained in he region Re(z) < σ From (a) and (3) Re(z) > Re( σ ) > σ Pole of F ( z) lie o he righ of he pah of inegraion Pole of F (z) are o he lef of he pah of inegraion Pole of F ( z) are funcion of wherea pole of F (z) are fixed in relaion o Example 537 Find he Laplace ranform of he funcion f() = f () f () = e e u() Soluion From Theorem 533 and he abolue convergence region for each funcion, we have F()=, σ > + F()=, σ > + Furher, f() = exp[ ( + )] u() implie ha σ f = σ + σ = 3 We now wrie F( z) F ( z )= z z = z + 3 z + + To carry ou he inegraion dicaed by equaion (39) we ue he conour hown in Figure 55 If we elec conour C and ue he reidue heorem, we obain ()= ( ) = ( ) F π j [ ] = + F z F z dz j F z F z π Re C z = 3 The invere of hi ranform i exp( 3) If we had eleced conour C, he reidue heorem give by CRC Pre LLC
17 FIGURE 55 The conour for Example 537 ()= ( ) = ( ) F π j F z F z dz j F z F z π Re = + C z = = The invere ranform of hi i alo exp( 3), a o be expeced [ ] Theorem 534 Iniial Value Theorem Le f() and f () () be Laplace ranformable funcion, hen for cae when lim F() a exi, lim F lim f ()= () + (3) Proof Begin wih equaion (36) and conider Becaue f(+) i independen of, and becaue he inegral vanihe for, hen Furhermore, f(+) = lim + f() o ha [ ] = () ( + ) df lim e d lim F f d () ( + ) lim F f [ ] = lim F lim f ()= () + If f() ha a diconinuiy a he origin, hi expreion pecifie he value of he impule f(+) If f() conain an impule erm, hen he lefhand ide doe no exi, and he iniial value propery doe no exi by CRC Pre LLC
18 Theorem 535 Final Value Theorem Le f() and f () () be Laplace ranformable funcion, hen for lim f lim F ()= () (3) Proof Begin wih equaion (36) and Le Thu, he expreion = () ( + ) df lim e d lim F f d [ ] Conider he quaniy on he lef Becaue and are independen and becaue e a, hen he inegral on he lef become, in he limi Combine he laer wo equaion o ge I follow from hi ha he final value of f() i given by Thi reul applie F() poee a imple pole a he origin, bu i doe no apply if F() ha imaginary axi pole, pole in he righ half plane, or higher order pole a he origin Example 538 Apply he final value heorem o he following wo funcion: Soluion For he fir funcion from F, For he econd funcion, df = () + d d lim f f lim f f lim F f F () ( + )= () ( + ) lim f lim F + a ()= + + a b ()= () +, F ()= + b + a lim = + a b F ()= + b by CRC Pre LLC
19 However, hi funcion ha ingulariie on he imaginary axi a = ±jb, and he Final Value Theorem doe no apply The imporan properie of he Laplace ranform are conained in Table 5 in he Appendix 54 The Invere Laplace Tranform We employ he ymbol {F()}, correponding o he direc Laplace ranform defined in (), o denoe a funcion f() whoe Laplace ranform i F() Thu, we have he Laplace pair F = { f ( )}, f ( ) = { F } (4) Thi correpondence beween F() and f() i called he invere Laplace ranformaion of f() Reference o Table 5 how ha F() i a raional funcion in if f() i a polynomial or a um of exponenial Furher, i appear ha he produc of a polynomial and an exponenial migh alo yield a raional F() If he quare roo of appear on f(), we do no ge a raional funcion in Noe alo ha a coninuou funcion f() may no have a coninuou invere ranform Oberve ha he F() funcion have been uniquely deermined for he given f() funcion by () A logical queion i wheher a given ime funcion in Table 5 i he only funcion ha will give he correponding F() Clearly, Table 5 i more ueful if here i a unique f() for each F() Thi i an imporan conideraion becaue he oluion of pracical problem uually provide a known F() from which f() mu be found Thi uniquene condiion can be eablihed uing he inverion inegral Thi mean ha here i a oneoone correpondence beween he direc and he invere ranform Thi mean ha if a given problem yield a funcion F(), he correponding f() from Table 5 i he unique reul In he even ha he available able do no include a given F(), we would eek o reolve he given F() ino form ha are lied in Table 5 Thi reoluion of F() i ofen accomplihed in erm of a parial fracion expanion A few example will how he ue of he parial fracion form in deducing he f() for a given F() Example 54 Find he invere Laplace ranform of he funcion F ()= (4) Soluion Oberve ha he denominaor can be facored ino he form ( + ) ( + 3) Thu, F() can be wrien in parial fracion form a F 3 A B ()= ( + ) = (43) where A and B are conan ha mu be deermined To evaluae A, muliply boh ide of (43) by ( + ) and hen e = Thi give A= F() ( + ) = 3 = + 3 and B( + )/( + 3) = i idenically zero In he ame manner, o find he value of B we muliply boh ide of (43) by ( + 3) and ge = = 5 by CRC Pre LLC
20 B= F() ( + ) = 3 3 = = 3 = 6 The parial fracion form of (43) i F ()= The invere ranform i given by where enry 8 in Table 5, i ued Example 54 Find he invere Laplace ranform of he funcion Soluion Thi funcion i wrien in he form f()= { F() }= = 5e + 6e F ()= + ( + ) F A 3 ()= + + B + C + = ( + ) + ( + ) The value of A i deduced by muliplying boh ide of hi equaion by ( + 3) and hen eing = 3 Thi give To evaluae B and C, combine he wo fracion and equae he coefficien of he power of in he numeraor Thi yield from which i follow ha 3+ = = 3 ( + ) + A= + 3 F + ( + ) B C ( + ) + ( + 3) = 3 + = ( + ) ( ) + B + (C + 3B ) + C = + by CRC Pre LLC
21 Combine likepowered erm o wrie ( + B ) + ( 4 + C + 3B ) + ( 5 + 3C ) = + Therefore, + B =, 4 + C + 3B =, 5 + 3C = From hee equaion we obain B =, C = The funcion F() i wrien in he equivalen form Now uing Table 5, he reul i F ()= f ( ) = e 3 + e co, > In many cae, F() i he quoien of wo polynomial wih real coefficien If he numeraor polynomial i of he ame or higher degree han he denominaor polynomial, fir divide he numeraor polynomial by he denominaor polynomial; he diviion i carried forward unil he numeraor polynomial of he remainder i one degree le han he denominaor Thi reul in a polynomial in plu a proper fracion The proper fracion can be expanded ino a parial fracion expanion The reul of uch an expanion i an expreion of he form A A A A A p p pr F = B+ B + L+ + + L+ + + L + r p ( p) ( p) (44) Thi expreion ha been wrien in a form o how hree ype of erm; polynomial, imple parial fracion including all erm wih diinc roo, and parial fracion appropriae o muliple roo To find he conan A, A, he polynomial erm are removed, leaving he proper fracion where F () (B + B + L) = F() (45) A A A A A A k p p pr F ()= + + L L + r k p ( p) ( p) To find he conan A k ha are he reidue of he funcion F() a he imple pole k, i i only neceary o noe ha a k he erm A k ( k ) will become large compared wih all oher erm In he limi A lim F k = ( k ) k (46) by CRC Pre LLC
22 Upon aking he invere ranform for each imple pole, he reul will be a imple exponenial of he form Ak k = Ak e k (47) Noe alo ha becaue F() conain only real coefficien, if k i a complex pole wih reidue A k, here will alo be a conjugae pole k wih reidue A k For uch complex pole Ak k A + k k = + k k Ae k Ae k Thee can be combined in he following way: ( k k) + ( k k) [ ] σ k e ( akcoωk bkinωk) σ k Ae k co( ωk + θk) ( σk+ jωk) ( σk jωk) repone = a + jb e a jb e σ k = e ( ak + jbk) ( coωk + jinωk)+ ( ak jbk) coωk + jinωk = = (48) where θ k = an (b k /a k ) and A k = a k /co θ k When he proper fracion conain a muliple pole of order r, he coefficien in he parialfracion expanion A p, A p,, A pr ha are involved in he erm A p A p ( p) + ( p) A pr + L+ ( p) mu be evaluaed A imple applicaion of (46) i no adequae Now he procedure i o muliply boh ide of (45) by ( p ) r, which give r r r A A ( p) F()= ( p) + A k + L+ k + A p r ( p) + L (49) + A ( ( )+ A ) p r p pr In he limi a = p all erm on he righ vanih wih he excepion of A pr Suppoe now ha hi equaion i differeniaed once wih repec o The conan A pr will vanih in he differeniaion bu A p(r ) will be deermined by eing = p Thi procedure will be coninued o find each of he coefficien A pk Specifically, he procedure i pecified by A pk d = ( r k)! d r k r F (), k=,, K, r r k p = p (4) by CRC Pre LLC
23 Example 543 Find he invere ranform of he following funcion: Soluion Thi i no a proper fracion The numeraor polynomial i divided by he denominaor polynomial by imple long diviion The reul i The proper fracion i expanded ino parial fracion form The value of A i deduced uing (46) To find A and A we proceed a pecified in (4) Therefore, A F A = p() = F ()= F ()= A A A Fp= + p = A = + F [ ] = = [ ] = d d F d p =! d = () = = = = + 3+ = = = = = 4 From Table 5 he invere ranform i F ()= f ( ) = δ ( ) e, for If he funcion F() exi in proper fracional form a he quoien of wo polynomial, we can employ he Heaviide expanion heorem in he deerminaion of f() from F() Thi heorem i an efficien mehod for finding he reidue of F() Le P () A A F ()= Q () = + Ak + L+ k by CRC Pre LLC
24 where P() and Q() are polynomial wih no common facor and wih he degree of P() le han he degree of Q() Suppoe ha he facor of Q() are diinc conan Then, a in (46) we find Alo, he limi P() i P( k ) Now, becaue A k k = lim P () k Q () hen Thu, k lim = lim, k Q () k Q () Q () A k Pk = Q () = ( k ) () ( k ) P () k P ( n) F ()= Q () = () Q n= ( n) n (4) From hi, he invere ranformaion become Thi i he Heaviide expanion heorem I can be wrien in formal form Theorem 54 Heaviide Expanion Theorem If F() i he quoien P()/Q() of wo polynomial in uch ha Q() ha he higher degree and conain imple pole he facor k, which are no repeaed, hen he erm in f() correponding o P k hi facor can be wrien e k Q Example 544 Repea Example 4 employing he Heaviide expanion heorem Soluion We wrie (4) in he form () ( k ) F f ()= ()= () () P Q Pn () Q k = () n= ( n) P Q () = = e n ( + ) by CRC Pre LLC
25 The derivaive of he denominaor i Q () = + 5 from which, for he roo of hi equaion, Q () ( ) =, Q () ( 3) = Hence, P ( ) = 5, P ( 3) = 6 The final value for f() i f ( ) = 5e + 6e 3 Example 545 Find he invere Laplace ranform of he following funcion uing he Heaviide expanion heorem: Soluion The roo of he denominaor are Tha i, he roo of he denominaor are complex The derivaive of he denominaor i Q () = + 4 We deduce he value P()/Q () () for each roo Then ( + ) = + + j 3 j 3 () For = j 3 Q = j 3 P= j 3 () For = + j 3 Q ( )=+ j 3 P( )= + j 3 f ()= = e = e j 3 e j 3 j e j 3 j 3 j 3 j 3 j e e j 3 j 3 j 3 j 3 ( e e ) + j e + e j 3 j 3 j 3 ( + ) 3 j 3 = e co 3 in 3 3 by CRC Pre LLC
26 55 Soluion of Ordinary Linear Equaion wih Conan Coefficien The Laplace ranform i ued o olve homogeneou and nonhomogeneou ordinary differen equaion or yem of uch equaion To underand he procedure, we conider a number of example Example 55 Find he oluion o he following differenial equaion ubjec o precribed iniial condiion: y(+); (dy/d) + ay = x() Soluion Laplace ranform hi differenial equaion Thi i accomplihed by muliplying each erm by e d and inegraing form o The reul of hi operaion i from which Y y (+) + ay = X, If he inpu x() i he uni ep funcion u(), hen X() = / and he final expreion for Y() i Upon aking he invere ranform of hi expreion wih he reul Y Y ()= () ()= Example 55 Find he general oluion o he differenial equaion X y + + a + a y ( + a) a y y ()= { Y ()}= a + a a y a a ( e )+ y( + ) e a ()= ubjec o zero iniial condiion dy dy y = d d Soluion Laplace ranform hi differenial equaion The reul i Y 5Y 5Y ()+ ()+ ()= by CRC Pre LLC
27 Solving for Y(), we ge Expand hi ino parialfracion form, hu Then Y ()= = + A Y ()= B ( + ) C and A= Y() + B= Y() + 4 = C= Y() = = = = = ( + ) + 4 = = 4 = = = = 4 3 The invere ranform i Y ()= ( + ) + ( + ) Example 553 Find he velociy of he yem hown in Figure 56a when he applied force i f() = e u() Aume zero iniial condiion Solve he ame problem uing convoluion echnique The inpu i he force and he oupu i he velociy Soluion The conrolling equaion i, from Figure 56b, 4 x ()= e + e dν + 5ν+ 4 νd = e u() d Laplace ranform hi equaion and hen olve for F() We obain by CRC Pre LLC
28 FIGURE 56 The mechanical yem and i nework equivalen Wrie hi expreion in he form V ()= ( + ) = + ( + ) where A B C V()= A = ( + ) = 4 4 = 9 d B = =! d + 4 = C = = = 4 9 The invere ranform of V() i given by ν()= e + e e, To find ν () by he ue of he convoluion inegral, we fir find h(), he impule repone of he yem The quaniy h() i pecified by dh d + 5h + 4 hd = δ () where he yem i aumed o be iniially relaxed The Laplace ranform of hi equaion yield by CRC Pre LLC
29 H ()= The invere ranform of hi expreion i eaily found o be + + = ( + ) = h ()= e e, 3 3 The oupu of he yem o he inpu e u() i wrien = ( τ ) 4 4τ τ ν()= h( τ) f τ dτ e e e dτ 3 3 τ 4 3 τ = e e dτ dτ = e 4 3 e = 9 e 4 4 e e +, 9 3 Thi reul i idenical wih ha found uing he Laplace ranform echnique Example 554 Find an expreion for he volage ν () for > in he circui of Figure 57 The ource ν (), he curren i L ( ) hrough L = H, and he volage ν c ( ) acro he capacior C = F a he wiching inan are all aumed o be known FIGURE 57 The circui for Example 554 Soluion Afer he wich i cloed, he circui i decribed by he loop equaion 3i di di + i d + d di di i + 3i d + + d + ν ()= i() = ν () id = by CRC Pre LLC
30 All erm in hee equaion are Laplace ranformed The reul i he e of equaion + 3+ I ( ) I= V()+ i( + ) i I 3 ()= [ ( + )+ ( + )] q + I i i V()= I() The curren hrough he inducor i i L ( ) = i ( ) i ( ) A he inan = + i L (+) = i (+) i (+) Alo, becaue C q ()= i () d C [ ] = () + () = + ( ) C lim i d i d ν c + C, hen q( + ) q = c + ( + ) ν = ν c( )= i ()= C The equaion e i olved for I (), which i wrien by Cramer rule I ()= + ()+ L( + ) 3 V i ν c + ( + ) il( + ) vc + ( 3+ ) il( + ) V i = + ( + ) ( 3+ ) ( + ) + [ L( + )] () 3 c 4iL V = + ν ( + ) by CRC Pre LLC
31 Furher V = I Then, upon aking he invere ranform ν ( ) = { I } If he circui conain no ored energy a =, hen i L (+) = ν c (+) = and now ()= ν For he paricular cae when ν = u() o ha V () = / ( + ) V() ()= ν = = The validiy of hi reul i readily confirmed becaue a he inan = + he inducor behave a an open circui and he capacior behave a a hor circui Thu, a hi inan, he circui appear a wo equal reior in a imple erie circui and he volage i hared equally Example 555 The inpu o he RL circui hown in Figure 58a i he recurren erie of impule funcion hown in Figure 58b Find he oupu curren 3 4, 3 4 = e ( + 34 ) FIGURE 58 (a) The circui, (b) he inpu pule rain Soluion The differenial equaion ha characerize he yem i by CRC Pre LLC
32 di () + i d ()= ν () For zero iniial curren hrough he inducor, he Laplace ranform of he equaion i ( + )I() = V() Now, from he fac ha {δ ()} = and he hifing propery of Laplace ranform, we can wrie he explici form for V(), which i 3 4 V()= + e + e + e + e + L ( L) = + e e e e = + e Thu, we mu evaluae i() from Expand hee expreion ino The invere ranform of hee expreion yield The reul ha been keched in Figure The Inverion Inegral e e I ()= + e + = e e I e e e e e ()= e e L L i ( ) = e u ( ) + e ( ) u ( ) + e ( 4) u ( 4) + L + e ( ) u ( ) + e ( 3) u( 3) + e ( 5) u( 5) + L The dicuion in Secion 53 relaed he invere Laplace ranform o he direc Laplace ranform by he expreion F = { f ( )} f ( ) = { F} (6a) (6b) The ubequen dicuion indicaed ha he ue of equaion (6b) uggeed ha he f() o deduced wa unique; ha here wa no oher f() ha yielded he pecified F() We found ha alhough f() repreen a real funcion of he poiive real variable, he ranform F() can aume a complex variable form Wha hi mean, of coure, i ha a mahemaical form for he invere Laplace ranform wa no eenial for linear funcion ha aified he Dirichle condiion In ome cae, Table 5 i no adequae by CRC Pre LLC
33 FIGURE 59 The repone of he RL circui o he pule rain for many funcion when i a complex variable and an analyic form for he inverion proce of (6b) i required To deduce he complex inverion inegral, we begin wih he Cauchy econd inegral heorem, which i wrien Fz z dz j π F = () where he conour encloe he ingulariy a The funcion F() i analyic in he halfplane Re() c If we apply he invere Laplace ranformaion o he funcion on boh ide of hi equaion, we can wrie { ()}= σ+ jω jπ F lim F z dz ω σ jω z Bu F() i he Laplace ranform of f(); alo, he invere ranform of /( z) i e z Then i follow ha f σ+ jω σ + j z z lim e F z dz e F z dz πj ω σ jω πj σ j ()= = (6) Thi equaion applie equally well o boh he oneided and he woided ranform I wa poined ou in Secion 5 ha he pah of inegraion (6) i rericed o value of σ for which he direc ranform formula converge In fac, for he woided Laplace ranform, he region of convergence mu be pecified in order o deermine uniquely he invere ranform Tha i, for he woided ranform, he region of convergence for funcion of ime ha are zero for >, zero for <, or in neiher caegory, mu be diinguihed For he oneided ranform, he region of convergence i given by σ, where σ i he abcia of abolue convergence The pah of inegraion in (6) i uually aken a hown in Figure 5 and coni of he raigh line ABC diplayed o he righ of he origin by σ and exending in he limi from j o +j wih connecing emicircle The evaluaion of he inegral uually proceed by uing he Cauchy inegral heorem, which pecifie ha by CRC Pre LLC
34 FIGURE 5 The pah of inegraion in he plane f π j lim R Γ F e ()= () d = reidue of F () e a he ingulariie o he lef of ABC; > (63) Bu he conribuion o he inegral around he circular pah wih R i zero, leaving he deired inegral along he pah ABC, and f π j lim R Γ F e ()= () d = reidue of F () e a he ingulariie o he righ of ABC; < (64) We will preen a number of example involving hee equaion Example 56 Ue he inverion inegral o find f() for he funcion Noe ha by enry 5 of Table 5, hi i in w /w Soluion The inverion inegral i wrien in a form ha how he pole of he inegrand f ()= F π j ()= The pah choen i Γ in Figure 5 Evaluae he reidue + w e + jw jw d ( ) by CRC Pre LLC
35 Re ( jw) Re ( + jw) e + w e + w e e = = + jw wj = jw = jw jw e e = = jw wj = jw = jw jw Therefore, f ()= Re = e jw e jw jw inw = w Example 56 Evaluae {/ } Soluion The funcion F() = / i a doublevalued funcion becaue of he quare roo operaion Tha i, if i repreened in polar form by re j θ, hen re j (θ +π ) i a econd accepable repreenaion, and re j ( θ+ π) jθ = = re, hu howing wo differen value for Bu a doublevalued funcion i no analyic and require a pecial procedure in i oluion The procedure i o make he funcion analyic by rericing he angle of o he range π < θ < π and by excluding he poin = Thi i done by conrucing a branch cu along he negaive real axi, a hown in Figure 5 The end of he branch cu, which i he origin in hi cae, i called a branch poin Becaue a branch cu can never be croed, hi eenially enure ha F() i ingle valued Now, however, he inverion inegral (63) become for > } FIGURE 5 The inegraion conour for { by CRC Pre LLC
36 σ + j f()= lim F () e d= F () e R πjgab πjσ j = π j BC Γ l γ l+ Γ3 FG d, (65) which doe no include any ingulariy Fir we will how ha for > he inegral over he conour BC and CD vanih a R, from which = = Γ = BC Γ Γ3 FG arc BC i, becaue e jθ =, = Noe from Figure 5 ha β = co (σ/r) o ha he inegral over he I σ jω π e e jθ σ σ π jre dθ = e R dθ = e R co BC β jθ R e = e σ Rin σ R σ R Bu for mall argumen in (σ /R) = σ /R, and in he limi a R, I By a imilar approach, we find ha he inegral over CD i zero Thu, he inegral over he conour Γ and Γ 3 are alo zero a R For evaluaing he inegral over γ, le = re jθ = r(co θ + j in θ ) and γ π r coθ+ jinθ e j () d= jre θ dθ = a r π jθ Fe re The remaining inegral in (65) are wrien f j F e d F e d ()= () + () π l l+ (66) Along pah l, le = ue jπ = u; = j u, and d = du, where u and u are real poiive quaniie Then () = = j u du j u u e e F e d l j u du Along pah l +, = ue jπ = u, = j u (no + j u ), and d = du Then Combine hee reul o find () = u u e e F e d = l+ j u du j j u du by CRC Pre LLC
37 u u f()= u e du u e du j j =, π π which i a andard form inegral wih he value f Example 563 Find he invere Laplace ranform of he funcion π π π, ()= = > Soluion F ()= ( e ) + e The inegrand in he inverion inegral poee imple pole a: = and = jnπ, n = ±, ( + e ) ±3, ±L (odd value) Thee are illuraed in Figure 5 We ee ha he funcion e /( + e ) i analyic in he plane excep a he imple pole a = and = jnπ Hence, he inegral i pecified in erm of he reidue in he variou pole We have, pecifically e Re + e jn e Re + e = = jn = for = = for = jnπ (67) FIGURE 5 The pole diribuion of he given funcion by CRC Pre LLC
38 The problem we now face in hi evaluaion i ha Re ( a) n d where he roo of d() are uch ha = a canno be facored However, we know from complex funcion heory ha [ ()] = dd d () () becaue d(a) = Combine hi reul wih he above equaion o obain = a = a d da lim a () = d = lim a a a () By combining (68) wih (67), we obain () () Re ( a) n d = a = () () n [ ] d d d = a (68) We obain, by adding all of he reidue, jnπ e e Re odd d = n jn ( + e ) π d = jnπ Thi can be rewrien a follow f ()= + n= jnπ e jnπ Thi aume he form j3π jπ jπ j3π e e e e f()= + L L j3π jπ jπ j3π = + n= n odd jinnπ jnπ f ()= + in( k ) π π k k = (69) by CRC Pre LLC
39 A a econd approach o a oluion o hi problem, we will how he deail in carrying ou he conour inegraion for hi problem We chooe he pah hown in Figure 5 ha include emicircular hook around each pole, he verical connecing line from hook o hook, and he emicircular pah a R Thu, we examine f ()= e π j + e d = + + BC A j verical connecing line Hook π I I I3 Re (6) We conider he everal inegral in hi equaion Inegral I By eing = re j θ and aking ino conideraion ha co θ = co θ for θ > π /, he inegral I a r Inegral I Along he Yaxi, = jy and Noe ha he inegrand i an odd funcion, whence I = Inegral I 3 Conider a ypical hook a = jnπ The reul i Thi expreion i evaluaed (a for (67)) and yield e jnπ /jnπ Thu, for all pole I jy e = j dy jy jy + e r jn e lim r jn ( + e ) = π I 3 π jnπ e jπ e = d = + j + e j jn = + π π n π π = π n Finally, he reidue encloed wihin he conour are n= odd n odd innπ n jnπ e = + jn = + π π n= n= n odd n odd e Re + e innπ, n which i een o be wice he value around he hook Then when all erm are included in (6), he final reul i innπ in k π f()= + = + π n π k n= k= n odd by CRC Pre LLC
40 We now hall how ha he direc and invere ranform pecified by (4) and lied in Table 5 coniue unique pair In hi connecion, we ee ha (6) can be conidered a proof of he following heorem: Theorem 56 Le F() be a funcion of a complex variable ha i analyic and of order O( k ) in he halfplane Re() c, where c and k are real conan, wih k > The inverion inegral (6) wrien {F()} along any line x = σ, wih σ c converge o he funcion f() ha i independen of σ, whoe Laplace ranform i F(), F = { f ( )}, Re c In addiion, he funcion f() i coninuou for > and f() =, and f() i of he order O(e c ) for all > Suppoe ha here are wo ranformable funcion f () and f () ha have he ame ranform { f ( )} = { f ( )} = F The difference beween he wo funcion i wrien φ () where φ() i a ranformable funcion Thu, Addiionally, { } f()= F() φ ( ) = f ( ) f ( ) { φ ( )} = F F = φ()= {}=, > Therefore, hi require ha f () = f () The reul how ha i i no poible o find wo differen funcion by uing wo differen value of σ in he inverion inegral Thi concluion can be expreed a follow: Theorem 56 Only a ingle funcion f() ha i ecionally coninuou, of exponenial order, and wih a mean value a each poin of diconinuiy, correpond o a given ranform F() 57 Applicaion o Parial Differenial Equaion The Laplace ranformaion can be very ueful in he oluion of parial differenial equaion A baic cla of parial differenial equaion i applicable o a wide range of problem However, he form of he oluion in a given cae i criically dependen on he boundary condiion ha apply in any paricular cae In conequence, he ep in he oluion ofen will call on many differen mahemaical echnique Generally, in uch problem he reuling invere ranform of more complicaed funcion of occur han hoe for mo linear yem problem Ofen he inverion inegral i ueful in he oluion of uch problem The following example will demonrae he approach o ypical problem by CRC Pre LLC
41 Example 57 Solve he ypical hea conducion equaion ϕ = ϕ, < x <, x (7) ubjec o he condiion C ϕ (x, ) = f(x), = C ϕ =, ϕ (x, ) = x x = Soluion Muliply boh ide of (7) by e x dx and inegrae from o Alo = x Φ, e ϕ x, dx Equaion (7) hu ranform, ubjec o C and zero boundary condiion, o The oluion o hi equaion i Φ = Ae By an applicaion of condiion C, in ranformed form, we have The oluion, ubjec o C, i hen ϕ = ( + ) ϕ ϕ ( + ) x e x dx Φ, x dφ Φ = d Φ= = λ A f e d Now apply he inverion inegral o wrie he funcion in erm of x from, ϕ x, λ λ = + λ Φ, e f λ e dλ π j π j λ x e f λ e dλ e d ( + )= = λ+ x f λ dλ e d by CRC Pre LLC
42 Noe ha we can wrie x ( λ)= ( x λ) ( x λ ) 4 Alo wrie ( x λ) = u Then x u du ϕ( x, )= f exp λ d e λ λ π j 4 Bu he inegral e du= u π Thu, he final oluion i ϕ x, Example 57 A emiinfinie medium, iniially a emperaure ϕ = hroughou he medium, ha he face x = mainained a emperaure ϕ Deermine he emperaure a any poin of he medium a any ubequen ime Soluion The conrolling equaion for hi problem i = π ( x λ) 4 f λ e dλ ϕ ϕ = x K (7) wih he boundary condiion: a ϕ = ϕ a x =, > b ϕ = a =, x > To proceed, muliply boh ide of equaion (7) by e d and inegrae from o The ranformed form of equaion (7) i d Φ Φ, K = > dx K by CRC Pre LLC
43 The oluion of hi differenial equaion i x K x K Φ= Ae + Be Bu Φ mu be finie or zero for infinie x ; herefore, B = and Apply boundary condiion (a) in ranformed form, namely Therefore, K x Φ(, x)= Ae Φ(, e ϕ )= ϕ d = for x = A = ϕ and he oluion in Laplace ranformed form i K x ϕ Φ(, x)= e (73) To find ϕ (x, ) require ha we find he invere ranform of hi expreion Thi require evaluaing he inverion inegral x σ + j K ϕ e e ϕ( x, )= d π jσ j (74) Thi inegral ha a branch poin a he origin (ee Figure 53) To carry ou he inegraion, we elec a pah uch a ha hown (ee alo Figure 5) The inegral in (74) i wrien A in Example 56 ϕ ϕ( x, )= π j BC Γ l γ l+ Γ3 FG For he egmen BC = = = = Γ Γ3 jπ, le = ρe and for, le = ρe FG l l+ jπ by CRC Pre LLC
44 FIGURE 53 The pah of inegraion Then for and +, wriing hi um I, Wrie I l = πj e e e d = π jx K jx K e d in x K u = K = ku, d = kudu Then we have Thi i a known inegral ha can be wrien I = e ux du Ku l in π u Finally, conider he inegral over he hook, x K u Il = e du π Le u wrie I y = e π jγ e x K d jθ j d = re, d = jre θ dθ, = jθ, by CRC Pre LLC
45 hen I γ j jθ jθ re x r K e = e e dθ π j For r, I γ = jπ πj =, hen I γ = Hence, he um of he inegral in (73) become πj πj x K u x ϕ()= ϕ e du= ϕ π erf K (75) Example 573 A finie medium of lengh l i a iniial emperaure ϕ There i no hea flow acro he boundary a x =, and he face a x = l i hen kep a ϕ (ee Figure 54) Deermine he emperaure ϕ () FIGURE 54 Deail for Example 573 Soluion Here we have o olve ubjec o he boundary condiion: a ϕ = ϕ = x l b ϕ = ϕ > x = l ϕ c = > x = x ϕ ϕ = x k Upon Laplace ranforming he conrolling differenial equaion, we obain d Φ Φ = dx k by CRC Pre LLC
46 The oluion i x x Φ= Ae + Be = A x + B x k k coh inh k k By condiion c d dx x Thi impoe he requiremen ha B =, o ha Φ=Acoh x k Now condiion b i impoed Thi require ha Thu, by b and c ϕ = Acoh l k cohx Φ=ϕ cohl k k Now, o aify c we have cohx ϕ ϕ Φ= cohl k k Thu, he final form of he Laplace ranformed equaion ha aifie all condiion of he problem i cohx ϕ ϕ ϕ Φ= + cohl k k To find he expreion for ϕ (x, ), we mu inver hi expreion Tha i, ϕ ϕ ϕ( x, )= ϕ + π j σ + j σ j e cohx cohl k k d (76) by CRC Pre LLC
47 n π The inegrand i a ingle valued funcion of wih pole a = and = k, n =,, l We elec he pah of inegraion ha i hown in Figure 55 Bu he inverion inegral over he pah BCA(=Γ ) = Thu, he inverion inegral become FIGURE 55 The pah of inegraion for Example 573 π j σ + j σ j e cohx cohl k k d By an applicaion of he Cauchy inegral heorem, we require he reidue of he inegrand a i pole There reul Re = = Re = k π n l π k n l πx e cohj n l = d cohl d k = k π n l Combine hee wih (75) o wrie finally ( ) n 4 ϕ ϕ k n π l ϕ( x, )= ϕ + l co n π x l π n n= (77) by CRC Pre LLC
48 Example 574 A circular cylinder of radiu a i iniially a emperaure zero The urface i hen mainained a emperaure ϕ Deermine he emperaure of he cylinder a any ubequen ime Soluion The hea conducion equaion in radial form i ϕ + ϕ = ϕ, r< a, > r r r k (78) And for hi problem he yem i ubjec o he boundary condiion C ϕ = = r < a C ϕ = ϕ > r = a To proceed, we muliply each erm in he parial differenial equaion by e d and inegrae We wrie Then (77) ranform o ϕ e = d Φ ( r, ) k d Φ dφ + dr r dr Φ =, which we wrie in he form d Φ dφ + µ Φ =, µ = dr r dr k Thi i he Beel equaion of order and ha he oluion Φ = AI ( µr ) + BN ( µr ) However, he Laplace ranformed form of C when z = impoe he condiion B = becaue N () i no zero Thu, Φ = AI ( µr ) The boundary condiion C require Φ(r, a) = when r = a, hence, o ha ϕ ϕ A = I µ a by CRC Pre LLC
49 ϕ I µ r Φ= I µ a To find he funcion ϕ(r, ) require ha we inver hi funcion By an applicaion of he inverion inegral, we wrie σ + j ϕ I ξr λ dλ λ ϕ( r, )= e, ξ = π jσ j I ξa λ k (79) Noe ha I (ξ r)/i (ξ a) i a inglevalued funcion of λ To evaluae hi inegral, we chooe a he pah for hi inegraion ha hown in Figure 56 The pole of hi funcion are a λ = and a he roo of he Beel funcion J (ξ a) (= I (jξ a)); hee occur when J (ξ a) =, wih he roo for J (ξ a) =, namely λ = kξ, kξ, The approximaion for I (ξ r) and I (ξ a) how ha when n he inegral over he pah BCA end o zero The reulan value of he inegral i wrien in erm of he reidue a zero and when λ = ξ Thee are k n Re Re Therefore, = = kξn λdi ξa = dλ kξn J ξ r kξn ( n ) ϕ( r, )= ϕ + e d n λ ξ d I ( a ) λ λ= kξn FIGURE 56 The pah of inegraion for Example 574 by CRC Pre LLC
50 d Furher, λ d I a ai () ξ = ξ ξ a n λ Hence, finally, kξ n ϕ()= ϕ + e a n= J ξ r ( n ) () ξnj ( ξna ) (7) Example 575 A emiinfinie reched ring i fixed a each end I i given an iniial ranvere diplacemen and hen releaed Deermine he ubequen moion of he ring Soluion Thi require olving he wave equaion a ϕ = ϕ x (7) ubjec o he condiion C ϕ (x, ) = f(x) =, ϕ (, ) = > C lim x ϕ (x, ) = To proceed, muliply boh ide of (7) by e d and inegrae The reul i he Laplaceranformed equaion > a d Φ = Φ ϕ +, x dx (7) C Φ(, ) = C lim x Φ(x, ) = To olve (7) we will carry ou a econd Laplace ranform, bu hi wih repec o x, ha i {Φ (x, )} = N(z, ) Thu, = zx N z, Φ x, e dx Apply hi ranformaion o boh member of (7) ubjec o Φ(, ) = The reul i ()= ()= { } Φ N z, Φ z a z N z,,, Φ z ϕ x We denoe Φ (, ) by C Then he oluion of hi equaion i x C N( z, )= Φ z () a z z a a by CRC Pre LLC
51 The invere ranformaion wih repec o z i, employing convoluion = ( ) x ac x Φ x, inh ϕξinh a a a x ξ d ξ To aify he condiion lim x Φ(x, ) = require ha he inh erm be replaced by heir exponenial form Thu, he facor Then we have he expreion Bu for hi funcion o be zero for x require ha Combine hi reul wih Φ(x, ) o ge Each inegral in hi expreion i inegraed by par Here we wrie The reuling inegraion lead o We noe by enry 6, Table 5 ha ξ a x inh, inh, a a x e ( ξ) x = x ac ξ a Φ x, ϕξe dξ a ac a ξ a ϕξe dξ, x = = + x x a x a x a aφ x, e ( ) ξ ( + ξ) ( ξ) ϕξ dξ ϕξe dξ ϕξe dξ ( ξ x) () a a u du d d e d e ( ξ k) a = ϕξ, = ϕ ( ξ) ξ; ν= ξ, ν= Φ x, ( ξ x) ( x () + ξ) a () a ϕ x ϕ ξ e dξ ϕ ξ e dξ x = + ( x ξ) ϕ ξ e a dξ () ( ξ x) = e a when a > ξ = when a < ξ x Thi funcion of ξ vanihe excep when ξ x + a Thu, by CRC Pre LLC
52 + () ( + ) x a ξ x a () ϕ ( ξ) e dξ = ϕ ( ξ) dξ = ϕ x + a ϕ x x x Proceed in he ame way for he erm () ( + ) x ξ a ϕ ( ξ) e dξ = when a > x + ξ = when a < x + ξ Thu, he econd erm become () ( + ) x+ a x ξ a () ϕ ( ξ) e dξ = ϕ ( ξ) dξ = ϕ( x a) x x The final erm become The final reul i x a x ϕ( x )+ ϕ( x a) when a < x ϕ when > [ ] > ϕ( x, )= f( a+ x) f( a x) when x a [ ] < = f ( x + a )+ f ( x a ) when x a (73) Example 576 A reched ring of lengh l i fixed a each end a hown in Figure 57 I i plucked a he midpoin and hen releaed a = The diplacemen i b Find he ubequen moion FIGURE 57 A reched ring plucked a i midpoin Soluion Thi problem require he oluion of y y = c y, < y< l, > (74) ubjec o by CRC Pre LLC
53 bx y = < x < l l = b y = ( l x) < x < l l 3 y x l 4 y = x = ; x = l > To proceed, muliply (73) by e d and inegrae in Thi yield Y y c dy ()= dx or c dy dx Y = y()= f ( x ) (75) ubjec o Y(, ) = Y(l, ) = To olve hi equaion, we proceed a in Example 575; ha i, we apply a ranformaion on x, namely {Y(x, )} = N(z, ) Thu, Thi equaion yield, wriing Y() a Φ(x, ), The invere ranform i where N z, Y c z N z, y, x ()= Φ z, N( z, )= z c Combine hee inegral wih he known form of f(x) in C and C Upon carrying ou he inegraion, he reuling form become, wih k =, c Φ x, Y( x, )= { N( x, ) }= c inh c l x x l ξ Φ x y c c d x l ξ (, )= inh ( ξ ) inh ξ+ inh y( ξ) inh dξ c c by CRC Pre LLC
54 ly x c inhkx =, x l b kl coh ly ( l x c k l x l = ) inh ( ), x l b kl coh To find y(), we mu inver hee expreion Noe ha ymmery exi and o we need conider only he fir erm We ue he inverion inegral on he erm inh kx kl coh Thu, we conider he inegral I = π j σ + j σ j e λb xλ inh c xλ coh c dλ λ We chooe he pah in he λplane a hown in Figure 58 The value of he inegral over pah Γ i zero Thu, he value of he inegral i given in erm of he reidue Thee occur a λ = and a he λ value for which coh =, which exi where c FIGURE 58 The pah of inegraion for Example 576 λ c n π = j l n or λ= ± j πc l Thu, we have, by he heory of reidue by CRC Pre LLC
55 Re Re λ= = x c n j π c l = e j( n ) πcx l πx inhj( n ) l d λ λ λl coh d c ( n ) j l π c Thee pole lead o πc Thu, he pole a ±j(n ) lead o l Then we have o ha finally = = n ly 4l = x x+ b π n= n 8b y = π n= n πx in( n l ) l e π c n l For he ring for which x < l, he correponding expreion i he ame excep ha (l x) replace x π Noe ha hi equaion can be wrien, wih η = (n ), l which how he raveling wave naure of he oluion n π c n j( n ) l πx πc in( n ) co( n 4l ) l l π c n n πx πc in( n ) co( n ) l l πx πc l in( n ) co ( n ), x l l inη( x c)+ inη x+ c in ηx co ηc =, by CRC Pre LLC
56 58 The Bilaeral or TwoSided Laplace Tranform In Secion 5 we dicued he fac ha he region of abolue convergence of he unilaeral or oneided Laplace ranform i he region o he lef of he abcia of convergence The iuaion for he woided Laplace ranform i raher differen; he region of convergence mu be pecified if we wih o inver a funcion F() ha wa obained uing he bilaeral Laplace ranform Thi requiremen i neceary becaue differen ime ignal migh have he ame Laplace ranform bu differen region of abolue convergence To eablih he region of convergence, wrie he bilaeral Laplace ranform in he form F e f d e f d e f d ()= () = () + () (8) If he funcion f() i of exponenial order (e σ ), hen he region of convergence for > i Re() > σ If he funcion f() for < i of exponenial order exp(σ ), hen he region of convergence i Re() < σ Hence, he funcion F () exi and i analyic in he verical rip defined by σ < Re() < σ, provided, of coure, ha σ < σ If σ > σ, no region of convergence would exi and he inverion proce could no be performed Thi region of convergence i hown in Figure 59 FIGURE 59 Region of convergence for he bilaeral ranform Example 58 Find he bilaeral Laplace ranform of he ignal f() = e a u() and f() = e a u() and pecify heir region of convergence Soluion Uing he baic definiion of he ranform (8), we obain and i region of convergence i Re() > a For he econd ignal, F e a a u e ( + ) d e d a ()= () = = + a + a F ()= e u( ) e d = e d = + a by CRC Pre LLC
57 and i region of convergence i Re() < a Clearly, he knowledge of he region of convergence i neceary o find he ime funcion unambiguouly Example 58 Find he funcion f(), if i Laplace ranform i given by F ()= ( 4) ( + ) ( + ) 3, < Re ()< Soluion The region of convergence and he pah of inegraion are hown in Figure 5 For > we cloe he conour o he lef and we obain f ()= 3e ( 4) + = = e, > FIGURE 5 Illuraing Example 58 For <, he conour cloe o he righ and now Appendix f ()= 3e ( 4) + 3e + ( + ) + = = e = e +, < 5 by CRC Pre LLC
58 TABLE 5 Laplace Tranform Pair by CRC Pre LLC
59 by CRC Pre LLC
60 by CRC Pre LLC
61 by CRC Pre LLC
6.003 Homework #4 Solutions
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