Laplace Transform. Chapter 8. Contents

Transcription

1 Chapter 8 Laplace Tranform Content 8.1 Introduction to the Laplace Method Laplace Integral Table Laplace Tranform Rule Heaviide Method Tranform Propertie Heaviide Step and Dirac Delta Laplace Table Derivation Modeling and Laplace Tranform The Laplace tranform can be ued to olve differential equation. Beide being a different and efficient alternative to variation of parameter and undetermined coefficient, the Laplace method i particularly advantageou for input term that are piecewie-defined, periodic or impulive. The Laplace method ha humble beginning a an extenion of the method of quadrature for higher order differential equation and ytem. The method i baed upon quadrature : Multiply the differential equation by the Laplace integrator dx e t dt and integrate from t to t. Iolate on the left ide of the equal ign the Laplace integral t t y(t)e t dt. Look up the anwer y(t) in a Laplace integral table. The Laplace integral or the direct Laplace tranform of a function f(t) defined for t < i the ordinary calculu integration problem f(t)e t dt, uccinctly denoted in cience and engineering literature by the ymbol L(f(t)).

2 8.1 Introduction to the Laplace Method 449 To the Student. When reading L in mathematical text, ay the word Laplace of. Think of the ymbol L( ) a tanding for E ( )dx where ( ) i the integrand to be ubtituted, E [, ) and dx e t dt i the Laplace integrator. For example, L(t 2 ) i horthand for (t2 )e t dt. The L notation recognize that integration alway proceed over t to t and that the integral involve a fixed integrator e t dt intead of the uual dt. Thee minor difference ditinguih Laplace integral from the ordinary integral found on the inide cover of calculu text. 8.1 Introduction to the Laplace Method The foundation of Laplace theory i Lerch cancelation law (1) y(t)e t dt f(t)e t dt implie y(t) f(t), or L(y(t) L(f(t)) implie y(t) f(t). In differential equation application, y(t) i the unknown appearing in the equation while f(t) i an explicit expreion extracted or computed from Laplace integral table. An Illutration. Laplace method will be applied to olve the initial value problem y 1, y(). No background in Laplace theory i aumed here, only a calculu background i ued. The Plan. The method obtain a relation L(y(t)) L( t), then Lerch cancelation law implie that the L-ymbol cancel, which give the differential equation olution y(t) t. The Laplace method i advertied a a table lookup method, in which the olution y(t) to a differential equation i found by looking up the anwer in a pecial integral table. In thi ene, the Laplace method i a generalization of the method of quadrature to higher order differential equation and ytem of differential equation. Laplace Integral. The integral g(t)e t dt i called the Laplace integral of the function g(t). It i defined by lim N N g(t)e t dt and depend on variable. The idea will be illutrated for g(t) 1, g(t) t and g(t) t 2, producing the integral formula in Table 1. (1)e t dt (1/)e t t t Laplace integral of g(t) 1. 1/ Aumed >. (t)e t dt d d (e t )dt Laplace integral of g(t) t.

3 45 Laplace Tranform d d (1)e t dt Ue d d F (t, )dt d d F (t, )dt. d d (1/) Ue L(1) 1/. 1/ 2 Differentiate. (t2 )e t dt d d (te t )dt Laplace integral of g(t) t 2. d d (t)e t dt d d (1/2 ) Ue L(t) 1/ 2. 2/ 3 Table 1. The Laplace integral g(t)e t dt for g(t) 1, t and t 2. (1)e t dt 1, (t)e t dt 1 2, (t2 )e t dt 2 3. In ummary, L(t n ) n! 1+n The Illutration. The idea of the Laplace method will be illutrated for the olution y(t) t of the problem y 1, y(). Laplace method, Table 2, which i entirely different from variation of parameter or undetermined coefficient, ue baic calculu and college algebra. In Table 3, a uccinct verion of Table 2 i given, uing L- notation. Table 2. Laplace method detail for the illutration y 1, y(). y (t)e t dt e t dt Multiply y 1 by e t dt. y (t)e t dt e t dt Integrate t to t. y (t)e t dt 1/ Ue Table 1 forward. y(t)e t dt y() 1/ Integrate by part on the left. y(t)e t dt 1/ 2 Ue y() and divide. y(t)e t dt ( t)e t dt Ue Table 1 backward. y(t) t Apply Lerch cancelation law. Solution found.

4 8.1 Introduction to the Laplace Method 451 Table 3. Laplace method L-notation detail for y 1, y() tranlated from Table 2. L(y (t)) L( 1) Apply L acro y 1, or multiply y 1 by e t dt, integrate t to t. L(y (t)) 1/ L(y(t)) y() 1/ L(y(t)) 1/ 2 L(y(t)) L( t) y(t) t Ue Table 1 forward. Integrate by part on the left. Ue y() and divide. Apply Table 1 backward. Invoke Lerch cancelation law. In Lerch law, the formal rule of eraing the integral ign i valid provided the integral are equal for large and certain condition hold on y and f ee Theorem 2. The illutration in Table 2 how that Laplace theory require an in-depth tudy of a pecial integral table, a table which i a true extenion of the uual table found on the inide cover of calculu book; ee Table 1 and ection 8.2, Table 4. The L-notation for the direct Laplace tranform produce briefer detail, a witneed by the tranlation of Table 2 into Table 3. The reader i advied to move from Laplace integral notation to the L notation a oon a poible, in order to clarify the idea of the tranform method. Some Tranform Rule. The formal propertie of calculu integral plu the integration by part formula ued in Table 2 and 3 lead to thee rule for the Laplace tranform: L(f(t) + g(t)) L(f(t)) + L(g(t)) L(cf(t)) cl(f(t)) L(y (t)) L(y(t)) y() L(y(t)) L(f(t)) implie y(t) f(t) The integral of a um i the um of the integral. Contant c pa through the integral ign. The t-derivative rule, or integration by part. See Theorem 3. Lerch cancelation law. See Theorem 2. The firt two rule are referenced a linearity of the tranform. The rule let u manipulate the ymbol L like it wa a matrix ubject to the rule of matrix algebra. In particular, Laplace method compare to multiplying a vector equation by a matrix.

5 452 Laplace Tranform Example 1 Example (Laplace method) Solve by Laplace method the initial value problem y 5 2t, y() 1 to obtain y(t) 1 + 5t t 2. Solution: Laplace method i outlined in Table 2 and 3. The L-notation of Table 3 will be ued to find the olution y(t) 1 + 5t t 2. L(y (t)) L(5 2t) Apply L acro y 5 2t. 5L(1) 2L(t) Linearity of the tranform Ue Table 1 forward. L(y(t)) y() Apply the part rule, Theorem 3. L(y(t)) Ue y() 1 and divide. L(y(t)) L(1) + 5L(t) L(t 2 ) Apply Table 1 backward. L(1 + 5t t 2 ) Linearity of the tranform. y(t) 1 + 5t t 2 Ue Lerch cancelation law. 2 Example (Laplace method) Solve by Laplace method the initial value problem y 1, y() y () to obtain y(t) 5t 2. Solution: The L-notation of Table 3 will be ued to find the olution y(t) 5t 2. L(y (t)) L(1) Apply L acro y 1. L(y (t)) y () L(1) Apply the part rule to y, that i, replace f by y in Theorem 3. [L(y(t)) y()] y () L(1) Repeat the part rule, on y. 2 L(y(t)) L(1) Ue y() y (). L(y(t)) 1 Ue Table 1 forward. Then divide. 3 L(y(t)) L(5t 2 ) y(t) 5t 2 Apply Table 1 backward. Invoke Lerch cancelation law. Exitence of the Tranform The Laplace integral e t f(t) dt i known to exit in the ene of the improper integral definition 1 N g(t)dt lim g(t)dt N 1 An advanced calculu background i aumed for the Laplace tranform exitence proof. Application of Laplace theory require only a calculu background.

6 8.1 Introduction to the Laplace Method 453 provided f(t) belong to a cla of function known in the literature a function of exponential order. For thi cla of function the relation (2) f(t) lim t e at i required to hold for ome real number a, or equivalently, for ome contant M and α, (3) f(t) Me αt. In addition, f(t) i required to be piecewie continuou on each finite ubinterval of t <, a term defined a follow. Definition 1 (piecewie continuou) A function f(t) i piecewie continuou on a finite interval [a, b] provided there exit a partition a t < < t n b of the interval [a, b] and function f 1, f 2,..., f n continuou on (, ) uch that for t not a partition point (4) f 1 (t) t < t < t 1, f(t).. f n (t) t n 1 < t < t n. The value of f at partition point are undecided by equation (4). In particular, equation (4) implie that f(t) ha one-ided limit at each point of a < t < b and appropriate one-ided limit at the endpoint. Therefore, f ha at wort a jump dicontinuity at each partition point. 3 Example (Exponential order) Show that f(t) e t co t + t i of exponential order. Solution: The proof mut how that f(t) i piecewie continuou and then find an α > uch that lim t f(t)/e αt. Already, f(t) i continuou, hence piecewie continuou. From L Hopital rule in calculu, lim t p(t)/e αt for any polynomial p and any α >. Chooe α 2, then f(t) co t t lim t e 2t lim t e t + lim. t e2t Theorem 1 (Exitence of L(f)) Let f(t) be piecewie continuou on every finite interval in t and atify f(t) Me αt for ome contant M and α. Then L(f(t)) exit for > α and lim L(f(t)).

7 454 Laplace Tranform Proof: It ha to be hown that the Laplace integral of f i finite for > α. Advanced calculu implie that it i ufficient to how that the integrand i abolutely bounded above by an integrable function g(t). Take g(t) Me ( α)t. Then g(t). Furthermore, g i integrable, becaue g(t)dt M α. Inequality f(t) Me αt implie the abolute value of the Laplace tranform integrand f(t)e t i etimated by f(t)e t Me αt e t g(t). The limit tatement lim L(f(t)) follow from L(f(t)) g(t)dt M, becaue the right ide of thi inequality ha limit zero at. The α proof i complete. Theorem 2 (Lerch) If f 1 (t) and f 2 (t) are continuou, of exponential order and f 1 (t)e t dt f 2 (t)e t dt for all >, then f 1 (t) f 2 (t) for t. Proof: See Widder [?]. Theorem 3 (Part Rule or t-derivative Rule) If f(t) i continuou, lim t f(t)e t for all large value of and f (t) i piecewie continuou and of exponential order, then L(f (t)) exit for all large and L(f (t)) L(f(t)) f(). Proof: See page 485. Theorem 4 (Atom have Laplace Integral) Let f(t) be the real or imaginary part of x n e ax+ibx, where b and a are real and n i an integer [f i an atom]. Then f i of exponential order and L(f(t)) exit. Further, if g(t) i a linear combination of atom, then L(g(t)) exit. Proof: By calculu, ln x 2x for x 1. Define c 2 n + a. Then f(t) e n ln x +ax e cx for x 1, which prove f i of exponential order. Becaue olution to undetermined coefficient problem are a linear combination of atom, then Laplace method applie to all uch differential equation. Thi i the cla of all contant-coefficient higher order linear differential equation, and all ytem of differential equation with contant coefficient, having a forcing term which i a linear combination of atom.

8 8.1 Introduction to the Laplace Method 455 Exercie 8.1 Laplace method. Solve the given initial value problem uing Laplace method. 1. y 2, y(). 2. y 1, y(). 3. y t, y(). 4. y t, y(). 5. y 1 t, y(). 6. y 1 + t, y(). 7. y 3 2t, y(). 8. y 3 + 2t, y(). 9. y 2, y() y (). 1. y 1, y() y (). 11. y 1 t, y() y (). 12. y 1 + t, y() y (). 13. y 3 2t, y() y (). 14. y 3 + 2t, y() y (). Exponential order. Show that f(t) i of exponential order, by finding a contant α in each cae uch that f(t) lim t e αt. 15. f(t) 1 + t 16. f(t) e t in(t) 17. f(t) N n c nx n, for any choice of the contant c,..., c N. 18. f(t) N n1 c n in(nt), for any choice of the contant c 1,..., c N. Exitence of tranform. Let f(t) te t2 in(e t2 ). Etablih thee reult. 19. The function f(t) i not of exponential order. 2. The Laplace integral of f(t), f(t)e t dt, converge for all >. Jump Magnitude. For f piecewie continuou, define the jump at t by J(t) lim f(t + h) lim f(t h). h + h + Compute J(t) for the following f. 21. f(t) 1 for t, ele f(t) 22. f(t) 1 for t 1/2, ele f(t) 23. f(t) t/ t for t, f() 24. f(t) in t/ in t for t nπ, f(nπ) ( 1) n Taylor erie. The erie relation L( n c nt n ) n c nl(t n ) often hold, in which cae the reult L(t n ) n! 1 n can be employed to find a erie repreentation of the Laplace tranform. Ue thi idea on the following to find a erie formula for L(f(t)). 25. f(t) e 2t n (2t)n /n! 26. f(t) e t n ( t)n /n!

9 456 Laplace Tranform 8.2 Laplace Integral Table The objective in developing a table of Laplace integral, e.g., Table 4 and 5, i to keep the table ize mall. Table manipulation rule appearing in Table 7, page 463, effectively increae the table ize manyfold, making it poible to olve typical differential equation from electrical and mechanical model. The combination of Laplace table plu the table manipulation rule i called the Laplace tranform calculu. Table 4 i conidered to be a table of minimum ize to be memorized. Table 5 add a number of pecial-ue entrie. For intance, the Heaviide entry in Table 5 might be memorized, but uually not the other. Derivation are potponed to page 492. The theory of the generalized factorial function, the gamma function Γ(x), appear on page 46. Table 4. A minimal Laplace integral table with L-notation (tn )e t dt n! 1+n L(t n ) n! 1+n (eat )e t dt 1 L(e at ) 1 a a (co bt)e t dt 2 + b 2 L(co bt) 2 + b 2 (in b b bt)e t dt 2 + b 2 L(in bt) 2 + b 2 Table 5. Laplace integral table extenion L(H(t a)) e a L(δ(t a)) e a L(floor(t/a)) (a ) Heaviide { unit tep, defined by 1 for t, H(t) otherwie. e a (1 e a ) Dirac delta, δ(t) dh(t). Special uage rule apply. Staircae function, floor(x) greatet integer x. L(qw(t/a)) 1 tanh(a/2) L(a trw(t/a)) 1 2 tanh(a/2) L(t α ) L(t 1/2 ) Γ(1 + α) 1+α π Square wave, qw(x) ( 1) floor(x). Triangular wave, trw(x) x qw(r)dr. Generalized power function, Γ(1 + α) e x x α dx. Becaue Γ(1/2) π.

10 8.2 Laplace Integral Table 457 Table 6. Minimal forward and backward Laplace integral table Forward Table f(t) L(f(t)) t n n! 1+n e at 1 a co bt 2 + b 2 in bt b 2 + b 2 Backward Table L(f(t)) f(t) 1 1+n t n n! 1 a e at 2 + b 2 co bt b 2 in bt b Example 4 Example (Laplace tranform) Let f(t) t(t 5) in 2t+e 3t. Compute L(f(t)) uing the forward Laplace table and tranform linearity propertie. Solution: L(f(t)) L(t 2 5t in 2t + e 3t ) Expand t(t 5). L(t 2 ) 5L(t) L(in 2t) + L(e 3t ) Linearity applied Forward Table. 5 Example (Invere Laplace tranform) Ue the backward Laplace table plu tranform linearity propertie to olve for f(t) in the equation Solution: L(f(t)) L(f(t)) Convert to table entrie. L(co 4t) + 2L(e 3t ) + L(t) L(t2 ) Backward Laplace table. L(co 4t + 2e 3t + t t2 ) Linearity applied. f(t) co 4t + 2e 3t + t t2 Lerch cancelation law. 6 Example (Heaviide) Find the Laplace tranform of f(t) in Figure Figure 1. A piecewie defined function f(t) on t < : f(t) except for 1 t < 2 and 3 t < 4.

11 458 Laplace Tranform Solution: The detail require the ue of the Heaviide function formula (1) H(t a) H(t b) { 1 a t < b, otherwie. The formula for f(t): f(t) 1 1 t < 2, 5 3 t < 4, otherwie { 1 1 t < 2, otherwie f 1 (t) + 5f 2 (t), { 1 3 t < 4, + 5 otherwie where relation (1) implie f 1 (t) H(t 1) H(t 2), f 2 (t) H(t 3) H(t 4). The extended Laplace table give L(f(t)) L(f 1 (t)) + 5L(f 2 (t)) Linearity. L(H(t 1)) L(H(t 2)) + 5L(f 2 (t)) Subtitute for f 1. e e 2 + 5L(f 2 (t)) Extended table ued. e e 2 + 5e 3 5e 4 Similarly for f 2. 7 Example (Dirac delta) A machine hop tool that repeatedly hammer a die i modeled by a Dirac impule model f(t) N n1 δ(t n). Verify the formula L(f(t)) N n1 e n. Solution: L(f(t)) L ( N n1 δ(t n) ) N n1 L(δ(t n)) Linearity. N n1 e n Extended Laplace table. 8 Example (Square wave) A periodic camhaft force f(t) applied to a mechanical ytem ha the idealized graph hown in Figure 2. Verify formula f(t) 1 + qw(t) and L(f(t)) 1 (1 + tanh(/2)) Figure 2. A periodic force f(t) applied to a mechanical ytem.

12 8.2 Laplace Integral Table 459 Solution: 1 + qw(t) { n t < 2n + 1, n, 1,..., 1 1 2n + 1 t < 2n + 2, n, 1,..., { 2 2n t < 2n + 1, n, 1,..., f(t). otherwie, By the extended Laplace table, L(f(t)) L(1) + L(qw(t)) 1 + tanh(/2). 9 Example (Sawtooth wave) Expre the P -periodic awtooth wave repreented in Figure 3 a f(t) ct/p c floor(t/p ) and obtain the formula c L(f(t)) c P 2 ce P e P. P 4P Figure 3. A P -periodic awtooth wave f(t) of height c >. Solution: The repreentation originate from geometry, becaue the periodic function f can be viewed a derived from ct/p by ubtracting the correct contant from each of interval [P, 2P ], [2P, 3P ], etc. The technique ued to verify the identity i to define g(t) ct/p c floor(t/p ) and then how that g i P -periodic and f(t) g(t) on t < P. Two P - periodic function equal on the bae interval t < P have to be identical, hence the repreentation follow. The fine detail: for t < P, floor(t/p ) and floor(t/p + k) k. Hence g(t + kp ) ct/p + ck c floor(k) ct/p g(t), which implie that g i P -periodic and g(t) f(t) for t < P. L(f(t)) c L(t) cl(floor(t/p )) P Linearity. c P 2 ce P e P Baic and extended table applied. 1 Example (Triangular wave) Expre the triangular wave f of Figure 4 in term of the quare wave qw and obtain L(f(t)) 5 π 2 tanh(π/2). 5 2π Figure 4. A 2π-periodic triangular wave f(t) of height 5.

13 46 Laplace Tranform Solution: The repreentation of f in term of qw i f(t) 5 t/π qw(x)dx. Detail: A 2-periodic triangular wave of height 1 i obtained by integrating the quare wave of period 2. A wave of height c and period 2 i given by c trw(t) c t qw(x)dx. Then f(t) c trw(2t/p ) c 2t/P qw(x)dx where c 5 and P 2π. Laplace tranform detail: Ue the extended Laplace table a follow. L(f(t)) 5 5 L(π trw(t/π)) π π 2 tanh(π/2). Gamma Function In mathematical phyic, the Gamma function or the generalized factorial function i given by the identity (2) Γ(x) e t t x 1 dt, x >. Thi function i tabulated and available in computer language like Fortran, C and C++. It i alo available in computer algebra ytem and numerical laboratorie. Some ueful propertie of Γ(x): (3) (4) Γ(1 + x) xγ(x) Γ(1 + n) n! for integer n 1. Detail for relation (3) and (4): Start with e t dt 1, which give Γ(1) 1. Ue thi identity and ucceively relation (3) to obtain relation (4). To prove identity (3), integration by part i applied, a follow: Γ(1 + x) e t t x dt Definition. t x e t t t + e t xt x 1 dt Ue u t x, dv e t dt. x e t t x 1 dt Boundary term are zero for x >. xγ(x). Exercie 8.2 Laplace Tranform. Uing the baic Laplace table and linearity propertie of the tranform, compute L(f(t)). Do not ue the direct Laplace tranform! 1. L(2t) 2. L(4t) 3. L(1 + 2t + t 2 ) 4. L(t 2 3t + 1) 5. L(in 2t) 6. L(co 2t) 7. L(e 2t ) 8. L(e 2t )

14 8.2 Laplace Integral Table L(t + in 2t) 1. L(t co 2t) 11. L(t + e 2t ) 12. L(t 3e 2t ) 13. L((t + 1) 2 ) 14. L((t + 2) 2 ) 15. L(t(t + 1)) 16. L((t + 1)(t + 2)) 17. L( 1 n tn /n!) 18. L( 1 n tn+1 /n!) 19. L( 1 n1 in nt) 2. L( 1 n co nt) Invere Laplace tranform. Solve the given equation for the function f(t). Ue the baic table and linearity propertie of the Laplace tranform. 21. L(f(t)) L(f(t)) L(f(t)) 1/ + 2/ 2 + 3/ L(f(t)) 1/ 3 + 1/ 25. L(f(t)) 2/( 2 + 4) 26. L(f(t)) /( 2 + 4) 27. L(f(t)) 1/( 3) 28. L(f(t)) 1/( + 3) 29. L(f(t)) 1/ + /( 2 + 4) 3. L(f(t)) 2/ 2/( 2 + 4) 31. L(f(t)) 1/ + 1/( 3) 32. L(f(t)) 1/ 3/( 2) 33. L(f(t)) (2 + ) 2 / L(f(t)) ( + 1)/ L(f(t)) (1/ 2 + 2/ 3 ) 36. L(f(t)) ( + 1)( 1)/ L(f(t)) 1 n n!/1+n 38. L(f(t)) 1 n n!/2+n 39. L(f(t)) 1 n n1 2 + n 2 4. L(f(t)) 1 n 2 + n 2 Laplace Table Extenion. Compute the indicated Laplace integral uing the extended Laplace table, page L(H(t 2) + 2H(t)) 42. L(H(t 3) + 4H(t)) 43. L(H(t π)(h(t) + H(t 1))) 44. L(H(t 2π) + 3H(t 1)H(t 2)) 45. L(δ(t 2)) 46. L(5δ(t π)) 47. L(δ(t 1) + 2δ(t 2)) 48. L(δ(t 2)(5 + H(t 1))) 49. L(floor(3t)) 5. L(floor(2t)) 51. L(5 qw(3t)) 52. L(3 qw(t/4)) 53. L(4 trw(2t)) 54. L(5 trw(t/2)) 55. L(t + t 3/2 + t 1/2 ) 56. L(t 3 + t 3/2 + 2t 1/2 ) Invere Laplace, Extended Table. Compute f(t), uing the extended Laplace integral table. 57. L(f(t)) e / 58. L(f(t)) 5e 2 /

15 462 Laplace Tranform 59. L(f(t)) e 2 6. L(f(t)) 5e L(f(t)) 62. L(f(t)) e /3 (1 e /3 ) e 2 (1 e 2 ) 63. L(f(t)) 4 tanh() 64. L(f(t)) 5 tanh(3) L(f(t)) 4 tanh() L(f(t)) 5 tanh(2) L(f(t)) L(f(t)) 1 3

16 8.3 Laplace Tranform Rule Laplace Tranform Rule In Table 7, the baic table manipulation rule are ummarized. tatement and proof of the rule appear in ection 8.5, page 484. Full The rule are applied here to everal key example. Partial fraction expanion do not appear here, but in ection 8.4, in connection with Heaviide coverup method. Table 7. Laplace tranform rule L(f(t) + g(t)) L(f(t)) + L(g(t)) L(cf(t)) cl(f(t)) L(y (t)) L(y(t)) y() Linearity. The Laplace of a um i the um of the Laplace. Linearity. Contant move through the L-ymbol. The t-derivative rule. Derivative L(y ) are replaced in tranformed equation. ( ) t L g(x)dx 1 L(g(t)) The t-integral rule. L(tf(t)) d d L(f(t)) The -differentiation rule. Multiplying f by t applie d/d to the tranform of f. L(e at f(t)) L(f(t)) ( a) L(f(t a)h(t a)) e a L(f(t)), L(g(t)H(t a)) e a L(g(t + a)) L(f(t)) P Firt hifting rule. Multiplying f by e at replace by a. Second hifting rule. Firt and econd form. f(t)e t dt 1 e P Rule for P -periodic function. Aumed here i f(t + P ) f(t). L(f(t))L(g(t)) L((f g)(t)) Convolution rule. t Define (f g)(t) f(x)g(t x)dx. Example 11 Example (Harmonic ocillator) Solve the initial value problem x +x, x(), x () 1 by Laplace method. Solution: The olution i x(t) in t. The detail: L(x ) + L(x) L() L(x ) x () + L(x) [L(x) x()] x () + L(x) Apply L acro the equation. Ue the t-derivative rule. Ue again the t-derivative rule. ( 2 + 1)L(x) 1 Ue x(), x () 1. L(x) Divide. L(in t) Baic Laplace table. x(t) in t Invoke Lerch cancellation law.

17 464 Laplace Tranform 12 Example (-differentiation rule) Show the tep for L(t 2 e 5t ) Solution: ( L(t 2 e 5t ) d ) ( d ) L(e 5t ) d d ( 1) 2 d ( ) d 1 d d 5 d ( ) 1 d ( 5) 2 2 ( 5) 3. Apply -differentiation. Baic Laplace table. Calculu power rule. 2 ( 5) 3 Identity verified. 13 Example (Firt hifting rule) Show the tep for L(t 2 e 3t ) Solution: 2 ( + 3) 3. L(t 2 e 3t ) L(t 2 ) ( 3) Firt hifting rule. ( ) 2 ( 3) 2+1 Baic Laplace table. 2 ( + 3) 3 Identity verified. 14 Example (Second hifting rule) Show the tep for L(in t H(t π)) e π Solution: The econd hifting rule i applied a follow. LHS L(in t H(t π)) Left ide of the identity. L(g(t)H(t a)) Chooe g(t) in t, a π. e a L(g(t + a) Second form, econd hifting theorem. e π L(in(t + π)) Subtitute a π. e π L( in t) Sum rule in(a + b) in a co b + in b co a plu in π, co π 1. e π Baic Laplace table. RHS Identity verified.

18 8.3 Laplace Tranform Rule Example (Trigonometric formula) Show the tep ued to obtain thee Laplace identitie: (a) L(t co at) 2 a 2 ( 2 + a 2 ) 2 (c) L(t 2 co at) 2(3 3a 2 ) ( 2 + a 2 ) 3 (b) L(t in at) Solution: The detail for (a): L(t co at) (d/d)l(co at) d ( ) d 2 + a 2 The detail for (c): 2a ( 2 + a 2 ) 2 (d) L(t 2 in at) 62 a a 3 ( 2 + a 2 ) 3 Ue -differentiation. Baic Laplace table. 2 a 2 ( 2 + a 2 ) 2 Calculu quotient rule. L(t 2 co at) (d/d)l(( t) co at) d ( 2 a 2 ) d ( 2 + a 2 ) 2 Ue -differentiation. Reult of (a). 23 6a 2 ) ( 2 + a 2 ) 3 Calculu quotient rule. The imilar detail for (b) and (d) are left a exercie. 16 Example (Exponential) Show the tep ued to obtain thee Laplace identitie: (a) L(e at a co bt) ( a) 2 + b 2 (c) L(te at co bt) ( a)2 b 2 (( a) 2 + b 2 ) 2 (b) L(e at b in bt) ( a) 2 + b 2 (d) L(te at 2b( a) in bt) (( a) 2 + b 2 ) 2 Solution: Detail for (a): L(e at co bt) L(co bt) a Firt hifting rule. ( ) 2 + b a 2 Baic Laplace table. a ( a) 2 + b 2 Verified (a). Detail for (c): L(te at co bt) L(t co bt) a ( d ) d a L(co bt) ( d ( )) d 2 + b a 2 Firt hifting rule. Apply -differentiation. Baic Laplace table.

19 466 Laplace Tranform ( ) 2 b 2 ( 2 + b 2 ) a 2 ( a)2 b 2 (( a) 2 + b 2 ) 2 Calculu quotient rule. Verified (c). Left a exercie are (b) and (d). 17 Example (Hyperbolic function) Etablih thee Laplace tranform fact about coh u (e u + e u )/2 and inh u (e u e u )/2. (a) L(coh at) 2 a 2 (c) L(t coh at) 2 + a 2 ( 2 a 2 ) 2 a 2a (b) L(inh at) 2 a 2 (d) L(t inh at) ( 2 a 2 ) 2 Solution: The detail for (a): L(coh at) 1 2 (L(eat ) + L(e at )) Definition plu linearity of L. 1 ( 1 2 a + 1 ) Baic Laplace table. + a 2 a 2 Identity (a) verified. The detail for (d): L(t inh at) d d ( a ) 2 a 2 Apply the -differentiation rule. a(2) ( 2 a 2 ) 2 Calculu power rule; (d) verified. Left a exercie are (b) and (c). 18 Example (-differentiation) Solve L(f(t)) Solution: The olution i f(t) t in t. The detail: 2 ( 2 for f(t). + 1) 2 2 L(f(t)) ( 2 + 1) 2 d ( ) 1 d Calculu power rule (u n ) nu n 1 u. d (L(in t)) d Baic Laplace table. L(t in t) Apply the -differentiation rule. f(t) t in t Lerch cancellation law. 19 Example (Firt hift rule) Solve L(f(t)) for f(t).

20 8.3 Laplace Tranform Rule 467 Solution: The anwer i f(t) e t co t + e t in t. The detail: L(f(t)) Signal for thi method: the denominator ha complex root. + 2 ( + 1) Complete the quare, denominator. S + 1 S Subtitute S for + 1. S S S Split into Laplace table entrie. L(co t) + L(in t) S+1 Baic Laplace table. L(e t co t) + L(e t in t) Firt hift rule. f(t) e t co t + e t in t Invoke Lerch cancellation law. 2 Example (Damped ocillator) Solve by Laplace method the initial value problem x + 2x + 2x, x() 1, x () 1. Solution: The olution i x(t) e t co t. The detail: L(x ) + 2L(x ) + 2L(x) L() Apply L acro the equation. L(x ) x () + 2L(x ) + 2L(x) The t-derivative rule on x. [L(x) x()] x () +2[L(x) x()] + 2L(x) The t-derivative rule on x. ( )L(x) 1 + Ue x() 1, x () L(x) Divide to iolate L(x). + 1 ( + 1) 2 Complete the quare in the denominator. + 1 L(co t) +1 Baic Laplace table. L(e t co t) Firt hifting rule. x(t) e t co t Invoke Lerch cancellation law. 21 Example (Rectified ine wave) Compute the Laplace tranform of the rectified ine wave f(t) in ωt and how that it can be expreed in the form L( in ωt ) ω coth ( ) π 2ω 2 + ω 2. Solution: The periodic function formula will be applied with period P 2π/ω. The calculation reduce to the evaluation of J P f(t)e t dt. Becaue in ωt on π/ω t 2π/ω, integral J can be written a J J 1 + J 2, where J 1 π/ω in ωt e t dt, J 2 2π/ω π/ω in ωt e t dt.

21 468 Laplace Tranform Integral table give the reult in ωt e t dt ωe t co(ωt) 2 + ω 2 e t in(ωt) 2 + ω 2. Then J 1 ω(e π /ω + 1) 2 + ω 2, J 2 ω(e 2π/ω + e π/ω ) 2 + ω 2, J ω(e π/ω + 1) ω 2. The remaining challenge i to write the anwer for L(f(t)) in term of coth. The detail: J L(f(t)) 1 e P Periodic function formula. J Apply 1 x 2 (1 x)(1 + x), (1 e P /2 )(1 + e P /2 ) x e P /2. ω(1 + e P /2 ) Cancel factor 1 + e P /2. (1 e P /2 )( 2 + ω 2 ) ep /4 P /4 + e ω e P /4 e P /4 2 + ω 2 2 coh(p /4) ω 2 inh(p /4) 2 + ω 2 Factor out e P /4, then cancel. Apply coh, inh identitie. ω coth(p /4) 2 + ω 2 Ue coth u coh u/ inh u. ω coth ( ) π 2ω 2 + ω 2 Identity verified. 22 Example (Half wave Rectification) Determine the Laplace tranform of the half wave rectification of in ωt, denoted g(t), in which the negative cycle of in ωt have been replaced by zero, to create g(t). Show in particular that L(g(t)) 1 ω ω 2 ( 1 + coth ( )) π 2ω Solution: The half wave rectification of in ωt i g(t) (in ωt + in ωt )/2. Therefore, the baic Laplace table plu the reult of Example 21 give Dividing by 2 produce the identity. L(2g(t)) L(in ωt) + L( in ωt ) ω 2 + ω 2 + ω coh(π/(2ω)) 2 + ω 2 ω 2 (1 + coh(π/(2ω)) + ω2 23 Example (Shifting Rule I) Solve L(f(t)) e for f(t).

22 8.3 Laplace Tranform Rule 469 Solution: The anwer i f(t) e 3 t co(t 3)H(t 3). The detail: L(f(t)) e ( + 1) Complete the quare. e 3 S S Replace + 1 by S. e 3S+3 (L(co t)) S+1 Baic Laplace table. e 3 ( e 3 L(co t) ) S+1 Regroup factor e 3S. e 3 (L(co(t 3)H(t 3))) S+1 Second hifting rule. e 3 L(e t co(t 3)H(t 3)) Firt hifting rule. f(t) e 3 t co(t 3)H(t 3) Lerch cancellation law. 24 Example (Shifting Rule II) Solve L(f(t) for f(t). Solution: The anwer i f(t) e 2t (co 2t in 2t). The detail: + 7 L(f(t)) ( + 2) S + 5 S S S S S+2 Complete the quare. Replace + 2 by S. Split into table entrie. Shifting rule preparation. L ( co 2t in 2t) S+2 Baic Laplace table. L(e 2t (co 2t in 2t)) Firt hifting rule. f(t) e 2t (co 2t in 2t) Lerch cancellation law. Exercie 8.3 Second Order Initial Value Problem. Diplay the Laplace method detail which verify the upplied anwer. 1. x + x 1, x() 1, x () ; x(t) x + 4x 4, x() 1, x () ; x(t) x + x, x() 1, x () 1; x(t) co t + in t. 4. x + x, x() 1, x () 2; x(t) co t + 2 in t. 5. x + 2x + x, x(), x () 1; x(t) te t. 6. x + 2x + x, x() 1, x () 1; x(t) e t. 7. x + 3x + 2x, x() 1, x () 1; x(t) e t. 8. x + 3x + 2x, x() 1, x () 2; x(t) e 2t. 9. x + 3x, x() 5, x () ; x(t) x + 3x, x() 1, x () 3; x(t) e 3t.

23 47 Laplace Tranform 11. x 2, x(), x () ; x(t) t x 6t, x(), x () ; x(t) t x 2, x(), x () 1; x(t) t + t x 6t, x(), x () 1; x(t) t + t x + x 6t, x(), x () 1; x(t) t + t x + x 6t, x(), x () 1; x(t) t + t 3.

24 8.4 Heaviide Method Heaviide Method The method olve an equation like L(f(t)) 2 ( + 1)( 2 + 1) for the t-expreion f(t) e t + co t + in t. The detail in Heaviide method involve a equence of eay-to-learn college algebra tep. Thi practical method wa popularized by the Englih electrical engineer Oliver Heaviide ( ). More preciely, Heaviide method tart with a polynomial quotient (1) a + a a n n b + b b m m and compute an expreion f(t) uch that a + a a n n b + b b m m L(f(t)) f(t)e t dt. It i aumed that a,..., a n, b,..., b m are contant and the polynomial quotient (1) ha limit zero at. Partial Fraction Theory In college algebra, it i hown that a rational function (1) can be expreed a the um of partial fraction, which are fraction with a contant in the numerator, and a denominator having jut one root. Such term have the form (2) A ( ) k. The numerator in (2) i a real or complex contant A and the denominator ha exactly one root. The power ( ) k mut divide the denominator in (1). Aume fraction (1) ha real coefficient. If in (2) i real, then A i real. If α + iβ in (2) i complex, then ( ) k alo divide the denominator in (1), where α iβ i the complex conjugate of. The correponding partial fraction ued in the expanion turn out to be complex conjugate of one another, which can be paired and re-written a a fraction (3) A ( ) k + A ( ) k Q() (( α) 2 + β 2 ) k, where Q() i a real polynomial. Thi jutifie the replacement of all partial fraction A/( ) k with complex by B + C (( )( )) k B + C (( α) 2 + β 2 ) k,

25 472 Laplace Tranform in which B and C are real contant. Thi real form i preferred over the um of complex fraction, becaue integral table and Laplace table typically contain only real formula. Simple Root. Aume that (1) ha real coefficient and the denominator of the fraction (1) ha ditinct real root 1,..., N and ditinct complex root α 1 ±iβ 1,..., α M ±iβ M. The partial fraction expanion of (1) i a um given in term of real contant A p, B q, C q by (4) a + a a n n N b + b b m m p1 A p p + M q1 B q + C q ( α q ) ( α q ) 2 + β 2 q. Multiple Root. Aume (1) ha real coefficient and the denominator of the fraction (1) ha poibly multiple root. Let N p be the multiplicity of real root p and let M q be the multiplicity of complex root α q + iβ q (β q > ), 1 p N, 1 q M. The partial fraction expanion of (1) i given in term of real contant A p,k, B q,k, C q,k by (5) N p1 1 k N p A M p,k ( p ) k + q1 B q,k + C q,k( α q ) (( α 1 k M q ) 2 + β 2 q q ) k. Summary. The theory for imple root and multiple root can be ditilled a follow. A polynomial quotient p/q with limit zero at infinity ha a unique expanion into partial fraction. A partial fraction i either a contant divided by a divior of q having exactly one real root, or ele a linear function divided by a real divior of q, having exactly one complex conjugate pair of root. The Sampling Method Conider the expanion in partial fraction (6) 1 ( + 1) 2 ( 2 + 1) A + B C ( + 1) 2 + D + E The five undetermined real contant A through E are found by clearing the fraction, that i, multiply (6) by the denominator on the left to obtain the polynomial equation (7) 1 A( + 1) 2 ( 2 + 1) + B( + 1)( 2 + 1) +C( 2 + 1) + (D + E)( + 1) 2.

26 8.4 Heaviide Method 473 Next, five different value of are ubtituted into (7) to obtain equation for the five unknown A through E. 2 We alway ue the root of the denominator to tart:, 1, i, i are the root of ( + 1) 2 ( 2 + 1). Each complex root reult in two equation, by taking real and imaginary part. The complex conjugate root i i not ued, becaue it duplicate equation already obtained from i. The three root, 1, i give only four equation, o we invent another value 1 to get the fifth equation: (8) 1 A ( ) 2 2C ( 1) i 1 (Di + E)i(i + 1) 2 ( i) 8A + 4B + 2C + 4(D + E) ( 1) Becaue D and E are real, the complex equation ( i) become two equation, a follow. i 1 (Di + E)i(i 2 + 2i + 1) Expand power. i 1 2Di 2E Simplify uing i D Equate imaginary part. 1 2E Equate real part. Solving the 5 5 ytem, the anwer are A 1, B 3/2, C 1, D 1/2, E 1/2. The Method of Atom Conider the expanion in partial fraction (9) 2 2 ( + 1) 2 ( 2 + 1) a + b c ( + 1) 2 + d + e Clearing the fraction in (9) give the polynomial equation (1) 2 2 a( + 1) 2 ( 2 + 1) + b( + 1)( 2 + 1) +c( 2 + 1) + (d + e)( + 1) 2. The method of atom expand all polynomial product and collect on power of (function 1,, 2,... are by definition called atom). The coefficient of the power are matched to give 5 equation in the five unknown a through e. Some detail: (11) 2 2 (a + b + d) 4 + (2a + b + c + 2d + e) 3 + (2a + b + d + 2e) 2 + (2a + b + c + e) + a 2 The value choen for are ample, that i, cleverly choen value. Generally, the number of value elected equal the number of ymbol A, B,... to be determined.

27 474 Laplace Tranform Matching power of implie the 5 equation a + b + d, 2a + b + c + 2d + e, 2a + b + d + 2e, 2a + b + c + e 2, a 2. Solving, the unique olution i a 2, b 3, c 2, d 1, e 1. Heaviide Coverup Method The method applie only to the cae of ditinct root of the denominator in (1). Extenion to multiple-root cae can be made; ee page 475. To illutrate Oliver Heaviide idea, conider the problem detail (12) ( 1)( + 1) A + B 1 + C + 1 L(A) + L(Be t ) + L(Ce t ) L(A + Be t + Ce t ) The firt line (12) ue college algebra partial fraction. The econd and third line ue the baic Laplace table and linearity of L. Myteriou Detail. Oliver Heaviide propoed to find in (12) the contant C 1 2 by a cover up method: C. ( 1) +1 The intruction are to cover up the matching factor ( + 1) on the left and right with box (Heaviide ued two fingertip), then evaluate on the left at the root which caue the box content to be zero. The other term on the right are replaced by zero. To jutify Heaviide cover up method, clear the fraction C/( + 1), that i, multiply (12) by the denominator + 1 of the partial fraction C/( + 1) to obtain the partially-cleared fraction relation (2 + 1) ( + 1) ( 1) ( + 1) A ( + 1) B ( + 1) C ( + 1) ( + 1). Set ( + 1) in the diplay. Cancelation left and right plu annihilation of two term on the right give Heaviide precription ( 1) C. + 1

28 8.4 Heaviide Method 475 The factor ( + 1) in (12) i by no mean pecial: the ame procedure applie to find A and B. The method work for denominator with imple root, that i, no repeated root are allowed. Heaviide method in word: A To determine A in a given partial fraction, multiply the relation by ( ), which partially clear the fraction. Subtitute from the equation into the partially cleared relation. Extenion to Multiple Root. Heaviide method can be extended to the cae of repeated root. The baic idea i to factor out the repeat. To illutrate, conider the partial fraction expanion detail R 1 ( + 1) 2 ( + 2) 1 ( ) 1 ( + 1)( + 2) ( ) ( + 1) ( + 1)( + 2) 1 ( + 1) A ample rational function having repeated root. Factor out the repeat. Apply the cover up method to the imple root fraction. Multiply. Apply the cover up method to the lat fraction on the right. Term with only one root in the denominator are already partial fraction. Thu the work center on expanion of quotient in which the denominator ha two or more root. Special Method. Heaviide method ha a ueful extenion for the cae of root of multiplicity two. To illutrate, conider thee detail: R 1 ( + 1) 2 ( + 2) A B ( + 1) 2 + C + 2 A ( + 1) ( + 1) A fraction with multiple root. 2 See equation (5), page Find B and C by Heaviide cover up method. 4 Detail below.

29 476 Laplace Tranform We dicu 4 detail. Multiply the equation 1 2 by +1 to partially clear fraction, the ame tep a the cover-up method: 1 ( + 1)( + 2) A + B C( + 1) We don t ubtitute + 1, becaue it give infinity for the econd term. Intead, et to get the equation A + C. Becaue C 1 from 3, then A 1. The illutration work for one root of multiplicity two, becaue will reolve the coefficient not found by the cover up method. In general, if the denominator in (1) ha a root of multiplicity k, then the partial fraction expanion contain term A 1 A 2 + ( ) A k ( ) k. Heaviide cover up method directly find A k, but not A 1 to A k 1. Conider the par- Cover-up Method and Complex Number. tial fraction expanion 1 ( + 1)( 2 + 9) A B + C The ymbol A, B, C are real. The value of A can be found directly by the cover-up method, giving A 1. To find B and C, multiply the fraction expanion by 2 + 9, in order to partially clear fraction, then formally et to obtain the two equation B + C, The method applie the identical idea ued for one real root. By clearing fraction in the firt, the equation become 1 B 2 + C + B + C, Subtitute 2 9 into the firt equation to give the linear equation 1 ( 9B + C) + (B + C). Becaue thi linear equation ha two complex root ±3i, then real contant B, C atify the 2 2 ytem Solving give B 1, C 1. 9B + C 1, B + C. The ame method applie epecially to fraction with 3-term denominator, like The only change made in the detail i the replacement 2 1. By repeated application of 2 1, the firt equation can be ditilled into one linear equation in with two root. A before, a 2 2 ytem reult.

30 8.4 Heaviide Method 477 Example 25 Example (Partial Fraction I) Show the detail of the partial fraction expanion ( 1)( 2 + 4)( ) 2/ / Solution: Background. The problem originate a equality 5 6 in the equence of Example 27, page 479, which olve for x(t) uing the method of partial fraction: L(x) ( 1)( 2 + 4)( ) 6 2/ / College algebra detail. College algebra partial fraction theory ay that there exit real contant A, B, C, D, E atifying the identity ( 1)( 2 + 4)( ) A 1 + B + C D + E A explained on page 471, the complex conjugate root ±2i and 1 ± i are not repreented a term c/( ), but in the combined real form een in the above diplay, which i uited for ue with Laplace table. The ampling method applie to find the contant. fraction are cleared to obtain the polynomial relation A( 2 + 4)( ) +(B + C)( 1)( ) +(D + E)( 1)( 2 + 4). In thi method, the The root of the denominator ( 1)( 2 + 4)( ) to be inerted into the previou equation are 1, 2i, 1+i. The conjugate root 2i and 1 i are not ued. Each complex root generate two equation, by equating real and imaginary part, therefore there will be 5 equation in 5 unknown. Subtitution of 1, 2i, 1 + i give three equation A, 2i 4i 3 (B + 2iC)(2i 1)( 4 + 4i + 2), 1 + i 5 (D E + Ei)( 2 + i)(2 2( 1 + i)). Writing each expanded complex equation in term of it real and imaginary part, explained in detail below, give 5 equation 1 2 5A, 2i 3 6B + 16C, 2i 4 8B 12C, 1 + i 5 6D 2E, 1 + i 8D 14E. The equation are olved to give A 2/5, B 1/2, C, D 7/1, E 2/5 (detail for B, C below).

31 478 Laplace Tranform Complex equation to two real equation. It i an algebraic mytery how exactly the complex equation 4i 3 (B + 2iC)(2i 1)( 4 + 4i + 2) get converted into two real equation. The proce i explained here. Firt, the complex equation i expanded, a though it i a polynomial in variable i, to give the tep 4i 3 (B + 2iC)(2i 1)( 2 + 4i) (B + 2iC)( 4i i 2 4i) Expand. (B + 2iC)( 6 8i) Ue i B 12iC 8Bi + 16C Expand, ue i 2 1. ( 6B + 16C) + ( 8B 12C)i Convert to form x + yi. Next, the two ide are compared. Becaue B and C are real, then the real part of the right ide i ( 6B + 16C) and the imaginary part of the right ide i ( 8B 12C). Equating matching part on each ide give the equation 6B + 16C 3, 8B 12C 4, which i a 2 2 linear ytem for the unknown B, C. Solving the 2 2 ytem. Such a ytem with a unique olution can be olved by Cramer rule, matrix inverion or elimination. The anwer: B 1/2, C. The eaiet method turn out to be elimination. Multiply the firt equation by 4 and the econd equation by 3, then ubtract to obtain C. Then the firt equation i 6B + 3, implying B 1/2. 26 Example (Partial Fraction II) Verify the partial fraction expanion ( + 1) 2 ( ) Solution: Baic partial fraction theory implie that there are unique real contant a, b, c, d, e, f atifying the equation (13) ( + 1) 2 ( ) 2 a b ( + 1) 2 + c + d e + f ( ) 2 The ampling method applie to clear fraction and replace the fractional equation by the polynomial relation a( + 1)( ) 2 +b( ) 2 +(c + d)( )( + 1) 2 +(e + f )( + 1) 2

32 8.4 Heaviide Method 479 However, the prognoi for the reultant algebra i grim: only three of the ix required equation can be obtained by ubtitution of the root ( 1, 1/2 + i 3/2) of the denominator. We abandon the idea, becaue of the complexity of the 6 6 ytem of linear equation required to olve for the contant a through f. Intead, the fraction R on the left of (13) i written with repeated factor extracted, a follow: ( ) 1 p(x) R ( + 1)( ) ( + 1)( 2, + + 1) p(x) Long diviion give the formula p(x) ( + 1)( ) Therefore, the fraction R on the left of (13) can be written a R p(x) ( + 1) 2 ( ) 2 ( + 3) 2 ( + 1)( ). The implified form of R ha a partial fraction expanion ( + 3) 2 ( + 1)( ) a b + c Heaviide cover-up method give a 4. Applying Heaviide method again to the quadratic factor implie the pair of equation ( + 3) b + c, Thee equation can be olved for b 5, c 3. The detail aume that i a root of , then ( + 3) 2 b + c + 1 The firt equation b + c + 1 Expand b + c + 1 Ue ( + 1)(b + c) Clear fraction b + c + b + c 2 Expand again b + c + b c c Ue The concluion 5 b and 8 b c follow becaue the lat equation i linear but ha two complex root. Then b 5, c Example (Third Order Initial Value Problem) Solve the third order initial value problem x x + 4x 4x 5e t in t, x(), x () x () 1.

33 48 Laplace Tranform Solution: The anwer i x(t) 2 5 et in 2t 3 1 e t in t 2 5 e t co t. Method. Apply L to the differential equation. In tep 1 to 3 the Laplace integral of x(t) i iolated, by applying linearity of L, integration by part L(f ) L(f) f() and the baic Laplace table. L(x ) L(x ) + 4L(x ) 4L(x) 5L(e t in t) 1 ( 3 L(x) 1) ( 2 L(x) 1) +4(L(x)) 4L(x) 5 2 ( + 1) ( )L(x) 5 ( + 1) Step 5 and 6 ue the college algebra theory of partial fraction, the detail of which appear in Example 25, page 477. Step 7 and 8 write the partial fraction expanion in term of Laplace table entrie. Step 9 convert the -expreion, which are baic Laplace table entrie, into Laplace integral expreion. Algebraically, we replace -expreion by expreion in ymbol L and t. 5 (+1) L(x) ( 1)( 2 + 4)( ) 2/ / / / / ( + 1) 2 1/1 + 4 ( + 1) / / /1 (2/5)( + 1) ( + 1) ( + 1) ( 2 L 5 et in 2t 3 1 e t in t 2 ) 5 e t co t The lat tep 1 applie Lerch cancelation theorem to the equation 4 9. x(t) 2 5 et in 2t 3 1 e t in t 2 5 e t co t 1 28 Example (Second Order Sytem) Solve for x(t) and y(t) in the 2nd order ytem of linear differential equation 2x x + 9x y y 3y, x() x () 1, 2 + x + 7x y + y 5y, y() y ().

34 8.4 Heaviide Method 481 Solution: The anwer i x(t) 1 3 et co(2 t) + 1 in(2 t), 3 y(t) 2 3 et 2 3 co(2 t) 1 in(2 t). 3 Tranform. The intent of tep 1 and 2 i to tranform the initial value problem into two equation in two unknown. Ued repeatedly in 1 i integration by part L(f ) L(f) f(). No Laplace table were ued. In 2 the ubtitution x 1 L(x), x 2 L(y) are made to produce two equation in the two unknown x 1, x 2. ( )L(x) + ( 2 3)L(y) 1 + 2, ( )L(x) + ( 2 + 5)L(y) 3 + 2, ( )x 1 + ( 2 3)x , ( )x 1 + ( 2 + 5)x Step 3 ue Cramer rule to compute the anwer x 1, x 2 to the equation ax 1 + bx 2 e, cx 1 + dx 2 f a the determinant fraction e b f d a e c f x 1 a b, x 2 c d a b. c d The variable name x 1, x 2 tand for the Laplace integral of the unknown x(t), y(t), repectively. The anwer, following a calculation: 2 + 2/3 x , 1/3 x Step 4 write each fraction reulting from Cramer rule a a partial fraction expanion uited for revere Laplace table look-up. Step 5 doe the table look-up and prepare for tep 6 to apply Lerch cancelation law, in order to diplay the anwer x(t), y(t). x 1 1/ x 2 2/ ( 1 L(x(t)) L L(y(t)) L , et co(2 t) + 1 ) in(2 t), 3 ( 2 3 et 2 3 co(2 t) 1 ) in(2 t). 3 x(t) 1 3 et co(2 t) + 1 in(2 t), 3 y(t) 2 3 et 2 3 co(2 t) in(2 t)

35 482 Laplace Tranform Partial fraction detail. We will how how to obtain the expanion 2 + 2/ / The denominator factor into 1 time Partial fraction theory implie that there i an expanion with real coefficient A, B, C of the form 2 + 2/3 ( 1)( 2 + 4) A 1 + B + C We will verify A 1/3, B 2/3, C 2/3. Clear the fraction to obtain the polynomial equation 2 + 2/3 A( 2 + 4) + (B + C)( 1). Intead of uing 1 and 2i, which are root of the denominator, we hall ue 1,, 1 to get a real 3 3 ytem for variable A, B, C: 1 : 1 + 2/3 A(1 + 4) +, : + 2/3 A(4) + C( 1), 1 : 1 + 2/3 A(1 + 4) + ( B + C)( 2). Write thi ytem a an augmented matrix G with variable A, B, C aigned to the firt three column of G: 5 5/3 G 4 1 2/ /3 Uing computer ait, calculate rref(g) 1 1/3 1 2/3 1 2/3 Then A, B, C are the lat column entrie of rref(g), which verifie the partial fraction expanion. Heaviide cover-up detail. It i poible to rapidly check that A 1/3 uing the cover-up method. Le obviou i that the cover-up method alo applie to the fraction with complex root. The idea i to multiply the fraction decompoition by to partially clear the fraction and then et Thi proce formally et equal to one of the two root ±2i. We avoid complex number entirely by olving for B, C in the pair of equation 2 + 2/3 1 A() + (B + C), Becaue /3 4, the firt equality i implified to B + C. Swap 1 ide of the equation, then cro-multiply to obtain B 2 +C B C 1/3 and then ue 2 4 again to implify to ( B + C) + ( 4B C) 1/3. Becaue thi linear equation in variable ha two olution, then B + C

36 8.4 Heaviide Method 483 and 4B C 1/3. B C 2/3. Solve thi 2 2 ytem by elimination to obtain We review the algebraic method. Firt, we found two equation in ymbol, B, C. Next, ymbol i eliminated to give two equation in ymbol B, C. Finally, the 2 2 ytem for B, C i olved.

37 484 Laplace Tranform 8.5 Tranform Propertie Collected here are the major theorem for the manipulation of Laplace tranform table, along with their derivation. Student who tudy in iolation are advied to dwell on the detail of proof and re-read the example of preceding ection. Theorem 5 (Linearity) The Laplace tranform ha thee inherited integral propertie: (a) (b) L(f(t) + g(t)) L(f(t)) + L(g(t)), L(cf(t)) cl(f(t)). Theorem 6 (The t-derivative Rule) Let y(t) be continuou, of exponential order and let y (t) be piecewie continuou on t. Then L(y (t)) exit and L(y (t)) L(y(t)) y(). Theorem 7 (The t-integral Rule) Let g(t) be of exponential order and continuou for t. Then ( ) L t g(x) dx 1 L(g(t)) or equivalently ( ) L(g(t)) L t g(x) dx Theorem 8 (The -Differentiation Rule) Let f(t) be of exponential order. Then L(tf(t)) d d L(f(t)). Theorem 9 (Firt Shifting Rule) Let f(t) be of exponential order and < a <. Then L(e at f(t)) L(f(t)) ( a). Theorem 1 (Second Shifting Rule) Let f(t) and g(t) be of exponential order and aume a. Then (a) (b) L(f(t a)h(t a)) e a L(f(t)), L(g(t)H(t a)) e a L(g(t + a)). Theorem 11 (Periodic Function Rule) Let f(t) be of exponential order and atify f(t + P ) f(t). Then L(f(t)) P f(t)e t dt 1 e P.

38 8.5 Tranform Propertie 485 Theorem 12 (Convolution Rule) Let f(t) and g(t) be of exponential order. Then ( t ) L(f(t))L(g(t)) L f(x)g(t x)dx. Proof of Theorem 5 (linearity): LHS L(f(t) + g(t)) Left ide of the identity in (a). (f(t) + g(t))e t dt Direct tranform. f(t)e t dt + g(t)e t dt Calculu integral rule. L(f(t)) + L(g(t)) Equal RHS; identity (a) verified. LHS L(cf(t)) Left ide of the identity in (b). cf(t)e t dt Direct tranform. c f(t)e t dt Calculu integral rule. cl(f(t)) Equal RHS; identity (b) verified. Proof of Theorem 6 (t-derivative rule): Already L(f(t)) exit, becaue f i of exponential order and continuou. On an interval [a, b] where f i continuou, integration by part uing u e t, dv f (t)dt give b a f (t)e t dt f(t)e t tb ta b a f(t)( )e t dt f(a)e a + f(b)e b + b a f(t)e t dt. On any interval [, N], there are finitely many interval [a, b] on each of which f i continuou. Add the above equality acro thee finitely many interval [a, b]. The boundary value on adjacent interval match and the integral add to give N N f (t)e t dt f()e + f(n)e N + f(t)e t dt. Take the limit acro thi equality a N. Then the right ide ha limit f() + L(f(t)), becaue of the exitence of L(f(t)) and lim t f(t)e t for large. Therefore, the left ide ha a limit, and by definition L(f (t)) exit and L(f (t)) f() + L(f(t)). Proof of Theorem 7 (t-integral rule): Let f(t) t g(x)dx. Then f i of exponential order and continuou. The detail: L( t g(x)dx) L(f(t)) By definition. 1 L(f (t)) Becaue f() implie L(f (t)) L(f(t)). 1 L(g(t)) Becaue f g by the Fundamental theorem of calculu.