1 Introduction. Let f(t) be a given function which is defined for all positive values of t, if. F(s) = A 0. Ae -st f(t) dt. L{f(t)} = F(s) = A 0

Transkript

1 LPLC TRNSFORMS Inroducion Le f() be a given funcion which i defined for all poiive value of, if F() = e - f() d exi, hen F() i called Laplace ranform of f() and i denoed by L{f()} = F() = e - f() d The invere ranform, or invere of L{f()} or F(), i f() = L - {F()} where i a poiive real number or a complex number wih poiive real par. Laplace -

2 ; e e ω where ( [xample] L{} = L{ e a } = - a L{ co ω } = e e - co ω d = - ( - co ω + ω in ω ) + = = + ω (Noe ha >, oherwie e - = diverge) L{ in ω } = - in ω d (inegraion by par) = e - in ω = + ω - co ω d = ω e - co ω d = ω L{ co ω } = ω + ω L{ n } = n e - d ( le = z/, d = dz/ ) = z n e -z dz = n+ z n e -z dz = Γ(n+) n+ α Recall ( ) Γ α = e d ) If n =,, 3,... Γ(n+) = n! L{ n } = n! n+ n i a poiive ineger Laplace -

3 =, L L [Theorem] Lineariy of he Laplace Tranform L{ a f() + b g() } = a L{ f() } + b L{ g() } where a and b are conan. [xample] If L{ e a } = - a hen L{ inh a } =?? Since L{ inh a } = L e a e -a = L{ e a } L{ e -a } = a + a = a a [xample] Find L - a L - a L - + a + a = - a a = e a + e e -a = a + e -a = coh a Laplace - 3

4 d d d d + xience of Laplace Tranform [xample] L{ / } From definiion, L {/ } = e - = e - + e - d Bu for in he inerval, e - e - (if > ), hen e - e - d e - d However, - d = lim - d = lim ln = lim ( ln ln ) = lim ( ln ) = e - diverge, no Laplace Tranform for /! Laplace - 4

5 e Piecewie Coninuou Funcion funcion i called piecewie coninuou in an inerval a b if he inerval can be ubdivided ino a finie number of inerval in each of which () he funcion i coninuou and () ha finie righ- and lef-hand limi. xience Theorem (Sufficien Condiion for xience of Laplace Tranform) Le f be piecewie coninuou on and aify he condiion f() M e γ for fixed non-negaive conan γ and M, hen L{ f() } exi for all > γ. [Proof] Since f() i piecewie coninuou, e - f() i inegraable over any finie inerval on >, L{ f() } = e - f() d - f() d M e γ e - d = M γ if Re() > γ L{ f() } exi. Laplace - 5

6 e π exi, + [xample] Do L{ n }, L{ e }, L{ -/ } exi? (i) e = + +! + 3 3! n n!... n n! e L { n } exi. (ii) > M e γ a approache infiniy L{ e } may no exi. (iii) L{ -/ } = bu noe ha -/ for! (See iem 4 on page 49 in exbook!) Laplace - 6

7 Some Imporan Properie of Laplace Tranform () Lineariy Properie L{ a f() + b g() } = a L{ f() } + b L{ g() } where a and b are conan. (i.e., Laplace ranform operaor i linear) () Laplace Tranform of Derivaive If f() i coninuou and f'() i piecewie coninuou for, hen [Proof] L { f'() } = L{ f() } f( + ) L{ f'() } = f'() e - d Inegraion by par by leing u = e - du = - e - d dv = f'() d v = f() L{ f'() } = e f() + e f() d = f() + L{ f ( ) } L { f'() } = L{ f() } f( + ) Laplace - 7

8 Theorem: f(), f'(),..., f (n-) () are coninuou funcion for, and f (n) () i piecewie coninuou funcion, hen L{ f (n) } = n L{ f } n- f() n- f'()... f (n-) () e.g, L{ f''() } = L{ f() } f() f'() and L{ f'''() } = 3 L{ f() } f() f'() f''() [xample] L{ e a } =?? f() = e a, f() = and f'() = a e a L{ f'() } = L{ f() } f() or L{ a e a } = L{ e a } or a L{ e a } = L{ e a } L{ e a } = a [xample] L{ in a } =?? f() = in a, f() = f'() = a co a, f'() = a f''() = a in a L{ f''() } = L{ f() } f() f'() L{ a in a } = L{ in a } a or a L{ in a } = L{ in a } a L{ in a } = a + a Laplace - 8

9 [xample] L{ in } = ( + 4) ( ) Known: in ; () ; ( ) in co in lo, f = f = f = = { } L in = + 4 { } = { f } = { f} f = { } Thu, L in L L () L in L{ in } = L{ in } = + ( 4) [xample] L{ f()} =L{ in ω } = ω ( + ω ) ( ) f = in ω, f ( ) = inω+ ωco ω, f () = f ( ) = ωcoω ω inω = ωcoω ω f( ) { } f () = f = ω ω ω f = f f = f L L{co } L{ ( )} L{ } f ( ) () L{ } ω + ω { } = ω { ω } = + ω ( + ω ) ( ) L f L in [xample] y'' 4 y =, y() =, y'() = (IVP!) [Soluion] Take Laplace Tranform on boh ide, L{ y'' 4 y } = L{ } or L{ y'' } 4 L{ y } = L{ y } y() y'() 4 L{ y } = or L{ y } 4 L{ y } = L{ y } = + 4 = y() = e Laplace - 9

10 ( ) Y = + + Laplace -

11 = = [xercie] y'' + 4 y =, y() =, y'() = (IVP!) y() = co + in [xercie] y'' 3 y' + y = 4 6, y() =, y'() = 3 (IVP!) ( y 3 ) 3 ( y ) + y = 4 6 y = + ( ) = + y = L - + ( ) L - + e + [xercie] y'' 5 y' + 4 y = e, y() =, y'() = (IVP!) y() = e e 6 e 4 Laplace -

12 or Queion: Can a boundary-value problem (BVP) be olved by Laplace Tranform mehod? [xample] y'' + 9 y = co, y() =, y(π/) = Le y'() = c L{ y'' + 9 y } = L{ co } y y() y'() + 9 y = + 4 y c + 9 y = + 4 y = + c ( + 9) ( + 4) = c ( + 4) y = L - { y } = 4 5 co 3 + c 3 in co Now ince y(π/) =, we have = c/3 /5 c = /5 y = 4 5 co in co [xercie] Find he general oluion o y'' + 9 y = co by Laplace Tranform mehod. Le y ( ) ( ) = c y = c Laplace -

13 = } L = ddiional Remark: Since L{ f'() } = L{ f() } f( + ) if f() i coninuou If f() = L - { f () = f'() (i.e., muliplied by ) [xample] If we know L - + in hen L - +?? [Soluion] Since in = L - + d = d - + d = d in = co Laplace - 3

14 } = e (divided L{ (3) Laplace Tranform of Inegral If f() i piecewie coninuou and f() M e γ, hen L{ a f(τ) dτ } = L{ f() } + a f(τ) dτ = e- a f(τ) dτ L{ a + f(τ) dτ = a f() e - d = f(τ) dτ a [Proof] - d (inegraion by par) f(τ) dτ + Special Cae: for a =, f() e - d = a f(τ) dτ + f() } L{ f(τ) dτ } = f () L{ f() } = Invere: L - f () f(τ) dτ by!) Laplace - 4

15 = = = [xample] If we know L = in, hen L - ( + 4) in τ dτ = co 4 [xercie] If we know L - + in, hen L - 3 ( + )?? in ddd ( co ) ( in ) = dd = d = + co = + co Laplace - 5

16 e = f() d L{ (4) Muliplicaion by n L{ n f() } = ( ) n d n f () n e.g., if n=, hen L{ f() } = f '() [Proof] f () L{ f() } = e - f() d df () d d = d - f() d = ( ) e- d (Leibniz formula)f = e - f() d = e - f() d = L{ f() } d L{ f() } = d f () d = d f() } Leibniz' Rule: d dα φ (α) F(x,α) dx = φ (α) φ (α) F α φ (α) dx + F(φ, α) dφ dα - F(φ, α) dφ dα Laplace - 6

17 + = = = = = ' = + [xample] L{ e } = d L{ e } = - d ( ) ( ) d L{ e } = d ( ) ( ) 3 [xercie] L{ in ω } =?? L{ co ω } =?? [xample] y'' y' y =, y() =, y'() = 3 [Soluion] Taking Laplace ranform of boh ide of he differenial equaion, we have L{ y'' y' y } = L{ } or L{ y'' } L{ y' } L{ y } = d d Noe ha L{ y'' } = d L{ y'' } = d ( y y() y'() ) y ' y dy y() = y d d d L{ y' } = d L{ y' } = d ( y y() ) y y = dy y d L{ y } = y y ' y y ' + y - y or y + y dy d y = Laplace - 7

18 indirecly or = = Solve he above equaion by eparaion of variable for y, we have y = c ( ) y = c e Bu y'() = 3, we have 3 = y'() = c (+) e = = c y() = 3 e [xample] valuae L - an - ( ) by (4), i.e., L{ f() } = - F'() and F()= an - (/). [Soluion] I i eaier o evaluae he inverion of he derivaive of an - ( ) wr. By making ue of he ideniy: ( an - ) = + +, one can obain F ()=( an - (/) ) = -/ (/) L d an = L- d in d - in = L an = d L- { F () } = - f() = L an, -in = - f(). L an = f() = in Laplace - 8

19 = } indirecly = [xample] valuae L ln( + - ) by (4) L - ln( + ) L - { f () = f() and f = d d (ln( + )) + + Since from (4) we have L - { f '() } = f() + e - = f() f() = e - Laplace - 9

20 d provided and exi (5) Diviion by L f() = f ( ~ ) ~ f() ha for. [xample] L{ in } = + I i known ha in lim = L in = d ~ ~ + = an = an - ( ) Laplace -

21 ln( = L{ d - ] = d = = (. ln d [xample] Deermine he Laplace Tranform of in L{ in } = L{ - co } = + 4 ( + 4 ) Thu, L{ in } = ( + 4 ) = = [ ln ) ] [ 4 ln ln + 4 lim ln =ln ) = + 4 [xample] valuae he inegral e - in d in } = e - in a =, Thu, e - in d = 4 ln + 4 = 4 5 Laplace -

22 ( ) } = } (6) Fir Tranlaion or Shifing Propery ( -Shifing ) If L{ f() } = f (), hen L{ e a f() } = f ( a) If L - { f () = f(), hen L - { f ( - a) = e a f() [xample] L{ co } = + 4 L{ e - co } = + ( + ) + 4 = [xercie] L{ e - in 4 } [xample] L L ( ) + 6 = L - 6 ( ) = 6L L + ( ) + 4 ( ) + 4 = 6 e co 4 + e in 4 = e ( 3 co 4 + in 4 ) Laplace -

23 and e (7) Second Tranlaion or Shifing Propery ( -Shifing) If L{ f() } = f () f( a) if > a g() = if < a L{ g() } = e -a f () [xample] L{ 3 } = 3! 4 = 6 4 ( ) 3 > g() = < L{ g() } = Laplace - 3

24 e e e and } (8) Sep Funcion, Impule Funcion and Periodic Funcion (a) Uni Sep Funcion (Heaviide Funcion) u( a) Definiion: < a u( a) = > a Thu, he funcion can be wrien a f( a) > a g() = < a g() = f( a) u( a) The Laplace ranform of g() can be calculaed a L{ f( a) u( a) } = e - f( a) u( a) d = a - f( a) d ( by leing x = a ) = -(x+a) f(x) dx = e -a -x f(x) dx = e -a L{ f() } = e -a f () L{ f( a) u( a) } = e -a L{ f() } = e -a f () L - { e -a f() = f( a) u( a) Laplace - 4

25 [xample] L{ in a( b) u( b) } = e -b L{ in a } = a e -b + a [xample] L{ u( a) } = e -a [xample] Calculae L{ f() } e where f() = e + co [Soluion] π > π Since he funcion u( π) co( π) = co( π) ( = co) < π > π he funcion f() can be wrien a f() = e + u( π) co( π) L{ f() } = L{ e } + L{ u( π) co( π) } = + e -π + Laplace - 5

26 =, [xample] L - = in u( e -π / + π ) in ( L - π + L - e -π / ) = in + u( + π ) co [xample] Recangular Pule f() = u( a) u( b) L{ f() } = L{ u( a) } L{ u( b) } = e -a e -b [xample] Saircae f() = u( a) + u( a) + u( 3a) +... L{ f() } = L{ u( a) } + L{ u( a) }+ L{ u( 3a) } +... = ( e-a + e -a + e -3a +... ) If a >, e -a <, and ha + x + x +... = n= x n = x x < hen, for >, L{ f() } = e -a e -a Laplace - 6

27 e } or wih ) [xample] Square Wave f() = u() u( a) + u( a) u( 3a) +... L{ f() } = = { ( e -a + e -a e -3a +... ) }= = ea/ e -a/ a/ + e -a/ = + e -a anh( a ( e-a + e -a e -3a +... ) = e-a + e -a [xample] Solve y' + y + 6 z d = u() y' + z' + z = y() = 5, z() = 6 [Soluion] We ake he Laplace ranform of he above e of equaion: ( L{ y } + 5 ) + L{ y } + 6 L{ z } = ( L{ y } + 5 ) + ( L{ z } 6 ) + L{ z } = ( + ) y + 6 z = 5 y + ( + ) z = y = = z y (3 + ) 4 = = = + + ( )( + 4) + 4 y = L - { y = u() 4 e 3 e -4 z = e + 4e 4 Laplace - 7

28 [xercie] y' + y + z' + 3 z = e - 3 y' y + 4 z' + z = y() =, z() = [xercie] y'' + y = f(), y() = y'() = where f() = > Laplace - 8

29 fk() (b) Uni Impule Funcion ( Dirac Dela Funcion ) δ( a) Definiion: /k a a+k Le fk() = oherwie and Ik = d = Define: δ( a) = lim k fk() From he definiion, we know Laplace - 9

30 δ( a) δ() δ( a) = a δ( a) = a and d = d = Noe ha d = δ() g() d = g() for any coninuou funcion g() δ( a) g() d = g(a) The Laplace ranform of δ() i L{ δ( b) } = e - δ( b) d = e -b [Queion] L{ e co δ( 3) } =?? Laplace - 3

31 = ( (+) = ) [xample] Find he oluion of y for y'' + y' + y = δ( ), y() =, y'() = 3 [Soluion] The Laplace ranform of he above equaion i ( y 3 ) + ( y ) + y e - or y e = = (+) + 5 (+) + e - (+) = ( + ) + e - ( + ) Since L{ e - } = ( + ) Recall L{ } = L - e - ( + ) ( ) e -( - ) u( ) y = e e - + ( ) e -( - ) u( ) = e - [ e ( ) u( ) ] Laplace - 3

32 e e (c) Periodic Funcion For all, f(+p) = f(), hen f() i aid o be periodic funcion wih period p. Theorem: The Laplace ranform of a piecewie coninuou periodic funcion f() wih period p i p L{ f } = e -p e - f() d [Proof] L{ f } = e - f() d p = p e - f() d + p e - f() d 3p + p - f() d +... (k+)p kp p - f() d = e -(u+kp) f(u+kp) du which i he (k+)h inegral! (where u = kp and <u<p) Laplace - 3

33 ( e e (Ue p = e -kp -u f(u) du...ince f(u+kp) = f(u) L{ f } = k= p -kp e -u f(u) du = p e -u f(u) du (e -p ) k k= = p e -u f(u) du e -p [xample] Find L{ in a }, a > [Soluion] p = π a due o ) L{ in a } = p e -u f() d e -p = π/a e - in a d e -π /a inegraion by par wice) Laplace - 33

34 (Thi ) = a + a + e -π /a e -π /a = π π a a e + e / a π π + a a a e e / = a + a coh( π a [xample] y'' + y' + 5 y= f(), y() = y'() = where f() = u() u( π) + u( π) u( 3π) +.. (Square Wave!) [Soluion] The Laplace ranform of he quare wave f() i L{ f() } = e -π + e -π wa derived previouly!) y + y + 5 y = e -π + e -π or y = e -π + e -π Now ( ) = = 5 + ( + ) + Laplace - 34

35 = u( kπ) (k=, in ( ) + = 5 ( + ) + ( + ) + and e -π + e -π ( e -π ) ( e -π + e -π e -3π +... ) = e -π + e -π e -3π +... (derived previouly) y = 5 + ( + ) + ( e -π + e -π e -3π + ) The invere Laplace ranform of y can be calculaed in he following way: L ( + ) + ( + ) ( ) ( ) = L = g( ) e co + in 5 = 5 (k=, he fir erm) L ( + ) + e -kπ, 3, ) = 5 ( g( kπ) ) Bu g( kπ) = e -(-kπ ) ( co ( kπ) + ( kπ) ) = e kπ g() = k e π e co + in Laplace - 35

36 5 u( π) u( π) u( 3π) y() = 5 ( g()) 5 ( eπ g()) + 5 ( eπ g()) - 5 ( e3π g()) + = 5 ( u( π) + u( π) u( 3π) +...) g() ( e π u( π) + e π u( π)...) = 5 ( f() g()( eπ u( π) + e π u( π) e 3π u( 3π) +...) ) Laplace - 36

37 e = = (9) Change of Scale Propery L{ f() } = f () hen L{ f(a) } = a f( a ) [Proof] L{ f(a) } = - f(a) d = e -u/a f(u) d(u/a) = a e -u/a f(u) du = a f( a ) [xercie] Given ha L in = an - (/) in a Find L?? in a Noe ha L a = a f (/a) = a an - (a/) L in a a L in a a = an - (a/) Laplace - 37

38 g( v) f(v) () Laplace Tranform of Convoluion Inegral Definiion of Convoluion If f and g are piecewie coninuou funcion, hen he convoluion of f and g, wrien a (f*g), i defined by (f*g)() f( τ) g(τ) dτ Properie (a) f*g = g*f (commuaive law) (f*g)() = = - f( τ) g(τ) dτ g( v) dv ( by leing v = τ) = f(v) dv = (g*f)() q.e.d. (b) f*(g + g) = f*g + f*g (lineariy) (c) (f*g)*v = f*(g*v) (d) f* = *f = Laplace - 38

39 = = (e) *f f in general Convoluion Theorem Le f () L{ f() } and g () L{ g() } hen L{ (f*g)() } = f () g () [Proof] Laplace - 39

40 = = f () g () e -τ f(τ) dτ e -v g(v) dv = e -(τ +v) f(τ) g(v) dv dτ Le = τ + v and conider inner inegral wih τ fixed (given), hen d = dv and f () g () τ e - f(τ) g( τ) d dτ τ d dτ = dτ d Laplace - 4

41 = = e } = f () g () τ e - f(τ) g( τ) d dτ = e - f(τ) g( τ) dτ d = e - g( τ) f(τ) dτ d = - (g*f)() d = e - (f*g)() d = L{ f*g } Corollary If f () L{ f() } and g () L{ g() }, hen L - { f() g () = (f*g)() Laplace - 4

42 L{ [xample] Find L - ( + ) Recall ha he Laplace ranform of co and in are L{ co } = + in } = + Laplace - 4

43 co coτ Thu, L - ( + ) = L = in * co Since in * co = in( τ) coτ dτ = ( in coτ co inτ ) dτ = in τ dτ co inτ coτ dτ co = in in co + + = in Noe, co(-b)=co co B + in in B Laplace - 43

44 [xample] Find he oluion of y o he differenial equaion y'' + y = f(), y() =, y'() = and < < f() = > [Soluion] The funcion f() can be wrien in erm of uni ep funcion: f() = u() u( ) Now ake he Laplace ranform on boh ide of he differenial equaion, we have y + y = e - or y + e = -e ( + ) = - + e - + y = co + in [in * u( ) ] Bu he convoluion in * u( ) = in( τ) u(τ ) dτ For <, u( ) =, in * u( ) = Laplace - 44

45 in( τ) and for >, u( ) =, u(τ ) dτ = in( τ) dτ Thu, in * u( ) = u( ) in( τ) dτ = u( ) co( τ) y = co + in u( ) [ co( ) ] = u( ) [ co( ) ] [xample] Volerra Inegral quaion y() = f() + g( τ) y(τ) dτ where f() and g() are coninuou. The oluion of y can eaily be obained by aking Laplace ranform of he above inegral equaion: Laplace - 45

46 = = + { y () f () g () y () y () f () - g () For example, o olve y() = + in( τ) y(τ) dτ y = y or y = L n! n} = n + y = + 4 Laplace - 46

47 = = = = = () Limiing Value (a) Iniial-Value Theorem lim f() = lim f () (b) Final-Value Theorem lim f() = lim f () [xample] f() = 3 e -, f() = 3, f() = f () L{ f() } = 3 + lim f () f() lim f () 3 + f() [xercie] Prove he above heorem Laplace - 47

48 = L { f'() } = L{ f() } f( + ) Parial Fracion - Pleae read Sec. 5.6 of he Texbook 3 L - F() G()?? where F() and G() are polynomial in. Cae G() = ha diinc real roo (i.e., G() conain unrepeaed facor ( a) ) Cae Laplace - 48

49 = e } } } eā x e L = L{ u 4 Laplace Tranform of Some Special Funcion () rror Funcion Definiion: erf() = π -x dx rror Funcion erfc() erf() = π -x dx Complemenary rror Funcion [xample] Find L{ erf erf π dx = π -/ e -u du ( by leing u = x ) L { erf = π L u -/ e -u du ( Recall ha L f(τ) dτ f() } ) L { erf = π { -/ e - } Laplace - 49

50 } = = { (by Bu L{ -/ } = Γ(/) π L α } ( α ) Γ + = α + we have L{ -/ e - } = π + -hif) L{ erf = + [xercie] Find L - ( )?? e erf () Beel Funcion [xample] Find L{ Jo() } Noe ha d y dy + + ( p ) y = d d d p J p p( ) = J p+ ( ) d [Soluion] Noe ha J() aifie he Beel' differenial equaion: J''() + J'() + J() = (p=) Laplace - 5

51 = = We now ake L on boh ide and noe ha J() = and J'() = J ( ) = d d ( J () ) + ( Jo dj ) d dj + J + J = d ( ) = J + By eparaion of variable Jo = c + Noe ha lim f () f() (Iniial Value Theorem) lim Jo = J() = we have c + = = c = Jo = L{ J() } = + Laplace - 5

52 Hin: d ) [xercie] Find L{ J(b) } =?? [xercie] Find L{ J() } if J () = - J() [xercie] Find L{ e -a J(b) } [xercie] Find L - J() + d = ln ( + + [xercie] Find J() d [xercie] Find L{ e - J() } [xercie] Find e - - J() Laplace - 5

53 } f(τ) ; = () ;. for ; SUMMRY L{ = L{ n } = n! n+ n N L{ e a } = a L{ in ω } = ω + ω L{ co ω } = + ω L{ a f() + b g() } = a L{ f() } + b L{ g() } L - { a f () + b g () } = a L - { f () } + b L - { g () } = a f() + b g() L { f () } = L{ f() } f( + ) Noe ha f() i coninuou for and f'() i piecewie coninuou. If f() = f () = f () =... = f (n-) () =, hen L - { n f () } = f (n) () 3 L{ dτ } = f () L{ f() } = Queion: wha if he inegraion ar from a inead of? 3 L - f () n.. f() d... d 4 L{ f() } = - f ; L{ n f() } = ( - ) n f (n) () Laplace - 53

54 d = = ; e d 6 if 7 exi f( ) 4 L - d n d n f () ( - ) n n f() 5 L f() = f ( ~ ) ~ f() for. 5 L f - ( ~ ) ~ f() 6. L{ e a f() } = f ( a) L - { f ( a) } = e a f() 7. L{ f( a) u( a) } = e -a f () L - { e -a f () } = f( a) u( a) 8. L{ u( a) } = e -a L{ δ( a) } = e -a ; L{ f } = p e -p - f() d where f() i a periodic funcion wih period p 9. L{ f(a) } = a f( a ) 9 L - { f (a) } = a a. L{ (f*g)() } = f () g () L - { f () g () } = f*g where (f*g)() f( τ) g(τ) dτ Laplace - 54

55 ;. lim f() = lim f () lim f() = lim f () Laplace - 55