LAPLACE TRANSFORMS. 1. Basic transforms

Transcription

1 LAPLACE TRANSFORMS. Bic rnform In hi coure, Lplce Trnform will be inroduced nd heir properie exmined; ble of common rnform will be buil up; nd rnform will be ued o olve ome dierenil equion by rnforming hem ino lgebric equion which cn be eily olved. Le f() be dened for. Then we dene he Lplce Trnform of f We lo dene he invere rnform L by F () L(f) f L F e f() d Remrk.. L i n operor on he funcion f. Where f depend on he vrible, F i independen of. Exmple. f(), F () e d e Thu we y h L().. Exponenil. Exmple f() e,, Lineriy of Lplce Trnform L(e ) e e d e ( ) d e ( ) if >. Le f() nd g() be dened for wih rnform F () L(f) nd G() L(g) repecively, nd le nd b be conn. Then L{f() + bg()} F () + bg()

2 LAPLACE TRANSFORMS Proof. L{f() + bg()} e f() + bg() d e f() d + b F () + bg() e g() d Exmple. coh e + e when > nd + >. We ummrie hi or.. Power of. L( ) L(coh ) {Le + Le } { + } + L(coh ) + where Γ() i he Gmm funcion dened by for >. Inegring by pr: Γ( + ) e d > ( e x x ) dx L( ) Γ() e x x dx e x x dx Γ( + ) + e x x dx eing x x e x + e x x dx Γ() Remrk.. Γ() e x dx e x Le n be poiive ineger. Then Γ(n + ) nγ(n) n(n )Γ(n ) n(n )(n )Γ(n ). n(n )(n )... (3)()()Γ() n!

3 . BASIC TRANSFORMS 3 Thu for poiive ineger n we hve L( n ) n! n+ We lo hve Γ(/) π. Even hough he Gmm funcion i o fr only dened for poiive vlue of, we cn exend he deniion o (ome) negive vlue of uing he propery Γ( + ) Γ(). For exmple, nd Γ( /) Γ(/) / π Γ( 3/) Γ( /) 3/ 4 π/3 However, Γ() Γ() nd o Γ() / which i unbounded. Similrly Γ(n) i unbounded for ll negive ineger n..3. Trigonomeric funcion. We lredy hve L(e ) nd o nd ince e iω co ω + i in ω we hve nd o, equing rel nd imginry pr, we hve L(e iω ) iω + iω + ω + ω + i ω + ω L(co ω + i in ω) + ω + i ω + ω L(co ω) + ω nd L(in ω) ω + ω.4. Sep funcion. k, < c; f(), c L{f()} c e f() d ke d k k e c e c

4 4 LAPLACE TRANSFORMS.5. Lplce rnform of derivive. Le f() be dierenible for wih rnform L{f()}. Then L{f ()} e f () d Inegring by pr, wih u e du e d dv f () d v f() Lf () f()e + e f() d Lf() f() Similrly, if f i wice dierenible, hen L{f ()} Lf () f () Lf() f() f () Lf() f() f () nd in generl L{f (n) ()} n L{f()} n f() n f ()... f (n ) () To verify hi, we cn ee for exmple h if f() 3, hen f() f () f () f () 6 Then L{f ()} L(6) 6/ 3 L{f()} f() f () f () 3 L( 3 ) nd o lredy een. Exmple (Iniil Vlue Problem) L( 3 ) 6/ 4 3!/ 4 y + 5y + 6y y(), y () 3 Le Y () L{y()} Then nd rnforming he equion give or L{y ()} Y y() Y L{y ()} Y y() y () Y 3 Y 3 + 5(Y ) + 6Y ( )Y + 3 which i clled he ubidiry equion. Thi h oluion Y which now mu be invered o obin he oluion y(). + 3 ( + 3)( + )

5 . BASIC TRANSFORMS 5 Uing pril frcion nd hu Y y L (Y ) 9L { + } 7L { + 3 } 9e 7e 3.6. The hifing heorem. To olve more compliced problem we need he following. Theorem.3. (Fir Shifing Theorem.) If L{f()} F () for > γ; hen L{e f()} F ( ) for > γ + Proof. F () e f() d F ( ) e ( ) f() d e e f() d L{e f()} Exmple. L() / F () for >. Thu L(e ) F ( ) /( ) for > een lredy. Exmple. y + y + 5y, y(), y () 4 Trnforming Y (Y ) + 5Y ( + + 5)Y Y ( + ) + + ( + ) + ( + ) + From before, we know h L { } co + nd L { } in + nd o, uing he r hifing heorem hi implie nd nd o y L (Y ) e co in L + { ( + ) + } e co L { ( + ) + } e in

6 6 LAPLACE TRANSFORMS The nlogue of he r hifing heorem for invere rnform i he econd hifing heorem. Thi mke ue of he uni ep funcion:, < u (), Thi i lo omeime wrien u( ) or he Heviide funcion H () or H( ). Theorem.4. (Second hifing heorem) If L (F ()) f(), hen L (e F ()) f(), < f( ), f( )u () Proof. e F () e e z f(z) dz + z e (+z) f(z) dz e f( ) d e f( )u () d Lf( )u () Exmple. L e 3 Since L + in, he econd hifing heorem implie h + Remrk.5. e L 3 in( 3)u 3 () +, < 3 in( 3), 3 L{u ()} e u () d e d e e Remrk.6. The del funcion δ () (lo wrien δ( )) which i innie he poin nd zero elewhere i he r derivive of he uni ep funcion u (). Thu L{δ ()} L{u ()} L{u ()} u () e

7 . BASIC TRANSFORMS 7.7. Trnform of inegrl. Theorem.7. If f h rnform L(f) F () hen Proof. Le g() L f F () f. Then g () f() nd g() nd o L{f()} L{g ()} L{g()} g() L{g()} nd o L{g()} L f L{f()} Exmple. Inver nd ( + ). L L ( + ) ( + ) L ( + ) L e + L e z dz e ( + ) L ( e z ) dz ( + ) L ( e ) ( + ) z + e z + e We cn ue hee reul o olve more compliced iniil vlue problem, which could no be olved wihou he ue of rnform: Exmple. Repone of n undmped yem o ingle qure wve. y + y r() y() y (), < r(), > Thu r() u () nd L{r()} e. Seing Y L(y) nd rnforming he equion we obin Y + Y ( e )

8 8 LAPLACE TRANSFORMS nd o To nd y we r inver nd o Y L + ( + ) e in We e f() L ( in z dz + ) co z ( co ) ( co ) nd herefore, uing he econd hifing heorem, L e ( u ()f( ) + ) co ( ) u () Thu For <, u () nd o y co co ( ) u () y co where, for, u () nd o y co ( ) co Remrk.8. Ined of uing inegrion, we could hve ued pril frcion: or ( + ) A + B + C + A( + ) + (B + C) For, nd o A /. Equing coecien of power of give : A + B B / : C Thu nd o before. ( + ) + L ( + ) co

9 . BASIC TRANSFORMS 9 Exmple. RC circui Ri() + i v() C, < where v() V, < < b, > b Th i, v() V u () u b () If we e I() Li() hen he rnformed equion i nd o RI + I C V e e b I F ()(e e b ) where F () V R + /C L V L V /R R + /C + /RC V R L + /RC Thu V R e /RC i() L I() L e F () L e b F () V { } e ( )/RC u () e ( b)/rc u b () R So, for <, u u b nd o i(). For < < b u bu u b ill nd o i() V R e/rc e /RC nd nlly, for > b, u u b nd hu V e /RC e b/rc e /RC R.8. Periodic funcion. A funcion f() h period T if f( + T ) f() for ll. For exmple, in nd co hve period π. We lredy know he rnform of hee funcion; bu o rnform more generl (including diconinuou) periodic funcion we do he following: Lf() T... + nt e f() d T e f() d + (n )T T e f() d T e f() d + e f() d +... T Seing u + T in he r inegrl, u + T in he econd, nd o on, wih u + (n )T in he nh, we obin

10 LAPLACE TRANSFORMS Lf() T... + T T e u f(u) du + e (u+(n )T ) f(u) du e T + e T +... T e u f(u) du T e (u+t ) f(u) du + e (u+t ) f(u) du+ (uing he periodiciy of f). Noing h he erm in qure brcke i geomeric erie, we cn um i o obin nlly Exmple. Periodic qure wve for n,,,..., which h period. Lf() T e T e f() d k, n < < (n + ) f() k, (n + ) < < (n + ) k -k Figure. f() Lf() e ke d k e e + e ( e ) k k e e + e e + e ( k nh ) ke d.9. Diereniion of rnform. If F () e f() d hen Hence, Lf() F () F () Exmple. Find he invere rnform of ( + ) e f() d Lf()

11 Noing h L{in } nd o Exmple. +, we hve. BASIC TRANSFORMS L in L ( + ) ( + ) in Diereniing: F () ln + ln( + ) ln( ) d d L F () e e ln + + inh Le e Linh.. Inegrion of rnform. From hi rule for diereniion of rnform, we cn obin he following rule for inegrion: f() Theorem.9. If F () Lf() nd lim + Proof. Exmple. We know h L{in } F (u) du exi, hen f() L F f() e u f() d du e u du d f() e d L in nd lim. Thu + f() L{ in } u + du co u co A hi ge we cn ummrie ll he rnform which hve been eblihed in he following ble.

12 LAPLACE TRANSFORMS Tble. Tble of Lplce Trnform f() F () k k n n! n+ α Γ(α + ) α+ e in + co + inh coh e b in b ( ) + b e co b ( ) + b e n n! ( ) n+ e f() F ( ) u ()f( ) in co f() n f() f () e F () ( + ) ( + ) F () ( ) n F (n) () F () f() f () F () f() f () f (n) () n F () n f()... f (n ) () F () f(u) du f() F (u) du.. Syem of liner dierenil equion. Thee preen no pecil diculie for Lplce rnform. The rnformed yem of liner lgebric equion i olved nd hen invered. Exmple. A circui i decribed by he yem of wo equion: L i + i R Mi + v() L i + i R Mi

13 If L L, M nd R R 3, hen hi become i + 3i i + v() i + 3i i Auming i () i (), he rnformed equion i hen. BASIC TRANSFORMS 3 or Equion nd o I I + 3I I + V () I + 3I I ( + 3)I I V () () I + ( + 3I ) () ( + 3) I V () V () ( + 3) V () 3( + )( + 3) Conider now he ce where v() E in for ome conn E. Then V () E nd o + I Uing pril frcion: or Then 4A nd o A /4 3 3 B B 3/ E 3 ( + )( + 3)( + ) ( + )( + 3)( + ) A + + B C + D + A( + 3)( + ) + B( + )( + ) + (C + D)( + )( + 3) Equing he coecien of 3 we hve: A + B + C nd o C A B (5 3)/ / nd nlly eing we obin 3A + B + 3D or D /3(3A + B) ( ) /5 3 Thu I E nd invering we obin i E 5e + 3e 3 + co + 4 in 6 Similrly, nd I + 3 I E ( + )( + 3)( + ) E i E 5e + 3e 3 8 co + 4 in 6

14 4 LAPLACE TRANSFORMS. Convoluion The convoluion of wo funcion f() nd g(), denoed f g i dened by (f g)() f(u)g( u) du Convoluion h mny of he properie of produc, bu no ll. For exmple, i i commuive: f g g f ince (f g)() f(u)g( u) du f( v)g(v) ( dv) However, noe h, ince Alo, f f cn be negive, e.g., if f() co nd wih v u du g(v)f( v) dv (g f)() co co Thi i negive, for exmple, 3π/. co u co( u) du co + co(u ) du u co + in(u ) { co + in } in( ) co + in The vlue of convoluion in he conex of Lplce rnform rie from he following: Theorem.. If f() nd g() hve Lplce rnform F () nd G() repecively, hen Proof. F ()G() L(f g)() F ()G() e u f(u) du e v f(v) dv e (u+v) f(u)g(v) dv du Now, le u + v nd o d dv (in he inner inegrl). Then F ()G() u e f(u)g( u) d du e f(u)g( u) du d (wiching he order of inegrion by king noe of he re of inegrion) Thu

15 . CONVOLUTION 5 F ()G()) e L(f g)() e (f g)() d f(u)g( u) du d Exmple. f() g() co F () G() + nd o F ()G() ( + ) Thu L ( + ) co co co + in Noe h we cn rrive hi reul in n lernive fhion by uing Thu L co d d + ( + ) + () ( + ) ( + ) L ( + ) L ( ) + ( + ) ( + ) L ( + ) + + co + in ( before).. Iniil vlue problem. Convoluion cn obviouly be ued n lernive wy of olving dierenil equion. One dvnge cn be een in he following: Exmple. Trnforming nd hu or nd hu y + ω y h() y() A, y () B Y Y A B + ω Y H() ( + ω )Y H() + A + B H() + ω + A + ω + B + ω y ω in ω h() + A co ω + B in ω ω Thu we cn obin oluion o problem in hi generl form uing convoluion wihou knowing h() nd if we do know h() we cn ue hi formul o clcule y wihou hving o rnform r.

16 6 LAPLACE TRANSFORMS.. Inegrl equion. Convoluion cn lo be ued ogeher wih Lplce rnform o olve ome inegrl equion very imply. Exmple. Trnforming y() in + y(u) in ( u) du nd hu y() in. Y ( + 4 Y Y ) Y +