1. Introduction: A Mixing Problem

Transcription

1 CHAPTER 7 Laplace Tranfrm. Intrductin: A Mixing Prblem Example. Initially, kg f alt are dilved in L f water in a tank. The tank ha tw input valve, A and B, and ne exit valve C. At time t =, valve A i pened, delivering 6 L/min f a brine lutin cntining.4 kg f alt per liter. At t = min, valve A i cled and valve B i pened, delivering 6 L/min f brine at a cncentratin f.2 kg/l.the exit valve C, which emptie the tank at 6 L/min, maintain the cntent f the tank at a cntant vlume. Auming the lutin i kept well-tirred, determine the amunt f alt in the tank at all time t >. Thi give u an input rate f g(t) = ( 2.4, < t <..2, t > 28

2 r. INTRODUCTION: A MIXING PROBLEM 29 Then x(t) change at a rate d x(t) x(t) = g(t) dt 5 dx dt + 5 x = g(t) with IC x() =. With the methd we have far, nting that g(t) i cntant n each f the tw interval it i defined fr, we have tw equatin that need t be lved by undetermined ce cient. But there i anther methd, uing the Laplace tranfrm. Thi i an alternate methd f lving linear IVP f all rder. The idea: We tranfrm an IVP in the variable t (the t-dmain) by integratin int an algebraic equatin in the variable (the -dmain), lve that equatin uing algebra, and then tranlate that anwer back t the riginal t-dmain. The fllwing chart cmpare the tw methd.

3 7. LAPLACE TRANSFORMS Fr nw, the lutin can be written a x(t) = 4 7e t/5 2 A graph f the lutin fllw: (, t apple [ e (t )/5 ], t.

4 2. DEFINITION OF THE LAPLACE TRANSFORM 2. Definitin f the Laplace Tranfrm Definitin ( The Laplace Tranfrm). Let f(t) be a functin whe dmain include all f t. The Laplace tranfrm f f i the functin F defined by the imprper integral F () := Z e t f(t) dt. The dmain f F i the et f -value fr which the imprper integral exit. Ntatin. f(t) L 7! F () () r a bit lppy, but ueful, Example. L{} = (2) Z L{e at } = L{f} = F r L{f}() = F () h e t dt = lim N! Z e t e at dt = h e (a lim N! a L{f(t)} = F (). Z )ti N e t i N e (a )t dt = = lim N! = lim N! h e (a a )N h a e N i + i =, >. = a, > a.

5 2 7. LAPLACE TRANSFORMS () h h lim N! e t L{in bt} = in bt b c bt Z b 2 in bt in bt be t c bt 2 e N in bn e t in bt dt = + & & e t e t + e t 2 i Z b b2 2 L{in bt} = be N 2 c bn + b 2 i 2 L{in bt} = b b2, > =) b L{in bt} = 2 + b2, >. e t in bt dt =) = b 2, > =)

6 (4) Let f(t) = 8 >< 2, < t < 5, 5 < t <. >: e 4t, t > Z 5 2. DEFINITION OF THE LAPLACE TRANSFORM L{f(t)} = e t 2 dt + h 2 2 Z 5 h 2 2 e 5 Z 5 Z e t f(t) dt = e t dt + e t dt + lim N! e t i 5 i + lim N! h e (4 + lim N! 2e 5 + Z N h e (4 Z e (4 )t dt = )ti N 4 )N e 4 4 ( 4) e 4, > 4. e t e 4t dt = = (4 ) Therem ( Linearity f the Laplace Tranfrm). Let f, f and f 2 be functin whe Laplace tranfrm exit fr > and let c be a cntant. Then, fr >, L{f + f 2 } = L{f } + L{f 2 }, L{cf} = cl{f}. The prf here fllw eaily frm the ame prpertie fr integral. i =

7 4 7. LAPLACE TRANSFORMS Example. L{ + 5e 4t 6 in 2t} = L{} + 5L{e 4t } 6L{in 2t} = = , > 4. Definitin. A functin f(t) i aid t have a jump dicntinuity at t 2 (a, b) if f(t) i dicntinuu at t, but the ne-ided limit exit a finite number. lim f(t) and lim t!t t!t + f(t) Definitin (2 Piecewie Cntinuu Functin). A functin f(t) i piecewie cntinuu n a finite interval [a, b] if it i cntinuu at each pint f [a, b] except fr at mt a finite number f jump dicntinuitie. A functin f(t) i piecewie cntinuu n the infinite interval [, ] if it i piecewie cntinuu n every finite interval [, T ]. There are functin fr which the integral fr the Laplace tranfrm de nt cnverge. Example are f(t) = t, which grw t fat near, and g(t) = et2, which increae t rapidly a t!. Hwever, rughly peaking, the Laplace tranfrm f a piecewie cntinuu functin exit, prvided that the functin de nt grw fater than an expnential.

8 2. DEFINITION OF THE LAPLACE TRANSFORM 5 Definitin ( Expnential Order). A functin f(t) i aid t be f expnential rder a t! if there are pitive cntant T and M uch that f(t) apple Me t fr all t > T. Ntatin. If f(t) i f expnential rder, we write O(e t ) i read a Big Oh f e t. Example. f(t) = O(e t ) (a t! ). () f(t) = e 5t in 2t. f(t) = e 5t in 2t apple e 5t =) f(t) = O(e 5t ) f(t) i expnential rder 5.

9 6 7. LAPLACE TRANSFORMS (2) f(t) = t n, t. e t = + t + t2 2! + t! + + tn n! + =) e t > tn n! =) tn = t n < n!e t =) f(t) = O(e t ) f(t) i expnential rder. Therem (Lng-Term Behavir f Functin f Expnential Order). If f(t) = O(e t ) a t!, then fr >, Prf. lim e t f(t) =. t! f(t) apple Me t =) Me t apple f(t) apple Me t =) Me ( )t apple e t f(t) apple Me ( )t t! =) apple lim t! e t f(t) apple =) lim t! e t f(t) =. Crllary (Lng-Term Behavir f Laplace Tranfrm). If a functin f(t) i bunded and integrable n every finite interval [, T ] and i f expnential rder, then lim L{f}() =.!

10 . PROPERTIES OF THE LAPLACE TRANSFORM 7 Therem (2 Cnditin fr Exitence f the Tranfrm). If f(t) i piecewie cntinuu n [, ) and i f expnential rder, then L{f}() exit fr >.. Prpertie f the Laplace Tranfrm Therem ( Tranlatin in ). If the Laplace tranfrm L{f}() = F () exit fr > a, then fr > + a. Prf. L{e at f(t)}() = Z Example. We knw Then, by tranlatin in, L{e at f(t)}() = F ( a) e t e at f(t) dt = Z L{in bt}() = F () = L{e at in bt}() = F ( a) = e ( a)t f(t) dt = F ( a). b 2 + b 2. b ( a) 2 + b 2. Therem (4 Laplace Tranfrm f the Derivative). Let f(t) be di erentiable n [, ) and f (t) be piecewie cntinuu n [, ), with bth f expnential rder. Then, fr >, L{f (t)}() = L{f(t)}() f().

11 8 7. LAPLACE TRANSFORMS b Example. L{in bt}() = 2 + b2, >. Find L{c bt}(). in bt apple = e t =) in bt = O(e t ) =) =. L{c bt}() = L{b c bt}() = b h i L{in bt}() in = b b b 2 + b 2 = 2 + b2, >. The fllwing therem are needed fr higher-rder linear DE. Therem (Laplace Tranfrm f the Secnd Derivative). Let f(t) and f (t) be cntinuu n [, ) and let f (t) be piecewie cntinuu n [, ), with all thee functin f expnential rder. Then, fr >, L{f (t)}() = 2 L{f(t)}() f() f (). Therem (5 Laplace Tranfrm f Higher-Order Derivative). Let f(t), f (t),..., f (n ) (t) be cntinuu n [, ) and let f (n) (t) be piecewie cntinuu n [, ), with all thee functin f expnential rder. Then, fr >, L{f (n) (t)}() = n L{f(t)}() (n ) f() (n 2) f () f (n ) ().

12 . PROPERTIES OF THE LAPLACE TRANSFORM 9 Therem (Tranfrm f an Integral). If f(t) i piecewie cntinuu n [, ) and f expnential rder, then ( Z ) t L f( ) d () = L{f(t)}() fr >. Prf. Z t Let g(t) = f( ) d. Nte that g() = and g (t) = f(t). Then ( Z ) t L{f(t)}() = L{g (t)}() = L f( ) d =) ( Z ) t L f( ) d = L{f(t)}() Example. ( Z ) t L e 4 d = L{e4t } = 4 = ( 4).

13 4 7. LAPLACE TRANSFORMS The fllwing therem hw that if F () i the Laplace tranfrm f f(t), then F () i al a Laplace tranfrm f me functin f t. In fact, Mre generally: F () = L{ tf(t)}(). Therem (Derivative f the Laplace Tranfrm). Suppe a functin f(t) i piecewie cntinuu n [, ) and f expnential rder. Let F () = L{f(t)}(). Then L{t n f(t)}() = ( ) ndn F d (). n Example. d 2 d 2 2 L{e 2t }() = 2. d = d ( 2) d 2 d = ( 2) 2. = 2( 2) ; d = 6( 2) 4 = d 2 6 ( 2) 4. Thu L{t e 2t }() = ( ) 6 ( 2) 4 = 6 ( 2) 4.

14 4. INVERSE LAPALCE TRANSFORM 4 Thu we have that multiplying a functin by t crrepnd t the negative f the derivative f the tranfrm. Similarly, the fllwing tate that dividing f by t crrepnd t an integral f it tranfrm. Therem (Integral f a Tranfrm). Suppe a functin f(t) i piecewie cntinuu n [, ) and f expnential rder. Furthermre, uppe exit and i finite. Then n f(t) L t where F () = L{f}(). f(t) lim t! + t () = Z F (u) du 4. Invere Lapalce Tranfrm Definitin (4 Invere Laplace Tranfrm). An invere Laplace tranfrm f a functin F () i the unique functin f(t) that i cntinuu n [, ) and atifie L{f(t)}() = F (). In cae all functin that atify L{f(t)}() = F () are dicntinuu n [, ), we elect a piecewie cntinuu functin that atifie L{f(t)}() = F (). In bth cae, we write L {F } = f. Therem (Therem 7 Linearity f the Invere Tranfrm). Aume that L {F }, L {F }, and L {F 2 } exit and are cntinuu n [, ) and let c be any cntant. Then L {F + F 2 } = L {F } + L {F 2 }, and L {cf } = cl {F }. In ther wrd, L i a linear peratr.

15 42 7. LAPLACE TRANSFORMS If L{f(t)}() = F (), then al L {F ()}(t) = f(t). L! f(t) F () L Example. n L n n n = L + L + 2L = + t + 2e t. n Prblem (Page 74 # 6). Find L. (2 + 5) Slutin. n L (2 + 5) = n 8 L ( + 5/2) = 6 t2 e 5t/2.

16 5. SOLVING INITIAL VALUE PROBLEMS 4 n Prblem (Page 75 # 24). Find {L} ( )( ) Slutin. Uing partial fractin, ( )( ) = 5 = ( 2) = ( 2) Then n L ( )( ) n n = 5L + 2L 2 ( 2) Slving Initial Value Prblem Slving IVP The Methd f Laplace Tranfrm () Take the Laplace tranfrm f bth ide f the DE. 5 ( 2) n 5L ( 2) = 5e t + 2e 2t c t 5e 2t in 2t. (2) Ue the initial cnditin and prpertie f the tranfrm t expre the algebraic equatin btained in Step in term f the tranfrm f the lutin. () Slve the equatin in Step 2 fr the tranfrm f the lutin. (4) Determine the lutin by finding a cntinuu (r piecewie cntinuu) functin with a tranfrm the ame a the ne btained in Step.

17 44 7. LAPLACE TRANSFORMS Therem (Lerch Therem Uniquene f Invere Laplace Tranfrm). Suppe f(t) and g(t) are cntinuu n [, ) and f expnential rder. If L{f(t)}() = L{g(t)}() fr all >, then f(t) = g(t) fr all t. Example. Slve y y = e 5t, y() =. L{y 5y}() = L{e 5t }() =) L{y }() 5L{y}() = L{e 5t }() =) L{y}() y() 5L{y}() = L{e 5t }() =) ( 5)L{y}() = 5 =) L{y}() = ( 5) =) 2 n y(t) = L =) ( 5) 2 y(t) = te 5t.

18 5. SOLVING INITIAL VALUE PROBLEMS 45 Example. Slve y + y = t, y() =. L{y + y}() = L{t}() =) L{y }() + L{y}() = L{t}() =) L{y}() y() + L{y}() = L{t}() =) L{y}() + L{y}() = 2 =) ( + )L{y}() = 2 + = =) L{y}() = ( + ) ( + ) = A + B + C 2 + = A2 + A + B + B + C 2 =) 2 ( + ) 8 >< A + C = A + B = =) A = =) C = 2. >: B = - Then L{y}() = =) n n n y(t) = L + L + 2L = + t + 2e t 2 + i the lutin.

19 46 7. LAPLACE TRANSFORMS Example. Slve y (t) + y(t) = in t, y() =. L{y (t)}() + L{y(t)}() = L{in t}() =) L{y(t)}() y() + L{y(t)}() = 2 + =) ( + )L{y(t)}() = 2 + =) ( + )L{y(t)}() = = =) L{y(t)}() = ( 2 + )( + ) ( 2 + )( + ) = A + B C + = A2 + A + B + B + C 2 + C ( 2 + )( + ) 8 >< A + C = A + B = =) A = >: 2, B = 2, C = 2. B + C = 2 - Then L{y(t)}() = =) =) y(t) = n 2 L + n L + n L + =) i the lutin. y(t) = 2 c t + 2 in t + 2 e t

20 Example. () Slve y + 4y =, x() = y () = 2. i the lutin. 5. SOLVING INITIAL VALUE PROBLEMS 47 L{y }() + 4L{y}() = L{}() =) 2 L{y}() x() x () + 4L{y}() = =) ( 2 + 4)L{y}() 2 2 = =) L{y}() = = =) n n 2 y(t) = 2L + L =) y(t) = 2 c 2t + in 2t

21 48 7. LAPLACE TRANSFORMS Example. Slve y 2y + 5y = 8e t, y() = 2, y () = 2. L{y }() 2L{y }() + 5L{y}() = 8L{e t }() =) 2 L{y}() y() y () 2 L{y}() y() 8 + 5L{y}() = + =) 2 L{y}() 2 2 2L{y}() L{y}() = ( )L{y}() = =) 8 + = 22 + =) L{y}() = ( )( + ) =) ( )( + ) = A + B C + = A 2 + A + B + B + C 2 2C + 5C =) ( )( + ) 8 8 A + C = 2 8 >< ()A + C = 2 >< [(2) ()]A 7C = >< C = (2)A + B 2C = =) =) A = >: ()B + 5C = >: >: B = 5 8C = 8 - L{y}() = = ( ) ( ) n y(t) = L n + 4L 2 ( ) ( ) y(t) = e t c 2t + 4e t in 2t e t i the lutin. L n + + =) =)

22 5. SOLVING INITIAL VALUE PROBLEMS 49 Example. Slve x + x = t 2 + 2, x() =, x () =. L{x }() + L{x}() = L{t 2 }() + 2L{}() =) 2 L{x}() x() x () + L{x}() = =) 2 L{x}() + + L{x}() = =) ( 2 + )L{x}() = = =) ( 2 + ) = A + B 2 + C + D + E 2 + = L{x}() = =) ( 2 + ) - A 4 + A 2 + B + B + C 2 + C + D 4 + E ( 2 + ) 8 A + D = >< B + E = A + C = 2 B = =) E = >: C = 2 =) A = =) D = - L{x} = =) n 2 n n x(t) = L + L L =) x(t) = t 2 + c t in t i the lutin. =)

23 5 7. LAPLACE TRANSFORMS 6. Tranfrm f Dicntinuu and Peridic Functin Definitin (5 Unit Step Functin). The unit tep (Heaviide) functin u(t) i defined by u(t) = (, t <, t >. Nte that thi functin i undefined at. Al, if the functin jump at t = a, (, t < a u(t a) =, t > a, and if the height f the jump i a cntant M, (, t < a Mu(t a) = M, t > a.

24 6. TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 5 Definitin (Rectangular Windw Functin). The rectangular windw functin a,b (t) i defined by a,b (t) = u(t a) u(t b) = 8 ><, t < a, a < t < b. >:, t > b We can al ue thi functin t che a windw fr a functin:

25 52 7. LAPLACE TRANSFORMS Any piecewie cntinuu functincan be expreed in term f windw and tep functin. Example. can be written a f(t) = 8 >< t, < t < t >: 2, < t < ln(t), t > f(t) = t, (t) + t 2, (t) + ln(t)u(t ) = tu(t) + (t 2 t)u(t ) + ( ln t t 2 )u(t ).

26 Example. Fr a, L{u(t a)} = 6. TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 5 Z e t u(t a) dt = Z a e t dt = h e t i N h e N lim N! = lim e a i = e a a N!, >. Al, n e L a (t) = u(t a) and n L n a,b (t) () = L u(t a) u(t b) = e a e b, < a < b. Jut a the hift in decribe the e ect n the Laplace tranfrm f multiplying a functin by e at, the next therem illutrate an analgud e ect f multiplying the Lapace tranfrm f a functin by e a. Therem (8 Tranlatin in t). Let F () = L{f}() exit fr >. If a i a pitive cntant, then L{f(t a)u(t a)}() = e a F (), and, cnverely, an invere Laplace tranfrm f e a F () i given by L ne a F () (t) = f(t a)u(t a). Prf. L{f(t a)u(t a)} = Z Z e t f(t a)u(t a) dt = v = t a, dv = dt Z Z e (v+a) f(v) dv = e a e v f(v) dv = e a F (). a e t f(t a) dt =

27 54 7. LAPLACE TRANSFORMS Crllary. Prf. L{g(t)u(t a)}() = e a L{g(t + a)}(). Identify g(t) with f(t a), f(t) = g(t + a). Example. Find L{(in t)u(t Let g(t) = in t and a =. Then Thu, )}(). g(t + a) = g(t + ) = in(t + ) = in t. L{g(t + a)}() = L{in t()} = 2 + =) L{(in t)u(t )}() = e 2 +. Prblem (Page 94 # 4). Find L n e (t) Slutin. We have F () = n 2 + 9, f(t) = L F () (t) = L n e (t) = f(t )u(t ) = in t. Thu in(t 9) u(t ).

28 6. TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 55 Example (Abrupt Change). We begin with a filled 5 L tank with 2 kg/l f alt. There are tw input valve. Valve A flw at 2 L/min with 4 kg/l f alt. Valve B flw at 2 L/min with 6 kg/l f alt. There i ne utput valve, Valve C, which al flw at 2 L/min. The inflw tart with Valve A fr minute, witche t valve B fr the next, then back t A fr the duratin. Valve C i pen cntinuuly during thi time. Find the amunt f alt in the tank at any time. Let t = time in minute. Let x(t) = kg f alt in tank at time t. ( dx dt = 2 4 fr apple t <, t > 2 6 < t < 2 ) 2 x(t) 5 x + 25 x = u(t ) 2u(t 2)

29 56 7. LAPLACE TRANSFORMS We need t lve x + x = u(t 25 ) 24(t 2), x() =. L{x }() + L{x}() = 48L{}() + 24L{u(t )}() 25 24L{u(t 2)}() =) L{x}() x() + L{x}() = 48L{}() + 24L{u(t )}() 25 + L{x}() = e L{x}() = ( + 25 ) + 24e ( + n x(t) = L + 25 ( + 24L e + 25 (t) + 48L ( + 25 ( )(t) 24L 24L{u(t 2)}() =) 24 e 2 25 ) =) 24e 2 ( + 25 ) =) ) (t) ) e 2 (t). + 25

30 B ( A + B = 25 A = =) A = TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 57 The fllwing i needed in three di erent term. - = A + = A + 25 A + B =) 25 and B = - Thi give u ( ) ( L 25 (t) = L + 25 ( ) 25 L (t) ) 25 + (t) = 25 ( ) 25 L + 25 Recalling ( ) n x(t) = L + (t) + 48L (t) ( ( + 24L e )(t) 24L + 25 the lutin i x(t) = e t e t h x(t) = 2 e 25 t + e 25 u(t (t ) ) (t) = 25 e e 25 t. ) (t), 25 e 25 (t ) u(t ) e 25 u(t (t 2) 2) =) i e 25 u(t (t 2) 2).

31 58 7. LAPLACE TRANSFORMS Maple. See abrupt-heaviide.mw r abrupt-heaviide.pdf. Definitin (Peridic Functin). A functin f(t) i aid t be peridic f perid T (6= ) if f(t + T ) = f(t) fr all t in the dmain f f. Example. f(t) = (, < t <, and f(t) ha perid 2., < t < 2 A ntatin fr the windwed verin f a peridic functin i f T (t) = f(t),t (t) = f(t) u(t) u(t T ) ( f(t), < t < T =, therwie.

32 6. TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 59 The Laplace tranfrm f f T (t) i given by F T () = Z e t f T (t) dt = Z T e t f(t) dt. Therem (9 Tranfrm f Peridic Functin). If f ha perid T and i piecewie cntinuu n [, T ], then the Laplace tranfrm are related by F () = Z e t f(t) dt and F T () = Z T e t f(t) dt F T () = F () e T r F () = F T() e T. Prblem (Page 95 # 8). Slve y + 5y + 6y = tu(t 2), y() =, y () =. Slutin. With Y () = L{y(t)}(), L{y (t)}() + 5L{y (t)}() + 6L{y(t)}() = L{tu(t 2)}() =) 2 Y () y() y () + 5 Y () y() + 6Y () L{(t 2)u(t 2)} + 2L{u(t 2)} =) 2 Y () y () + 5 Y () + 6Y () = e e 2 =) 2 ( )Y () = + e 2 (2 + ) =) 2 Y () = ( + 2)( + ) + e 2 (2 + ) =) (uing partial fractin) 2 ( + 2)( + )

33 6 7. LAPLACE TRANSFORMS Y () = + 2 ( y(t) = L + 2 y(t) = e 2t e t + apple + + e apple + + e 2 apple t e 4( + 2) + 5 9( + ) 2(t 2) 4 4( + 2) + 5 9( + ) (t 2) 5e Impule and the Dirac Delta Functin =) ) u(t 2). T deal with vilent frce f hrt duratin, we intrduce the Dirac Delta Functin. Definitin. The Dirac Delta functin (t) i characterized the fllwing tw prpertie: (, t 6= ( ) (t) =, t =, and ( ) Z f(t) (t) dt = f() fr any functin f(t) that i cntinuu n an pen interval cntaining t =. Nte. () By hifting the argument f (t), we have ( (, t 6= a ) (t a) =, t = a, and ( ) Z f(t) (t a) dt = f(a) fr any functin f(t) that i cntinuu n an pen interval cntaining t = a. =)

34 8. IMPULSES AND THE DIRAC DELTA FUNCTION 6 (2) (t a) i nt a functin in the uual ene, but i called a generalized functin r ditributin. () Z Z t (4) (t a) =. Thi i a pecial cae f ( ) by taking f(t) =. (t a) dt = (, t < a, t > a = u(t a). (5) Di erentiating the abve with repect t t, and uing the Fundamental Therem f Calculu, (t a) = u (t a). Example. Fr a >, L{ (t a)} = Z e t (t a) dt = e t (t a) + & e t u(t a) h i Z e t u(t a) + e t u(t a) dt = h i lim e T u(t a) + e a = e a, >. T! Maple. See abrupt-heaviide.mw r abrupt-heaviide.pdf.

35 62 7. LAPLACE TRANSFORMS Example. A rancher ha 6 lnghrn cattle, with a per capita grwth rate f 8% in the herd. The rancher decide t get rid f thi herd in tw year time by elling the ame number every 6 mnth tarting 6 mnth frm nw. Hw many huld be ld at a time? Let p(t) = # f cattle at time t (in year). The input r grwth rate i.8p. Let N = # ld each 6 mnth. Then the ttal number ld at time t i n(t) = Nu t + Nu(t ) + Nu t 2 Then the utput r elling rate i n (t) = N t + N (t ) + N t 2 + Nu(t 2). 2 + N (t 2). 2 The DE that then mdel thi i dp dt =.8p N t 2 Then ur IVP i N (t ) N t 2 N (t 2).

36 8. IMPULSES AND THE DIRAC DELTA FUNCTION 6 p.8p = N t 2 N (t )+N t 2 N (t 2). p() = 6 =) L{p }() L{.8p}() = h n n n N L t ()+L t ()+L t 2 n i ()+L t 2 () 2 =) L{p}() p().8l{p}() = h n n n n i N L t ()+L t ()+L t ()+L t 2 () 2 2 i =) (.8)L{p}() 6 = N he 2 + e + e 2 + e 2 =) h e 2 N L{p}() = e n p(t) = 6L (t) N.8 n +L e n (t)+l e p(t) = 6e.8t h N e.8 t 2 u t.8 + e e 2 i =).8 (t).8 n (t)+l e 2 i (t).8 hl n e 2 2 e.8 t 2 u T find N, we ue the fact that p(2 + ) =. + e.8 t u 6e.6 N = e.2 + e.8 + e Maple. See abrupt-dirac.mw r abrupt-dirac.pdf. t 2 =) t + i + e.8 t 2 u t 2

37 64 7. LAPLACE TRANSFORMS Example. An undergrund trage tank cntain gal f galine. At time t =, a pump i turned n and ga i pumped int the tank at 5 gal/min. The pump turn autmatically when the 6 gal capacity f the tank i reached. A leak in the tank develp at t = 4 minute with a mdel fr the galln leaked being g(t) = t 2 8t + 7, t 4. (a) Expre the galln leaked a a tep functin. (, apple t < 4 g(t) = t 2 8t + 7, t > 4 = (t 2 8t + 7)u(t 4). (b) Expre the number f galln in the tank in term f t. Let x(t) = the number f galln in the tank at time t. x(t) = + 5t (t 2 8t + 7)u(t 4), apple t < 4, t > 4. (c) Will the pump hut autmatically? Fr t > 4, x(t) = t t 2. x (t) = 58 2t = () t = 29. x (t) = 2 =) x(29) = 824 i the maximum number f gal. Thu the pump never turn. (d) When de the tank becme empty? t t 2 = () t = 29 ± 4 p 4 () t 7.7 r t.7. The tank will empty in 7.7 minute.

38 8. IMPULSES AND THE DIRAC DELTA FUNCTION 65 (e) Mdel the rate f change f galln in the tank. We have g(t) = (t 2 8t + 7)u(t 4), apple t <, t 6= 4 =) g (t) = (2t 8)u(t 4) + (t 2 8t + 7) (t 4) dx h i dt = 5 (2t 8)u(t 4) + (t 2 8t + 7) (t 4), x() = x = 5 (2t 8)u(t 4) (t 2 8t + 7) (t 4), x() = (f) Fr what time i the equatin in part (e) valid? The equatin i valid until the tank verflw r becme empty. (g) Find a frmula fr the number f galln f galine in the tank. lim N! L{(t 2 8t + 7) (t 4)}() = Z e t (t 2 8t + 7) (t 4) dt = " Z # N e t (t 2 8t + 7) (t 4) dt = u = e t (t 2 8t + 7) dv = (t 4) dt du = ( e t (t 2 8t + 7) + e t (2t 8)) dt v = u(t 4) lim N! + " N e t (t 2 8t + 7)u(t 4) Z N e t (t 2 8t + 7)u(t 4) dt Z T e t (2t 8)u(t 4) dt # =

39 66 7. LAPLACE TRANSFORMS lim N! " + lim N! Z N " e t (t 2 8t+7)u(t 4) dt Z N 4 e t (t 2 8t + 7) dt Z N Z N 4 e 4 + e t (2t 8)u(t 4) dt e t (2t 2e 4 Maple. See Dirac integral.mw r Dirac integral.pdf. Then 2 8) dt # = 2e 4 # 2 = e 4. x = 5 (2t 8)u(t 4) (t 2 8t + 7) (t 4), x() = =) L{x }() = L{5}() L{(2t 8)u(t 4)}() L{(t 2 8t+7) (t 4)}() =) L{x}() x() = 5L{}() 2L{(t 4)u(t 4)}() e 4 =) L{x}() = 5 e 4 =) L{x}() = + 5 L{x}() = e 4 L{t}() {z } hift in t 2e 4 2 e 4 =) 2e 4 e 4 =) n n x = L (t) + 5L (t) L ne 4 2 n e (t) L 4 2 x = 5t (t 4) 2 u(t 4) u(t 4) =) x(t) = + 5t (t 2 8t + 7)u(t 4) = (t) =) i the number f galln in the tank after t minute until the tank i empty.