CHAPTER 6 The Laplace Transform
Many practical engineering problems involve mechanical or electrical systems acted on by discontinuous or impulsive forcing terms. For such problems the methods described in Chapter 3 are often rather awkward to use. Another method that is especially well suited to these problems, although useful much more generally, is based on the Laplace transform. In this chapter we describe how this important method works, emphasizing problems typical of those arising in engineering applications.
6.1 Definition of the Laplace Transform
Among the tools that are very useful for solving linear differential equations are integral transforms. An integral transform is a relation of the form β F(s) K(s,t)f(t)dt, (1) = �α whereK(s,t) is a given function, called the kernel of the transformation, and the limits of integrationα andβ are also given. It is possible thatα orβ , or both. The relation (1) transforms the functionf into another function= −∞ F, which=∞ is called the transform off . The general idea in using an integral transform to solve a differential equation is as follows: Use the relation (1) to transform a problem for an unknown functionf into a simpler problem forF, then solve this simpler problem to findF, and finally recover the desired functionf from its transformF. This last step is known as “inverting the transform.” 293 294 Chapter 6. The Laplace Transform
There are several integral transforms that are useful in applied mathematics, but in this chapter we consider only the Laplace1 transform. This transform is defined in the following way. Letf(t) be given fort 0, and suppose thatf satisfies certain conditions to be stated a little later. Then≥ the Laplace transform off , which we will denote byL f(t) or byF(s), is defined by the equation { } ∞ st L f(t) F(s) e− f(t)dt. (2) { } = = �0 st The Laplace transform makes use of the kernelK(s,t) e − . Since the solutions of linear differential equations with constant coefficients= are based on the exponential function, the Laplace transform is particularly useful for such equations. Since the Laplace transform is defined by an integral over the range from zero to infinity, it is useful to review some basic facts about such integrals. In the first place, an integral over an unbounded interval is called an improper integral, and is defined as a limit of integrals over finite intervals; thus
A ∞ f(t)dt lim f(t)dt, (3) = A �a →∞ �a whereA is a positive real number. If the integral froma toA exists for eachA>a, and if the limit asA exists, then the improper integral is said to converge to →∞ that limiting value. Otherwise the integral is said to diverge, or to fail to exist. The following examples illustrate both possibilities.
Letf(t) e ct ,t 0, wherec is a real nonzero constant. Then EXAMPLE = ≥ A ct A ∞ e 1 ect dt lim ect dt lim = A = A c �0 →∞ �0 →∞ �0 1 � lim (ecA 1). � = A − � →∞ c It follows that the improper integral converges ifc< 0, and diverges ifc> 0. Ifc 0, the integrandf(t) is the constant function with value 1, and the integral again diverges.=
Letf(t) 1/t,t 1. Then EXAMPLE = ≥ A ∞ dt dt 2 lim lim lnA. t = A t = A �1 →∞ �1 →∞ Since lim lnA , the improper integral diverges. A =∞ →∞
1The Laplace transform is named for the eminent French mathematician P. S. Laplace, who studied the relation (2) in 1782. However, the techniques described in this chapter were not developed until a century or more later. They are due mainly to Oliver Heaviside (1850–1925), an innovative but unconventional English electrical engineer, who made significant contributions to the development and application of electromagnetic theory. 6.1 Definition of the Laplace Transform 295
p Letf(t) t − ,t 1, wherep is a real constant andp 1; the casep 1 was EXAMPLE considered= in Example≥ 2. Then �= =
3 A ∞ p p 1 1 p t− dt lim t− dt lim (A − 1). = A = A 1 p − �1 →∞ �1 →∞ − 1 p 1 p ∞ p AsA ,A − 0 ifp> 1, butA − ifp< 1. Hence t− dt con- →∞ → → ∞ 1 verges forp> 1, but (incorporating the result of Example 2) diverges for�p 1. These ≤ ∞ p results are analogous to those for the infinite series n− . n 1 �=
Before discussing the possible existence of ∞ f(t)dt, it is helpful to define certain a terms. A functionf is said to be piecewise continuous on an intervalα t β if the interval can be partitioned by a finite number of pointsα t 1.f is continuous on each open subintervalt i 1 y α t1 t2 β t FIGURE 6.1.1 A piecewise continuous function. 296 Chapter 6. The Laplace Transform Theorem 6.1.1 Iff is piecewise continuous fort a,if f(t) g(t) whent M for some positive ≥ | | ≤ ≥ constantM, and if ∞ g(t)dt converges, then ∞ f(t)dt also converges. On M a the other hand, iff(t) g(t) 0 fort M, and if ∞ g(t)dt diverges, then �≥ ≥ ≥ � M ∞ f(t)dt also diverges. a � � The proof of this result from the calculus will not be given here. It is made plausible, however, by comparing the areas represented by ∞ g(t)dt and ∞ f(t) dt. The M M | | ct p functions most useful for comparison purposes aree andt − , which were considered in Examples 1, 2, and 3. � � We now return to a consideration of the Laplace transformL f(t) orF(s), which is defined by Eq. (2) whenever this improper integral converges.{ In} general, the pa- rameters may be complex, but for our discussion we need consider only real values ofs. The foregoing discussion of integrals indicates that the Laplace transformF of a functionf exists iff satisfies certain conditions, such as those stated in the following theorem. Theorem 6.1.2 Suppose that 1.f is piecewise continuous on the interval 0 t A for any positiveA. ≤ ≤ 2. f(t) Ke at whent M. In this inequalityK,a, andM are real constants,K | and| ≤M necessarily positive.≥ Then the Laplace transformL f(t) F(s), defined by Eq. (2), exists fors>a. { } = To establish this theorem it is necessary to show only that the integral in Eq. (2) converges fors>a. Splitting the improper integral into two parts, we have M ∞ st st ∞ st e− f(t)dt e− f(t)dt e− f(t)dt. (4) = + �0 �0 �M The first integral on the right side of Eq. (4) exists by hypothesis (1) of the theorem; hence the existence ofF(s) depends on the convergence of the second integral. By hypothesis (2) we have, fort M, ≥ st st at (a s)t e− f(t) Ke − e Ke − , | | ≤ = ∞ (a s)t and thus, by Theorem 6.1.1,F(s) exists provided that e − dt converges. Refer- M ring to Example 1 withc replaced bya s, we see that this latter integral converges − whena s< 0, which establishes Theorem 6.1.2. � Unless− the contrary is specifically stated, in this chapter we deal only with functions satisfying the conditions of Theorem 6.1.2. Such functions are described as piecewise continuous, and of exponential order ast . The Laplace transforms of some important elementary functions are given in→∞ the following examples. 6.1 Definition of the Laplace Transform 297 Letf(t) 1,t 0. Then EXAMPLE = ≥ ∞ st 1 4 L 1 e− dt ,s>0. { } = = s �0 Letf(t) e at ,t 0. Then EXAMPLE = ≥ at ∞ st at ∞ (s a)t L e e− e dt e− − dt 5 { } = = �0 �0 1 ,s>a. = s a − Letf(t) sinat,t 0. Then EXAMPLE = ≥ ∞ st L sinat F(s) e− sinat dt,s>0. 6 { }= = �0 Since A st F(s) lim e− sinat dt, = A →∞ �0 upon integrating by parts we obtain st A A e− cosat s st F(s) lim e− cosat dt = A − a − a →∞ � �0 �0 � � 1 s ∞ st � e− cosat� dt. = a − a �0 A second integration by parts then yields 2 1 s ∞ st F(s) e− sinat dt = a − a2 �0 1 s2 F(s). = a − a2 Hence, solving forF(s), we have a F(s) ,s>0. = s2 a 2 + Now let us suppose thatf 1 andf 2 are two functions whose Laplace transforms exist fors>a 1 ands>a 2, respectively. Then, fors greater than the maximum ofa 1 anda 2, ∞ st L c f (t) c f (t) e− [c f (t) c f (t)]dt { 1 1 + 2 2 } = 1 1 + 2 2 �0 ∞ st ∞ st c e− f (t)dt c e− f (t)dt = 1 1 + 2 2 ; �0 �0 298 Chapter 6. The Laplace Transform hence L c f (t) c f (t) c L f (t) c L f (t) . (5) { 1 1 + 2 2 } = 1 { 1 } + 2 { 2 } Equation (5) is a statement of the fact that the Laplace transform is a linear operator. This property is of paramount importance, and we make frequent use of it later. PROBLEMS In each of Problems 1 through 4 sketch the graph of the given function. In each case determine whetherf is continuous, piecewise continuous, or neither on the interval 0 t 3. ≤ ≤ t2,0 t 1 t2,0 t 1 ≤ ≤ ≤ ≤1 1.f(t) 2 t,1 ibt ibt ibt In each of Problems 11 through 14 recall that cosbt (e e − )/2 and sinbt (e ibt = + = − e− )/2i. Assuming that the necessary elementary integration formulas extend to this case, find the Laplace transform of the given function;a andb are real constants. 11. sinbt 12. cosbt 13.e at sinbt 14.e at cosbt In each of Problems 15 through 20, using integration by parts, find the Laplace transform of the given function;n is a positive integer anda is a real constant. 15. teat 16.t sinat 17.t coshat 18.t neat 19.t 2 sinat 20.t 2 sinhat In each of Problems 21 through 24 determine whether the given integral converges or diverges. ∞ 2 1 ∞ t 21. (t 1) − dt 22. te− dt + �0 �0 ∞ 2 t ∞ t 23. t− e dt 24. e− cost dt �1 �0 25. Suppose thatf andf � are continuous fort 0, and of exponential order ast . Show by integration by parts that ifF(s) L f(t)≥, then lim F(s) 0. The result→∞ is actually = { } s = true under less restrictive conditions, such as those of→∞ Theorem 6.1.2. 26. The Gamma Function. The gamma function is denoted by�(p) and is defined by the integral ∞ x p �(p 1) e− x dx. (i) + = �0 6.2 Solution of Initial Value Problems 299 The integral converges asx for allp. Forp< 0 it is also improper because the integrand becomes unbounded→∞ asx 0. However, the integral can be shown to converge atx 0 forp> 1. → (a)= Show that− forp>0 �(p 1) p�(p). + = (b) Show that�(1) 1. (c) Ifp is a positive= integern, show that �(n 1) n!. + = Since�(p) is also defined whenp is not an integer, this function provides an extension of the factorial function to nonintegral values of the independent variable. Note that it is also consistent to define 0! 1. (d) Show that forp>0= p(p 1)(p 2) (p n 1) �(p n)/�(p). + + ··· + − = + Thus�(p) can be determined for all positive values ofp if�(p) is known in a single 1 3 interval of unit length, say, 0 1. (a) Referring to Problem 26, show that − L p ∞ st p 1 ∞ x p t e− t dt p 1 e− x dx { } = 0 = s + 0 � p 1 � �(p 1)/s + ,s>0. = + (b) Letp be a positive integern in (a); show that n n 1 L t n!/s + ,s>0. { } = (c) Show that 1/2 2 ∞ x 2 L t − e− dx,s>0. { } = √s �0 It is possible to show that ∞ x 2 √π e− dx = 2 ; �0 hence 1/2 L t− π/s,s>0. { } = (d) Show that � L t 1/2 √π/2s3/2,s>0. { } = 6.2 Solution of Initial Value Problems In this section we show how the Laplace transform can be used to solve initial value problems for linear differential equations with constant coefficients. The usefulness of the Laplace transform in this connection rests primarily on the fact that the transform 300 Chapter 6. The Laplace Transform off � is related in a simple way to the transform off . The relationship is expressed in the following theorem. Theorem 6.2.1 Suppose thatf is continuous andf � is piecewise continuous on any interval 0 t A. Suppose further that there exist constantsK,a, andM such that f(t) ≤at≤ | | ≤ K e fort M. ThenL f �(t) exists fors>a, and moreover ≥ { } L f �(t) sL f(t) f(0). (1) { } = { } − To prove this theorem we consider the integral A st e− f �(t)dt. �0 Iff � has points of discontinuity in the interval 0 t A, let them be denoted by ≤ ≤ t1,t 2,...,t n. Then we can write this integral as A t1 t2 A st st st st e− f �(t)dt e− f �(t)dt e− f �(t)dt e− f �(t)dt. = + t +···+ t �0 �0 � 1 � n Integrating each term on the right by parts yields A t t A st st 1 st 2 st e− f �(t)dt e − f(t) e − f(t) e − f(t) 0 = 0 + t + · · · + t � � � 1 � n t1 � � t2 � A st st st s �e− f(t)dt � e− f(t)dt � e− f(t)dt . + + t +···+ t ��0 � 1 � n � Sincef is continuous, the contributions of the integrated terms att 1,t 2,...,t n cancel. Combining the integrals gives A A st s A st e− f �(t)dt e − f(A) f(0) s e− f(t)dt. = − + �0 �0 s A AsA ,e − f(A) 0 whenevers>a. Hence, fors>a, →∞ → L f �(t) sL f(t) f(0), { } = { } − which establishes the theorem. Iff � andf �� satisfy the same conditions that are imposed onf andf �, respectively, in Theorem 6.2.1, then it follows that the Laplace transform off �� also exists fors>a and is given by 2 L f ��(t) s L f(t) sf(0) f �(0). (2) { } = { } − − Indeed, provided the functionf and its derivatives satisfy suitable conditions, an expression for the transform of thenth derivativef (n) can be derived by successive applications of this theorem. The result is given in the following corollary. (n 1) (n) Corollary 6.2.2 Suppose that the functionsf,f �,...,f − are continuous, and thatf is piecewise continuous on any interval 0 t A. Suppose further that there exist constantsK,a, ≤ ≤ 6.2 Solution of Initial Value Problems 301 at at (n 1) at andM such that f(t) Ke , f �(t) Ke ,..., f − (t) Ke fort M. ThenL f (n)(t) | exists| ≤ fors>a and| is given| ≤ by | | ≤ ≥ { } (n) n n 1 (n 2) (n 1) L f (t) s L f(t) s − f(0) sf − (0) f − (0). (3) { } = { } − −···− − We now show how the Laplace transform can be used to solve initial value problems. It is most useful for problems involving nonhomogeneous differential equations, as we will demonstrate in later sections of this chapter. However, we begin by looking at some homogeneous equations, which are a bit simpler. For example, consider the differential equation y�� y � 2y 0 (4) − − = and the initial conditions y(0) 1,y �(0) 0. (5) = = This problem is easily solved by the methods of Section 3.1. The characteristic equation is r 2 r 2 (r 2)(r 1) 0, (6) − − = − + = and consequently the general solution of Eq. (4) is t 2t y c e− c e . (7) = 1 + 2 To satisfy the initial conditions (5) we must havec c 1 and c 2c 0; 1 + 2 = − 1 + 2 = hencec 2 andc 1 , so that the solution of the initial value problem (4) and (5) is 1 = 3 2 = 3 2 t 1 2t y φ(t) e− e . (8) = = 3 + 3 Now let us solve the same problem by using the Laplace transform. To do this we must assume that the problem has a solutiony φ(t), which with its first two derivatives satisfies the conditions of Corollary 6.2.2.= Then, taking the Laplace transform of the differential equation (4), we obtain L y �� L y � 2L y 0, (9) { } − { } − { }= where we have used the linearity of the transform to write the transform of a sum as the sum of the separate transforms. Upon using the corollary to expressL y �� andL y � in terms ofL y , we find that Eq. (9) becomes { } { } { } 2 s L y sy(0) y �(0) [sL y y(0)] 2L y 0, { }− − − { }− − { }= or 2 (s s 2)Y(s) (1 s)y(0) y �(0) 0, (10) − − + − − = whereY(s) L y . Substituting fory(0) andy �(0) in Eq. (10) from the initial condi- tions (5), and= then{ } solving forY(s), we obtain s 1 s 1 Y(s) − − . (11) = s2 s 2 = (s 2)(s 1) − − − + We have thus obtained an expression for the Laplace transformY(s) of the solution y φ(t) of the given initial value problem. To determine the functionφ we must find the= function whose Laplace transform isY(s), as given by Eq. (11). 302 Chapter 6. The Laplace Transform This can be done most easily by expanding the right side of Eq. (11) in partial fractions. Thus we write s 1 a b a(s 1) b(s 2) Y(s) − + + − , (12) = (s 2)(s 1) = s 2 + s 1 = (s 2)(s 1) − + − + − + where the coefficientsa andb are to be determined. By equating numerators of the second and fourth members of Eq. (12), we obtain s 1 a(s 1) b(s 2), − = + + − an equation that must hold for alls. In particular, if we sets 2, then it follows that 1 = 2 a 3 . Similarly, if we sets 1, then we find thatb 3 . By substituting these values= fora andb, respectively,=− we have = 1/3 2/3 Y(s) . (13) = s 2 + s 1 − + 1 2t Finally, if we use the result of Example 5 of Section 6.1, it follows that 3 e has the 1 1 2 t 2 1 transform 3 (s 2) − ; similarly, 3 e− has the transform 3 (s 1) − . Hence, by the linearity of the− Laplace transform, + 1 2t 2 t y φ(t) e e− = = 3 + 3 has the transform (13) and is therefore the solution of the initial value problem (4), (5). Of course, this is the same solution that we obtained earlier. The same procedure can be applied to the general second order linear equation with constant coefficients, ay�� by � cy f(t). (14) + + = Assuming that the solutiony φ(t) satisfies the conditions of Corollary 6.2.2 for n 2, we can take the transform= of Eq. (14) and thereby obtain = 2 a[s Y(s) sy(0) y �(0)] b[sY(s) y(0)] cY(s) F(s), (15) − − + − + = whereF(s) is the transform off(t). By solving Eq. (15) forY(s) we find that (as b)y(0) ay �(0) F(s) Y(s) + + . (16) = as2 bs c + as2 bs c + + + + The problem is then solved, provided that we can find the functiony φ(t) whose transform isY(s). = Even at this early stage of our discussion we can point out some of the essential features of the transform method. In the first place, the transformY(s) of the unknown functiony φ(t) is found by solving an algebraic equation rather than a differential equation,= Eq. (10) rather than Eq. (4), or in general Eq. (15) rather than Eq. (14). This is the key to the usefulness of Laplace transforms for solving linear, constant coefficient, ordinary differential equations—the problem is reduced from a differential equation to an algebraic one. Next, the solution satisfying given initial conditions is automatically found, so that the task of determining appropriate values for the arbitrary constants in the general solution does not arise. Further, as indicated in Eq. (15), nonhomogeneous equations are handled in exactly the same way as homogeneous ones; it is not necessary to solve the corresponding homogeneous equation first. Finally, the method can be applied in the same way to higher order equations, as long as we assume that the solution satisfies the conditions of the corollary for the appropriate value ofn. 6.2 Solution of Initial Value Problems 303 Observe that the polynomialas 2 bs c in the denominator on the right side of Eq. (16) is precisely the characteristic+ polynomial+ associated with Eq. (14). Since the use of a partial fraction expansion ofY(s) to determineφ(t) requires us to factor this polynomial, the use of Laplace transforms does not avoid the necessity of finding roots of the characteristic equation. For equations of higher than second order this may be a difficult algebraic problem, particularly if the roots are irrational or complex. The main difficulty that occurs in solving initial value problems by the transform technique lies in the problem of determining the functiony φ(t) corresponding to the transformY(s). This problem is known as the inversion= problem for the Laplace transform;φ(t) is called the inverse transform corresponding toY(s), and the process of findingφ(t) fromY(s) is known as inverting the transform. We also use the notation 1 L− Y(s) to denote the inverse transform ofY(s). There is a general formula for the inverse{ Laplace} transform, but its use requires a knowledge of the theory of functions of a complex variable, and we do not consider it in this book. However, it is still possible to develop many important properties of the Laplace transform, and to solve many interesting problems, without the use of complex variables. In solving the initial value problem (4), (5) we did not consider the question of whether there may be functions other than the one given by Eq. (8) that also have the transform (13). In fact, it can be shown that iff is a continuous function with the Laplace transformF, then there is no other continuous function having the same transform. In other words, there is essentially a one-to-one correspondence between functions and their Laplace transforms. This fact suggests the compilation of a table, such as Table 6.2.1, giving the transforms of functions frequently encountered, and vice versa. The entries in the second column of Table 6.2.1 are the transforms of those in the first column. Perhaps more important, the functions in the first column are the inverse transforms of those in the second column. Thus, for example, if the transform of the solution of a differential equation is known, the solution itself can often be found merely by looking it up in the table. Some of the entries in Table 6.2.1 have been used as examples, or appear as problems in Section 6.1, while others will be developed later in the chapter. The third column of the table indicates where the derivation of the given transforms may be found. While Table 6.2.1 is sufficient for the examples and problems in this book, much larger tables are also available (see the list of references at the end of the chapter). Transforms and inverse transforms can also be readily obtained electronically by using a computer algebra system. Frequently, a Laplace transformF(s) is expressible as a sum of several terms, F(s) F (s) F (s) F (s). (17) = 1 + 2 +···+ n 1 1 Suppose thatf (t) L − F (s) ,...,f (t) L − F (s) . Then the function 1 = { 1 } n = { n } f(t) f (t) f (t) = 1 +···+ n has the Laplace transformF(s). By the uniqueness property stated previously there is no other continuous functionf having the same transform. Thus 1 1 1 L− F(s) L − F (s) L − F (s) (18) { } = { 1 } + · · · + { n }; that is, the inverse Laplace transform is also a linear operator. In many problems it is convenient to make use of this property by decomposing a given transform into a sum of functions whose inverse transforms are already known or can be found in the table. Partial fraction expansions are particularly useful in this 304 Chapter 6. The Laplace Transform TABLE 6.2.1 Elementary Laplace Transforms 1 f(t) L − F(s) F(s) L f(t) Notes = { } = { } 1 1. 1 ,s>0 Sec. 6.1; Ex. 4 s 1 2.e at ,s>a Sec. 6.1; Ex. 5 s a − n! n , > 3.t n positive integer n 1 s 0 Sec. 6.1; Prob. 27 ; = s + �(p 1) p, > , > 4.t p 1 p+ 1 s 0 Sec. 6.1; Prob. 27 − s + a 5. sinat ,s> 0 Sec. 6.1; Ex. 6 s2 a 2 + s 6. cosat ,s> 0 Sec. 6.1; Prob. 6 s2 a 2 + a 7. sinhat ,s> a Sec. 6.1; Prob. 8 s2 a 2 | | − s 8. coshat ,s> a Sec. 6.1; Prob. 7 s2 a 2 | | − b 9.e at sinbt ,s>a Sec. 6.1; Prob. 13 (s a) 2 b 2 − + s a 10.e at cosbt − ,s>a Sec. 6.1; Prob. 14 (s a) 2 b 2 − + n! n at , , > 11.t e n positive integer n 1 s a Sec. 6.1; Prob. 18 = (s a) + − cs e− 12.u (t) ,s> 0 Sec. 6.3 c s cs 13.u (t)f(t c)e − F(s) Sec. 6.3 c − 14.e ct f(t)F(s c) Sec. 6.3 − 1 s 15.f(ct) F ,c> 0 Sec. 6.3; Prob. 19 c c t � � 16. f(t τ)g(τ)dτF(s)G(s) Sec. 6.6 0 − cs 17.δ(t c)e − Sec. 6.5 � − (n) n n 1 (n 1) 18.f (t)s F(s) s − f(0) f − (0) Sec. 6.2 − −···− 19.( t) n f(t)F (n)(s) Sec. 6.2; Prob. 28 − 6.2 Solution of Initial Value Problems 305 connection, and a general result covering many cases is given in Problem 38. Other useful properties of Laplace transforms are derived later in this chapter. As further illustrations of the technique of solving initial value problems by means of the Laplace transform and partial fraction expansions, consider the following examples. Find the solution of the differential equation EXAMPLE y�� y sin 2t, (19) 1 + = satisfying the initial conditions y(0) 2,y �(0) 1. (20) = = We assume that this initial value problem has a solutiony φ(t), which with its first two derivatives satisfies the conditions of Corollary 6.2.2.= Then, taking the Laplace transform of the differential equation, we have 2 2 s Y(s) sy(0) y �(0) Y(s) 2/(s 4), − − + = + where the transform of sin 2t has been obtained from line 5 of Table 6.2.1. Substituting fory(0) andy �(0) from the initial conditions and solving forY(s), we obtain 2s3 s 2 8s 6 Y(s) + + + . (21) = (s2 1)(s 2 4) + + Using partial fractions we can writeY(s) in the form as b cs d (as b)(s 2 4) (cs d)(s 2 1) Y(s) + + + + + + + . (22) = s2 1 + s2 4 = (s2 1)(s 2 4) + + + + By expanding the numerator on the right side of Eq. (22) and equating it to the numerator in Eq. (21) we find that 2s3 s 2 8s 6 (a c)s 3 (b d)s 2 (4a c)s (4b d) + + + = + + + + + + + for alls. Then, comparing coefficients of like powers ofs, we have a c 2,b d 1, + = + = 4a c 8,4b d 6. + = + = Consequently,a 2,c 0,b 5 , andd 2 , from which it follows that = = = 3 =− 3 2s 5/3 2/3 Y(s) . (23) = s2 1 + s2 1 − s2 4 + + + From lines 5 and 6 of Table 6.2.1, the solution of the given initial value problem is y φ(t) 2cost 5 sint 1 sin 2t. (24) = = + 3 − 3 Find the solution of the initial value problem EXAMPLE yiv y 0, (25) 2 − = y(0) 0,y �(0) 1,y ��(0) 0,y ���(0) 0. (26) = = = = 306 Chapter 6. The Laplace Transform In this problem we need to assume that the solutiony φ(t) satisfies the conditions of Corollary 6.2.2 forn 4. The Laplace transform of the= differential equation (25) is = 4 3 2 s Y(s) s y(0) s y�(0) sy ��(0) y ���(0) Y(s) 0. − − − − − = Then, using the initial conditions (26) and solving forY(s), we have s2 Y(s) . (27) = s4 1 − A partial fraction expansion ofY(s) is as b cs d Y(s) + + , = s2 1 + s2 1 − + and it follows that (as b)(s 2 1) (cs d)(s 2 1) s 2 (28) + + + + − = for alls. By settings 1ands 1, respectively, in Eq. (28) we obtain the pair of equations = =− 2(a b) 1,2( a b) 1, + = − + = 1 1 and thereforea 0 andb 2 . If we sets 0 in Eq. (28), thenb d 0, sod 2 . Finally, equating= the coefficients= of the cubic= terms on each side− of= Eq. (28),= we find thata c 0, soc 0. Thus + = = 1/2 1/2 Y(s) , (29) = s2 1 + s2 1 − + and from lines 7 and 5 of Table 6.2.1 the solution of the initial value problem (25), (26) is sinht sint y φ(t) + . (30) = = 2 The most important elementary applications of the Laplace transform are in the study of mechanical vibrations and in the analysis of electric circuits; the governing equations were derived in Section 3.8. A vibrating spring–mass system has the equation of motion d2u du m γ ku F(t), (31) dt2 + dt + = wherem is the mass,γ the damping coefficient,k the spring constant, andF(t) the applied external force. The equation describing an electric circuit containing an inductanceL, a resistanceR, and a capacitanceC (an LRC circuit) is d2 Q d Q 1 L R Q E(t), (32) dt2 + dt + C = whereQ(t) is the charge on the capacitor andE(t) is the applied voltage. In terms of the currentI(t) dQ(t)/dt we can differentiate Eq. (32) and write = d2 I d I 1 d E L R I (t). (33) dt2 + dt + C = dt Suitable initial conditions onu,Q, orI must also be prescribed. 6.2 Solution of Initial Value Problems 307 We have noted previously in Section 3.8 that Eq. (31) for the spring–mass system and Eq. (32) or (33) for the electric circuit are identical mathematically, differing only in the interpretation of the constants and variables appearing in them. There are other physical problems that also lead to the same differential equation. Thus, once the mathematical problem is solved, its solution can be interpreted in terms of whichever corresponding physical problem is of immediate interest. In the problem lists following this and other sections in this chapter are numerous initial value problems for second order linear differential equations with constant coefficients. Many can be interpreted as models of particular physical systems, but usually we do not point this out explicitly. PROBLEMS In each of Problems 1 through 10 find the inverse Laplace transform of the given function. 3 4 1. 2. s2 4 (s 1) 3 + − 2 3s 3. 4. s2 3s 4 s2 s 6 + − − − 2s 2 2s 3 5. + 6. − s2 2s 5 s2 4 + + − 2s 1 8s2 4s 12 7. + 8. − + s2 2s 2 s(s 2 4) − + + 1 2s 2s 3 9. − 10. − s2 4s 5 s2 2s 10 + + + + In each of Problems 11 through 23 use the Laplace transform to solve the given initial value problem. 11.y �� y � 6y 0 y(0) 1,y �(0) 1 − − = ; = =− 12.y �� 3y � 2y 0 y(0) 1,y �(0) 0 + + = ; = = 13.y �� 2y � 2y 0 y(0) 0,y �(0) 1 − + = ; = = 14.y �� 4y � 4y 0 y(0) 1,y �(0) 1 − + = ; = = 15.y �� 2y � 2y 0 y(0) 2,y �(0) 0 − − = ; = = 16.y �� 2y � 5y 0 y(0) 2,y �(0) 1 iv + + = ; = =− 17.y 4y ��� 6y �� 4y � y 0 y(0) 0,y �(0) 1,y ��(0) 0,y ���(0) 1 iv − + − + = ; = = = = 18.y y 0 y(0) 1,y �(0) 0,y ��(0) 1,y ���(0) 0 iv − = ; = = = = 19.y 4y 0 y(0) 1,y �(0) 0,y ��(0) 2,y ���(0) 0 − 2= ; = 2 = =− = 20.y �� ω y cos 2t, ω 4 y(0) 1,y �(0) 0 + = �= ; = = 21.y �� 2y � 2y cost y(0) 1,y �(0) 0 − + = ;t = = 22.y �� 2y � 2y e − y(0) 0,y �(0) 1 − + = t ; = = 23.y �� 2y � y 4e − y(0) 2,y �(0) 1 + + = ; = =− In each of Problems 24 through 26 find the Laplace transformY(s) L y of the solution of the given initial value problem. A method of determining the inverse= transform{ } is developed in Section 6.3. 1,0 t<π, 24.y �� 4y ≤ y(0) 1,y �(0) 0 + = 0, π t< = = � ≤ ∞; t,0 t<1, 25.y �� y ≤ y(0) 0,y �(0) 0 + = 0,1 t< = = � ≤ ∞; 308 Chapter 6. The Laplace Transform t,0 t<1, 26.y �� 4y ≤ y(0) 0,y �(0) 0 + = 1,1 t< = = � ≤ ∞; 27. The Laplace transforms of certain functions can be found conveniently from their Taylor series expansions. (a) Using the Taylor series for sint, n 2n 1 ∞ ( 1) t + sint − , ( ) = n 0 2n 1 ! �= + and assuming that the Laplace transform of this series can be computed term by term, verify that 1 L sint ,s>1. { }= s2 1 + (b) Let (sint)/t,t 0, f(t) �= = 1,t 0. � = Find the Taylor series forf aboutt 0. Assuming that the Laplace transform of this function can be computed term by term,= verify that L f(t) arctan(1/s),s>1. { }= (c) The Bessel function of the first kind of order zeroJ 0 has the Taylor series (see Section 5.8) n 2n ∞ ( 1) t ( ) . J0 t −2n 2 = n 0 2 (n!) �= Assuming that the following Laplace transforms can be computed term by term, verify that 2 1/2 L J (t) (s 1) − ,s>1, { 0 }= + and 1 1/4s L J (√t) s − e− ,s>0. { 0 }= Problems 28 through 36 are concerned with differentiation of the Laplace transform. 28. Let ∞ st F(s) e− f(t)dt. = �0 It is possible to show that as long asf satisfies the conditions of Theorem 6.1.2, it is legitimate to differentiate under the integral sign with respect to the parameters when s>a. (a) Show thatF �(s) L t f(t) . (b) Show thatF (n)(=s) {−L ( t)} n f(t) ; hence differentiating the Laplace transform cor- responds to multiplying= the{ − original function} by t. − In each of Problems 29 through 34 use the result of Problem 28 to find the Laplace transform of the given function;a andb are real numbers andn is a positive integer. 29. teat 30.t 2 sinbt 31.t n 32.t neat 33. teat sinbt 34. teat cosbt 35. Consider Bessel’s equation of order zero ty�� y � ty 0. + + = 6.2 Solution of Initial Value Problems 309 Recall from Section 5.4 thatt 0 is a regular singular point for this equation, and therefore solutions may become unbounded= ast 0. However, let us try to determine whether there are any solutions that remain finite att→ 0 and have finite derivatives there. Assuming that there is such a solutiony φ(t), letY(s=) L φ(t) . (a) Show thatY(s) satisfies= = { } 2 (1 s )Y �(s) sY(s) 0. + + = 2 1/2 (b) Show thatY(s) c(1 s )− , wherec is an arbitrary constant. = 2 +1/2 (c) Expanding(1 s )− in a binomial series valid fors> 1 and assuming that it is permissible to take+ the inverse transform term by term, show that n 2n ∞ ( 1) t ( ), y c −2n 2 cJ 0 t = n 0 2 (n!) = �= whereJ is the Bessel function of the first kind of order zero. Note thatJ (0) 1, and that 0 0 = J0 has finite derivatives of all orders att 0. It was shown in Section 5.8 that the second solution of this equation becomes unbounded= ast 0. 36. For each of the following initial value problems→ use the results of Problem 28 to find the differential equation satisfied byY(s) L φ(t) , wherey φ(t) is the solution of the given initial value problem. = { } = (a)y �� ty 0 y(0) 1,y �(0) 0 (Airy’s equation) − 2= ; = = (b)(1 t )y �� 2ty � α(α 1)y 0 y(0) 0,y �(0) 1 (Legendre’s equation) − − + + = ; = = Note that the differential equation forY(s) is of first order in part (a), but of second order in part (b). This is due to the fact thatt appears at most to the first power in the equation of part (a), whereas it appears to the second power in that of part (b). This illustrates that the Laplace transform is not often useful in solving differential equations with variable coefficients, unless all the coefficients are at most linear functions of the independent variable. 37. Suppose that t g(t) f(τ)dτ. = �0 IfG(s) andF(s) are the Laplace transforms ofg(t) andf(t), respectively, show that G(s) F(s)/s. = 38. In this problem we show how a general partial fraction expansion can be used to calculate many inverse Laplace transforms. Suppose that F(s) P(s)/Q(s), = whereQ(s) is a polynomial of degreen with distinct zerosr 1,...,r n andP(s)isa polynomial of degree less thann. In this case it is possible to show thatP(s)/Q(s) has a partial fraction expansion of the form P(s) A A 1 n , (i) Q(s) = s r + · · · + s r − 1 − n where the coefficientsA 1,...,A n must be determined. (a) Show that A P(r )/Q �(r ),k 1,...,n. (ii) k = k k = Hint: One way to do this is to multiply Eq. (i) bys r and then to take the limit ass r . − k → k 310 Chapter 6. The Laplace Transform (b) Show that n 1 P(rk ) r t L− F(s) e k . (iii) ( ) { }= k 1 Q� rk �= 6.3 Step Functions In Section 6.2 we outlined the general procedure involved in solving initial value problems by means of the Laplace transform. Some of the most interesting elemen- tary applications of the transform method occur in the solution of linear differential equations with discontinuous or impulsive forcing functions. Equations of this type frequently arise in the analysis of the flow of current in electric circuits or the vibra- tions of mechanical systems. In this section and the following ones we develop some additional properties of the Laplace transform that are useful in the solution of such problems. Unless a specific statement is made to the contrary, all functions appearing below will be assumed to be piecewise continuous and of exponential order, so that their Laplace transforms exist, at least fors sufficiently large. To deal effectively with functions having jump discontinuities, it is very helpful to introduce a function known as the unit step function, or Heaviside function. This function will be denoted byu c, and is defined by 0,t y y 1 1 c t c t FIGURE 6.3.1 Graph ofy u c(t). FIGURE 6.3.2 Graph of = y 1 u (t). = − c Sketch the graph ofy h(t), where EXAMPLE = h(t) u (t) u (t),t 0. 1 = π − 2π ≥ From the definition ofu c(t) in Eq. (1) we have 0 0 0,0 t<π, h(t) 1−0=1,≤ π t<2π, = 1−1=0,2π t< ≤. − = ≤ ∞ 6.3 Step Functions 311 Thus the equationy h(t) has the graph shown in Figure 6.3.3. This function can be thought of as a rectangular= pulse. y 1 π 2π 3π t FIGURE 6.3.3 Graph ofy u (t) u (t). = π − 2π The Laplace transform ofu c is easily determined: L ∞ st ∞ st uc(t) e− uc(t)dt e− dt { } = 0 = c � cs � e− ,s>0. (2) = s For a given functionf , defined fort 0, we will often want to consider the related functiong defined by ≥ 0,t y y f (0) f (0) t c t (a) (b) FIGURE 6.3.4 A translation of the given function. (a)y f(t);(b)y u (t)f(t c). = = c − Theorem 6.3.1 IfF(s) L f(t) exists fors>a 0, and ifc is a positive constant, then = { } ≥ cs cs L u (t)f(t c) e − L f(t) e − F(s),s>a. (3) { c − } = { } = 312 Chapter 6. The Laplace Transform 1 Conversely, iff(t) L − F(s) , then = { } 1 cs u (t)f(t c) L − e− F(s) . (4) c − = { } Theorem 6.3.1 simply states that the translation off(t) a distancec in the positivet cs direction corresponds to the multiplication ofF(s) bye − . To prove Theorem 6.3.1 it is sufficient to compute the transform ofu (t)f(t c): c − ∞ st L u (t)f(t c) e− u (t)f(t c)dt { c − } = c − �0 ∞ st e− f(t c)dt. = − �c Introducing a new integration variableξ t c, we have = − L ∞ (ξ c)s cs ∞ sξ uc(t)f(t c) e− + f (ξ)dξ e − e− f (ξ)dξ { − } = 0 = 0 � cs � e − F(s). = Thus Eq. (3) is established; Eq. (4) follows by taking the inverse transform of both sides of Eq. (3). A simple example of this theorem occurs if we takef(t) 1. Recalling thatL 1 L =cs { } = 1/s, we immediately have from Eq. (3) that u c(t) e − /s. This result agrees with that of Eq. (2). Examples 2 and 3 illustrate further{ } how = Theorem 6.3.1 can be used in the calculation of transforms and inverse transforms. If the functionf is defined by EXAMPLE sint,0 t<π/4, f(t) ≤ 2 = sint cos(t π/4),t π/4, � + − ≥ findL f(t) . The graph ofy f(t) is shown in Figure 6.3.5. Note{ that} f(t) sint g(t)=, where = + 0,t<π/4, g(t) = cos(t π/4),t π/4. � − ≥ Thus g(t) u (t) cos(t π/4), = π/4 − and L f(t) L sint L u (t) cos(t π/4) { } = { }+ { π/4 − } πs/4 L sint e − L cost . = { }+ { } Introducing the transforms of sint and cost, we obtain πs/4 1 πs/4 s 1 se − L f(t) e − + . { } = s2 1 + s2 1 = s2 1 + + + You should compare this method with the calculation ofL f(t) directly from the definition. { } 6.3 Step Functions 313 y 2 1.5 1 0.5 π y = sin t + uπ/4(t)cos(t – 4 ) 0.5π 1 1.5 2 2.5 3 t 4 FIGURE 6.3.5 Graph of the function in Example 2. Find the inverse transform of EXAMPLE 2s 1 e − 3 F(s) − . = s2 From the linearity of the inverse transform we have 2s 1 1 1 1 e− f(t) L − F(s) L − L − = { } = 2 − 2 �s � � s � t u (t)(t 2). = − 2 − The functionf may also be written as t,0 t<2, f(t) ≤ = 2,t 2. � ≥ The following theorem contains another very useful property of Laplace transforms that is somewhat analogous to that given in Theorem 6.3.1. Theorem 6.3.2 IfF(s) L f(t) exists fors>a 0, and ifc is a constant, then = { } ≥ L ect f(t) F(s c),s>a c. (5) { } = − + 1 Conversely, iff(t) L − F(s) , then = { } ct 1 e f(t) L − F(s c) . (6) = { − } According to Theorem 6.3.2, multiplication off(t) bye ct results in a translation of the transformF(s) a distancec in the positives direction, and conversely. The proof of this theorem requires merely the evaluation ofL e ct f(t) . Thus { } ct ∞ st ct ∞ (s c)t L e f(t) e− e f(t)dt e− − f(t)dt { } = = �0 �0 F(s c), = − 314 Chapter 6. The Laplace Transform which is Eq. (5). The restrictions>a c follows from the observation that, according + at ct (a c)t to hypothesis (ii) of Theorem 6.1.2, f(t) Ke ; hence e f(t) Ke + . Equa- tion (6) follows by taking the inverse| transform| ≤ of Eq. (5),| and the| proof≤ is complete. The principal application of Theorem 6.3.2 is in the evaluation of certain inverse transforms, as illustrated by Example 4. Find the inverse transform of EXAMPLE 1 G(s) . 4 = s2 4s 5 − + By completing the square in the denominator we can write 1 G(s) F(s 2), = (s 2) 2 1 = − − + 2 1 1 whereF(s) (s 1) − . SinceL − F(s) sint, it follows from Theorem 6.3.2 that = + { } = 1 2t g(t) L − G(s) e sint. = { } = The results of this section are often useful in solving differential equations, partic- ularly those having discontinuous forcing functions. The next section is devoted to examples illustrating this fact. PROBLEMS In each of Problems 1 through 6 sketch the graph of the given function on the intervalt 0. ≥ 1.u 1(t) 2u 3(t) 6u 4(t) 2.(t 3)u 2(t) (t 2)u 3(t) 3.f(t π)+u (t),− wheref(t) t 2 4.f(−t 3)u (−t), −wheref(t) sint − π = − 3 = 5.f(t 1)u 2(t), wheref(t) 2t 6.f(t)− (t 1)u (t) 2(t 2)u= (t) (t 3)u (t) = − 1 − − 2 + − 3 In each of Problems 7 through 12 find the Laplace transform of the given function. 0,t<2 0,t<1 7.f(t) 8.f(t) = (t 2) 2,t 2 = t2 2t 2,t 1 � − ≥ � − + ≥ 0,t<π 9.f(t) t π, π t<2π 10.f(t) u (t) 2u (t) 6u (t) = = 1 + 3 − 4 0−,t 2π ≤ ≥ 11.f(t) (t 3)u (t) (t 2)u (t) 12.f(t) t u (t)(t 1),t 0 = − 2 − − 3 = − 1 − ≥ In each of Problems 13 through 18 find the inverse Laplace transform of the given function. 2s 3! e− 13.F(s) 14.F(s) = (s 2) 4 = s2 s 2 − + − 2s 2s 2(s 1)e − 2e− 15.F(s) − 16.F(s) = s2 2s 2 = s2 4 − + − s s 2s 3s 4s (s 2)e − e− e − e − e − 17.F(s) − 18.F(s) + − − = s2 4s 3 = s − + 6.3 Step Functions 315 19. Suppose thatF(s) L f(t) exists fors>a 0. (a) Show that ifc= is a{ positive} constant,≥ then 1 s L f(ct) F ,s>ca. { }= c c � � (b) Show that ifk is a positive constant, then 1 1 t L− F(ks) f . { }= k k � � (c) Show that ifa andb are constants witha> 0, then 1 1 bt/a t L− F(as b) e− f . { + }= a a � � In each of Problems 20 through 23 use the results of Problem 19 to find the inverse Laplace transform of the given function. n 1 2 + n! 2s 1 ( ) ( ) 20.F s n 1 21.F s 2 + = s + = 4s 4s 5 + + 2 4s 1 e e− 22.F(s) 23.F(s) = 9s2 12s 3 = 2s 1 − + − In each of Problems 24 through 27 find the Laplace transform of the given function. In Problem 27 assume that term-by-term integration of the infinite series is permissible. 1,0 t<1 1,0 t<1 0,1≤t<2 24.f(t) ≤ 25.f(t) ≤ = 0,t 1 = 1,2 t<3 � ≥ 0,t ≤3 ≥ 2n 1 + k 26.f(t) 1 u 1(t) u 2n(t) u 2n 1(t) 1 ( 1) uk (t) = − +···+ − + = + k 1 − �= ∞ k 27.f(t) 1 ( 1) uk (t). See Figure 6.3.6. = + k 1 − �= y 1 1 2 3 4 5 t FIGURE 6.3.6 A square wave. 28. Letf satisfyf(t T) f(t) for allt 0 and for some fixed positive numberT;f is said to be periodic+ = with periodT≥ on0 t< . Show that ≤ ∞ T st e− f(t)dt L ( ) 0 . f t � sT { }= 1 e − − 316 Chapter 6. The Laplace Transform In each of Problems 29 through 32, use the result of Problem 28 to find the Laplace transform of the given function. 1,0 t<1, 1,0 t<1, 29.f(t) ≤ 30.f(t) ≤ = 0,1 t<2 = 1,1 t<2 � ≤ ; � − ≤ ; f(t 2) f(t). f(t 2) f(t). + = + = Compare with Problem 27. See Figure 6.3.7. y 1 1 2 3 4 5 t –1 FIGURE 6.3.7 A square wave. 31.f(t) t,0 t<1 32.f(t) sint,0 t<π f(=t 1)≤ f(t).; f(=t π) f≤(t). ; + = + = See Figure 6.3.8. See Figure 6.3.9. y y 1 1 1 2 3 4 t π 2π 3π t FIGURE 6.3.8 A sawtooth wave. FIGURE 6.3.9 A rectified sine wave. L 33. (a) Iff(t) 1 u 1(t), find f(t) ; compare with Problem 24. Sketch the graph of y f(t). = − { } = t (b) Letg(t) f(ξ)dξ, where the functionf is defined in part (a). Sketch the graph = 0 ofy g(t) and findL g(t) . = ( ) ( ) { } ( ) ( ) (c) Leth t g t� u 1 t g t 1 , whereg is defined in part (b). Sketch the graph of y h(t) and= findL−h(t) . − 34. Consider= the function{ p} defined by t,0 t<1, p(t) ≤ p(t 2) p(t). = 2 t,1 t<2 + = � − ≤ ; (a) Sketch the graph ofy p(t). (b) FindL p(t) by noting= thatp is the periodic extension of the functionh in Prob- lem 33(c) and{ then} using the result of Problem 28. 6.4 Differential Equations with Discontinuous Forcing Functions 317 (c) FindL p(t) by noting that { } t p(t) f(t)dt, = �0 wheref is the function in Problem 30, and then using Theorem 6.2.1. 6.4 Differential Equations with Discontinuous Forcing Functions In this section we turn our attention to some examples in which the nonhomogeneous term, or forcing function, is discontinuous. Find the solution of the differential equation EXAMPLE 2y �� y � 2y g(t), (1) 1 + + = where 1,5 t< 20, g(t) u (t) u (t) ≤ (2) = 5 − 20 = 0,0 t<5 andt 20. � ≤ ≥ Assume that the initial conditions are y(0) 0,y �(0) 0. (3) = = This problem governs the charge on the capacitor in a simple electric circuit with a unit voltage pulse for 5 t< 20. Alternatively,y may represent the response of a damped oscillator subject≤ to the applied forceg(t). The Laplace transform of Eq. (1) is 2 2s Y(s) 2sy(0) 2y �(0) sY(s) y(0) 2Y(s) L u (t) L u (t) − − + − + = { 5 } − { 20 } 5s 20s (e − e − )/s. = − Introducing the initial values (3) and solving forY(s), we obtain 5s 20s e− e − Y(s) − . (4) = s(2s 2 s 2) + + To findy φ(t) it is convenient to writeY(s)as = 5s 20s Y(s) (e − e − )H(s), (5) = − where H(s) 1/s(2s 2 s 2). (6) = + + 1 Then, ifh(t) L − H(s) , we have = { } y φ(t) u (t)h(t 5) u (t)h(t 20). (7) = = 5 − − 20 − 318 Chapter 6. The Laplace Transform 5s Observe that we have used Theorem 6.3.1 to write the inverse transforms ofe − H(s) 20s ande − H(s), respectively. Finally, to determineh(t) we use the partial fraction expansion ofH(s): a bs c H(s) + . (8) = s + 2s2 s 2 + + Upon determining the coefficients we find thata 1 ,b 1, andc 1 . Thus = 2 =− =− 2 1/2 s 1 H(s) + 2 = s − 2s2 s 2 + + 1/2 1 (s 1 ) 1 + 4 + 4 , (9) = s − 2 (s 1 )2 15 � � + 4 + 16 so that, by referring to lines 9 and 10 of Table 6.2.1, we obtain 1 1 t/4 t/4 h(t) [e− cos(√15t/4) ( √15/15)e− sin(√15t/4)]. (10) = 2 − 2 + In Figure 6.4.1 the graph ofy φ(t) from Eqs. (7) and (10) shows that the solution consists of three distinct parts.= For 0 2y �� y � 2y 0 (11) + + = and the initial conditions are given by Eq. (3). Since the initial conditions impart no energy to the system, and since there is no external forcing, the system remains at rest; that is,y 0 for 0 y(5) 0,y �(5) 0. (12) = = Oncet> 5, the differential equation becomes 2y �� y � 2y 1, (13) + + = whose solution is the sum of a constant (the response to the constant forcing function) and a damped oscillation (the solution of the corresponding homogeneous equation). The plot in Figure 6.4.1 shows this behavior clearly for the interval 5 t 20. An expression for this portion of the solution can be found by solving the≤ ≤ differential equation (13) subject to the initial conditions (12). Finally, fort> 20 the differential equation becomes Eq. (11) again, and the initial conditions are obtained by evaluating the solution of Eqs. (13), (12) and its derivative att 20. These values are = y(20) 0.50162,y �(20) 0.01125. (14) =∼ =∼ The initial value problem (11), (14) contains no external forcing, so its solution is a damped oscillation abouty 0, as can be seen in Figure 6.4.1. = 6.4 Differential Equations with Discontinuous Forcing Functions 319 y 0.8 0.6 0.4 0.2 10 20 4030 t – 0.2 FIGURE 6.4.1 Solution of the initial value problem (1), (2), (3). While it may be helpful to visualize the solution shown in Figure 6.4.1 as composed of solutions of three separate initial value problems in three separate intervals, it is somewhat tedious to find the solution by solving these separate problems. Laplace transform methods provide a much more convenient and elegant approach to this problem and to others having discontinuous forcing functions. The effect of the discontinuity in the forcing function can be seen if we examine the solutionφ(t) of Example 1 more closely. According to the existence and uniqueness Theorem 3.2.1 the solutionφ and its first two derivatives are continuous except possibly at the pointst 5andt 20 whereg is discontinuous. This can also be seen at once = = from Eq. (7). One can also show by direct computation from Eq. (7) thatφ andφ � are continuous even att 5andt 20. However, if we calculateφ ��, we find that = = lim φ��(t) 0, lim φ��(t) 1/2. t 5 = t 5 = → − → + Consequently,φ ��(t) has a jump of 1/2 att 5. In a similar way one can show that = φ��(t) has a jump of 1/2 att 20. Thus the jump in the forcing termg(t) at these − = points is balanced by a corresponding jump in the highest order term 2y �� on the left side of the equation. Consider now the general second order linear equation y�� p(t)y � q(t)y g(t), (15) + + = wherep andq are continuous on some intervalα Describe the qualitative nature of the solution of the initial value problem EXAMPLE y�� 4y g(t), (16) 2 + = y(0) 0,y �(0) 0, (17) = = where 0,0 t<5, g(t) (t ≤5)/5,5 t< 10, (18) = 1,−t 10, ≤ ≥ and then find the solution. In this example the forcing function has the graph shown in Figure 6.4.2 and is known as ramp loading. It is relatively easy to identify the general form of the solution. Fort< 5 the solution is simplyy 0. On the other hand, fort> 10 the solution has the form = y c cos 2t c sin 2t 1/4. (19) = 1 + 2 + The constant 1/4 is a particular solution of the nonhomogeneous equation while the other two terms are the general solution of the corresponding homogeneous equation. Thus the solution (19) is a simple harmonic oscillation abouty 1/4. Similarly, in the intermediate range 5 y 1 y = g(t) 0.5 5 10 15 20 t FIGURE 6.4.2 Ramp loading;y g(t) from Eq. (18). = To solve the problem it is convenient to write g(t) [u (t)(t 5) u (t)(t 10)]/5, (20) = 5 − − 10 − as you may verify. Then we take the Laplace transform of the differential equation and use the initial conditions, thereby obtaining 2 5s 10s 2 (s 4)Y(s) (e − e − )/5s , + = − or 5s 10s Y(s) (e − e − )H(s)/5, (21) = − where H(s) 1/s 2(s2 4). (22) = + 6.4 Differential Equations with Discontinuous Forcing Functions 321 Thus the solution of the initial value problem (16), (17), (18) is y φ(t) [u (t)h(t 5) u (t)h(t 10)]/5, (23) = = 5 − − 10 − whereh(t) is the inverse transform ofH(s). The partial fraction expansion ofH(s) is 1/4 1/4 H(s) , (24) = s2 − s2 4 + and it then follows from lines 3 and 5 of Table 6.2.1 that h(t) 1 t 1 sin 2t. (25) = 4 − 8 The graph ofy φ(t) is shown in Figure 6.4.3. Observe that it has the qualitative form that we indicated= earlier. To find the amplitude of the eventual steady oscillation it is suf- ficient to locate one of the maximum or minimum points fort> 10. Setting the deriva- tive of the solution (23) equal to zero, we find that the first maximum is located approxi- mately at(10.642,0.2979), so the amplitude of the oscillation is approximately 0.0479. y 0.30 0.20 0.10 5 10 15 20 t FIGURE 6.4.3 Solution of the initial value problem (16), (17), (18). Note that in this example the forcing functiong is continuous butg � is discontinuous att 5 andt 10. It follows that the solutionφ and its first two derivatives are = = continuous everywhere, butφ ��� has discontinuities att 5 and att 10 that match = = the discontinuities ing � at those points. PROBLEMS In each of Problems 1 through 13 find the solution of the given initial value problem. Draw the graphs of the solution and of the forcing function; explain how they are related. 1,0 t<π/2 � 1.y �� y f(t) y(0) 0,y �(0) 1 f(t) ≤ + = ; = = ; = 0, π/2 t< � ≤ ∞ 1, π t<2π � 2.y �� 2y � 2y h(t) y(0) 0,y �(0) 1 h(t) ≤ + + = ; = = ; = 0,0 t<π andt 2π � ≤ ≥ 322 Chapter 6. The Laplace Transform � 3.y �� 4y sint u (t)sin(t 2π) y(0) 0,y �(0) 0 + = − 2π − ; = = � 4.y �� 4y sint u (t)sin(t π) y(0) 0,y �(0) 0 + = + π − ; = = 1,0 t< 10 � 5.y �� 3y � 2y f(t) y(0) 0,y �(0) 0 f(t) ≤ + + = ; = = ; = 0,t 10 � ≥ � 6.y �� 3y � 2y u (t) y(0) 0,y �(0) 1 + + = 2 ; = = � 7.y �� y u (t) y(0) 1,y �(0) 0 + = 3π ; = = 5 � 8.y �� y � y t u (t)(t π/2) y(0) 0,y �(0) 0 + + 4 = − π/2 − ; = = t/2,0 t<6 � 9.y �� y g(t) y(0) 0,y �(0) 1 g(t) ≤ + = ; = = ; = 3,t 6 � ≥ 5 sint,0 t<π � 10.y �� y � y g(t) y(0) 0,y �(0) 0 g(t) ≤ + + 4 = ; = = ; = 0,t π � ≥ � 11.y �� 4y u (t) u (t) y(0) 0,y �(0) 0 + = π − 3π ; = = � iv 12.y y u 1(t) u 2(t) y(0) 0,y �(0) 0,y ��(0) 0,y ���(0) 0 iv − = − ; = = = = � 13.y 5y �� 4y 1 u (t) y(0) 0,y �(0) 0,y ��(0) 0,y ���(0) 0 + + = − π ; = = = = 14. Find an expression involvingu (t) for a functionf that ramps up from zero att t to c = 0 the valueh att t 0 k. 15. Find an expression= involving+ u (t) for a functiong that ramps up from zero att t to the c = 0 valueh att t 0 k and then ramps back down to zero att t 0 2k. � 16. A certain= spring–mass+ system satisfies the initial value problem= + 1 u�� u� u kg(t),u(0) 0,u �(0) 0, + 4 + = = = whereg(t) u (t) u (t) andk> 0 is a parameter. = 3/2 − 5/2 (a) Sketch the graph ofg(t). Observe that it is a pulse of unit magnitude extending over one time unit. (b) Solve the initial value problem. (c) Plot the solution fork 1/2,k 1, andk 2. Describe the principal features of the solution and how they depend= on=k. = (d) Find, to two decimal places, the smallest value ofk for which the solutionu(t) reaches the value 2. (e) Supposek 2. Find the timeτ after which u(t) <0.1 for allt>τ. � 17. Modify the problem= in Example 2 in the text by| replacing| the given forcing functiong(t) by f(t) [u 5(t)(t 5) u 5 k (t)(t 5 k)]/k. = − − + − − (a) Sketch the graph off(t) and describe how it depends onk. For what value ofkisf(t) identical tog(t) in the example? (b) Solve the initial value problem y�� 4y f(t),y(0) 0,y �(0) 0. + = = = (c) The solution in part (b) depends onk, but for sufficiently larget the solution is always a simple harmonic oscillation abouty 1/4. Try to decide how the amplitude of this eventual oscillation depends onk. Then= confirm your conclusion by plotting the solution for a few different values ofk. � 18. Consider the initial value problem 1 y�� y� 4y f (t),y(0) 0,y �(0) 0, + 3 + = k = = where 1/2k,4 k t<4 k f (t) − ≤ + k = 0,0 t<4 k andt 4 k � ≤ − ≥ + and 0 (a) Sketch the graph off k (t). Observe that the area under the graph is independent ofk. Iff k (t) represents a force, this means that the product of the magnitude of the force and the time interval during which it acts does not depend onk. (b) Writef k (t) in terms of the unit step function and then solve the given initial value problem. (c) Plot the solution fork 2,k 1, andk 1 . Describe how the solution depends onk. = = = 2 Resonance and Beats. In Section 3.9 we observed that an undamped harmonic oscillator (such as a spring–mass system), with a sinusoidal forcing term experiences resonance if the frequency of the forcing term is the same as the natural frequency. If the forcing frequency is slightly different from the natural frequency, then the system exhibits a beat. In Problems 19 through 23 we explore the effect of some nonsinusoidal periodic forcing functions. � 19. Consider the initial value problem y�� y f(t),y(0) 0,y �(0) 0, + = = = where n k f(t) u 0(t) 2 ( 1) ukπ (t). = + k 1 − �= (a) Draw the graph off(t) on an interval such as 0 t 6π. (b) Find the solution of the initial value problem.≤ ≤ (c) Letn 15 and plot the graph of the solution for 0 t 60. Describe the solution and explain= why it behaves as it does. ≤ ≤ (d) Investigate how the solution changes asn increases. What happens asn ? � 20. Consider the initial value problem →∞ y�� 0.1y � y f(t),y(0) 0,y �(0) 0, + + = = = wheref(t) is the same as in Problem 19. (a) Plot the graph of the solution. Use a large enough value ofn and a long enought- interval so that the transient part of the solution has become negligible and the steady state is clearly shown. (b) Estimate the amplitude and frequency of the steady-state part of the solution. (c) Compare the results of part (b) with those from Section 3.9 for a sinusoidally forced oscillator. � 21. Consider the initial value problem y�� y g(t),y(0) 0,y �(0) 0, + = = = where n k g(t) u 0(t) ( 1) ukπ (t). = + k 1 − �= (a) Draw the graph ofg(t) on an interval such as 0 t 6π. Compare the graph with that off(t) in Problem 19(a). ≤ ≤ (b) Find the solution of the initial value problem. (c) Letn 15 and plot the graph of the solution for 0 t 60. Describe the solution and explain= why it behaves as it does. Compare it with≤ the≤ solution of Problem 19. (d) Investigate how the solution changes asn increases. What happens asn ? � 22. Consider the initial value problem →∞ y�� 0.1y � y g(t),y(0) 0,y �(0) 0, + + = = = whereg(t) is the same as in Problem 21. 324 Chapter 6. The Laplace Transform (a) Plot the graph of the solution. Use a large enough value ofn and a long enough t-interval so that the transient part of the solution has become negligible and the steady state is clearly shown. (b) Estimate the amplitude and frequency of the steady-state part of the solution. (c) Compare the results of part (b) with those from Problem 20 and from Section 3.9 for a sinusoidally forced oscillator. � 23. Consider the initial value problem y�� y h(t),y(0) 0,y �(0) 0, + = = = where n k f(t) u 0(t) 2 ( 1) u11k/4 (t). = + k 1 − �= Observe that this problem is identical to Problem 19 except that the frequency of the forcing term has been increased somewhat. (a) Find the solution of this initial value problem. (b) Letn 33 and plot the solution for 0 t 90 or longer. Your plot should show a clearly recognizable≥ beat. ≤ ≤ (c) From the graph in part (b) estimate the “slow period” and the “fast period” for this oscillator. (d) For a sinusoidally forced oscillator in Section 3.9 it was shown that the “slow fre- quency“ is given by ω ω /2, whereω is the natural frequency of the system andωis | − 0| 0 the forcing frequency. Similarly the “fast frequency” is(ω ω 0)/2. Use these expressions to calculate the “fast period” and the “slow period” for the+ oscillator in this problem. How well do the results compare with your estimates from part (c)? 6.5 Impulse Functions In some applications it is necessary to deal with phenomena of an impulsive nature, for example, voltages or forces of large magnitude that act over very short time intervals. Such problems often lead to differential equations of the form ay�� by � cy g(t), (1) + + = whereg(t) is large during a short intervalt 0 τ In particular, let us suppose thatt 0 is zero, and thatg(t) is given by 1/2τ, τ lim dτ (t) 0,t 0. (5) τ 0 = �= → Further, sinceI (τ) 1 for eachτ 0, it follows that = �= lim I (τ) 1. (6) τ 0 = → Equations (5) and (6) can be used to define an idealized unit impulse functionδ, which imparts an impulse of magnitude one att 0, but is zero for all values oft other than zero. That is, the “function”δ is defined to= have the properties δ(t) 0,t 0 (7) = �= ; ∞ δ(t)dt 1. (8) = �−∞ y 1 2τ –τ τ t FIGURE 6.5.1 Graph ofy d (t). = τ y t FIGURE 6.5.2 Graphs ofy d (t) asτ 0. = τ → 326 Chapter 6. The Laplace Transform There is no ordinary function of the kind studied in elementary calculus that satisfies both of Eqs. (7) and (8). The “function”δ, defined by those equations, is an example of what are known as generalized functions and is usually called the Dirac2 delta function. Sinceδ(t) corresponds to a unit impulse att 0, a unit impulse at an arbitrary point t t is given byδ(t t ). From Eqs. (7) and= (8) it follows that = 0 − 0 δ(t t ) 0,t t (9) − 0 = �= 0; ∞ δ(t t )dt 1. (10) − 0 = �−∞ The delta function does not satisfy the conditions of Theorem 6.1.2, but its Laplace transform can nevertheless be formally defined. Sinceδ(t) is defined as the limit of dτ (t) asτ 0, it is natural to define the Laplace transform ofδ as a similar limit of → L the transform ofd τ . In particular, we will assume thatt 0 > 0, and define δ(t t 0) by the equation { − } L L δ(t t ) lim dτ (t t ) . (11) { − 0 } = τ 0 { − 0 } → To evaluate the limit in Eq. (11) we first observe that ifτ sinhsτ st L d (t t ) e− 0 . (12) { τ − 0 } = sτ The quotient(sinhsτ)/sτ is indeterminate asτ 0, but its limit can be evaluated by L’Hospital’s rule. We obtain → sinhsτ s coshsτ lim lim 1. τ 0 τ = τ 0 = → s → s Then from Eq. (11) it follows that st L δ(t t ) e − 0 . (13) { − 0 } = 2Paul A. M. Dirac (1902–1984), English mathematical physicist, received his Ph.D. from Cambridge in 1926 and was professor of mathematics there until 1969. He was awarded the Nobel prize in 1933 (with Erwin Schrodinger)¨ for fundamental work in quantum mechanics. His most celebrated result was the relativistic equation for the electron, published in 1928. From this equation he predicted the existence of an “anti-electron,” or positron, which was first observed in 1932. Following his retirement from Cambridge, Dirac moved to the United States and held a research professorship at Florida State University. 6.5 Impulse Functions 327 L Equation (13) defines δ(t t 0) for anyt 0 > 0. We extend this result, to allowt 0 to be zero, by lettingt { 0 on− the right} side of Eq. (13); thus 0 → st L δ(t) lim e− 0 1. (14) { } = t 0 = 0→ In a similar way it is possible to define the integral of the product of the delta function and any continuous functionf . We have ∞ ∞ δ(t t )f(t)dt lim dτ (t t )f(t)dt. (15) − 0 = τ 0 − 0 �−∞ → �−∞ Using the definition (4) ofd τ (t) and the mean value theorem for integrals, we find that t τ ∞ 1 0+ dτ (t t 0)f(t)dt f(t)dt − = 2τ t τ �−∞ � 0− 1 2τ f(t ∗) f(t ∗), = 2τ · · = wheret 0 τ Find the solution of the initial value problem EXAMPLE 2y �� y � 2y δ(t 5), (17) 1 + + = − y(0) 0,y �(0) 0. (18) = = This initial value problem arises from the study of the same electrical circuit or me- chanical oscillator as in Example 1 of Section 6.4. The only difference is in the forcing term. To solve the given problem we take the Laplace transform of the differential equation and use the initial conditions, obtaining 2 5s (2s s 2)Y(s) e − . + + = Thus 5s 5s e− e− 1 Y(s) . (19) = 2s2 s 2 = 2 (s 1 )2 15 + + + 4 + 16 By Theorem 6.3.2 or from line 9 of Table 6.2.1 1 L 1 4 t/4 √15 . − 1 2 15 √ e− sin 4 t (20) �(s ) � = 15 + 4 + 16 328 Chapter 6. The Laplace Transform Hence, by Theorem 6.3.1, we have 1 2 (t 5)/4 √15 y L − Y(s) u (t)e − − sin (t 5), (21) = { } = √15 5 4 − which is the formal solution of the given problem. It is also possible to writey in the form 0, t<5, y 2 (t 5)/4 √15 (22) = e− − sin (t 5),t 5. � √15 4 − ≥ The graph of Eq. (22) is shown in Figure 6.5.3. Since the initial conditions att 0 are homogeneous, and there is no external excitation untilt 5, there is no response= in the interval 0 y 0.3 0.2 0.1 5 10 15 20 t – 0.1 FIGURE 6.5.3 Solution of the initial value problem (17), (18). PROBLEMS In each of Problems 1 through 12 find the solution of the given initial value problem and draw its graph. � 1.y �� 2y � 2y δ(t π) y(0) 1,y �(0) 0 + + = − ; = = � 2.y �� 4y δ(t π) δ(t 2π) y(0) 0,y �(0) 0 + = − − − ; = = � 3.y �� 3y � 2y δ(t 5) u (t) y(0) 0,y �(0) 1/2 + + = − + 10 ; = = � 4.y �� y 20δ(t 3) y(0) 1,y �(0) 0 − =− − ; = = � 5.y �� 2y � 3y sint δ(t 3π) y(0) 0,y �(0) 0 + + = + − ; = = � 6.y �� 4y δ(t 4π) y(0) 1/2,y �(0) 0 + = − ; = = � 7.y �� y δ(t 2π) cost y(0) 0,y �(0) 1 + = − ; = = � 8.y �� 4y 2δ(t π/4) y(0) 0,y �(0) 0 + = − ; = = 6.5 Impulse Functions 329 � 9.y �� y u (t) 3δ(t 3π/2) u (t) y(0) 0,y �(0) 0 + = π/2 + − − 2π ; = = � 10. 2y �� y � 4y δ(t π/6)sint y(0) 0,y �(0) 0 + + = − ; = = � 11.y �� 2y � 2y cost δ(t π/2) y(0) 0,y �(0) 0 iv + + = + − ; = = � 12.y y δ(t 1) y(0) 0,y �(0) 0,y ��(0) 0,y ���(0) 0 − = − ; = = = = � 13. Consider again the system in Example 1 in the text in which an oscillation is excited by a unit impulse att 5. Suppose that it is desired to bring the system to rest again after exactly one cycle,= that is, when the response first returns to equilibrium moving in the positive direction. (a) Determine the impulsekδ(t t ) that should be applied to the system in order to − 0 accomplish this objective. Note thatk is the magnitude of the impulse andt 0 is the time of its application. (b) Solve the resulting initial value problem and plot its solution to confirm that it behaves in the specified manner. � 14. Consider the initial value problem y�� γy � y δ(t 1),y(0) 0,y �(0) 0, + + = − = = whereγ is the damping coefficient (or resistance). (a) Letγ 1 . Find the solution of the initial value problem and plot its graph. = 2 (b) Find the timet 1 at which the solution attains its maximum value. Also find the maximum valuey 1 of the solution. (c) Letγ 1 and repeat parts (a) and (b). = 4 (d) Determine howt 1 andy 1 vary asγ decreases. What are the values oft 1 andy 1 when γ 0? � 15. Consider= the initial value problem y�� γy � y kδ(t 1),y(0) 0,y �(0) 0, + + = − = = wherek is the magnitude of an impulse att 1 andγ is the damping coefficient (or resistance). = (a) Letγ 1 . Find the value ofk for which the response has a peak value of 2; call this = 2 valuek 1. (b) Repeat part (a) forγ 1 . = 4 (c) Determine howk 1 varies asγ decreases. What is the value ofk 1 whenγ 0? � 16. Consider the initial value problem = y�� y f (t),y(0) 0,y �(0) 0, + = k = = wheref k (t) [u 4 k (t) u 4 k (t)]/2k with0 20 20 k 1 � 17.f(t) δ(t kπ) � 18.f(t) ( 1) + δ(t kπ) = k 1 − = k 1 − − �= �= 20 20 k 1 � 19.f(t) δ(t kπ/2) � 20.f(t) ( 1) + δ(t kπ/2) = k 1 − = k 1 − − �= �= 15 40 k 1 � 21.f(t) δ[t (2k 1)π] � 22.f(t) ( 1) + δ(t 11k/4) = k 1 − − = k 1 − − �= �= � 23. The position of a certain lightly damped oscillator satisfies the initial value problem 20 k 1 y�� 0.1y � y ( 1) + δ(t kπ),y(0) 0,y �(0) 0. + + = k 1 − − = = �= Observe that, except for the damping term, this problem is the same as Problem 18. (a) Try to predict the nature of the solution without solving the problem. (b) Test your prediction by finding the solution and drawing its graph. (c) Determine what happens after the sequence of impulses ends. � 24. Proceed as in Problem 23 for the oscillator satisfying 15 y�� 0.1y � y δ[t (2k 1)π],y(0) 0,y �(0) 0. + + = k 1 − − = = �= Observe that, except for the damping term, this problem is the same as Problem 21. 25. (a) By the method of variation of parameters show that the solution of the initial value problem y�� 2y � 2y f(t) y(0) 0,y �(0) 0 + + = ; = = is t (t τ) y e− − f(τ)sin(t τ)dτ. = − �0 (b) Show that iff(t) δ(t π), then the solution of part (a) reduces to = − (t π) y u (t)e − − sin(t π). = π − (c) Use a Laplace transform to solve the given initial value problem withf(t) δ(t π) and confirm that the solution agrees with the result of part (b). = − 6.6 The Convolution Integral Sometimes it is possible to identify a Laplace transformH(s) as the product of two other transformsF(s) andG(s), the latter transforms corresponding to known functionsf andg, respectively. In this event, we might anticipate thatH(s) would be the transform of the product off andg. However, this is not the case; in other words, the Laplace transform cannot be commuted with ordinary multiplication. On the other hand, if an appropriately defined “generalized product” is introduced, then the situation changes, as stated in the following theorem. 6.6 The Convolution Integral 331 Theorem 6.6.1 IfF(s) L f(t) andG(s) L g(t) both exist fors>a 0, then = { } = { } ≥ H(s) F(s)G(s) L h(t) ,s>a, (1) = = { } where t t h(t) f(t τ)g(τ)dτ f (τ)g(t τ)dτ. (2) = − = − �0 �0 The functionh is known as the convolution off andg; the integrals in Eq. (2) are known as convolution integrals. The equality of the two integrals in Eq. (2) follows by making the change of variable t τ ξ in the first integral. Before giving the proof of this theorem let us make some observations− = about the convolution integral. According to this theorem, the transform of the convolution of two functions, rather than the transform of their ordinary product, is given by the product of the separate transforms. It is conventional to emphasize that the convolution integral can be thought of as a “generalized product” by writing h(t) (f g)(t). (3) = ∗ In particular, the notation(f g)(t) serves to indicate the first integral appearing in Eq. (2). ∗ The convolutionf g has many of the properties of ordinary multiplication. For example, it is relatively∗ simple to show that f g g f (commutative law) (4) ∗ = ∗ f (g g ) f g f g (distributive law) (5) ∗ 1 + 2 = ∗ 1 + ∗ 2 (f g) h f (g h) (associative law) (6) ∗ ∗ = ∗ ∗ f 0 0 f 0. (7) ∗ = ∗ = The proofs of these properties are left to the reader. However, there are other properties of ordinary multiplication that the convolution integral does not have. For example, it is not true in general thatf 1 is equal tof . To see this, note that ∗ t t (f 1)(t) f(t τ) 1dτ f(t τ)dτ. ∗ = − · = − �0 �0 If, for example,f(t) cost, then = t τ t (f 1)(t) cos(t τ)dτ sin(t τ) = ∗ = 0 − =− − τ 0 � � = sin 0 sint � = − + � sint. = Clearly,(f 1)(t) f(t). Similarly, it may not be true thatf f is nonnegative. See Problem 3∗ for an�= example. ∗ Convolution integrals arise in various applications in which the behavior of the system at timet depends not only on its state at timet, but on its past history as well. Systems of this kind are sometimes called hereditary systems and occur in such diverse fields as neutron transport, viscoelasticity, and population dynamics. 332 Chapter 6. The Laplace Transform Turning now to the proof of Theorem 6.6.1, we note first that if ∞ sξ F(s) e− f (ξ)dξ = �0 and ∞ sη G(s) e− g(η)dη, = �0 then ∞ sξ ∞ sη F(s)G(s) e− f (ξ)dξ e− g(η)dη. (8) = �0 �0 Since the integrand of the first integral does not depend on the integration variable of the second, we can writeF(s)G(s) as an iterated integral, ∞ ∞ s(ξ η) F(s)G(s) g(η)dη e− + f (ξ)dξ. (9) = �0 �0 This expression can be put into a more convenient form by introducing new variables of integration. First letξ t η, for fixedη. Then the integral with respect toξ in Eq. (9) is transformed into= − one with respect tot; hence ∞ ∞ st F(s)G(s) g(η)dη e− f(t η)dt. (10) = − �0 �η Next letη τ; then Eq. (10) becomes = ∞ ∞ st F(s)G(s) g(τ)dτ e− f(t τ)dt. (11) = − �0 �τ The integral on the right side of Eq. (11) is carried out over the shaded wedge-shaped region extending to infinity in thetτ-plane shown in Figure 6.6.1. Assuming that the order of integration can be reversed, we finally obtain t ∞ st F(s)G(s) e− dt f(t τ)g(τ)dτ, (12) = − �0 �0 τ τ = t t =τ t→ ∞ τ = 0 t FIGURE 6.6.1 Region of integration inF(s)G(s). 6.6 The Convolution Integral 333 or ∞ st F(s)G(s) e− h(t)dt = �0 L h(t) , (13) = { } whereh(t) is defined by Eq. (2). This completes the proof of Theorem 6.6.1. Find the inverse transform of EXAMPLE a H(s) . (14) 1 = s2(s2 a 2) + 2 2 2 It is convenient to think ofH(s) as the product ofs − anda/(s a ), which, according to lines 3 and 5 of Table 6.2.1, are the transforms oft and sinat,+ respectively. Hence, by Theorem 6.6.1, the inverse transform ofH(s)is t at sinat h(t) (t τ)sinaτdτ − . (15) = − = a2 �0 You can verify that the same result is obtained ifh(t) is written in the alternate form t h(t) τ sina(t τ)dτ, = − �0 which confirms Eq. (2) in this case. Of course,h(t) can also be found by expanding H(s) in partial fractions. Find the solution of the initial value problem EXAMPLE y�� 4y g(t), (16) 2 + = y(0) 3,y �(0) 1. (17) = =− By taking the Laplace transform of the differential equation and using the initial conditions, we obtain s2Y(s) 3s 1 4Y(s) G(s), − + + = or 3s 1 G(s) Y(s) − . (18) = s2 4 + s2 4 + + Observe that the first and second terms on the right side of Eq. (18) contain the dependence ofY(s) on the initial conditions and forcing function, respectively. It is convenient to writeY(s) in the form s 1 2 1 2 Y(s) 3 G(s). (19) = s2 4 − 2 s2 4 + 2 s2 4 + + + Then, using lines 5 and 6 of Table 6.2.1 and Theorem 6.6.1, we obtain t y 3 cos 2t 1 sin 2t 1 sin 2(t τ)g(τ)dτ. (20) = − 2 + 2 − �0 334 Chapter 6. The Laplace Transform If a specific forcing functiong is given, then the integral in Eq. (20) can be evaluated (by numerical means, if necessary). Example 2 illustrates the power of the convolution integral as a tool for writing the solution of an initial value problem in terms of an integral. In fact, it is possible to proceed in much the same way in more general problems. Consider the problem consisting of the differential equation ay�� by � cy g(t), (21) + + = wherea,b, andc are real constants andg is a given function, together with the initial conditions y(0) y ,y �(0) y � . (22) = 0 = 0 The transform approach yields some important insights concerning the structure of the solution of any problem of this type. The initial value problem (21), (22) is often referred to as an input–output problem. The coefficientsa,b, andc describe the properties of some physical system, andg(t) is the input to the system. The valuesy 0 andy 0� describe the initial state, and the solution y is the output at timet. By taking the Laplace transform of Eq. (21) and using the initial conditions (22), we obtain 2 (as bs c)Y(s) (as b)y ay � G(s). + + − + 0 − 0 = If we let (as b)y ay � G(s) �(s) + 0 + 0 ,�(s) , (23) = as2 bs c = as2 bs c + + + + then we can write Y(s) �(s) �(s). (24) = + Consequently, y φ(t) ψ(t), (25) = + 1 1 whereφ(t) L − �(s) andψ(t) L − �(s) . Observe thaty φ(t) is a solution of the initial= value{ problem} = { } = ay�� by � cy 0,y(0) y ,y �(0) y � , (26) + + = = 0 = 0 obtained from Eqs. (21) and (22) by settingg(t) equal to zero. Similarly,y ψ(t) is the solution of = ay�� by � cy g(t),y(0) 0,y �(0) 0, (27) + + = = = in which the initial valuesy 0 andy 0� are each replaced by zero. 1 Once specific values ofa,b, andc are given, we can findφ(t) L − �(s) by using Table 6.2.1, possibly in conjunction with a translation or a partial= fraction{ expansion.} 1 To findψ(t) L − �(s) it is convenient to write�(s) as = { } �(s) H(s)G(s), (28) = 6.6 The Convolution Integral 335 2 1 3 whereH(s) (as bs c) − . The functionH is known as the transfer function and depends= only on+ the+ properties of the system under consideration; that is,H(s) is determined entirely by the coefficientsa,b, andc. On the other hand,G(s) depends only on the external excitationg(t) that is applied to the system. By the convolution theorem we can write t 1 ψ(t) L − H(s)G(s) h(t τ)g(τ)dτ, (29) = { } = − �0 1 whereh(t) L − H(s) , andg(t) is the given forcing function. To obtain= a better{ understanding} of the significance ofh(t), we consider the case in whichG(s) 1; consequently,g(t) δ(t) and�(s) H(s). This means that y h(t) is the solution= of the initial value= problem = = ay�� by � cy δ(t),y(0) 0,y �(0) 0, (30) + + = = = obtained from Eq. (27) by replacingg(t) byδ(t). Thush(t) is the response of the system to a unit impulse applied att 0, and it is natural to callh(t) the impulse = response of the system. Equation (29) then says thatψ(t) is the convolution of the impulse response and the forcing function. Referring to Example 2, we note that in that case the transfer function isH(s) 1/(s2 4), and the impulse response ish(t) (sin 2t)/2. Also, the first two terms on= the right+ side of Eq. (20) constitute the function= φ(t), the solution of the corresponding homogeneous equation that satisfies the given initial conditions. PROBLEMS 1. Establish the commutative, distributive, and associative properties of the convolution integral. (a)f g g f ∗ = ∗ (b)f (g 1 g 2) f g 1 f g 2 (c)f∗(g h) +(f g) =h ∗ + ∗ 2. Find∗ an∗ example= ∗ ∗ different from the one in the text showing that(f 1)(t) need not be equal tof(t). ∗ 3. Show, by means of the examplef(t) sint, thatf f is not necessarily nonnegative. = ∗ In each of Problems 4 through 7 find the Laplace transform of the given function. t t 2 (t τ) 4.f(t) (t τ) cos 2τdτ 5.f(t) e− − sinτdτ = − = �0 �0 t t 6.f(t) (t τ)e τ dτ 7.f(t) sin(t τ) cosτdτ = − = − �0 �0 In each of Problems 8 through 11 find the inverse Laplace transform of the given function by using the convolution theorem. 1 s 8.F(s) 9.F(s) = s4(s2 1) = (s 1)(s 2 4) + + + 1 G(s) 10.F(s) 11.F(s) = (s 1) 2(s2 4) = s2 1 + + + 3This terminology arises from the fact thatH(s) is the ratio of the transforms of the output and the input of the problem (27). 336 Chapter 6. The Laplace Transform In each of Problems 12 through 19 express the solution of the given initial value problem in terms of a convolution integral. 2 12.y �� ω y g(t) y(0) 0,y �(0) 1 + = ; = = 13.y �� 2y � 2y sinαt y(0) 0,y �(0) 0 + + = ; = = 14. 4y �� 4y � 17y g(t) y(0) 0,y �(0) 0 + +5 = ; = = 15.y �� y � y 1 u (t) y(0) 1,y �(0) 1 + + 4 = − π ; = =− 16.y �� 4y � 4y g(t) y(0) 2,y �(0) 3 + + = ; = =− 17.y �� 3y � 2y cosαt y(0) 1,y �(0) 0 iv + + = ; = = 18.y y g(t) y(0) 0,y �(0) 0,y ��(0) 0,y ���(0) 0 iv − = ; = = = = 19.y 5y �� 4y g(t) y(0) 1,y �(0) 0,y ��(0) 0,y ���(0) 0 20. Consider+ the+ equation= ; = = = = t φ(t) k(t ξ)φ(ξ)dξ f(t), + − = �0 in whichf andk are known functions, andφ is to be determined. Since the unknown functionφ appears under an integral sign, the given equation is called an integral equation; in particular, it belongs to a class of integral equations known as Volterra integral equations. Take the Laplace transform of the given integral equation and obtain an expression for L φ(t) in terms of the transformsL f(t) andL k(t) of the given functionsf andk. The{ inverse} transform ofL φ(t) is the{ solution} of{ the} original integral equation. 21. Consider the Volterra integral{ equation} (see Problem 20) t φ(t) (t ξ)φ(ξ)dξ sin 2t. + − = �0 (a) Show that ifu is a function such thatu ��(t) φ(t), then = u��(t) u(t) tu �(0) u(0) sin 2t. + − − = (b) Show that the given integral equation is equivalent to the initial value problem u��(t) u(t) sin2t u(0) 0,u �(0) 0. + = ; = = (c) Solve the given integral equation by using the Laplace transform. (d) Solve the initial value problem of part (b) and verify that the solution is the same as that obtained in part (c). 22. The Tautochrone. A problem of interest in the history of mathematics is that of finding the tautochrone—the curve down which a particle will slide freely under gravity alone, reaching the bottom in the same time regardless of its starting point on the curve. This problem arose in the construction of a clock pendulum whose period is independent of the amplitude of its motion. The tautochrone was found by Christian Huygens (1629– 1695) in 1673 by geometrical methods, and later by Leibniz and Jakob Bernoulli using analytical arguments. Bernoulli’s solution (in 1690) was one of the first occasions on which a differential equation was explicitly solved. The geometrical configuration is shown in Figure 6.6.2. The starting pointP(a,b) is joined to the terminal point(0,0) by the arcC. Arc lengths is measured from the origin, andf(y) denotes the rate of change ofs with respect toy: 1/2 ds dx 2 f(y) 1 . (i) = dy = + dy � � � � Then it follows from the principle of conservation of energy that the timeT(b) required for a particle to slide fromP to the origin is 1 b f(y) T(b) dy. (ii) = √2g √b y �0 − 6.6 The Convolution Integral 337 y P(a, b) C s x FIGURE 6.6.2 The tautochrone. (a) Assume thatT(b) T 0, a constant, for eachb. By taking the Laplace transform of Eq. (ii) in this case and= using the convolution theorem, show that 2g T F(s) 0 (iii) = � π √s ; then show that 2g T f(y) 0 . (iv) = � π √y Hint: See Problem 27 of Section 6.1. (b) Combining Eqs. (i) and (iv), show that dx 2α y − , (v) dy = � y whereα gT 2/π 2. = 0 (c) Use the substitutiony 2αsin 2(θ/2) to solve Eq. (v), and show that = x α(θ sinθ),y α(1 cosθ). (vi) = + = − Equations (vi) can be identified as parametric equations of a cycloid. Thus the tautochrone is an arc of a cycloid. REFERENCES The books listed below contain additional information on the Laplace transform and its applications: Churchill, R. V., Operational Mathematics (3rd ed.) (New York: McGraw-Hill, 1971). Doetsch, G., Nader, W. (tr.), Introduction to the Theory and Application of the Laplace Transform (New York: Springer-Verlag, 1974). Kaplan, W., Operational Methods for Linear Systems (Reading, MA: Addison-Wesley, 1962). Kuhfittig, P. K. F., Introduction to the Laplace Transform (New York: Plenum, 1978). Miles, J. W., Integral Transforms in Applied Mathematics (London: Cambridge University Press, 1971). Rainville, E. D., The Laplace Transform: An Introduction (New York: Macmillan, 1963). Each of the books just mentioned contains a table of transforms. Extensive tables are also available; see, for example: Erdelyi, A. (ed.), Tables of Integral Transforms (Vol. 1) (New York: McGraw-Hill, 1954). 338 Chapter 6. The Laplace Transform Roberts, G. E., and Kaufman, H., Table of Laplace Transforms (Philadelphia: Saunders, 1966). A further discussion of generalized functions can be found in: Lighthill, M. J., Fourier Analysis and Generalized Functions (London: Cambridge University Press, 1958).