CHAPTER 8 Laplace Transforms

CHAPTER Laplace Transforms IN THIS CHAPTER we study the method of Laplace transforms, which illustrates one of the basic problem solving techniques in mathematics: transform a difficult problem into an easier one, solve the latter, and then use its solution to obtain a solution of the original problem. The method discussed here transforms an initial value problem for a constant coefficient equation into an algebraic equation whose solution can then be used to solve the initial value problem. In some cases this method is merely an alternative procedure for solving problems that can be solved equally well by methods that we considered previously; however, in other cases the method of Laplace transforms is more efficient than the methods previously discussed. This is especially true in physical problems dealing with discontinuous forcing functions. SECTION. defines the Laplace transform and developes its properties. SECTION. deals with the problem of finding a function that has a given Laplace transform. SECTION.3 applies the Laplace transform to solve initial value problems for constant coefficient second order differential equations on.; /. SECTION.4 introduces the unit step function. SECTION.5 uses the unit step function to solve constant coefficient equations with piecewise continuous forcing functions. SECTION.6 deals with the convolution theorem, an important theoretical property of the Laplace transform. SECTION.7 introduces the idea of impulsive force, and treats constant coefficient equations with impulsive forcing functions. SECTION. is a brief table of Laplace transforms. 393

394 Chapter Laplace Transforms. INTRODUCTION TO THE LAPLACE TRANSFORM Definition of the Laplace Transform To define the Laplace transform, we first recall the definition of an improper integral. If g is integrable over the interval Œa; T for every T>a, then the improper integral of g over Œa; / is defined as Z Z T g.t/ dt D lim g.t/ dt: (..) a T! a We say that the improper integral converges if the limit in (..) exists; otherwise, we say that the improper integral diverges or does not exist. Here s the definition of the Laplace transform of a function f. Definition.. Let f be defined for t and let s be a real number: Then the Laplace transform of f is the function F defined by Z F.s/ D e st f.t/dt; (..) for those values of s for which the improper integral converges: It is important to keep in mind that the variable of integration in (..) ist, while s is a parameter independent of t. We use t as the independent variable for f because in applications the Laplace transform is usually applied to functions of time. The Laplace transform can be viewed as an operator L thattransforms the functionf D f.t/intothe function F D F.s/. Thus, (..) can be expressed as F D L.f /: The functions f and F form a transform pair, which we ll sometimes denote by f.t/ $ F.s/: It can be shown that if F.s/ is defined for s D s then it s defined for all s>s (Exercise 4(b)). Computation of Some Simple Laplace Transforms Example.. Find the Laplace transform of f.t/ D. Solution From (..) withf.t/ D, If s then Therefore F.s/ D Z T Z Z T e st dt D lim e st dt: T! e st dt D s e stˇˇˇt D e st s Z ( T lim e st dt D s ; s > ; T! ; s < : : (..3) (..4)

Section. Introduction to the Laplace Transform 395 If s D the integrand reduces to the constant, and Z T lim T! Therefore F./ is undefined, and dt D lim T! F.s/ D Z This result can be written in operator notation as or as the transform pair Z T dt D lim T! T D: e st dt D s ; s > : L./ D s ; s > ; $ s ; s > : REMARK: It is convenient to combine the steps of integrating from to T and letting T!. Therefore, instead of writing (..3) and(..4) as separate steps we write Z e st dt D ( e stˇˇˇ s D s ; s > ; ; s < : We ll follow this practice throughout this chapter. Example.. Find the Laplace transform of f.t/ D t. Solution From (..) withf.t/ D t, F.s/ D Z If s, integrating by parts yields Z e st tdt D te st s ˇ C s ( D s ; s > ; ; s<: If s D, the integral in (..5) becomes Therefore F./ is undefined and This result can also be written as or as the transform pair Z e st tdt: (..5) Z tdtd t ˇ t e st dt D s C e stˇˇˇˇ s D: F.s/ D s ; s > : L.t/ D s ; s > ; t $ s ; s > :

396 Chapter Laplace Transforms Example..3 Find the Laplace transform of f.t/ D e at,wherea is a constant. Solution From (..) with f.t/ D e at, Combining the exponentials yields F.s/ D F.s/ D However, we know from Example.. that Replacing s by s a here shows that This can also be written as Z Z Z e st e at dt: e.s a/t dt: e st dt D s ; s > : F.s/ D s a ; s > a: L.e at / D s a ; s > a; or eat $ s a ; s > a: Example..4 Find the Laplace transforms of f.t/ D sin!t and g.t/ D cos!t,where! is a constant. Solution Define and F.s/ D G.s/ D If s>, integrating (..6) by parts yields so F.s/ D e st s If s>, integrating (..7) by parts yields so G.s/ D e st cos!t s Now substitute from (..) into this to obtain Z Z ˇ sin!tˇ C! s e st sin!t dt (..6) e st cos!t dt: (..7) Z e st cos!t dt; F.s/ D! G.s/: (..) s ˇ ˇ! s Z G.s/ D s! s F.s/: e st sin!t dt; G.s/ D s! s G.s/:

Section. Introduction to the Laplace Transform 397 Solving this for G.s/ yields This and (..) imply that G.s/ D F.s/ D s s C! ; s > :! s C! ; s > : Tables of Laplace transforms Extensive tables of Laplace transforms have been compiled and are commonly used in applications. The brief table of Laplace transforms in the Appendix will be adequate for our purposes. Example..5 Use the table of Laplace transforms to find L.t 3 e 4t /. Solution The table includes the transform pair Setting n D 3 and a D 4 here yields L.t 3 e 4t / D t n e at $ nš.s a/ nc : 3Š.s 4/ D 6 4.s 4/ : 4 We ll sometimes write Laplace transforms of specific functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms. Linearity of the Laplace Transform The next theorem presents an important property of the Laplace transform. Theorem.. ŒLinearity Property Suppose L.f i / is defined for s > s i ; i n/: Let s be the largest of the numbers s, s ;...,s n ; and let c, c,..., c n be constants: Then L.c f C c f CCc n f n / D c L.f / C c L.f / CCc n L.f n / for s>s : Proof We give the proof for the case where n D. Ifs>s then L.c f C c f / D Z e st.c f.t/ C c f.t/// dt Z Z D c e st f.t/ dt C c e st f.t/ dt D c L.f / C c L.f /: Example..6 Use Theorem.. and the known Laplace transform to find L.cosh bt/.b /. L.e at / D s a

39 Chapter Laplace Transforms Solution By definition, Therefore cosh bt D ebt C e bt : L.cosh bt/ D L ebt C e bt D L.ebt / C L.e bt / (linearity property) D s b C s C b ; (..9) where the first transform on the right is defined for s>band the second for s> b; hence, both are defined for s>jbj. Simplifying the last expression in (..9) yields L.cosh bt/ D s s b ; s > jbj: The First Shifting Theorem The next theorem enables us to start with known transform pairs and derive others. (For other results of this kind, see Exercises 6 and 3.) Theorem..3 ŒFirst Shifting Theorem If F.s/ D Z e st f.t/dt (..) is the Laplace transform of f.t/ for s > s,thenf.s a/ is the Laplace transform of e at f.t/ for s>s C a. PROOF. Replacing s by s a in (..) yields F.s a/ D Z if s a>s ;thatis,ifs>s C a. However,(..) can be rewritten as e.s a/t f.t/dt (..) which implies the conclusion. F.s a/ D Z e st e at f.t/ dt; Example..7 Use Theorem..3 and the known Laplace transforms of, t, cos!t, andsin!t to find L.e at /; L.te at /; L.e t sin!t/;and L.e t cos!t/: Solution In the following table the known transform pairs are listed on the left and the required transform pairs listed on the right are obtained by applying Theorem..3.

Section. Introduction to the Laplace Transform 399 f.t/$ F.s/ e at f.t/ $ F.s a/ $ s ; s > eat $.s a/ ; s > a t $ s ; s > teat $.s a/ ; s > a sin!t $! s C! ; s > et sin!t $ cos!t $ s s C! ; s > et sin!t $!.s / C! ;s> s.s / C! ;s> Existence of Laplace Transforms Not every function has a Laplace transform. For example, it can be shown (Exercise 3) that Z e st e t dt D for every real number s. Hence, the function f.t/ D e t does not have a Laplace transform. Our next objective is to establish conditions that ensure the existence of the Laplace transform of a function. We first review some relevant definitions from calculus. Recall that a limit lim f.t/ t!t exists if and only if the one-sided limits both exist and are equal; in this case, lim f.t/ and lim f.t/ t!t t!t C lim t!t f.t/d lim f.t/d t!t lim f.t/: t!t C Recall also that f is continuous at a point t in an open interval.a; b/ if and only if lim f.t/ D f.t /; t!t which is equivalent to For simplicity, we define lim f.t/ D lim f.t/d f.t /: (..) t!t C t!t f.t C/ D lim f.t/ and f.t / D lim f.t/; t!t C t!t so (..) can be expressed as f.t C/ D f.t / D f.t /: If f.t C/ and f.t / have finite but distinct values, we say that f has a jump discontinuity at t,and f.t C/ f.t /

4 Chapter Laplace Transforms y f (t +) f (t ) t x Figure.. A jump discontinuity is called the jump in f at t (Figure..). If f.t C/ and f.t / are finite and equal, but either f isn t defined at t or it s defined but f.t / f.t C/ D f.t /; we say that f has a removable discontinuity at t (Figure..). This terminolgy is appropriate since a function f with a removable discontinuity at t can be made continuous at t by defining (or redefining) f.t / D f.t C/ D f.t /: REMARK: We know from calculus that a definite integral isn t affected by changing the values of its integrand at isolated points. Therefore, redefining a function f to make it continuous at removable discontinuities does not change L.f /. Definition..4 (i) A function f is said to be piecewise continuous on a finite closed interval Œ; T if f.c/ and f.t / are finite and f is continuous on the open interval.; T / except possibly at finitely many points, where f may have jump discontinuities or removable discontinuities. (ii) A function f is said to be piecewise continuous on the infinite interval Œ; / if it s piecewise continuous on Œ; T for every T>. Figure..3 shows the graph of a typical piecewise continuous function. It is shown in calculus that if a function is piecewise continuous on a finite closed interval then it s integrable on that interval. But if f is piecewise continuous on Œ; /, thensoise st f.t/, and therefore Z T e st f.t/dt

Section. Introduction to the Laplace Transform 4 y f(t ) f(t ) = f(t +) t x a b x Figure.. Figure..3 A piecewise continuous function on Œa; b exists for every T>. However, piecewise continuity alone does not guarantee that the improper integral Z Z T e st f.t/dt D lim e st f.t/dt (..3) T! converges for s in some interval.s ; /. For example, we noted earlier that (..3) diverges for all s if f.t/ D e t. Stated informally, this occurs because e t increases too rapidly as t!. The next definition provides a constraint on the growth of a function that guarantees convergence of its Laplace transform for s in some interval.s ; /. Definition..5 A function f is said to be of exponential order s if there are constants M and t such that jf.t/jme st ; t t : (..4) In situations where the specific value of s is irrelevant we say simply that f is of exponential order. The next theorem givesuseful sufficient conditionsfora functionf to have a Laplace transform. The proof is sketched in Exercise. Theorem..6 If f is piecewise continuous on Œ; / and of exponential order s ; then L.f / is defined for s>s. REMARK: We emphasize that the conditions of Theorem..6 are sufficient, but not necessary, forf to have a Laplace transform. For example, Exercise 4(c) shows that f may have a Laplace transform even though f isn t of exponential order. Example.. If f is bounded on some interval Œt ; /, say jf.t/jm; t t ; then (..4) holds with s D, sof is of exponential order zero. Thus, for example, sin!t and cos!t are of exponential order zero, and Theorem..6 implies that L.sin!t/ and L.cos!t/ exist for s>. This is consistent with the conclusion of Example..4.

4 Chapter Laplace Transforms Example..9 It can be shown that if lim t! e s t f.t/ exists and is finite then f is of exponential order s (Exercise 9). If is any real number and s >then f.t/ D t is of exponential order s,since lim t! e s t t D ; by L Hôpital s rule. If, f is also continuous on Œ; /. Therefore Exercise 9 and Theorem..6 imply that L.t / exists for s s. However, since s is an arbitrary positive number, this really implies that L.t / exists for all s>. This is consistent with the results of Example.. and Exercises 6 and. Example.. Find the Laplace transform of the piecewise continuous function ; t<; f.t/ D 3e t ; t : Solution Since f is defined by different formulas onœ; / and Œ; /, we write Since F.s/ D Z e st f.t/dt D Z Z e st./ dt C Z < e s e st ; s ; dt D s : ; s D ; e st. 3e t /dt: and Z e st. 3e t /dt D 3 it follows that ˆ< e s F.s/ D s ˆ: This is consistent with Theorem..6, since Z e.sc/t dt D 3e.sC/ s C ; s > ; 3 e.sc/ ; s > ; s ; s C 3 ; s D : e jf.t/j3e t ; t ; and therefore f is of exponential order s D. REMARK: In Section.4 we ll develop a more efficient method for finding Laplace transforms of piecewise continuous functions. Example.. We stated earlier that Z e st e t dt D for all s, so Theorem..6 implies that f.t/d e t is not of exponential order, since so lim t! e t Me s t D lim t! M et s t D; e t >Me s t for sufficiently large values of t, for any choice of M and s (Exercise 3).

Section. Introduction to the Laplace Transform 43. Exercises. Find the Laplace transforms of the followingfunctions by evaluating the integral F.s/ D R e st f.t/dt. (a) t (b) te t (c) sinh bt (d) e t 3e t (e) t. Use the table of Laplace transforms to find the Laplace transforms of the following functions. (a) cosh t sin t (b) sin t (c) cos t (d) cosh t (e) t sinh t (f) sin t cos t (g) sin t C (h) cos t cos 3t (i) sin t C cos 4t 4 3. Show that Z e st e t dt D for every real number s. 4. Graph the following piecewise continuous functions and evaluate f.tc/, f.t /, andf.t/at each point of discontinuity. < t; t<; < t C ; t<; (a) f.t/ D t 4; t<3; (b) f.t/ D 4; t D ; : : ; t 3: t; t > : < (c) f.t/d : sin t; t<=; sin t; = t<; cos t; t : 5. Find the Laplace transform: e (a) f.t/ D t ; t<; e t ; t : t; t<; (c) f.t/d ; t : ˆ< (d) f.t/ D ˆ: t; t<; ; t D ; t; t<; 3; t D ; 6; t > : ; t<4; (b) f.t/ D t; t 4: te (d) f.t/ D t ; t<; e t ; t : 6. Prove that if f.t/ $ F.s/ then t k f.t/ $. / k F.k/.s/. HINT: Assume that it s permissible to differentiate the integral R e st f.t/dt with respect to s under the integral sign. 7. Use the known Laplace transforms L.e t sin!t/ D! and L.e t s cos!t/ D.s / C!.s / C! and the result of Exercise 6 to find L.te t cos!t/ and L.te t sin!t/.. Use the known Laplace transform L./ D =s and the result of Exercise 6 to show that L.t n / D nš s nc ; n D integer: 9. (a) Show that if lim t! e st f.t/exists and is finite then f is of exponential order s. (b) Show that if f is of exponential order s then lim t! e st f.t/d for all s>s.

44 Chapter Laplace Transforms (c) Show that if f is of exponential order s and g.t/ D f.t C / where >,theng is also of exponential order s.. Recall the next theorem from calculus. THEOREM A. Let g be integrable on Œ; T for every T>:Suppose there s a function w defined on some interval Œ; / (with ) such that jg.t/j w.t/ for t and R w.t/dt converges. Then R g.t/ dt converges. Use Theorem A to show that if f is piecewise continuous on Œ; / and of exponential order s, then f has a Laplace transform F.s/ defined for s>s.. Prove: If f is piecewise continuous and of exponential order then lim s! F.s/ D.. Prove: If f is continuous on Œ; / and of exponential order s >,then L f./d D s L.f /; s > s : HINT: Use integration by parts to evaluate the transform on the left. 3. Suppose f is piecewise continuous and of exponential order, and that lim t!c f.t/=t exists. Show that Z f.t/ L D F.r/dr: t s HINT: Use the results of Exercises 6 and. 4. Suppose f is piecewise continuous on Œ; /. (a) Prove: If the integral g.t/ D R t e s f./d satisfies the inequality jg.t/j M.t/, then f has a Laplace transform F.s/ defined for s>s.hint: Use integration by parts to show that Z T Z T e st f.t/dt D e.s s/t g.t / C.s s / e.s s/t g.t/ dt: (b) Show that if L.f / exists for s D s then it exists for s>s. Show that the function f.t/d te t cos.e t / has a Laplace transform defined for s>, even though f isn t of exponential order. (c) Show that the function f.t/d te t cos.e t / has a Laplace transform defined for s>, even though f isn t of exponential order. 5. Use the table of Laplace transforms and the result of Exercise 3 to find the Laplace transforms of the following functions. sin!t cos!t (a).! > / (b).! > / (c) eat e bt t t t cosh t (d) (e) sinh t t t 6. The gamma function is defined by. / D which can be shown to converge if >. Z x e x dx;

Section. The Inverse Laplace Transform 45 (a) Use integration by parts to show that. C / D. /; > : (b) Show that.n C / D nš if n D,, 3,... (c) From (b) and the table of Laplace transforms,. C / L.t / D ; s > ; s C if is a nonnegative integer. Show that this formula is valid for any >. HINT: Change the variable of integration in the integral for. C /. 7. Suppose f is continuous on Œ; T and f.tc T/D f.t/for all t.(wesayinthiscasethatf is periodic with period T.) (a) Conclude from Theorem..6 that the Laplace transform of f is defined for s>.hint: Since f is continuous on Œ; T and periodic with period T, it s bounded on Œ; /. (b) (b) Show that HINT: Write Then show that F.s/ D Z.nC/T nt e st F.s/ D X nd Z T Z.nC/T nt e st f.t/dt; s >: e st f.t/dt: Z T e st f.t/dt D e nst e st f.t/dt; and recall the formula for the sum of a geometric series.. Use the formula given in Exercise 7(b) to find the Laplace transforms of the given periodic functions: t; t<; (a) f.t/d f.t C / D f.t/; t t; t<; ; t< (b) f.t/d ; ; t<; f.t C / D f.t/; t (c) f.t/djsin tj sin t; t<; (d) f.t/d f.t C / D f.t/ ; t<;. THE INVERSE LAPLACE TRANSFORM Definition of the Inverse Laplace Transform In Section. we defined the Laplace transform of f by F.s/ D L.f / D Z e st f.t/dt: We ll also say that f is an inverse Laplace Transform of F, and write f D L.F /:

46 Chapter Laplace Transforms To solve differential equations with the Laplace transform, we must be able to obtain f from its transform F. There s a formula for doing this, but we can t use it because it requires the theory of functions of a complex variable. Fortunately, we can use the table of Laplace transforms to find inverse transforms that we ll need. Example.. Use the table of Laplace transforms to find (a) L s SOLUTION(a) Setting b D in the transform pair shows that SOLUTION(b) Setting! D 3 in the transform pair and (b) L s s C 9 sinh bt $ b s b L D sinh t: s cos!t $ s s C! shows that s L D cos 3t: s C 9 The next theorem enables us to find inverse transforms of linear combinations of transforms in the table. We omit the proof. Theorem.. ŒLinearity Property If F ;F ;...;F n are Laplace transforms and c ;c ;..., c n are constants; then L.c F C c F CCc n F n / D c L.F / C c L.F / CCc n L F n : : Example.. Find L s C 5 C 7 : s C 3 Solution From the table of Laplace transforms in Section. e at $ s a Theorem.. with a D 5and! D p 3 yields L s C 5 C 7 s C 3 and sin!t $! s C! : D L s C 5 C 7L s C 3 D L C 7 p! 3 p L s C 5 3 s C 3 D e 5t C 7 p 3 sin p 3t:

Section. The Inverse Laplace Transform 47 Example..3 Find 3s C L : s C s C 5 Solution Completing the square in the denominator yields 3s C s C s C 5 D 3s C.s C / C 4 : Because of the form of the denominator, we consider the transform pairs and write e t cos t $ s C.s C / C 4 L 3s C.s C / C 4 D D and e t sin t $.s C / C 4 ; L 3s C 3 C L 5.s C / C 4.s C / C 4 3L s C C 5.s C / C 4 L.s C / C 4 D e t.3 cos t C 5 sin t/: REMARK: We ll often write inverse Laplace transforms of specific functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms in Section.. Inverse Laplace Transforms of Rational Functions Using the Laplace transform to solve differential equations often requires finding the inverse transform of a rational function F.s/ D P.s/ Q.s/ ; where P and Q are polynomials in s with no common factors. Since it can be shown that lim s! F.s/ D if F is a Laplace transform, we need only consider the case where degree.p / < degree.q/. To obtain L.F /, we find the partial fraction expansion of F, obtain inverse transforms of the individual terms in the expansion from the table of Laplace transforms, and use the linearity property of the inverse transform. The next two examples illustrate this. Example..4 Find the inverse Laplace transform of Solution (METHOD ) Factoring the denominator in (..) yields F.s/ D 3s C s 3s C : (..) F.s/ D The form for the partial fraction expansion is 3s C.s /.s / : (..) 3s C.s /.s / D A s C B s : (..3)

4 Chapter Laplace Transforms Multiplying this by.s /.s / yields 3s C D.s /A C.s /B: Setting s D yields B D and setting s D yields A D 5. Therefore and F.s/ D 5 s C s L.F / D 5L C L D 5e t C e t : s s Solution (METHOD ) We don t really have to multiply (..3) by.s /.s / to compute A and B. We can obtain A by simply ignoring the factor s in the denominator of (..) and setting s D elsewhere; thus, A D 3s C ˇ D 3 C D 5: (..4) s ˇsD Similarly, we can obtain B by ignoring the factor s in the denominator of (..) and setting s D elsewhere; thus, B D 3s C ˇ D 3 C D : (..5) s ˇsD To justify this, we observe that multiplying (..3)bys yields 3s C B D A C.s / s s ; and setting s D leads to (..4). Similarly, multiplying (..3)bys yields 3s C A D.s / s s C B and setting s D leads to (..5). (It isn t necesary to write the last two equations. We wrote them only to justify the shortcut procedure indicated in (..4)and(..5).) The shortcut employed in the second solution of Example..4 is Heaviside s method. The next theorem states this method formally. For a proof and an extension of this theorem, see Exercise. Theorem.. Suppose P.s/ F.s/ D.s s /.s s /.s s n / ; (..6) where s, s ;...;s n are distinct and P is a polynomial of degree less than n: Then F.s/ D A C A CC A n ; s s s s s s n where A i can be computed from (..6) by ignoring the factor s s i and setting s D s i elsewhere. Example..5 Find the inverse Laplace transform of F.s/ D 6 C.s C /.s 5s C / : (..7) s.s /.s /.s C /

Section. The Inverse Laplace Transform 49 Solution The partial fraction expansion of (..7) isoftheform F.s/ D A s C B s C C s C D s C : (..) To find A, we ignore the factor s in the denominator of (..7)andsets D elsewhere. This yields A D 6 C././. /. /./ D 7 : Similarly, the other coefficients are given by and Therefore and L.F / D 7 L B D 6 C./.7/./. /./ D ; D D F.s/ D 7 C D 6 C 3.5/./.3/ D 7 ; 6. /. /. 3/ D : s s C 7 s s C L C 7 s s L L s s C D 7 et C 7 et e t : REMARK: We didn t multiply out the numerator in (..7) before computing the coefficients in (..), since it wouldn t simplify the computations. Example..6 Find the inverse Laplace transform of F.s/ D.s C /.4s C /.s C /.s C / : (..9) Solution The form for the partial fraction expansion is F.s/ D A s C C B s C C C.s C / : (..) Because of the repeated factor.s C / in (..9), Heaviside s method doesn t work. Instead, we find a common denominator in (..). This yields F.s/ D A.s C / C B.s C /.s C / C C.s C / : (..).s C /.s C / If (..9)and(..) are to be equivalent, then A.s C / C B.s C /.s C / C C.s C / D.s C /.4s C /: (..)

4 Chapter Laplace Transforms The two sides of this equation are polynomials of degree two. From a theorem of algebra, they will be equal for all s if they are equal for any three distinct values of s. We may determine A, B and C by choosing convenient values of s. The left side of (..) suggests that we take s D toobtain C D,ands D toobtain A D. We can now choose any third value of s to determine B. Takings D yields 4A C B C C D. Since A D and C D this implies that B D 6. Therefore F.s/ D s C 6 s C.s C / and L.F / D L 6L L s C s C.s C / D e t 6e t te t : Example..7 Find the inverse Laplace transform of F.s/ D s 5s C 7.s C / 3 : Solution The form for the partial fraction expansion is F.s/ D A s C C B.s C / C C.s C / : 3 The easiest way to obtain A, B, and C is to expand the numerator in powers of s C. This yields Therefore s 5s C 7 D Œ.s C / 5Œ.s C / C 7 D.s C / 9.s C / C : F.s/ D.s C / 9.s C / C.s C / 3 D s C 9.s C / C.s C / 3 and L.F / D L 9L s C.s C / C L.s C / 3 D e t 9t C t : Example.. Find the inverse Laplace transform of F.s/ D s.5 C 3s/ sœ.sc / C : (..3)

Section. The Inverse Laplace Transform 4 Solution One form for the partial fraction expansion of F is F.s/ D A s C Bs C C.s C / C : (..4) However, we see from the table of Laplace transforms that the inverse transform of the second fraction on the right of (..4) will be a linear combination of the inverse transforms e t cos t and e t sin t of s C and.s C / C respectively. Therefore, instead of (..4) we write.s C / C F.s/ D A s C B.s C / C C.s C / C : (..5) Finding a common denominator yields If (..3)and(..6) are to be equivalent, then F.s/ D A.s C / C C B.s C /s C Cs : (..6) sœ.sc/ C A.s C / C C B.s C /s C Cs D s.5 C 3s/: This is true for all s if it s true for three distinct values of s. Choosing s D,,and yields the system A D A C D 3 5A C B C C D 7: Solving this system yields A D ; B D 7 ; C D 5 : Hence, from (..5), Therefore F.s/ D s 7 s C.s C / C 5.s C / C : L.F / D L 7 s C s L 5.s C / C L.s C / C D 7 e t cos t 5 e t sin t: Example..9 Find the inverse Laplace transform of F.s/ D C 3s.s C /.s C 4/ : (..7)

4 Chapter Laplace Transforms Solution The form for the partial fraction expansion is F.s/ D A C Bs s C C C C Ds s C 4 : The coefficients A, B, C and D can be obtained by finding a common denominator and equating the resulting numerator to the numerator in (..7). However, since there s no first power of s in the denominator of (..7), there s an easier way: the expansion of F.s/ D.s C /.s C 4/ can be obtained quickly by using Heaviside s method to expand and then setting x D s to obtain Multiplying this by C 3s yields F.s/ D.x C /.x C 4/ D 3.s C /.s C 4/ D 3 C 3s.s C /.s C 4/ D 3 x C x C 4 s C : s C 4 C 3s s C C 3s : s C 4 Therefore L.F / D 3 sin t C cos t 4 sin t cos t: 3 USING TECHNOLOGY Some software packages that do symbolic algebra can find partial fraction expansions very easily. We recommend that you use such a package if one is available to you, but only after you ve done enough partial fraction expansions on your own to master the technique.. Exercises. Use the table of Laplace transforms to find the inverse Laplace transform. 3 s 4 (a) (b) (c).s 7/ 4 s 4s C 3 s C 4s C (d) (g) s C 9 (e) s 4 (h).s 4s C 5/.s 3/ 9 s (f).s C /.s / 4 (i) s 4s C 3.s 4s C 5/. Use Theorem.. and the table of Laplace transforms to find the inverse Laplace transform.

(a) s C 3.s 7/ 4 (b) (d) s C s C 9 (e) s (c).s / 6 s s C s C Section. The Inverse Laplace Transform 43 s C 5 s C 6s C (f) s C s 9 (g) s3 C s s 3 s C 3 (h) (i).s C / 4.s / C 4 s s s C (j) 3s C 4 3 (k) s s C 4s C 3 s C 6 (l) s C 9.s C / s C 4 3. Use Heaviside s method to find the inverse Laplace transform. 3.s C /.s / 7 C.s C 4/. 3s/ (a) (b).s C /.s C /.s /.s 3/.s /.s C 4/ (c) C.s /.3 s/.s /.s C /.s 3/ 3 C.s /. s s / (e).s /.s C /.s /.s C 3/ 4. Find the inverse Laplace transform. (a) (c) C 3s.s C /.s C /.s C / 3s C.s /.s C s C 5/ (d) (f) (b) (d) 3.s /.s C /.s C 4/.s /.s / 3 C.s 3/.s C s /.s 3/.s /.s C 4/.s / 3s C s C.s C /.s C s C / 3s C s C.s /.s C /.s C 3/ s C s C 3 3s C (e) (f).s /.s C /.s C /.s / 5. Use the method of Example..9 to find the inverse Laplace transform. 3s C 4s C 5s C 3 (a) (b) (c).s C 4/.s C 9/.s C /.s C 6/.s C /.s C 4/ s C 7s 34 s (d) (e) (f).4s C /.s C /.s C 6/.6s C /.4s C /.9s C / 6. Find the inverse Laplace transform. 7s 5 s C 56 (a) (b).s s C 5/.s C s C /.s 6s C 3/.s C s C 5/ s C 9 3s (c) (d).s C 4s C 5/.s 4s C 3/.s 4s C 5/.s 6s C 3/ 3s s C 4 (e) (f).s s C /.s C s C 5/.4s 4s C 5/.4s C 4s C 5/ 7. Find the inverse Laplace transform. (a) (b) s.s C /.s /.s s C 7/ 3s C 34 7s (c) (d).s /.s C s C /.s /.s s C 5/ s C s (e) (f).s 3/.s C s C 5/.s /.s C s C /. Find the inverse Laplace transform.

44 Chapter Laplace Transforms s C s C (a) (b).s C /.s /.s 3/.s C s C /.s / s s 6 (c) (d).s s C /.s C /.s /.s /.s C 4/ s 3 5s 5 (e) (f) s.s /.s s C 5/.s 4s C 3/.s /.s / 9. Given that f.t/ $ F.s/, find the inverse Laplace transform of F.as b/, wherea>.. (a) If s, s,..., s n are distinct and P is a polynomial of degree less than n, then P.s/.s s /.s s /.s s n / D A C A CC A n : s s s s s s n Multiply through by s s i to show that A i can be obtained by ignoring the factor s s i on the left and setting s D s i elsewhere. (b) Suppose P and Q are polynomials such that degree.p / degree.q / and Q.s /. Show that the coefficient of =.s s / in the partial fraction expansion of P.s/ F.s/ D.s s /Q.s/ is P.s /=Q.s /. (c) Explain how the results of (a) and (b) are related..3 SOLUTION OF INITIAL VALUE PROBLEMS Laplace Transforms of Derivatives In the rest of this chapter we ll use the Laplace transform to solve initial value problems for constant coefficient second order equations. To do this, we must know how the Laplace transform of f is related to the Laplace transform of f. The next theorem answers this question. Theorem.3. Suppose f is continuous on Œ; / and of exponential order s, and f is piecewise continuous on Œ; /: Then f and f have Laplace transforms for s>s ; and L.f / D sl.f / f./: (.3.) Proof We know from Theorem..6 that L.f / is defined for s>s. We first consider the case where f is continuous on Œ; /. Integration by parts yields Z T Z T e st f.t/ dt D e st ˇ f.t/ ˇT C s e st f.t/dt Z T (.3.) D e st f.t/ f./c s e st f.t/dt for any T>.Sincef is of exponential order s, lim T! e st f.t/d and the last integral in (.3.) converges as T!if s>s. Therefore Z e st f.t/ dt D f./c s Z D f./c sl.f /; e st f.t/dt

Section.3 Solution of Initial Value Problems 45 which proves (.3.). Now suppose T > and f is only piecewise continuous on Œ; T, with discontinuities at t <t < <t n. For convenience, let t D and t n D T. Integrating by parts yields i i e st f.t/ dt D e st ˇ f.t/ ˇti C s e st f.t/dt t t i i t i D e st i f.t i / e st i f.t i / C s e st f.t/dt: t i Summing both sides of this equation from i D to n and noting that e st f.t / e st f.t / C e st f.t / e st f.t / CC e st N f.t N / e st N f.t N / i D e st N f.t N / e st f.t / D e st f.t/ f./ yields (.3.), so (.3.) follows as before. Example.3. In Example..4 we saw that L.cos!t/ D Applying (.3.) with f.t/d cos!t shows that s s C! : s! L.! sin!t/ D s D s C! s C! : Therefore! L.sin!t/ D s C! ; which agrees with the corresponding result obtained in..4. In Section. we showed that the solution of the initial value problem y D ay; y./ D y ; (.3.3) is y D y e at. We ll now obtain this result by using the Laplace transform. Let Y.s/ D L.y/ be the Laplace transform of the unknown solution of (.3.3). Taking Laplace transforms of both sides of (.3.3) yields L.y / D L.ay/; which, by Theorem.3., can be rewritten as sl.y/ y./ D al.y/; or sy.s/ y D ay.s/: Solving for Y.s/ yields Y.s/ D y s a ; so y D L.Y.s// D L y D y L D y e at ; s a s a which agrees with the known result. We need the next theorem to solve second order differential equations using the Laplace transform.

46 Chapter Laplace Transforms Theorem.3. Suppose f and f are continuous on Œ; / and of exponential order s ; and that f is piecewise continuous on Œ; /: Then f, f, and f have Laplace transforms for s>s, L.f / D sl.f / f./; (.3.4) and L.f / D s L.f / f./ sf./: (.3.5) Proof Theorem.3. implies that L.f / exists and satisfies (.3.4) fors>s. To prove that L.f / exists and satisfies (.3.5) fors>s, we first apply Theorem.3. to g D f. Since g satisfies the hypotheses of Theorem.3., we conclude that L.g / is defined and satisfies L.g / D sl.g/ g./ for s>s. However, since g D f, this can be rewritten as L.f / D sl.f / f./: Substituting (.3.4) into this yields (.3.5). Solving Second Order Equations with the Laplace Transform We ll now use the Laplace transform to solve initial value problems for second order equations. Example.3. Use the Laplace transform to solve the initial value problem y 6y C 5y D 3e t ; y./ D ; y./ D 3: (.3.6) Solution Taking Laplace transforms of both sides of the differential equation in (.3.6) yields L.y 6y C 5y/ D L 3e t D 3 s ; which we rewrite as L.y / 6L.y / C 5L.y/ D 3 s : (.3.7) Now denote L.y/ D Y.s/. Theorem.3. and the initial conditions in (.3.6) imply that L.y / D sy.s/ y./ D sy.s/ and L.y / D s Y.s/ y./ sy./ D s Y.s/ 3 s: Substituting from the last two equations into (.3.7) yields Therefore so s Y.s/ 3 s 6.sY.s/ / C 5Y.s/ D 3 s :.s 6s C 5/Y.s/ D 3 C.3 C s/ C 6. /; (.3.) s.s 5/.s /Y.s/ D 3 C.s /.s 9/ ; s

Section.3 Solution of Initial Value Problems 47 and 3 C.s /.s 9/ Y.s/ D.s /.s 5/.s / : Heaviside s method yields the partial fraction expansion Y.s/ D s C s 5 C 5 s ; and taking the inverse transform of this yields y D e t C e5t C 5 et as the solution of (.3.6). It isn t necessary to write all the steps that we used to obtain (.3.). To see how to avoid this, let s apply the method of Example.3. to the general initial value problem ay C by C cy D f.t/; y./d k ; y./ D k : (.3.9) Taking Laplace transforms of both sides of the differential equation in (.3.9) yields al.y / C bl.y / C cl.y/ D F.s/: (.3.) Now let Y.s/ D L.y/. Theorem.3. and the initial conditions in (.3.9) imply that Substituting these into (.3.) yields L.y / D sy.s/ k and L.y / D s Y.s/ k k s: a s Y.s/ k k s C b.sy.s/ k / C cy.s/ D F.s/: (.3.) The coefficient of Y.s/ on the left is the characteristic polynomial p.s/ D as C bs C c of the complementary equation for (.3.9). Using this and moving the terms involving k and k to the right side of (.3.) yields p.s/y.s/ D F.s/C a.k C k s/ C bk : (.3.) This equation corresponds to (.3.) ofexample.3.. Having established the form of this equation in the general case, it is preferable to go directly from the initial value problem to this equation. You may find it easier to remember (.3.) rewritten as p.s/y.s/ D F.s/C a y./ C sy./ C by./: (.3.3) Example.3.3 Use the Laplace transform to solve the initial value problem y C 3y C y D e t ; y./ D 4; y./ D : (.3.4) Solution The characteristic polynomial is p.s/ D s C 3s C D.s C /.s C /

4 Chapter Laplace Transforms and so (.3.3) becomes F.s/ D L.e t / D s C ;.s C /.s C /Y.s/ D C. 4s/ C 3. 4/: s C Solving for Y.s/ yields 4..s C /.s C // Y.s/ D.s C =/.s C /.s C / : Heaviside s method yields the partial fraction expansion so the solution of (.3.4) is Y.s/ D 4 3 s C = s C C 3 s C ; (Figure.3.). y D L.Y.s// D 4 3 e t= e t C 3 e t y y.5 3 4 5 6 7 t 3 4 5 6 7 t 3 3 4 4 Figure.3. y D 4 3 e t= e t C 3 e t Figure.3. y D 7 e t cos t 5 e t sin t Example.3.4 Solve the initial value problem y C y C y D ; y./ D 3; y./ D : (.3.5) Solution The characteristic polynomial is p.s/ D s C s C D.s C / C and F.s/ D L./ D s ;

Section.3 Solution of Initial Value Problems 49 so (.3.3) becomes.s C / C Y.s/ D s C. 3s/ C. 3/: Solving for Y.s/ yields s.5 C 3s/ Y.s/ D sœ.sc / C : In Example.. we found the inverse transform of this function to be y D 7 e t cos t 5 e t sin t (Figure.3.), which is therefore the solution of (.3.5). REMARK: In our examples we applied Theorems.3. and.3. without verifying that the unknown function y satisfies their hypotheses. This is characteristic of the formal manipulative way in which the Laplace transform is used to solve differential equations. Any doubts about the validity of the method for solving a given equation can be resolved by verifying that the resulting function y is the solution of the given problem..3 Exercises In Exercises 3 use the Laplace transform to solve the initial value problem.. y C 3y C y D e t ; y./ D ; y./ D 6. y y 6y D ; y./ D ; y./ D 3. y C y y D e 3t ; y./ D ; y./ D 4 4. y 4y D e 3t ; y./ D ; y./ D 5. y C y y D e 3t ; y./ D ; y./ D 6. y C 3y C y D 6e t ; y./ D ; y./ D 7. y C y D sin t; y./ D ; y./ D. y 3y C y D e 3t ; y./ D ; y./ D 9. y 3y C y D e 4t ; y./ D ; y./ D. y 3y C y D e 3t ; y./ D ; y./ D 4. y C 3y C y D e t ; y./ D ; y./ D. y C y y D 4; y./ D ; y./ D 3 3. y C 4y D 4; y./ D ; y./ D 4. y y 6y D ; y./ D ; y./ D 5. y C 3y C y D e t ; y./ D ; y./ D 6. y y D ; y./ D ; y./ D 7. y C 4y D 3 sin t; y./ D ; y./ D. y C y D e 3t ; y./ D ; y./ D 4 9. y C y D ; y./ D ; y./ D. y C y D t; y./ D ; y./ D

4 Chapter Laplace Transforms. y C y D t 3 sin t; y./ D ; y./ D 3. y C 5y C 6y D e t ; y./ D ; y./ D 3 3. y C y C y D 6 sin t 4 cos t; y./ D ; y./ D 4. y y 3y D cos t; y./ D ; y./ D 7 5. y C y D 4 sin t C 6 cos t; y./ D 6; y./ D 6. y C 4y D sin t C 9 cos t; y./ D ; y./ D 7. y 5y C 6y D e t cos t; y./ D ; y./ D. y C y C y D t; y./ D ; y./ D 7 9. y y C y D 5 sin t C cos t; y./ D ; y./ D 3. y C 4y C 3y D e t 36e t ; y./ D ; y./ D 6 3. y C 4y C 5y D e t.cos t C 3 sin t/; y./ D ; y./ D 4 3. y 3y y D 4e t ; y./ D ; y./ D 33. 6y y y D 3e t ; y./ D ; y./ D 34. y C y C y D t; y./ D ; y./ D 35. 4y 4y C 5y D 4 sin t 4 cos t; y./ D ; y./ D =7 36. 4y C 4y C y D 3 sin t C cos t; y./ D ; y./ D 37. 9y C 6y C y D 3e 3t ; y./ D ; y./ D 3 3. Suppose a; b,andc are constants and a. Let y D L as C b and y as D L a : C bs C c as C bs C c Show that y./ D ; y./ D and y./ D ; y./ D : HINT: Use the Laplace transform to solve the initial value problems ay C by C cy D ; y./ D ; y./ D ay C by C cy D ; y./ D ; y./ D :.4 THE UNIT STEP FUNCTION In the next section we ll consider initial value problems ay C by C cy D f.t/; y./ D k ; y./ D k ; where a, b, andc are constants and f is piecewise continuous. In this section we ll develop procedures for using the table of Laplace transforms to find Laplace transforms of piecewise continuous functions, and to find the piecewise continuous inverses of Laplace transforms.

Section.4 The Unit Step Function 4 Example.4. Use the table of Laplace transforms to find the Laplace transform of ( t C ; t<; f.t/ D 3t; t (.4.) (Figure.4.). Solution Since the formula for f changes at t D, we write L.f / D D Z Z e st f.t/dt e st.t C / dt C Z To relate the first term to a Laplace transform, we add and subtract e st.3t/ dt: (.4.) in (.4.) to obtain Z e st.t C / dt L.f / D D D Z Z e st.t C / dt C e st.t C / dt C L.t C / C Z Z Z e st.t / dt: e st.3t t / dt e st.t / dt (.4.3) To relate the last integral to a Laplace transform, we make the change of variable x D t and rewrite the integral as Z e st.t / dt D Z e s.xc/.x C / dx Z D e s e sx.x C / dx: Since the symbol used for the variable of integration has no effect on the value of a definite integral, we can now replace x by the more standard t and write Z Z e st.t / dt D e s e st.t C / dt D e s L.t C /: This and (.4.3) imply that L.f / D L.t C / C e s L.t C /: Now we can use the table of Laplace transforms to find that L.f / D s C s C e s s C : s

4 Chapter Laplace Transforms y y 9 7 6 5 4 3 3 4 t τ t Figure.4. The piecewise continuous function (.4.) Figure.4. y D u.t / Laplace Transforms of Piecewise Continuous Functions We ll now develop the method of Example.4. into a systematic way to find the Laplace transform of a piecewise continuous function. It is convenient to introduce the unit step function,definedas u.t/ D ; t < ; t : (.4.4) Thus, u.t/ steps from the constant value to the constant value at t D. If we replace t by t in (.4.4), then ; t < ; u.t / D I ; t that is, the step now occurs at t D (Figure.4.). The step function enables us to represent piecewise continuous functions conveniently. For example, consider the function ( f.t/; t<t ; f.t/d (.4.5) f.t/; t t ; whereweassumethatf and f are defined on Œ; /, even though they equal f only on the indicated intervals. This assumption enables us to rewrite (.4.5)as f.t/d f.t/ C u.t t /.f.t/ f.t// : (.4.6) To verify this, note that if t<t then u.t t / D and (.4.6) becomes If t t then u.t t / D and (.4.6) becomes f.t/ D f.t/ C./.f.t/ f.t// D f.t/: f.t/ D f.t/ C./.f.t/ f.t// D f.t/: We need the next theorem to show how (.4.6) can be used to find L.f /.

Section.4 The Unit Step Function 43 Theorem.4. Let g be defined on Œ; /: Suppose and L.g.t C // exists for s>s : Then L.u.t /g.t// exists for s>s, and Proof By definition, L.u.t /g.t// D e s L.g.t C //: L.u.t /g.t// D From this and the definition of u.t /, L.u.t /g.t// D Z Z e st u.t /g.t/dt: e st./ dt C Z e st g.t/ dt: The first integral on the right equals zero. Introducing the new variable of integration x D t in the second integral yields Z Z L.u.t /g.t// D e s.xc/ g.x C /dx D e s e sx g.x C /dx: Changing the name of the variable of integration in the last integral from x to t yields Z L.u.t /g.t// D e s e st g.t C /dt D e s L.g.t C //: Example.4. Find L u.t /.t C / : Solution Here D and g.t/ D t C, so g.t C / D.t C / C D t C t C : Since Theorem.4. implies that L.g.t C // D s 3 C s C s ; L u.t /.t C / D e s s C 3 s C : s Example.4.3 Use Theorem.4. to find the Laplace transform of the function ( t C ; t<; f.t/ D 3t; t ; from Example.4.. Solution We first write f in the form (.4.6)as f.t/ D t C C u.t /.t /:

44 Chapter Laplace Transforms Therefore L.f / D L.t C / C L.u.t /.t // D L.t C / C e s L.t C / (from Theorem.4./ D s C s C e s s C ; s which is the result obtained in Example.4.. Formula (.4.6) can be extended tomore general piecewise continuousfunctions. For example, we can write ˆ< f.t/; t<t ; f.t/d f.t/; t t<t ; ˆ: f.t/; t t ; as f.t/d f.t/ C u.t t /.f.t/ f.t// C u.t t /.f.t/ f.t// if f, f,andf are all defined on Œ; /. Example.4.4 Find the Laplace transform of ; t<; ˆ< t C ; t<3; f.t/d 3t; 3 t<5; ˆ: t ; t 5 (.4.7) (Figure.4.3). Solution In terms of step functions, f.t/ D C u.t /. t C / C u.t 3/.3t C t / Cu.t 5/.t 3t/; or f.t/d u.t /t C u.t 3/.5t / u.t 5/.t C /: Now Theorem.4. implies that L.f / D L./ e s L.t C / C e 3s L.5.t C 3/ / e 5s L..t C 5/ C / D L./ e s L.t C / C e 3s L.5t C 4/ e 5s L.t C / D s e s s C 5 C e 3s s s C 4 e 5s s s C : s The trigonometric identities sin.a C B/ D sin A cos B C cos A sin B (.4.) cos.a C B/ D cos A cos B sin A sin B (.4.9) are useful in problems that involve shifting the arguments of trigonometric functions. We ll use these identities in the next example.

Section.4 The Unit Step Function 45 y 6 4 6 4 4 6 3 4 5 6 t Figure.4.3 The piecewise contnuous function (.4.7) Example.4.5 Find the Laplace transform of sin t; t< ˆ< ; f.t/ D cos t 3 sin t; ˆ: t<; 3 cos t; t (Figure.4.4). Solution In terms of step functions, f.t/d sin t C u.t =/.cos t 4 sin t/ C u.t /.cos t C 3 sin t/: Now Theorem.4. implies that L.f / D L.sin t/ C e s L cos t C 4 sin t C Ce s L.cos.t C /C 3 sin.t C //: Since cos t C 4 sin t C D sin t 4 cos t and cos.t C /C 3 sin.t C / D cos t 3 sin t; we see from (.4.)that L.f / D L.sin t/ e s= L.sin t C 4 cos t/ e s L. cos t C 3 sin t/ D s e s C 4s 3 C s e s : C s C s C (.4.) (.4.)

46 Chapter Laplace Transforms y 3 3 4 5 6 t 3 Figure.4.4 The piecewise continuous function (.4.) The Second Shifting Theorem Replacing g.t/ by g.t / in Theorem.4. yields the next theorem. Theorem.4. ŒSecond Shifting Theorem If and L.g/ exists for s>s then L.u.t /g.t // exists for s>s and L.u.t /g.t // D e s L.g.t//; or, equivalently, if g.t/ $ G.s/; then u.t /g.t / $ e s G.s/: (.4.) REMARK: Recall that the First Shifting Theorem (Theorem..3 states that multiplyinga function by e at corresponds to shifting the argument of its transform by a units. Theorem.4. states that multiplying a Laplace transform by the exponential e s corresponds to shifting the argument of the inverse transform by units. Example.4.6 Use (.4.)to find L e s s : Solution To apply (.4.)welet D and G.s/ D =s.theng.t/ D t and (.4.) implies that e L s D u.t /.t /: s

Section.4 The Unit Step Function 47 Example.4.7 Find the inverse Laplace transform h of H.s/ D e s s s C s and find distinct formulas for h on appropriate intervals. C e 4s 4 s 3 C s ; Solution Let G.s/ D s ; G.s/ D s C s ; G.s/ D 4 s C 3 s : Then g.t/ D t; g.t/ D t C ; g.t/ D t C : Hence, (.4.) and the linearity of L imply that h.t/ D L.G.s// L.e s G.s// C L e 4s G.s/ D t u.t / Œ.t / C C u.t 4/.t 4/ C D t u.t /.t C / C u.t 4/.t 6t C 33/; which can also be written as ˆ< t; t<; h.t/ D ; t<4; ˆ: t 6t C 3; t 4: Example.4. Find the inverse transform of H.s/ D s s C 4 e s 3s C s C 9 C s C e s s C 6s C : Solution Let and Then G.s/ D G.s/ D s s C 4 ; G.3s C /.s/ D s C 9 ; s C.s C 3/ D s C 6s C.s C 3/ C : g.t/ D cos t; g.t/ D 3cos 3t sin 3t; 3 and g.t/ D e 3t.cos t sin t/: Therefore (.4.) and the linearity of L imply that h.t/ D cos t u.t =/ 3 cos 3.t =/ C 3 sin 3 t Cu.t /e 3.t / Œcos.t / sin.t / :

4 Chapter Laplace Transforms Using the trigonometric identities (.4.)and(.4.9), we can rewrite this as h.t/ D cos t C u.t =/ 3 sin 3t cos 3t 3 u.t /e 3.t /.cos t sin t/ (.4.3) (Figure.4.5). y 5 4 3 3 4 5 6 7 3 4 5 6 t Figure.4.5 The piecewise continouous function (.4.3).4 Exercises In Exercises 6 find the Laplace transform by the method of Example.4.. Then express the given function f in terms of unit step functions as in Eqn. (.4.6), and use Theorem.4. to find L.f /. Where indicated by C/G,graphf.. f.t/d ( ; t<4; t; t 4:. f.t/d ( t; t<; ; t : ( ( t ; t<; 3. C/G f.t/ D 4. C/G f.t/d t; t : ; t<; t C ; t : 5. f.t/d ( t ; t<; 4; t : 6. f.t/d ( t ; t<; ; t :

Section.4 The Unit Step Function 49 In Exercises 7 express the given function f in terms of unit step functions and use Theorem.4. to find L.f /. Where indicated by C/G, graph f. 7. f.t/d ( ; t<; t C 3t; t :. f.t/d ( t C ; t<; t; t : 9. f.t/d ( te t ; t<; e t ; t : t; t<; ˆ<. f.t/d t 4; t<3; ˆ: ; t 3:. f.t/d ( e t ; t<; e t ; t : ; t<; ˆ<. f.t/d t; t<; ˆ: ; t : t; t<; t; t<; ˆ< ˆ< 3. f.t/d t ; t<; 4. f.t/d t; t<; ˆ: ˆ: ; t : 6; t > : sin t; t< ˆ< ; 5. C/G f.t/ D sin t; ˆ: t<; cos t; t : ; t<; ˆ< 6. C/G f.t/ D t C ; t<3; ˆ: 3t; t 3: 3; t<; ˆ< 7. C/G f.t/ D 3t C ; t<4; ˆ: 4t; t 4: (.t C / ; t<;. C/G f.t/ D.t C / ; t : In Exercises 9 use Theorem.4. to express the inverse transforms in terms of step functions, and then find distinct formulas the for inverse transforms on the appropriate intervals, as in Example.4.7. Where indicated by C/G, graph the inverse transform. 9. H.s/ D e s s. C/G H.s/ D e s. C/G H.s/ D s 3 C e s s s C s. H.s/ D e s s.s C / C e s 3s s C e 3s s C s

43 Chapter Laplace Transforms 3. H.s/ D 5 s 6 C e 3s s s C 7 C 3e 6s s s 3 4. H.s/ D e s. s/ s C 4s C 5 5. C/G H.s/ D s s C e 3s s s C s C 6. H.s/ D e s 3.s 3/.s C /.s / s C.s /.s / 7. H.s/ D s C 3 s C e s s C 4 C e 3s s s C 3 s. H.s/ D s 3 s C 3 e s s C e 4s s 3 s 9. Find L.u.t //. 3. Let ft m g md be a sequence of points such that t D, t mc >t m, and lim m! t m D.For each nonnegative integer m, letf m be continuous on Œt m ; /, andletf be defined on Œ; / by f.t/d f m.t/; t m t<t mc.m D ;;:::/: Show that f is piecewise continuouson Œ;/ and that it has the step function representation f.t/d f.t/ C X u.t t m /.f m.t/ f m.t// ; t<: md How do we know that the series on the right converges for all t in Œ; /? 3. In addition to the assumptions of Exercise 3, assume that jf m.t/j Me s t ;t t m ;md ;;:::;.A/ and that the series X md e tm converges for some >. Using the steps listed below, show that L.f / is defined for s>s and.b/ X L.f / D L.f / C e stm L.g m / md.c/ for s>s C,where g m.t/ D f m.t C t m / f m.t C t m /: (a) Use (A) and Theorem..6 to show that is defined for s>s. X L.f / D md mc t m e st f m.t/ dt.d/

Section.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 43 (b) (c) Show that (D) can be rewritten as L.f / D X md Z t m Z! e st f m.t/ dt e st f m.t/ dt :.E/ t mc Use (A), the assumed convergence of (B), and the comparison test to show that the series X md Z t m e st f m.t/ dt and X md Z t mc e st f m.t/ dt (d) both converge (absolutely) if s>s C. Show that (E) can be rewritten as L.f / D L.f / C X md Z t m e st.f m.t/ f m.t// dt if s>s C. (e) Complete the proof of (C). 3. Suppose ft m g md and ff mg md satisfy the assumptions of Exercises 3 and 3, and there s a positive constant K such that t m Km for m sufficiently large. Show that the series (B) of Exercise 3 converges for any >, and conclude from this that (C) of Exercise 3 holds for s>s. In Exercises 33 36 find the step function representation of f and use the result of Exercise 3 to find L.f /. HINT: You will need formulas related to the formula for the sum of a geometric series. 33. f.t/d m C ; m t<mc.md ; ; ; : : :/ 34. f.t/d. / m ;mt<mc.md ; ; ; : : :/ 35. f.t/d.m C / ;mt<mc.md ; ; ; : : :/ 36. f.t/d. / m m; m t<mc.md ; ; ; : : :/.5 CONSTANT COEEFFICIENT EQUATIONS WITH PIECEWISE CONTINUOUS FORCING FUNC- TIONS We ll now consider initial value problems of the form ay C by C cy D f.t/; y./ D k ; y./ D k ; (.5.) where a, b, andc are constants (a ) andf is piecewise continuous on Œ; /. Problems of this kind occur in situations where the input to a physical system undergoes instantaneous changes, as when a switch is turned on or off or the forces acting on the system change abruptly. It can be shown (Exercises 3 and 4) that the differential equation in (.5.) has no solutions on an open interval that contains a jump discontinuity of f. Therefore we must define what we mean by a solution of (.5.) onœ; / in the case where f has jump discontinuities. The next theorem motivates our definition. We omit the proof.

43 Chapter Laplace Transforms Theorem.5. Suppose a; b, and c are constants.a /; and f is piecewise continuous on Œ; /: with jump discontinuities at t ;..., t n ; where <t < <t n : Let k and k be arbitrary real numbers. Then there is a unique function y defined on Œ; / with these properties: (a) y./ D k and y./ D k. (b) y and y are continuous on Œ; /. (c) y is defined on every open subinterval of Œ; / that does not contain any of the points t ;..., t n, and ay C by C cy D f.t/ on every such subinterval. (d) y has limits from the right and left at t ;...;t n. We define the function y of Theorem.5. to be the solution of the initial value problem (.5.). We begin by considering initial value problems of the form ( ay C by f.t/; t<t ; C cy D f.t/; t t ; y./ D k ; y./ D k ; (.5.) where the forcing function has a single jump discontinuity at t. We can solve (.5.) by the these steps: Step. Find the solution y of the initial value problem Step. Compute c D y.t / and c D y.t /. ay C by C cy D f.t/; y./ D k ; y./ D k : Step 3. Find the solution y of the initial value problem ay C by C cy D f.t/; y.t / D c ; y.t / D c : Step 4. Obtain the solution y of (.5.) as ( y.t/; t<t y D y.t/; t t : ItisshowninExercise3 that y exists and is continuous at t. The next example illustrates this procedure. Example.5. Solve the initial value problem y C y D f.t/; y./ D ; y./ D ; (.5.3) where < ; t< f.t/ D ; : ; t :