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Lecture notes on Laplace transform

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Chapter 1 The Laplace Transform DEFINITION OF THE LAPLACE TRANSFORM Let F(t) be a function of t specified for t 0. Then the Laplace transform of F(t), denoted by 4 (F(t)), is defined by f e-St F(t)) = f(s) = F(t) dt (1) 0 where we assume at present that'the parameter s is real. Later it will be found useful to consider s complex. The Laplace transform of F(t) is said to exist if the integral (1) converges for some value of s; otherwise it does not exist. For sufficient conditions under which the Laplace transform does exist, see Page 2. NOTATION If a function of t. is indicated in terms of a capital letter, such as F(t), G(t), Y(t), etc., the Laplace transform of the function is denoted by the corresponding lower case letter, i.e. f (s), g(s), y(s), etc. In other cases, a tilde (-) can be used to denote the Laplace trans- form. Thus, for example, the Laplace transform of u(t) is is (s). LAPLACE TRANSFORMS OF SOME ELEMENTARY FUNCTIONS F(t) -C F(t) = f(8) 1. 1 1 80 s 2. t s0 s2 3. to 8 0 The adjacent table shows sn Laplace transforms of various n = 0, 1, 2, ... Note. Factorial n = n = 12 n elementary functions. For de- Also, by definition 0 = 1. tails of evaluation using defini- tion (1), see Problems-1 and 2. 4. eat 1 s a For a more extensive table see s-a Appendix B, Pages 245 to 254. a 5. sin at s 0 82 _ +a2 8 6. cos at 8 0 82 a2 7. sinh at a 8 jai 82 - a2 8. cosh at 8 lat 82 a2 1THE LAPLACE TRANSFORM CHAP. 1 2 SECTIONAL OR PIECEWISE CONTINUITY A function is called sectionally continuous or piecewise continuous in an interval c t- a if the interval can be subdivided into a finite number of intervals in each of which the function is continuous and has finite right and left hand limits. F(t) I/ j i t a ti t2 1t3 R Fig. 1-1 An example of a function which is sectionally continuous is shown graphically in Fig. 1-1 above. This function has discontinuities at ti, t2 and t3. Note that the right and left hand limits at t2, for example, are represented by lim F(t2 + E) = F(t2 + 0) = F(t2+) 0 e and lim F(t2 - E) = F(t2 - 0) = F(t2-) respectively, where c is positive. E-+0 FUNCTIONS OF EXPONENTIAL ORDER If real constants M 0 and y exist such that for all t N I e-It F(t) I M or I F(t) 1 Melt we say that F(t) is a function of exponential order y as t- - or, briefly, is of exponential order. Example 1. F(t) = t2 is of exponential order 3 (for example), since ;t2j = t2 eat for all t 0. et3 Example 2. F(t) = is not of exponential order since I e-vt et' 1 = et3-yt can be made larger than any given constant by increasing t. Intuitively, functions of exponential order cannot "grow" in absolute value more rapidly than Me"' as t increases. In practice, however, this is no restriction since M and y can be as large as desired. Bounded functions, such as sin at or cos at, are of exponential order. SUFFICIENT CONDITIONS FOR EXISTENCE OF LAPLACE TRANSFORMS Theorem 1-1. If F(t) is sectionally continuous in every finite interval 0 t N and of exponential order y for t N, then its Laplace transform f (s) exists for all s y. For a proof of this see Problem 47. It must be emphasized that the stated conditions are sufficient to guarantee the existence of the Laplace transform. If the conditions are not satisfied, however, the Laplace transform may or may not exist see Problem 32. Thus the conditions are not necessary for the existence of the Laplace transform. For other sufficient conditions, see Problem 145.CHAP. 11 THE LAPLACE TRANSFORM 3 SOME IMPORTANT PROPERTIES OF LAPLACE TRANSFORMS In the following list of theorems we assume, unless otherwise stated, that all functions satisfy the conditions of Theorem 1-1 so that their Laplace transforms exist. 1. Linearity property. Theorem 1-2. If c1 and C2 are any constants while F1(t) and F2(t) are functions with Laplace transforms f i (s) and f2 (s) respectively, then = C14 Fi(t) + c2a(F2(t) (2) .(C1F1(t)+C2F2(t)I = clfl(s) + C2f2(s) The result is easily extended to more than two functions. = 4.C t2 - 3.C cos 2t + 5.4 e-t Example. C (4t2 - 3 cos 2t + 5e-1 4(83) 3(s2+4)+5Cs+1) 8 _ 3s + 5 s3 s2+4 s+1 The symbol C, which transforms F(t) into f (s), is often called the Laplace trans- formation operator. Because of the property of t expressed in this theorem, we say that e( is a linear operator or that it has the linearity property. 2. First translation or shifting property. Theorem 1-3. If aC F(t) = f(s) then a( eal F(t) = f(s - a) (3) S Example. Since e cos 2t = we have .32+ 4 , s+1 = s+1 .Ce-tcos2t = (s+1)2+4 s2+2s+5 3. Second translation or shifting property. Theorem 1-4. If (F(t)) s and G(t) = td(t - a) t a , then .(G(t)) = e-as f(s) (4) Example. Since i t3 = = s4 , the Laplace transform of the function s4 j(t - 2)3 t2 G(t) = 10 t2 is 6e-2s/g4 4. Change of scale property. Theorem 1-5. If .( F(t) = f(s), then .t F(at) (5) .( sin t = s2 + 1 we have Example. Since 1 _ 3 sin 3t = 1 3 (s/3)2 + 1 s2 + 94 THE LAPLACE TRANSFORM CHAP. 1 5. Laplace transform of derivatives. Theorem 1-6. If t F(t) = f(s), then i (F'(t)) = s f(s) - F(O) (6) if F(t) is continuous for 0 t N and of exponential order for t N while F'(t) is sectionally continuous for 0 t N. r F(t) = Example. If F(t) = cos 3t, then and we have 82 + 9 F'(t) = -3 sin 3t s (82+9 ) 82+ - 1 9 The method is useful in finding Laplace `transforms without integration see Problem 151. Theorem 1-7. If in Theorem 1-6, F(t) fails to be continuous at t = 0 but lim F(t) = F(0+) exists but is not equal to F(0), which may or may not exist, then t-.o .C F'(t)) = sf(s) - F(0+) (7) Theorem 1-8. If in Theorem 1-6, F(t) fails to be continuous at t = a, then J (F'(t)) = s f (s) - F(O) - e -°s F(a+) - F(a-) (8) where F(a+) - F(a-) is sometimes called the jump at the discontinuity t = a. For more than one discontinuity, appropriate modifications can be made. Theorem 1-9. If F(t) = f (s), then F"(t) = s2 f(s) - sF(0) - F'(0) (9) if F(t) and F'(t) are continuous for 0 = t N and of exponential order for t N while F"(t) is sectionally continuous for 0 t-:5 N. If. F(t) and F'(t) have discontinuities, appropriate modification of (9) can be made as in Theorems 1-7 and 1-8. Theorem 1-10. If (' F(t) = f(s), then 2 F'(0) - sn-' F(0) - sn Fcn-2,(0) = sn f(s) - s - Fcn 1)(0) 4 (Fcn)(t) (10) if F(t), F'(t), ..., F`n-"(t) are continuous for 0 t N and of exponential order for t N while F(n)(t) is sectionally continuous for 0 t N. 6. Laplace transform of integrals. If Theorem 1-11. C F(t) = f(s), then foF(u) du1 of if +4 , we have Example. Since t sin 2t = - (J J sin 2u du - 2 S(82+4) r 1 U as can be verified directly.CHAP. 1) THE LAPLACE TRANSFORM 5 7. Multiplication by tn. Theorem 1-12. If . F(t) = f(s), then = (-1)n f(n)(s) -(' tn F(t) _ (-1)n f(s) dsn _1_ = Example. Since C e2t we have s - 2 ' d (s-2 1 1 1 .e te2t ds (s-2)2 d( 1 ) _ 2 2 .e t2e2t d82 s-2 (s-2) 8. Division by t. F(t) = f(s), Theorem 1-13. If . then f f (u) du (13) s provided lim F(t)/t exists. t-.o 1 sin t Exam le. p Since (sin t - 82 + 1 and E = 1, we have o t sin t l __ du f x tan' 1(11s) - j J t u2 + 1 9. Periodic functions. Theorem 1-14. Let F(t) have period T 0 so that F(t + T) = F(t) see Fig. 1-2. T f° e-" F(t) dt Then F'(t) _ (14) 1 - e-ST F(t) Fig. 1-2 10. Behavior of f (s) as s - co . Theorem 1-15. If e (F(t) = f(s), then lim f (s) = 0 (15) S -. 'Jo 11. Initial-value theorem. Theorem 1-16. If the indicated limits exist, then lim F(t) (16) = lim s f (s) t0 s-CHAP. 1 6 THE LAPLACE TRANSFORM 12. Final-value theorem. Theorem 1-17. If the indicated limits exist, then lim F(t) = lim s f (s) (17) t-. o s-.o Generalization of initial-value theorem. 13. then we say that for values of t near t = 0 small t, F(t) is If urn F(t)/G(t) = 1, t-.o close to G(t) and we write F(t) - G(t) as t- 0. lim f(s)/g(s) = 1, then we say that for large values of s, f (s) is Similarly if -4 00 close to g(s) and we write f (s) - g(s) as s With this notation we have the following generalization of Theorem 1-16. Theorem 1-18. If F(t) - G(t) as t - 0, then f(s) - g(s) as s - oo where f (s) .C (F(t)) and g(s) _ .( G(t). 14. Generalization of final-value theorem. Similarly if lim f(s)/g(s) = 1, If lim F(t)/G(t) = 1, we write F(t) - G(t) as t tW So we write f (s) g(s) as s - 0. Then we have the following generalization of Theorem 1-17. _ Theorem 1-19. If F(t) G(t) as t - oc, then f (s) - g(s) as s - 0 where f (s) .C F(t) g(s) _ .(G(t). and METHODS OF FINDING LAPLACE TRANSFORMS Various means are available for determining Laplace transforms as indicated in the following list. This involves direct use of definition (1). 1. Direct method. 2. Series method. If F(t) has a power series expansion given by (18) F(t) = ao + ait + a2t2 + =I ante n-o its Laplace transform can be obtained by taking the sum of the Laplace transforms of each term in the series. Thus = ao a, 2-a2 n an .C F(t) + + + _ (19) n+1 s'3 S S- n-U A condition under which the result is valid is that the series (19) be convergent for s y. See Problems 34, 36, 39 and 48. 3. Method of differential equations. This involves finding a differential equation satis- See Problems 34 and 48. fied by F(t) and then using the above theorems. 4. Differentiation with respect to a parameter. See Problem 20. Miscellaneous methods involving special devices such as indicated in the above theo- 5. rems, for example Theorem 1-13. 6. Use of Tables (see Appendix).CHAP. 1 THE LAPLACE TRANSFORM 7 EVALUATION OF INTEGRALS If f (s) = C F(t) , then f0 e-StF(t) dt = f(s) (20) Taking the limit as s - 0, we have f F(t) dt = f(0) (21) 0 assuming the integral to be convergent. The results (20) and (21) are often useful in evaluating various integrals. See Problems 45 and 46. SOME SPECIAL FUNCTIONS 1. The Gamma function. If n 0, we define the gamma function by r(n) = f un- ' e-u du (22) 0 The following are some important properties of the gamma function. 1. r(n + 1) = n r(n), n 0 Thus since r(1) 1, we have I'(2) = 1, r(3) = 2 = 2, or(4) = 3 and in general r(n + 1) = n, if n is a positive integer. For this reason the function is some- times called the factorial function. 2. r(:) = N5 3. r(p) r(1- p) 0 p 1 - sin per ' 4. For large n, - r(n+1) 27rnn11 e-n Here - means "approximately equal to for large n". More exactly, we write F(n) G(n) if lim F(n)/G(n) = 1. This is called Stirling's formula. n 0 we can define r(n) by r(n + 1) = r(n) n II. Bessel functions. We define a Bessel function of order n by _ t2 0 _ ... (t) = (23) 2(2n+2) 1 2nr(n+1) 2.4(2n + 2)(2n + 4) Some important properties are 1. J-n (t) (-1)n Jn (t) if n is a positive integer 2t 2. Jn+Z (t) = J, (t) - (t) 3. to Jn-i (t). If n = 0, we have Jo(t) = -J1(t). dt 1' 4. e 'h"-1/u) = Jn(t)un 9=-a This is called the generating function for the Bessel functions.8 THE. LAPLACE TRANSFORM CHAP. 1 5. J. (t) satisfies Bessel's differential . equation. t2 Y11(t) + t Y'(t) + (t2 - n2) Y(t) _ 0 J0(it) = i-nIn(t) where I0(t) It is convenient to define is called the modified Bessel function of order n. III. The Error function is defined as e-u2 du 2 erf (t) = (24) 0 IV. The Complementary Error function is defined as ft 5e_u2du erfc (t) = erf (t). 2 e-u2du 1 1 - (25) 0 V. The Sine and Cosine integrals are defined by t sin u du Si (t) (26) = f 0 u Cos u Ci (t) = du (27) f u t The Exponential integral is defined as VI. 00 eu Ei (t) = du (28) 5 VII. The Unit Step function, also called Heaviside's unit function, is defined as 0 t a _ 'U(t - a) (29) 1 t a See Fig. 1-3. u(t - a) Fe(t) i I 1/E t. t a Fig. 1-3 Fig. 1-4 VIII. The Unit Impulse function or Dirac delta function... Consider the function 1/E FE (t) (30) 0 t where e 0, whose graph appears in Fig. 1-4.CHAP. 1 THE LAPLACE TRANSFORM 9 It is geometrically evident that as e - 0 the height of the rectangular shaded region increases indefinitely and the width decreases in such a way that the area is always equal to 1, i.e. f F,(t) dt = 1. 0 This idea has led some engineers and physicists to think of a limiting function, denoted by S(t), approached by FE(t) as E-0. This limiting function they have called the unit impulse function or Dirac delta function. Some of its properties are 1. 58(t)dt = 1 0 2. f 3(t) G(t) dt = G(0) for any continuous function G(t). 0 8(t - a) G(t) dt = 3. f G(a) for any continuous function G(t). 0 Although mathematically speaking such a function does not exist, manipulations or operations using it can be made rigorous. IX. Null functions. If N(t) is a function of t such that for all t 0 t f '((u) du = 0 we call N (t) a null function. 1 1 t - 1/2 Example. The function F(t) 1 t = 1 is a null function. 0 otherwise In general, any function which is zero at all but a countable set of points i.e. a set of points which can be put into one-to-one correspondence with the natural numbers 1, 2, 3, ... is a null function. LAPLACE TRANSFORMS OF SPECIAL FUNCTIONS In the following table we have listed Laplace transforms of various special functions. For a more extensive list see Appendix B, Page 245. Table of Laplace transforms of special functions F(t) f(s) = C F(t) r(n + 1) 1. to gn+1 Note that if n = 0,1,2,... this reduces to entry 3, Page 1. 1 J0(at) 2. s2 -+a2 s2 + a2 - 8)n ( 3. Jn(at) an s2 + a2 V-7 0-114s 4. sin vt- 20/2 COS X Ft e-1/4s 5.THE LAPLACE TRANSFORM CHAP. 1 10 Table of Laplace transforms of special functions (cont.) F(t) f (s) _ i F(t) 6. erf (t) e32/4 erfc (s/2) 8 1 7. e r f (Vi) 8 a+1 8. Si (t) tan-1 s 8 In (82 + 1) 9. Ci (t) 2s In (s + 1) 10. Ei (t) 8 ns a 11. 1f(t - a) 8 12. S (t) 1 a-as 13. S(t - a) 14. 0 `Nf t) Solved Problems LAPLACE TRANSFORMS OF SOME ELEMENTARY FUNCTIONS eat = 1. Prove that: (a) ( (1) = 1, s 0; (b) t = 1, s 0; (c) 1 , s a. S s s-a p (a) .i 1 = e-st (1) dt = lim e-st dt P.4,0 J fo 0 0 e-st IP lim if s0 -.m -3 to oa (b) e-st (t) dt = lim t e-st dt -C (t) P...m o J p Pe-sP\ e-SP P = lim 82 82 lim (t) (e-8t) - (1)(e824t ) P-.w 8 /) (1 = 1 if 80 82 where we have used integration by parts.CHAP. 1 THE LAPLACE TRANSFORM 11 P = e-t (eat) dt eat = e (s-a)t dt (c) lim f °° 0 foP" IP 1- e-s 011 1 e-(s-a)t lim if s a = lira P-+oo -(s -a) 0 P-.m s-a s-a For methods not employing direct integration, see Problem 15. sin at) _ (b) . cos at = s2+t12 if s0. 2. Prove that (a) . s2+a2, x P (a) sin at e-st sin at dt = f e-st sin at dt J plim 0 e-st (- s sin at - a cos at) (P lim 82 + a2 P- 0 e-sP (s sin aP + a cos aP)1 J a lim 1 S2 + a2 S2 + a2 P-. J a if 80 S2 + a2 P f e-st e-st cos at dt = lim cos at dt (b) cos at P-. .e 0 0 e`st (- s cos at + a sin at) IP lim P-. 82 + a2 0 e-sP (a cos aP - a sin aP) s f lim 1s2+a2 - S2+a2 P-.x 8 if s0 S2+ a2 We have used here the results eat (a sin /3t - /3 cos /3t) f eat sin /it dt (1) a2 + /32 eat (a cos /3t + /3 sin /3t) eat cos /3t dt f (2) a2+/32 Another method. Assuming that the result of Problem 1(c) holds for complex numbers (which can be proved), we have 1 s + ia .C etat (S) a - ta s2 + a2 eiat But = cos at + i sin at. Hence a .C etat e-st (cos at + i sin at) J , e-st cos at dt + if e-st sin at dt .C cos at + i .e sin at f. 0 From (3) and (4) we have on equating real and imaginary parts, a = s2 C sin at = 82 + a2 .C cos at a2 0 +CHAP. 1 THE LAPLACE TRANSFORM 12 S a if s jal. (b) . cosh at) = s2 a2 (a) ( (sinh at) = a2, 3. Prove that s2 2e-atl eat 2e-atl -e-It /eat dt = f (a) .e (sinh at) 1\ J t f e-st e-at dt e-st eat dt 2 0 0 2 e-at 2 .e eat a _ 1 1 _ 1 - for s Ial 2 s-a s-+a j s2-a2 Using the linearity property of the Laplace transformation, we have at once Another method. eat 2e-at eat - 2 .t; e-at .e sinh at) . 2 1 1 _ a = for s laj g2-a2 2 s-a +a) (b) As in part (a), = ( eat + e-at _ 1 C eat + C e-at e cosh at l 2 - : 2 2 If I - s Ial for s s2-a2 -" + s+a j 2 13-a 5 0t3 Find C F(t) if 4. F(t) _ 0 t 3 By definition, 3 m = o f e-st (0) dt e-st (5) dt + = e-stF(t) dt .i F(t) J0 0 3 3 5 e_ tit 3 - 5(1 - e-35) 5 0 a-st dt S -s to 0 THE LINEARITY PROPERTY Prove the linearity property Theorem 1-2, Page 3. 5. , e-stF2(t) dt. Then if F1(t) dt and ,.C F2(t)) = f2(s) = f e Fi(t) = f1(s) = f, e -st Let o ° ci and c2 are any constants, M e-st c1 F1(t) + c. F2(t) dt .( c1 F1(t) + c2 F2 (t) J. a-t F2(t) dt = cifx e-st F1(t) dt + c2J 0 0 c1.C Fl(t) + c2.i F2(t) = clfl(8) + c2f2(e) The result is easily generalized see Problem 61.CHAP. 1 THE LAPLACE TRANSFORM 13 6. Find C (4e5t + 6t3 - 3 sin 4t + 2 cos 2t . By the linearity property Problem 6 we have 4 4e5L + 6t3 - 3 sin 4t + 2 cos 2t = 4.C e5t + 6.C t3 - 3.4 sin 4t + 2.( cos 2t 1 3' 4 8 4(s-5)+6(84) -3 32+16)+2`82+4 4 36 12 2s s2+16 + 82+4 s-5 + s4 where s 6. TRANSLATION AND CHANGE OF SCALE PROPERTIES Prove the first translation or shifting property: If .1 F(t) = f(s), then ., (eatF(t)) _ 7. f(s - a). C F(t) = f e-atF(t) dt = f(s) We have 0 = f Then e eat F(t) 0 e-at eat F(t) dt 0 W e-(s-a)tF(t) dt = f(s-a) so 8. Find (a) .t t2e3t, (b) .e (e-2t sin 4t), (c) 4 e4t cosh 5t, (d) I e-2t(3 cos 6t - 5 sin 6t). 2 2 t2 = 83 = Then C t2e3t = s . (8-3)3' 4 4 4 - C sin 4t = 82 16 . (b) Then 4 e-2t sin 4t = (8 + 2)2 + 16 s2 + -4s+ 20 ' + s 8-4 = s-4 (c) . cosh 5t = Then C e4t cosh 5t = 8225 ' 82-8s-9 (a-4)2-25 Another method fe4t /e5t+e-5t) 14 e9t+e-t .( e4t cosh 5t 1I\\ 2 2 l = 1 J 1 + 1 _ s-4 18-9 8+1 82-8s-9 e 3 cos 6t - 5 sin 6t = 3.C cos 6t - 5 -C sin 6t (d) _ s 38 - 30 6 (82+36 3(82+36 - 5 = 82+36 3(8+2)-30 = 3s - 24 = . e-2t (3 cos 6t - 5 sin 6t) Then (8+2)2+86 s2+4s+40THE LAPLACE TRANSFORM CHAP.1 14 9. Prove the second translation or shifting property: o (t - a) t a If C F(t) = f (s) and G(t) = then r X G(t) = e-as f (s). o0 w a .t G(t) = e-st G(t) dt = f e-st G(t) dt + f e-st G(t) dt f0 a f a e- st (0) dt + f e-st F(t - a) dt 0 e-st F(t -a) dt a f oo e-s(u+a) F(u) du e-as f eF(u) du 0 e -as f (s) where we have used the substitution t = u + a. r cos (t - 27r/3) t 2zr/3 F(t) = j 0 10. Find e (F(t)) if t27r/3 2rr/3 0 Method 1. e-st (0) dt + e-st cos (t - 2rr/3) dt C F(t) = f 0 KF/3 00 e-s(u+27r/3) cos u du fo se - 2vs/3 e-21rs/3 e-su cos = u du = S2+1 f 0 C (cos t = + it follows from Problem 9, with a = 27r/3, that Method 2. Since s2 se -gas/3 t F(t) 82+1 11. Prove the change of scale property: If e F(t) = f(s), then f e-st F(at) dt 0 f e-s(u/a) F(u) d(u/a) 0 1 f e-su/a F(u) du a o(1) of using the transformation t = u/a.CHAP. 1 1 THE LAPLACE TRANSFORM 15 (siat lsin t 12. Given that 1 = tan- 1(1/s), find C By Problem 11, a-C Isi tat I tan-1 1/(sla) (als) a tan at a = sinat). Then tan-1(a/8). tI LAPLACE TRANSFORM OF DERIVATIVES F(t) = f (s), 13. Prove Theorem 1-6: If . then Z (F'(t) = s f (s) - F(O). Using integration by parts, we have 00 P .C F'(t) e-st F'(t) dt = lim f e -st F'(t) dt I P-.00 0 P P lim j e-st F(t) + s f e-st F(t) dtl P- .o l 0 0 P e_SPF(P) lim - F(0) + s f e-st F(t) dt P -. oO o s fo, e-s1 F(t) dt - F(O) a f(8) - F(O) for s y. using the fact that F(t) is of exponential order y as t so that lim a-4P F(P) = 0 P-' For cases where F(t) is not continuous at t = 0, see Problem 68. 14. Prove Theorem 1-9, Page 4: If C F(t) = f (s) then . F"(t) = s2 f (S) - s F(0) - F'(0). By Problem 13, = ., G'(t) s .1 G(t) - G(0) = s g(s) - G(0) Let G(t) = F'(t). Then .C F"(t) s.i F'(t) - F'(0) s s C F(t) - F(0) - F'(0) 82.C F(t) - s F(O) - F'(0) 82 f(8) - s F(0) - F'(0) The generalization to higher order derivatives can be proved by using mathematical inductiofi see Problem 651. 15. Use Theorem 1-6, Page 4, to derive each of the following Laplace transforms: 1 1 (a) -C (1) = , (b) t = s , _ (c) -((eat) s-a Theorem 1-6 states, under suitable conditions given on Page 4, that .. F'(t) = 8 F(t) - F(0) (1)THE LAPLACE TRANSFORM CHAP. 1 16 (a) Let F(t) = 1. Then F'(t) = 0, F(0) = 1, and (1) becomes .C 1 = 1/s ,C 0 = 0 = s .I 1 - 1 or (2) (b) Let F(t) = t. Then F'(t) = 1, F(0) = 0, and (1) becomes using part (a) of t = 1/32 .C 1 = 1/s = s C t - 0 or (3) By using mathematical induction we can similarly show that t') = n/sii+1 for any positive integer n. (c) Let F(t) = eat. Then F'(t) = aeat, F(O) = 1, and (1) becomes .e aeat = s p eat - 1, a C (eat) = s .1 eat - 1 .1 eat = 1/(s - a) i.e. or e (sin at) = a 2. 16. Use Theorem 1-9 to show that s2 +a Let F(t) = sin at. Then F'(t) = a cos at, F"(t) = -a2 sin at, F(O) = 0, F'(0) = a. Hence from the result J F"(t) 92.C F(t) - s F(0) - F'(0) we have X - a2 sin at) = 92 e sin at - s (0) - a = S2.( sin at - a i.e. -a2.C sin at) a = or X sin at s2 + a2 LAPLACE TRANSFORM OF INTEGRALS t 17. Prove Theorem 1-11: If C F(t) = f(s), then .1 F(u) du = f(s)/s. fo t f F(u)dn. Then G'(t) = F(t) and G(0) = 0. Taking the Laplace transform Let G(t) _ 0 of both sides, we have .t G'(t) = s.C G(t) - G(0) = s.C G(t) = f(s) t f ) .( G(t) = or .e F(u) du ( = f S ) Thus o J 18. Find ,e Jo t siu u du j . We have by the Example following Theorem 1-13 on Page 5, s1t t tan-l s Thus by Problem 17, 1 rt sin u dul = 1 tan-1 ,J0 8 8 u JCHAP. 11 THE LAPLACE TRANSFORM 17 MULTIPLICATION BY POWERS OF t 19. Prove Theorem 1-12, Page 5: n If F(t) = f (s), then to F(t) = (-1)n dsn AS) = (-1)n f n)(s) where n = 1, 2, 3, .... We have f(s) _ e-st F(t) dt 00 Then by Leibnitz's rule for differentiating under the integral sign, f « s a-st F(t) dt e xt F(t) dt TI; d = J 0 0 -te-3t F(t) dt J : 0 f 'e-sttF(t) _ - dt J0 = -.C It F(t)) C It F(t) dsf = -f'(s) Thus (1) which proves the theorem for n = 1. To establish the theorem in general, we use mathematical induction. Assume the theorem true for n = k, i.e. assume f(k)(s) (-1)k i e- t tk F(t) dt (2) Then d (_1)k f(k+1)(s) e-St tk F(t) dt asf or by Leibnitz's rule, - f (_1)k f(k+1)(s) e-st tk+1 F(t) dt 0 i.e. W (_1)k+1 f(k+1)(s) e-st tk+1 F(t) dt = (3) J: I It follows that if (2) is true, i.e. if the theorem holds for n = k, then (3) is true, i.e. the theorem holds for n = k + 1. But by (1) the theorem is true for n = 1. Hence it is true for n = 1 + 1 = 2 and n = 2 + 1 = 3, etc., and thus for all positive integer values of n. To be completely rigorous, it is necessary to prove that Leibnitz's rule can be applied. For this, see Problem 166. 20. Find (a) e t sin at, (b) (' t2 cos at). a (a) Since sin at) = , we have by Problem 19 s2 + a2 .( It sin at = (32 + a2)218 THE LAPLACE TRANSFORM CHAP. 1 Another method. s Since tcos at - T e It cos at dt = 82 + a2 o we have by differentiating with respect to the parameter a using Leibnitz's rule, x e-st - t sin at dt = qt cos at dt -.C t sin at f 2as do( 82Ta2) (82 + a2)2 from which 2as X t sin at = (S2+0)2 d (cos at) _ . as cos at Note that the result is equivalent to a S (b) Since ,t cos at we have by Problem 19 32+a2' 2s3 - 6a2s ( t2 cos at 2 s2+a2) -j1- ( S (82 + a2)3 We can also use the second method of part (a) by writing d2 f d2 1 .1 t = e 1- (cos at) - Cla2 .( cos at) ' cos at dal which gives the same result. DIVISION BY t f 21. Prove Theorem 1-13, Page 5: If ;F(t) = f(s), then f(u) du. = F(t) Let G(t) = Flt) . Then F(t) = t G(t). Taking the Laplace transform of both sides and using Problem 19, we have G(t) F(t) or f(s) = do ds Then integrating, we have g(s) = -J f (u) du = f f (u) du (1) f I JF(t) f(u) du I t r Js Note that in (1) we have chosen the "constant of integration" so that lim g(s) = 0 see Theorem 1-15, Page 51. x 22. (a) Prove that f Ft t) dt = f f (u) du provided that the integrals converge. sin t dt = 7T (b) Show that fo, (a) From Problem 21, F(t) f:0 a-3t dt = f x f(u) du t . i sCHAP. 1 THE LAPLACE TRANSFORM 19 Then taking the limit as s - 0+, assuming the integrals converge, the required result is obtained. Then (b) Let F(t) = sin t so that f (S) = 1/(82 + 1) in part (a). x x du _ sin t f = dt tan-I u J0 t u2 + 1 2 0 0 PERIODIC FUNCTIONS 23. Prove Theorem 1-14, Page 5: If F(t) has period T 0 then fT e-8t F(t) dt a .C F(t) = vT 1 - e We have = o .i F(t) e-st F(t) dt 0 fT 2T IT e-st F(t) dt + f e-st F(t) dt 4 ` e It F(t) dt + T 2T t = n + T, in the third integral let t = u + 2T, etc. Then In the second integral )et T T T J F(t) f e-s(u+T) F(u + T) du + e-;('" 42T) F(u + 2T) du + . f e-su F(u) du + 0 0 fo T T e-su 2., 1' e-su F(u) du + e-sT F(u) du + e JoT esu F(v) du + .. . of J0 T (1 + e`sT + e-28T + ...) e-su F(u) du J. f T e-sn F(u) du 0 1 - e-sT where we have used the periodicity to write F(u + T) = F(u), F(?+ 2T) = F(u), .... and the fact that 1 1+r+r2+r3+ = Irl1 1-r 24. (a) Graph the function f0 sin t 0 t 71- = F(t) ;r ?r t 2 extended periodically with period 27- (b) Find t F(t). (a) The graph appears in Fig. 1-5. F(t) t 0 ,r 2rr 3s 4a Fig. 1-5THE LAPLACE TRANSFORM CHAP. 1 20 (b) By Problem 23, since T = 21r, we have .2,r e st F(t) dt - (F(t)) = 1 - e- 2,rs J 0 1 "` J e -5t sin t dt 1 - e-2,rs 0 Sr 1 J e - st (- s sin t - cos t) a2+1 0 - I J1+e-,rsl 1 e-l 2rrs 1 - s2 + 1 (1 - e Ts)(32 + 1) using the integral (1) of Problem 2, Page 11. The graph of the function F(t) is often called a half wave rectified sine curve. INITIAL AND FINAL VALUE THEOREMS 25. Prove the initial-value theorem: lim F(t) = lim s f (s). t -+ 0 s - x By Problem 13, .C F'(t) st F'(t) dt = s f(s) - F(0) (1) = f 0 But if F'(t) is sectionally continuous and of exponential order, we have lim a st F'(t) dt = 0 (2) s-.m 0 Then taking the limit as s oo in (1), assuming F(t) continuous at t = 0, we find that lim s f(s) = F(0) = lim F(t) 0 = lim s f(s) - F(O) or a- S- t-.0 If F(t) is not continuous at t = 0, the required result still holds but we must use Theorem 1-7, Page 4. a 26. Prove the final-value theorem: lim F(t) = lim s f(s). t-.00 S-0 By Problem 13, .e F'(t) f e-st F'(t) dt = s f (s) - F(0) 0 The limit of the left hand side as s 0 is f r lim f e-st F'(t) dt = F'(t) dt = lim F'(t) dt J P-.. s-.o J 0 0 0 lira F(P) - F(0) lira F(t) - F(0) P-.oo t -coo The limit of the right hand side as s - 0 is lira s f(s) - F(0) s-.o Thus lim F(t) - F(0) = lim s f(s) - F(0) t-.oo s-.O or, as required, lim F(t) = lim s f(s) t-aao s-»0 If F(t) is not continuous, the result still holds but we must use Theorem 1-7, Page 4.