The Laplace Transform

Transcription

1 Chapter 7 The Laplace Tranform 85 In thi chapter we will explore a method for olving linear differential equation with contant coefficient that i widely ued in electrical engineering. It involve the tranformation of an initial-value problem into an algebraic equation, which i eaily olved, and then the invere tranformation back to the olution of the original problem, thereby bypaing the need to olve for arbitrary contant in the general olution. The technique i epecially well uited for finding generalized olution to ytem driven by impule or by dicontinuou or periodic forcing function. 7. Definition and Baic Propertie Given a function f(t) defined for t, it Laplace tranform F () i defined a F () = e t f(t)dt. () Notice that the variable appear a a parameter in an improper integral. We ay that the Laplace tranform exit if thi improper integral converge for all ufficiently large. The notational convention ued here i common, though not univeral: The uppercae verion of the function name denote it Laplace tranform, and i ued for the tranform independent variable. Before we go further, let illutrate thi definition and notation with a couple of imple example. Example Conider a contant function f(t) = c, t. It Laplace tranform i F () = e t c dt = c e t = c ( lim t= t e t e ) = c ( ) = c, provided that >. Note that the improper integral diverge when. Example 2 Let f(t) = e at, t. It Laplace tranform i F () = = a e t e at dt = ( e ( a)t dt lim t e ( a)t e ) = ( ) = a a, provided that > a. Note that the improper integral diverge when a.

2 86 Chapter 7. The Laplace Tranform The Laplace tranform F () of a function f(t) i the reult of applying a linear operator to f. Denoting thi linear operator by L, we can write Lf = F, or L[f]() = F (). where F i given by (). Uing thi notation, the reult of Example 2, for intance, i that L[e at ]() = a, > a. It i eay to check that the operator L i linear: L[cf]() = L[f + g]() = = e t cf(t)dt = c e t (f(t) + g(t))dt e t f(t)dt + e t f(t)dt = c F () e t g(t)dt = F () + G(). Indeed, the linearity of L i a imple conequence of the linearity propertie of integration. Our tudy of Laplace tranform i primarily for the purpoe of olving initialvalue problem y + py + qy = f, y() = y, y () = v, (2) where p and q are contant. Later (tarting in Section 3) we will conider generalized olution that can arie when the nonhomogeneou term f i piecewie continuou on [, ). A function f i piecewie continuou on [, ) if (i) f ha at mot finitely many dicontinuitie in any bounded interval [, T ], (ii) lim f(t) exit, and t + (iii) at every number a > both lim f(t) and lim f(t) exit. t a t a + Note that condition (iii) i automatically true at any point where f i continuou, and it ay that all of f dicontinuitie in (, ) are imple jump dicontinuitie. Alo, piecewie continuity guarantee integrability on any bounded interval [, T ]. We alo need a condition that will guarantee the exitence of a function Laplace tranform. We ay that a function f i exponentially bounded on [, ) if there are contant M and r uch that f(t) Me rt for all t.

3 7.. Definition and Baic Propertie 87 If f i piecewie continuou, then T e t f(t) dt will exit for all T > and all. If f i alo exponentially bounded, then T T e t f(t) dt M e (r )t dt = M ( ) e (r )T ; r thu T e t f(t) dt converge a T, provided that > r. Thi mean that the Laplace tranform of f exit, which implie that the Laplace tranform of f exit. Throughout the ret of thi chapter we will be concerned almot excluively with function defined on [, ). The phrae piecewie continuou and exponentially bounded hould alway be undertood to mean piecewie continuou on [, ) and exponentially bounded on [, ). We alo emphaize that throughout thi chapter we will deal only with exponentially bounded function, ince only thoe have Laplace tranform. Alo, in thi ection and the next, all of our example will involve ordinary elementary function polynomial, inuoidal function, exponential function, etc. that are clearly continuou everywhere. Not until Section 3 will we begin to conider differential equation with nonhomogeneou term that are piecewiecontinuou. The Firt Differentiation Theorem The reaon why Laplace tranform are ueful in olving differential equation i embodied in the following theorem, which (together with the corollary that follow) we will refer to a the firt differentiation theorem. THEOREM Suppoe that y i continuou and exponentially bounded and that y i piecewie continuou. Then L[y ] exit, and L[y ]() = Y () y(). Proof. Firt, for T > we ue integration by part to compute T e t y (t)dt = e t y(t) T T ( )e t y(t)dt t= = e T y(t ) y() + T e t y(t)dt. Thi computation i valid becaue of the piecewie continuity of y and the continuity of y (at and T ). Now, ince y i aumed to be exponentially bounded, it follow that, for ufficiently large, lim T e T y(t ) =, and Y () = lim T T e t y(t)dt exit.

4 88 Chapter 7. The Laplace Tranform Therefore, L[y ] exit, and L[y ]() = y() + Y (). By repeated application of Theorem, we arrive at the following corollary. COROLLARY Suppoe that y and y are continuou and exponentially bounded and that y i piecewie continuou. Then L[y ] exit, and L[y ]() = 2 Y () y() y (). Moreover, if y, y,..., y (n ) are continuou and exponentially bounded, and if y (n) i piecewie continuou, then L[y (n) ] exit, and L[y (n) ]() = n Y () n y() n 2 y () y (n ) (). Becaue of the propertie tated in Theorem and Corollary, the Laplace tranform i particularly well uited for olving linear initial-value problem with contant coefficient. The following are two example that indicate the baic idea. Example 3 Conider the firt-order initial-value problem y + 2y = 4, y() =, and let Y denote the tranform of the olution y. By applying the operator L to both ide of the differential equation and uing the reult of Example on the right ide, we find that Y () y() + 2Y () = 4. Now we ue the given initial value and olve for Y (): ( + 2)Y () = 4 + ; o Y () = + 4 ( + 2). Now our job i to find the function y(t) that ha thi tranform. A partial fraction expanion (conult your calculu book) reveal that Y () = The reult of Example and 2 now tell u that Y () i the tranform of y(t) = 2 e 2t. Example 4 Conider the initial-value problem y + 3y + 2y =, y() =, y () =,

5 7.. Definition and Baic Propertie 89 and let Y denote the tranform of the olution y. By applying the operator L to both ide of the differential equation, we find that 2 Y () y() y () + 3( Y () y()) + 2Y () =. Uing the given initial value and rearranging, thi become Now we olve for Y (), finding ( )Y () 3 =. Y () = = + 3 ( + )( + 2). Now we look for the function y(t) that ha thi tranform. A partial fraction expanion reveal that Y () = From the reult of Example 2 above, we ee that thi i the tranform of The General Procedure y(t) = 2e t e 2t. Example 3 and 4 each illutrate a general procedure for olving initial value problem with the help of Laplace tranform: () Tranform each ide of the differential equation, uing the given initial value. (2) Solve the reulting algebraic equation for the tranform Y () of the olution y(t). (3) Find the olution y by identifying the tranform Y () with known tranform. The final tep amount to finding the invere tranform of Y (). An important quetion to ak i whether the operator L i actually invertible. In other word, given a tranform Y (), i there a unique function y(t), t, uch that Ly = Y? The anwer to thi i no, unle we require the function y to be continuou. However, thi i not a difficulty in the context of olving differential equation, ince olution will be continuou. Let now conider what happen in general when we apply the Laplace tranform technique to a linear econd-order initial-value problem with contant coefficient. Suppoe that the differential equation i and we have initial condition P (D)y = f, with P (D) = D 2 + p D + q I, y() = y, y () = v.

6 9 Chapter 7. The Laplace Tranform The Laplace tranform of the nonhomogeneou term i imply Lf = F. Tranforming the left ide produce LP (D)y = LD 2 y + p LD y + q Ly = 2 Y () y v + p( Y () y ) + q Y () = P () Y () ( y + v + p y ). The tranformed initial-value problem therefore become and olving for Y () give u P () Y () ( y + v + p y ) = F (), Y () = F () P () + γ() P (), (3) where γ() = y + v + p y. The term F ()/P () in (3) i the tranform of the ret olution of P (D)y = f, and the term γ()/p () in (3) i the tranform of the olution of the homogeneou equation P (D)y = that atifie the given initial condition. The ame tatement are true for all linear differential equation with contant coefficient, regardle of order. Partial Fraction Expanion Step and 2 of the general procedure jut decribed are eay, provided that the tranform of the nonhomogeneou term i known. The challenging part of olving any problem in thi way lie in tep 3. Extenive table of known tranform are available to help in thi tak. A in Example 3 and 4, partial fraction expanion i often a ueful tool for expreing Y () i term of known tranform. Computer algebra ytem uch a Mathematica and Maple have the ability to compute partial fraction expanion a well a all of the tandard Laplace tranform and invere Laplace tranform. (You can even compute partial fraction expanion with the expand() function on the TI-89 and Voyage calculator.) Neverthele, it i beneficial to have ome kill in efficiently computing partial fraction expanion by hand. A rational function N() P () with degree(n) < degree(p ) may be expanded into a um of rational function with denominator correponding to the linear and irreducible quadratic factor of P () and with numerator that are either contant or linear. A imple linear factor a of P () reult in a ingle imple term of the form A a, and repeated linear factor ( a) k can give rie to term A a + A 2 ( a) A k ( a) k.

7 7.. Definition and Baic Propertie 9 A imple irreducible quadratic factor 2 + b + c of P () reult in a term of the form B+C 2 +b+c, and repeated quadratic factor (2 + b + c) k can give rie to term B + C 2 + b + c + B 2 + C 2 ( 2 + b + c) B k + C k ( 2 + b + c) k. The number of undetermined contant needed in any partial fraction expanion i equal to the degree of the denominator P (). Example 5 For any numerator N() with degree le than, there are contant A, A 2, A 3, B, C, D, C 2, D 2, E, and F uch that i equal to N() 3 ( + )( 2 + ) 2 ( ) A + A A B + + C + D 2 + C 2 + D 2 + ( 2 + ) 2 + E + F The problem of computing the contant in a partial fraction expanion ultimately conit of olving a ytem of linear equation. There are variou way of aembling a uitable ytem of equation, one of which i to clear fraction, expand product, and equate coefficient. Thi approach i intereting from a theoretical point of view but uually lead to a ytem of equation that require coniderable effort to olve. On the other hand, one can uually aemble a much impler et of equation with the help of a few imple algebraic manipulation and evaluation at wiely choen value of. The following example illutrate thee technique. Example 6 Let find the partial fraction expanion of 2 ( + )( + 2)( + 3), which i typical of cae involving denominator with ditinct linear factor. We eek contant A, A 2, and A 3 uch that Multiplying both ide by + give 2 ( + )( + 2)( + 3) = A + + A A ( + 2)( + 3) = A + + A A , into which we ubtitute =, obtaining 2 = A. Multiplying both ide intead by + 2 give 2 ( + )( + 3) = A A A 3 + 3,

8 92 Chapter 7. The Laplace Tranform into which we ubtitute = 2, obtaining Multiplying both ide by + 3 give 4 = A 2. 2 ( + )( + 2) = A A A 3, into which we ubtitute = 3, obtaining Thu our partial fraction expanion i 9 2 = A 3. 2 ( + )( + 2)( + 3) = / / Now, looking back over the preceding calculation, it i evident that an elegant hortcut amount to evaluating the left-hand ide at each root of the denominator after removing the correponding factor from the denominator; that i, 2 A = = /////////( + ) + 2)( + 3) = 2, A 2 = A 3 = 2 = 4, ( + )( /////////( + 2) + 3) = 2 2 = 9 ( + )( + 2)( ///////// + 3) = 3 2. Example 7 Let find the partial fraction expanion of ( + 4) 3, which i typical of cae involving denominator with repeated linear factor. We eek contant A, A 2, and A 3 uch that 2 ( + 4) 3 = A A 2 ( + 4) 2 + A 3 ( + 4) 3. We can find A 3 a in the preceding example. Multiplying both ide by ( + 4) 3 give into which we ubtitute = 4, obtaining 2 = A ( + 4) 2 + A 2 ( + 4) + A 3, 6 = A 3. To find A, we ll multiply both ide by + 4, obtaining 2 ( + 4) 2 = A + A A 3 ( + 4) 2.

9 7.. Definition and Baic Propertie 93 Now letting give So far we have = A. 2 ( + 4) 3 = A 2 ( + 4) ( + 4) 3. At thi point we can imply evaluate both ide at = 3 (o that + 4 = ) and olve for A 2 : Thu our expanion i 9 = + A 2 + 6; o A 2 = 8. 2 ( + 4) 3 = ( + 4) ( + 4) 3. Example 8 Let find the partial fraction expanion of 2 ( + 2)( ), in which the factor i aid to irreducible, ince it ha no real zero. We eek contant A, B, and C uch that Multiplying both ide by + 2 give 2 ( + 2)( ) = A B + C into which we ubtitute = 2, obtaining 4 3 = A. Now, letting in the ame equation give (B + C)( + 2) = A + 2, + + = A + B; o B = A = 3. With only C left to find, we can imply evaluate at =, which give Thu our expanion i = A 2 + C; o C = A 2 = ( + 2)( ) = 3 Example 9 Let find the partial fraction expanion of 5 3 ( 2 + 4)( ), ( )

10 94 Chapter 7. The Laplace Tranform whoe denominator conit of two nonrepeated, irreducible quadratic. We eek contant A, B, C, and D uch that 5 3 ( 2 + 4)( ) = A + B C + D Let firt evaluate each ide at =, which give u a imple equation involving only B and D: = B/4 + D/2, i.e., B = 2D. Next, let multiply both ide by and let. That give u 5 = A + C. Now we ll multiply both ide by 2 + 4, obtaining = A + B + (C + D)(2 + 4) 2, into which we ll ubtitute = 2i: 4i = 2Ai + B i + 2 Simplifying thi give from which we conclude that B + 2Ai = 4 ( 2 + i), A = 2 and B = 8. Now returning to the relationhip 5 = A + C and B = 2D, we obtain C = 3 and D = 4. Finally, the deired expanion i 5 3 2( 4) ( 2 + 4)( 2 = ) Example Let find the partial fraction expanion of 2 ( ) ( ) 2, whoe denominator conit of a repeated irreducible quadratic. We eek contant A, B, C, and D uch that 2 ( ) ( ) 2 = A + B C + D ( ) 2. A in the previou example, we begin by evaluating both ide at = and then by multiplying both ide by and letting, which give = B/2 + D/4 and = A +.

11 7.. Definition and Baic Propertie 95 Next, let clear fraction: 2 ( ) = (A + B)( ) + C + D. The root of are ± i, o we ll evaluate both ide at = + i: Simplifying thi give from which we conclude that ( + i) 2 (2 i) = C ( + i) + D. D C + C i = 2 4i, C = 4 and D = 6. Now returning to the relationhip B/2 + D/4 =, we find that B = 3, and ince we have already that A =, our expanion i 2 ( ) ( ) 2 = ( + 3) ( ) 2. Problem In Problem through 4, ue the reult of Example and 2 to write down the Laplace tranform of the given function. Expre the reult a a ingle quotient.. f(t) = 3 5e t 2. f(t) = 2e t + 3e 2t 3. f(t) = coh bt 4. f(t) = inh bt In Problem 5 through, find the Laplace tranform Y () of the olution of the given initial-value problem. 5. y + 3y = e t, y() = 6. y + y =, y() =, y () = 7. y + y + y =, y() =, y () = 8. y + 3y + 2y =, y() =, y () = 9. y 4y = e t, y() =, y () =. y y =, y() = y () = y () = In Problem through 4, uppoe that f ha the Laplace tranform F (), and write down by inpection the tranform Y () of the differential equation ret olution.. y + y + y = f 2. y + ω 2 y = f 3. y + y = f 4. y y + 2y = f 5. Show that y = te at atifie y 2ay + a 2 y =, y() =, y () =, and ue that fact to find the Laplace tranform of te at. 6. (a) Ue the fact that y = co ωt i the olution of y + ω 2 y =, y() =, y () =, to find the Laplace tranform of co ωt. (b) Similarly find the Laplace tranform of in ω t.

12 96 Chapter 7. The Laplace Tranform 7. (a) Ue the fact that y = t in ωt atifie y + ω 2 y = 2ω co ωt, y() =, y () =, to find the Laplace tranform of t in ωt. Make ue of the known tranform of co ωt from Problem 6. (b) Similarly find the Laplace tranform of t co ωt. Auming that degree(n) < degree(p ), write down the form of the partial fraction expanion of each of the rational function in Problem 8 through N() ( ) 2 9. N() 2 ( ) 2. N() ( 2 + )( 2 + 4) 2 2. N() 4 Ue the method decribed at the end of Example 6 to find the partial fraction expanion of the rational function in Problem 22 through ( )( 2) ( + )( 3) ( )( + ) Ue the method of Example 7 to help find the partial fraction expanion of the rational function in each of Problem 25 through ( ) ( + 2) ( + 3) 4 Ue the method of Example 8 to find the partial fraction expanion of the rational function in Problem 28 through ( + )( 2 + ) ( + 2)( 2 + ) ( + )( ) Ue the method of Example 9 find the partial fraction expanion of the rational function in Problem 3 and ( 2 + )( ) ( )( ) Ue the method of Example find the partial fraction expanion of the rational function in Problem 33 through ( ) ( ) Find general formula for the partial fraction expanion of (a) 2 ( 2 + a 2 ) 2 (b) 3 ( 2 + a 2 ) 2 Solve Problem 37 through 43 a in Example 3 and 4. 3 ( ) y + y =, y() = 38. y y = e t, y() = 39. y y =, y() =, y () = 4. y 4y = 8, y() = y () = 4. y + 4y + 3y = 6, y() = y () = 42. y + 3y + 2y = e 3t, y() = y () =

13 7.2. More Tranform and Further Propertie y + 2y y 2y = 6, y() = y () = y () = 44. Suppoe that L[f(t)]() = F (). Prove the following caling theorem. (a) L[cf(ct)] = F (/c) b) L[f(t/c)] = c F (c) 45. Ue a comparion tet to how that if the improper integral in () converge for ome =, then it converge for all >. 46. (a) Argue that f(t) = e t2 doe not have a Laplace tranform (i.e., that the defining improper integral i divergent for all ). (b) Argue that f(t) = t t doe not have a Laplace tranform. (Hint: t t = e t ln t.) 47. Let a < b. Aume that the fundamental theorem of calculu, b a f (t) dt = f(b) f(a), i true when f i continuou on [a, b]. Prove that the ame i true if f i continuou and f i piecewie continuou on [a, b]. 7.2 More Tranform and Further Propertie In thi ection we will find Laplace tranform for more of the elementary function that commonly arie a olution of linear differential equation with contant coefficient. We will alo derive additional propertie of Laplace tranform that will help u in finding both tranform and invere tranform. Firt, let recall the reult of Example and 2 in Section : L[c ]() = c, > ; L[eat ]() =, > a. (, 2) a Note that the firt of thee i actually a conequence of the econd (with a = ) and linearity. Alo, the computation that produced the econd of thee remain valid if a i a complex contant, provided that > Re(a). That i, L[e (α+iβ)t ]() = α + iβ = So by the Euler-DeMoivre formula we have L[e αt (co βt + i in βt)]() = α iβ ( α) 2 + β 2, > α. α iβ ( α) 2 + β 2, > α. By equating real and imaginary part we arrive at L[e αt α co βt]() = ( α) 2, > α, (3) + β2 L[e αt in βt]() = β ( α) 2, > α, (4) + β2

14 98 Chapter 7. The Laplace Tranform and, in particular, L[co βt] = The Firt Shift Theorem 2 + β 2, >, L[in βt] = β 2, >. (5, 6) + β2 Notice that the tranform of e αt co βt and e αt in βt in (3) and (4) can be viewed a hifted tranform of co βt and in βt obtained by replacing with α. It turn out that the ame thing happen whenever any function i multiplied by e αt. To ee thi, uppoe that f(t) ha a known tranform F () and conider the tranform of e αt f(t): L[e αt f(t)]() = e t e αt f(t)dt = Thu we have what we will call the firt hift theorem: L[e αt f(t)]() = F ( α). Example Let find the invere tranform of F () = e ( α)t f(t)dt = F ( α). Since the denominator i an irreducible quadratic, the key i to complete the quare and then ue the firt hift theorem. Completing the quare how that the denominator i (+2) Now in order take advantage of the hift theorem, we mut expre the numerator in term of + 2 a well: F () = ( + 2) = + 2 ( + 2) ( + 2) The firt term on the right i the tranform of e 2t co 3t by (3), and the econd i the tranform of 2 3 e 2t in 3t by (4). Therefore, f(t) = e 2t( co 3t 2 3 in 3t ). Example 2 Suppoe that we wih to find the ret olution of the equation y + 2y + y =. Tranforming each ide of the equation (and uing y() = y () = ) give u ( ) Y () =. So the tranform of the olution i Y () = ( ) =

15 7.2. More Tranform and Further Propertie 99 The denominator in the econd term i ( + ) To take advantage of thi, we need the numerator in term of +. So we write Y () = + ( + ) ( + ) and oberve (by (), (3), and (4)) that Y () i the Laplace tranform of y = e t co 3t 3 e t in 3t. The Second Differentiation Theorem The firt hift theorem decribe the tranform of the product e αt f(t) in term of the tranform of f. A imilar reult decribing the tranform of the product t n f(t) would alo be ueful. To that end, let u conider a function f(t) with known tranform F (): e t f(t)dt = F (), >. Differentiating with repect to, we obtain e t ( tf(t))dt = d d F (), >. Since the left ide here i the tranform of tf(t), we have found we hall call the econd differentiation theorem: Repeated application of thi reult eaily produce L[ tf(t)]() = d d F (). L[ t n f(t)]() = ( ) n dn F (), n =, 2, 3,.... dn The econd differentiation theorem i ueful for the derivation of a number of tranform. For intance, with the contant function f(t) =, t, we find that L[t ]() = d ( d ) ; that i, L[t ]() = 2, >. Repeating the ame procedure produce the tranform of t 2, t 3, and o on: L[t n ]() = n!, >. (7) n+ * Here we have made the quetionable move of differentiating under the integral ign; however, it can be jutified in thi ituation by a theorem from advanced calculu.

16 2 Chapter 7. The Laplace Tranform Combining (7) with the firt hift theorem yield L[t n e at ]() = The econd differentiation theorem alo produce n!, > a. (8) ( a) n+ L[t co βt] = 2 β 2 ( 2 + β 2, >, (9) ) 2 L[t in βt] = 2β ( 2 + β 2, >. () ) 2 To put (9) into a form more ueful for finding invere tranform, we note that 2 + β 2 2 β 2 ( 2 + β 2 ) 2 = 2β 2 ( 2 + β 2 ) 2. Now we combine (6) with (9) to obtain [ ] in βt L t co βt = β 2β 2 ( 2 + β 2, >. () ) 2 The following two example illutrate the ue of (8) and () in the context of olving initial-value problem. Example 3 Conider the initial-value problem y + 4y + 4y =, y() =, y () =. Tranforming each ide of the equation give u 2 Y () + 4(Y () ) + 4 Y () =, which implifie to ( ) Y () = + 5. So the tranform of the olution i + 5 Y () = = ( + 2) 2 = ( + 2) 2. Therefore, becaue of (2) and (8), we ee that Y i the tranform of y = e 2t + 3te 2t = ( + 3t)e 2t, which i eaily verified a the olution of the initial-value problem. Example 4 Conider the initial-value problem y + 4y = in 2t, y() =, y () =.

17 7.2. More Tranform and Further Propertie 2 Tranforming each ide of the equation give u 2 Y () + 4 Y () = So the tranform of the olution i 2 Y () = = 2 ( 2 + 4) By (5) and (), we ee that y = ( ) in 2t t co 2t + co 2t. 4 2 Problem In Problem through 3, find the invere tranform (a) by partial fraction expanion and (b) by completing the quare and uing the firt hift theorem with the baic tranform of coh and inh: L[inh bt] = b 2 b and L[coh bt] = 2 2 b Find the invere tranform of each expreion in Problem 4 through ( + 2) ( 2 + 4) ( + ) 3. 6 ( 2 + 4) ( + 3) ( 2 + 4) ( ) 3 3 ( 2 + 4) 2 Ue formula () and () and the firt hift theorem to find the invere tranform of each expreion in Problem 6 through ( ) 2 7. ( ) ( ) 2 In Problem 9 through 2, ue the Laplace tranform to olve the initial-value problem. 9. y + 4y + 3y =, y() =, y () = 4 2. y + 6y + 3y =, y() =, y () = 2 2. y + 4y + 4y =, y() =, y () = In Problem 22 through 27, ue the Laplace tranform to find the ret olution of the differential equation. 22. y + 4y + 3y = 4e t 23. y + 6y + 3y = 69t 24. y + 4y + 4y = e 2t 25. y + 4y = 6 in t 26. y + 9y = 9 in 3t 27. y + y = co t

18 22 Chapter 7. The Laplace Tranform 28. Thi purpoe of thi problem i to derive the Laplace tranform of t. (a) Ue the definition of the Laplace tranform and integration by part to how that L[ t ]() = 2 (b) Ue the ubtitution t = x 2 to arrive at (c) After oberving that k 2 = L[ t ]() = k 3/2 where k = e x2 dx e y2 dy = e t t dt. e x2 dx. e (x2 +y 2) dx dy, ue polar coordinate to how that k 2 = π/4. Finally conclude that L[ π t ]() = 2. 3/2 29. Let f be continuou on [, ) with Laplace tranform F (). Apply the firt differentiation theorem to the antiderivative t f(τ)dτ to obtain the firt integration theorem: [ t ] L f(u)du () = F (). 3. In thi problem we will derive the econd integration theorem: [ ] L t f(t) () = F (σ)dσ. (a) Suppoe that f ha the tranform F () and that t f(t) ha the tranform Φ(). Apply the econd differentiation theorem to how that Φ () = F (). Hence Φ i ome antiderivative of F. (b) By the reult of Problem 36, we know that Φ() a. Show that F (σ)dσ ha that property, a well a being an antiderivative of F (), and therefore mut be the deired tranform Φ(). In Problem 3 through 34, ue the econd integration theorem (ee Problem 3) to find the Laplace tranform of the given function. e at co βt in βt inh bt t t t t 35. Ue the econd integration theorem (ee Problem 3) and the reult of Problem 28 to find the Laplace tranform of f(t) = / t. 36. Aymptotic Behavior of the Laplace Tranform. Suppoe that f(t) i bounded on every bounded interval [a, b], a. Suppoe further that the Laplace tranform L[f]() = F () exit for all >, where. By the following equence of tep, prove that lim F () =.

19 7.2. More Tranform and Further Propertie 23 (a) Argue that e t f(t) a t for any >. (b) Argue that becaue of (a) there i a time T > uch that e t f(t) < for all t T, and conequently f(t) < e t for all t T. (c) Ue the reult of (b) to obtain F () < T e t f(t) dt + T e t e t dt. (d) Ue the reult of (c) and the fact that f i bounded on [, T ] to obtain F () < M + for all >, where M i a bound on f(t) for all t T. (e) Finally, conclude that F () a. Alo oberve the tronger fact that F () i bounded for large. 37. Combine the reult of Problem 36 with the firt differentiation theorem to how that if f and f have Laplace tranform and f i continuou, then lim F () = f(). 38. The error function erf and the complementary error function erfc are erf(t) = 2 t e x2 dx and erfc(t) = erf(t) = 2 e x2 dx π π (a) Show that y = e t2 /4 atifie the initial-value problem y + t y =, y() =. 2 Then how that the tranform Y of y atifie Y () 2 Y () =. (b) Ue the appropriate integrating factor (ee Section 2.) to olve for Y () ubject to the condition Y () a (cf. Problem 36). Conclude that L[e t2 /4 ]() = π e 2 erfc(). (c) Ue the firt integration theorem (ee Problem 29) and the reult of part (b) to how that L [ erf(t/2)] () = e2 erfc(). (d) Finally, ue the caling theorem L[f(t/k)] () = k F (k) (ee Problem 43 in Section 7.) to how that, for k >, L[e t2 /(4k 2) ]() = k [ ( π e k2 2 t )] erfc(k) and L erf () = k 2 2k ek2 erfc(k). t

20 24 Chapter 7. The Laplace Tranform In Problem 39 and 4, ue the firt integration theorem (ee Problem 29) to help in finding the olution of the given integro-differential initial-value problem. 39. y + t y(τ) dτ = t, y() = 4. y + t y(τ) dτ =, y() = 3, y () = 7.3 Heaviide Function and Piecewie-Defined Input The Laplace tranform i particularly ueful for olving initial-value problem in which the differential equation ha a piecewie-defined nonhomogeneou term. In order to expre uch piecewie-defined function in a convenient manner, we will ue the Heaviide unit tep function, which i defined by {, if t < ; h(t) = (), if t. Thi function may be thought of a a witch that turn on at time t =. A witch that turn on at time t = c may be obtained by a imple hift: {, if t < c; h(t c) = (2), if t c. Figure a c how the graph of h(t), h(t 3), and h(t 3) h(t 5) Figure a Figure b Figure c For further illutration, Figure 2 how the graph of (a) h(t) + h(t 2), (b) h(t) 2h(t ) + h(t 2), (c) h(t) + h(t ) + h(t 2) 3h(t 3) Figure 2a Figure 2b Figure 2c

21 7.3. Heaviide Function and Piecewie-Defined Input 25 Example Let look cloely at the function f(t) = t h(t) + (2 2t)h(t ) + (t 2)h(t 2) with the goal of expreing it in piecewie form. When t <, all three of the Heaviide function are off, and o f(t) =. When t <, only the Heaviide function h(t) in the firt term i on, and o f(t) = t. When t < 2, both Heaviide function in the firt two term are on, and o f(t) = t + (2 2t) = 2 t. When t 2, all three Heaviide function are on, and o f(t) = t + (2 2t) + (t 2) =. Thu, f ha the piecewie form, t < ; t, t < ; f(t) = 2 t, t < 2; 2 3 4, 2 t. It graph i hown on the right. A witch that i on during a given time interval a t < b i epecially ueful and can be contructed a {, if t < a; h(t a) h(t b) =, if a t < b; (3), if b t. The product of any function g(t) with h(t a) h(t b) i a function that agree with g(t) when a t < b and i zero elewhere. For intance, Figure 3 how the graph of (a) t (h(t) h(t )), (b) (in t) (h(t) h(t π)), (c) (2 t)(h(t ) h(t 3)) Π 2 Π 3 Π 2 Figure 3a Figure 3b Figure 3c When a piecewie-defined function conit of multiple nonzero piece, it may be viewed a a um of function of thi type. Thi provide a convenient way of expreing the function in term of Heaviide function, a i illutrated in the next example.

22 26 Chapter 7. The Laplace Tranform Example 2 Conider the function, t < ; t f(t) = 2, t < ; (t ) 2, t < 2;, 2 t. We can expre f in term of Heaviide function a follow. We begin by thinking of f 2 a the um of impler function f and f 2, where { { f (t) = t 2, t < ; and f, elewhere; 2 (t) = (t ) 2, t < 2;, elewhere. Thee function can be expreed a f (t) = t 2 (h(t) h(t )) and f 2 (t) = (t ) 2 (h(t ) h(t 2)). So we have f(t) = t 2 (h(t) h(t )) + (t ) 2 (h(t ) h(t 2)), which we rearrange and write a f(t) = t 2 h(t) + ( (t ) 2 t 2) h(t ) (t ) 2 h(t 2). Another way of obtaining an expreion of a piecewie-defined function in term of Heaviide function i often convenient a well. Suppoe that f i given in piecewie form a, t < ; f (t), t < t ; f f(t) = (t), t t < t 2 ; f 2 (t), t 2 t < t 3 ;.. The repreentation of f in term of Heaviide function i f(t) = f (t)h(t) + (f (t) f (t))h(t t ) + (f 2 (t) f (t))h(t t 2 ) +, the form of the typical term being (f new f prev )h(t c). Thi i illutrated in the next example. Example 3 The function, t < ; t, t < ; f(t) =, t < 2; 3 t, 2 t < 3; 3 t; 2 3 4

23 7.3. Heaviide Function and Piecewie-Defined Input 27 may be written in term of Heaviide function a f(t) = t h(t) + ( t) h(t ) + (3 t ) h(t 2) + ( (3 t)) h(t 3), which after implification become f(t) = t h(t) + ( t) h(t ) + (2 t) h(t 2) + (t 3) h(t 3). The Second Shift Theorem Given a function f(t) with Laplace tranform F (), conider the function h(t c)f(t c) where c. The graph of thi function i the ame a that of f hifted to the right by c unit and zeroed out for t < c. Thi i illutrated in Figure 5. Figure 5 The Laplace tranform of h(t c)f(t c) i computed a follow: L[h(t c)f(t c)]() = = c e t h(t c)f(t c)dt e t f(t c)dt = Thi computation give u the econd hift theorem: e (t+c) f(t)dt = e c F (). L[h(t c)f(t c)]() = e c F (). A a imple conequence (with f(t c) = ), we obtain the tranform of any imple Heaviide function: L[h(t c)]() = e c. (4) Note that with c = thi give L[h(t)]() = /, which we have already een a the tranform of the function f(t) =, t.

24 28 Chapter 7. The Laplace Tranform Example 4 Suppoe that we want to find the invere tranform of F () = 2e + e 2. By plitting thi into three term, we recognize that F () = 2 e + e 2, f(t) = h(t) 2h(t ) + h(t 2). Thi i the function hown in Figure 2b. Example 5 Suppoe that we want to find the invere tranform of F () = 2e + e 2 2. After plitting thi into three term and recalling that / 2 = L[t](), the econd hift theorem tell u that f(t) = t h(t) 2(t )h(t ) + (t 2)h(t 2). Thi i the function in Example. Example 6 Let find the invere tranform of F () = π ( + e ) 2 + π 2. After plitting thi into two term and recalling that π/( 2 + π 2 ) = L[in πt](), the econd hift theorem tell u that f(t) = h(t) in πt + h(t ) in(π(t )). Since in(π(t )) = in πt, thi become f(t) = (h(t) h(t )) in πt = { in πt, t <,, t. A we aw in Example 4 6, the econd hift theorem make it very traightforward to write down the invere tranform of the product of e c and a known tranform F (). However, a lightly different form of the econd hift theorem i often more convenient for the purpoe of computing tranform. If we let g(t) = f(t c), then f(t) = g(t + c) and o F () = L[g(t + c)](). Thu the econd hift theorem i equivalent to L[h(t c)g(t)]() = e c L[g(t + c)](). (5)

25 7.3. Heaviide Function and Piecewie-Defined Input 29 Example 7 Conider the function f(t) = t h(t) t h(t ), whoe graph i hown in Figure 3a. To find the tranform of f we apply the econd hift theorem in the form of (5) a follow: F () = L[t h(t)]() L[t h(t )]() = L[t]() e L[t + ]() = ( 2 e 2 + ) = e ( + ) 2. Example 8 Let find the tranform of f(t) = e t h(t) + (e t e t )h(t ). Uing (5), we have F () = L[e t h(t)]() + L[(e t e t )h(t )]() = + e L[e (t+) e t+ ]() = + e Initial Value Problem ( e + e ) = e + e +. Our main purpoe here, of coure, i to ue all of thi to help u olve initial-value problem with piecewie-defined input. We will illutrate with the following example. Example 9 Conider an undamped pring-ma ytem with a natural frequency of ω 2π = cycle per unit time, o that ω = 2π. Suppoe that the ma i initially at ret and the platform to which the pring i attached undergoe motion decribed by the function een in Figure 2b. Thu we wih to find the ret olution of the equation y + 4π 2 y = 4π 2 (h(t) 2h(t ) + h(t 2)). Formally taking Laplace tranform, we find Y () = 2e + e 2 4π 2 ( 2 + 4π 2 ). A partial fraction expanion how that Y () = ( 2e + e 2) ( ) 2 + 4π 2. The econd factor i the tranform of co 2πt; therefore y(t) = ( co 2πt)h(t) 2( co 2π(t ))h(t ) + ( co 2π(t 2))h(t 2)

26 2 Chapter 7. The Laplace Tranform by the econd hift theorem. Since co 2πt = co 2π(t k) for all integer k and all t, the olution ha the imple piecewie form, t < ; co 2πt, t < ; y(t) = co 2πt, t < 2;, 2 t. The graph of thi olution and the forcing function are een in Figure 6. 2 Figure 6 We point out that in Example 9 we applied the firt differentiation theorem to tranform each ide of the differential equation without taking care to check the hypothee of that theorem. However, it i traightforward to verify a poteriori that the computation wa jutified. It i clear that y i continuou and exponentially bounded, and eay computation reveal that y i alo continuou and exponentially bounded and that y i piecewie continuou. The olution in Example 9 i not a olution i the normal ene, becaue y fail to exit at the point where f i dicontinuou. Such a olution i called a generalized olution. Definition. Let p and q be contant, and let f be piecewie continuou on [, ). A generalized olution of y + py + qy = f i a continuou function y on [, ) uch that (i) y i continuou and (ii) the differential equation i atified at each point where f i continuou. The following theorem on generalized olution jutifie the computation in Example 9 and imilar problem. We omit the proof. THEOREM Let p and q be contant, and let f be piecewie continuou on [, ). Then for any number y and v, the initial-value problem y + py + qy = f, y() = y, y () = v, ha a unique generalized olution y on [, ). Moreover, y i piecewie continuou, and if f i exponentially bounded, then o are y, y, and y. Problem In Problem 3, write the given function f in piecewie form. Then ketch the graph.. f(t) = h(t) + h(t ) + h(t 2) h(t 3) h(t 4) 2. f(t) = 4t( t) ( h(t) h(t ) ) 3. f(t) = t h(t) + ( t)h(t ) h(t 2)

27 7.3. Heaviide Function and Piecewie-Defined Input 2 In Problem 4 6, graph the given function without writing it piecewie form. 4. f(t) = 3h(t) h(t ) h(t 2) h(t 3) 5. f(t) = h(t) + ( t)h(t ) + (t 2)h(t 2) 6. f(t) = h(t) 2h(t ) + h(t 2) + h(t 3) 2h(t 4) + h(t 5) In Problem 7 9, expre the given function in term of Heaviide function Let x denote the floor (or greatet integer) function; that i, x = n, where n x < n + with n an integer Write each of the function in Problem through 3 in term of Heaviide function. Alo ketch the graph. { { ( ). f(t) = t, t < 3 t t, t < 3. f(t) =, elewhere, elewhere { { t t +.5, t < 2 3 t, t < 3 2. f(t) = 3. f(t) =, elewhere, elewhere Problem n = 4,..., 26: Find the Laplace tranform of the function in Problem n 3. In Problem 27 through 32, find the invere Laplace tranform and ketch it graph e 2 + e 3 2e 2 + e e + e 2 4 ( 2e π + e 2π ) 2 + 4π 2 3. ( 2 + 4) + e 2 2e 3 e e π + e 2π 2 + In Problem 33 and 34, olve the initial-value problem; then graph the olution and the nonhomogeneou term on the right ide of the differential equation. 33. y y = e h(t ), y() = 34. y + y = (t + )( h(t 2)), y() = In Problem 35 and 36, olve the initial-value problem; then graph the olution and the forcing function. 35. y + π 2 y = π 2( h(t 2) h(t 4) ), y() =, y () = 36. y + y = t h(t) t h(t 2π), y() =, y () = In Problem 37 39, obtain the tranform of the given function by direct application of the definition of the tranform, and then check by finding the tranform with the econd hift theorem. 37. f(t) = h(t ) h(t 2) 38. f(t) = h(t) 2h(t ) + h(t 2)

28 22 Chapter 7. The Laplace Tranform 39. f(t) = e t( h(t) h(t ) ) 4. A piecewie contant function that i right-continuou for all t, zero on (, t ), and dicontinuou at t, t, t 2,... can be expreed a f(t) = j h(t t ) + j h(t t ) + j 2 h(t t 2 ) + where each j i i a contant. Show that j i = lim t t + i jump of f(t) at t i.) Show further that f(t) lim t t i F () = j e t + j e t + j 2 e t2 +. Uing thee reult, redo Problem 9 and 22. f(t). (Thu j i the 7.4 Periodic Input A function f defined on [, ) i periodic if there i a number p > uch that f(t + p) = f(t) for all t. The leat uch p i the function period. Such a function may be regarded a the periodic extenion of the function f (t) = f(t)(h(t) h(t p)), which agree with f on [, p) and i zero elewhere. We will ee that the Laplace tranform of the periodic extenion f of f can be eaily expreed in term of the tranform of f. Since f(t + p) = f(t) for all t, it follow that f(t) = f(t p) for all t p. Thu we can expre f a f (t) = h(t)f(t) h(t p)f(t p), which allow ue of the econd hift theorem, producing the tranform F () = F () e p F (), where F = Lf a uual. Now we olve for F (), finding F () = F (). () e p Thu the tranform of the periodic extenion of f i imply the tranform of f divided by e p. The tranform F = Lf in the numerator can be obtained either by expreing f in term of Heaviide function and applying the econd hift theorem or by evaluating the definite integral F () = p e t f()d. (2)

29 7.4. Periodic Input 23 Example The function f(t) = ( ) t, t i the periodic extenion, with period 2, of f (t) = h(t) 2h(t ) + h(t 2). Graph of f and f are hown below. The tranform of f i F () = 2e + e 2 ; therefore, the tranform of f i Example 2 The function F () = 2e + e 2 ( e 2 ) = e + e = tanh(/2). f(t) = in t, t can be viewed a the periodic extenion of f (t) = in t (h(t) h(t π)) = h(t) in t + h(t π) in(t π) with period π. Thee function are plotted below. The Laplace tranform of f i F () = ( + e π ) * Recall that tanh x = ex e x = e 2x. e x +e x +e 2x 2 + ;

30 24 Chapter 7. The Laplace Tranform conequently, the tranform of f i Invere tranform F () = + e π e π 2 + = coth(π/2) 2. + The key to finding the invere tranform of a given tranform with e p a a factor in it denominator i the geometric erie: x = + x + x2 + x 3 +, if x <. Since < e p < for all > (and p > ), we have the expanion e p = + e p + e 2p + e 3p +, >. Although the tranform of any periodic function can be expreed in the form (), a factor of e p in the denominator doe not guarantee that the invere tranform will be periodic, a we hall ee in the firt of the following example. Example 3 The tranform can be written a the erie F () = F () = + e ( e ) + e 2 + e 3 Taking invere tranform term by term, we find that +. f(t) = h(t) + h(t ) + h(t 2) + h(t 3) +, which i eaily recognizable a the floor function f(t) = t, t. Example 4 The tranform F () = can be written a the erie F () = e e ( e 2 ) = ( + e ) + e 2 e 3 Taking invere tranform term by term, we find that +. f(t) = h(t) h(t ) + h(t 2) h(t 3) +, which can be expreed compactly a f(t) = ( + ( ) t ) /2, t.

31 7.4. Periodic Input 25 Example 5 The tranform can be written a F () = ( 2 + )( e π ) F () = 2 + ( + e π + e 2π + e 3π + ). Since the firt factor i the tranform of in t, we apply ucceive hift to obtain f(t) = h(t) in t + h(t π) in(t π) + h(t 2π) in(t 2π) +. Since in(t kπ) = ( ) k in t for any integer k, we can write thi function a f(t) = (h(t) h(t π) + h(t 2π) h(t 3π) + ) in t, which i the ame a f(t) = ( in t + in t ). 2 Example 6 Let f be a in Example ; that i, let f be the periodic extenion of f (t) = h(t) 2h(t ) + h(t 2) with period 2. Let find the ret olution of the equation The tranform of f i o the tranform of the olution i y + 4π 2 y = 4π 2 f(t). F () = 2e + e 2 ( e 2 ) Y () = 2e + e 2 e 2 4π 2 ( 2 + 4π 2 ). The firt factor ha the erie expanion 2e + e 2 e 2 = ( 2e + e 2 )( + e 2 + e 4 + e 6 + ) = 2e + 2e 2 2e 3 +, and the econd factor ha the partial fraction expanion So the tranform can be written a 4π 2 ( 2 + 4π 2 ) = 2 + 4π 2. Y () = ( 2e + 2e 2 2e 3 + ) ; ( ) 2 + 4π 2.

32 26 Chapter 7. The Laplace Tranform Recognizing the econd factor a the tranform of co 2πt, we apply ucceive hift, obtaining the invere tranform y(t) = ( co 2πt)h(t) 2( co 2π(t ))h(t ) + 2( co 2π(t 2))h(t 2) 2( co 2π(t 3))h(t 3) +. Since co 2π(t k) = co 2πt for any integer k, we finally arrive at the olution y(t) = ( co 2πt)(h(t) 2h(t ) + 2h(t 3) ) = ( ) t ( co 2πt). Notice that thi i preciely the periodic extenion with period 2 of the olution in Example 8 in Section 7.3. Problem The function in Problem through 3 are the ame a thoe in Problem 7 through 9 in Section 7.3. For each of them, find the Laplace tranform of the periodic extenion with period p = Let f (t) = h(t) h(t ) h(t 2) + h(t 3). (a) Sketch the graph of the periodic extenion of f with period p = 3 and find it Laplace tranform. (b) Sketch the graph of the periodic extenion of f with period p = 4 and find it Laplace tranform. 5. Let f(t) = t t, t. Graph f, decribe it a a periodic extenion of ome repreentative function f, and then find it Laplace tranform. 6. Let f(t) = (t t ) 2, t. Graph f, decribe it a a periodic extenion of ome repreentative function f, and then find it Laplace tranform. In Problem 7 through, find the invere Laplace tranform and ketch it graph ( 2 + )( e π ) e 2 ( e 2 ) 8.. e 2π ( 2 + )( e 4π ) e 2 ( + e 2 )

33 7.4. Periodic Input 27 In Problem and 2, find the ret olution of the equation and plot it graph. Note that each equation ha a forcing function imilar to t t from Problem 5.. y + 4π 2 y = 4π 2( (t t ) ) 2. y + π 2 y = π 3( t t ) 3. Let f be the periodic extenion of f (t) = h(t) 2h(t ) + h(t 2) with period p = 4. Find the ret olution of the equation y + 4π 2 y = 4π 2 f(t) and ketch it graph along with the graph of f. A function f defined on [, ) i aid to be antiperiodic with antiperiod k if f(t + k) = f(t) for all t. For example, in t i antiperiodic with antiperiod π, and ( ) t i antiperiodic with antiperiod. An antiperiodic function with antiperiod k can be decribed a the antiperiodic extenion of f (t) = f(t)(h(t) h(t k)), which vanihe outide the interval [, k]. Since f(t + k) = f(t) for all t, it follow that f (t) can be written a f (t) = f(t)h(t) + h(t k)f(t k). In Problem 4 through 6, aume that f i the antiperiodic extenion of f with antiperiod k. 4. Show that the tranform of f can be expreed in term of the tranform of f a F () = F () + e k = + e k k e t f(t)dt. 5. Let φ(t) = f(t) for all t. To electrical engineer, f(t) i known a the full-wave rectification of f(t). (a) Show that φ i periodic with period k and thu Φ() = e k k e t f(t) dt. (b) Conclude that if f(t) for t < k, then the following relationhip hold between the Laplace tranform of f(t) and φ(t) = f(t) : Φ() = + e k F () = coth(k/2)f (). e k 6. The function g(t) = 2( f(t) + f(t) ) i the half-wave rectification of f(t). (a) Sketch the graph of the half-wave rectification of in 2π t. (b) Show that if f(t) for t < k, then the following relationhip ) hold between the Laplace tranform of f(t) and g(t) = 2( f(t) + f(t) : G() = e 2k k e t f(t) dt = F () e k.

34 28 Chapter 7. The Laplace Tranform 7. Ue the reult of Problem 4 to rework Example. 8. Ue the reult of Problem 5 to rework Example Ue the reult of Problem 5 in a backward fahion to obtain the reult of Example from the tranform L[h(t)] = /. In Problem 2 and 2, ketch a graph of the given function, and ue the appropriate reult from either Problem 5 or 6, along with the firt hift theorem, to find it Laplace tranform. 2. g(t) = e t in πt 2. g(t) = 2 e t( in πt + in πt ). 7.5 Impule and the Dirac Ditribution Conider a pring-ma ytem being acted upon by ome external force. By Newton econd law, force i the rate of change in momentum: F (t) = dµ dt, where µ = mv. Thu, a illutrated in Figure, a force acting on the ma over a time interval [t, t ] reult in a change in momentum given by µ = µ(t ) µ(t ) = t t F (t)dt. Momentum Figure A force acting on the ma over a very mall time interval [ t, t + t] might reult, for intance, from tiking the ma with a hammer. It will be ueful to imagine an approximation to uch a force in the form of an impule an idealized force concentrated at a ingle point t in time and reulting in an intantaneou change in momentum. Our goal here i to develop a model for uch a force. Conider a unit change in momentum caued by a force F (t) whoe value i a contant b between time t = and t = t and zero at all other time; that i, t F (t) = (h(t) h(t t)) b and µ = b dt =. Thi arrangement i alway poible no matter how hort the time interval i. In fact, b and t are related by b t = ; therefore, b = / t, and o h(t) h(t t) F (t) =. () t The limit of uch force a t define a unit impule at time t = and may be thought of a generalized derivative of the Heaviide function h(t). Force

35 7.5. Impule and the Dirac Ditribution 29 Clearly thi will not be a function in the uual ene, but we will ue notation that eem to ugget that it i. To capture the limiting behavior of F a t, we define the ymbol δ(t), with unit of time in the current context, by the following propertie: (i) δ(t) = for all t ; (ii) δ(t)dt =. We emphaize that δ(t) i not a function of t in the uual ene, becaue (i) and (ii) are contradictory in the realm of ordinary calculu. Intead, δ(t) i a peculiar invention that erve a a limit for a equence of function whoe value converge to zero except at a ingle point, but whoe integral do not converge to zero. It i an example of a cla of generalized function called ditribution. Thi particular ditribution δ(t) i known a the Dirac ditribution, or the Dirac delta function, after the great phyicit Paul Dirac. Let g(t) be a continuou function on (, ). Since δ(t) repreent the limit a t of the function F (t) in (), we define g(t)δ(t) dt = lim t g(t) h(t) h(t t) dt = lim t t t t g(t)dt. Thi lat quantity i the limit a t of the average value of g(t) over [, t]. Since g i continuou, thi limit mut be g(). Therefore, it i conitent with (i) and (ii) above to define (iii) g(t)δ(t)dt = g() for any continuou g on (, ). A unit impule at time t = c i repreented by δ(t c), the unit impule at t = hifted to occur at t = c. A imple change of variable applied to (iii) give (iv) g(t)δ(t c)dt = g(c) for any continuou g on (, ). Thee lat two propertie put u in a poition to tate the Laplace tranform of δ(t) and δ(t c). Applying (iii) and (iv) with g(t) = e t, we find that Initial-Value Problem L[δ(t)]() = and L[δ(t c)]() = e c. (2) Impule can caue dicontinuitie in the derivative of the olution of a differential equation, or in the olution itelf in the cae of a firt-order equation. Therefore, initial value hould be undertood in the ene of left-ided limit. For example, in a econd-order initial-value problem, the initial condition y() = y and y () = v are undertood to mean y( ) = lim t y(t) = y and y ( ) = lim t y (t) = v. (3)

36 22 Chapter 7. The Laplace Tranform The ret olution of a econd-order equation i undertood to be the olution y for which y(t) = for all t, which implie that each of the tatement in (3) hold with y = v =. Example Suppoe that an undamped pring-ma ytem with with ma m, tiffne k, and natural angular frequency ω = k/m i initially at ret. At time t =, the ma i truck with a hammer, imparting an intantaneou tranfer of momentum µ. The ubequent motion may be modeled with the equation m y + k y = µ δ(t). (Note that all term have unit of force.) With zero initial value, the tranformed equation i m 2 Y () + k Y () = µ. Thu the tranform of the ret olution i Y () = µ m 2 + k = µ/m 2 + ω 2, and conequently the olution i y(t) = µ h(t) in ωt. m ω For t, thi i the ame a the olution of m y + k y = with y() = and initial velocity y () = µ/m. Alo note that, if viewed on (, ), the olution i continuou, but it derivative ha a jump dicontinuity at t =. Example 2 Suppoe that an undamped pring-ma ytem with ma m, tiffne k, and natural angular frequency ω = k/m i initially in motion with poition y() = and velocity y () = v. At a later time t = t, the ma i truck with a hammer, imparting an intantaneou tranfer of momentum µ. The motion of the ma may be modeled by The tranformed equation i m y + k y = µ δ(t t ), y() =, y () = v. o the tranform of the olution i m ( 2 Y () v ) + k Y () = µ e t ; Y () = µ e t + mv m 2 + k ( µ e t = m ω + v ) ω ω 2 + ω 2. Since the econd factor i the tranform of in ωt, we apply the econd hift theorem to obtain y(t) = µ m ω h(t t ) in (ω(t t )) + v h(t) in ωt. ω