Laplace Transform. f(t)e st dt,

Transcription

1 Chapter 7 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 direct Laplace tranform or the Laplace integral of a function f(t) defined for t < i the ordinary calculu integration problem f(t)e t dt, uccinctly denoted L(f(t)) in cience and engineering literature. The L notation recognize that integration alway proceed over t to t and that the integral involve an 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. 7.1 Introduction to the Laplace Method The foundation of Laplace theory i Lerch cancellation 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 ought-after unknown while f(t) i an explicit expreion taken from integral table. Below, we illutrate Laplace method by olving the initial value problem y 1, y(). The method obtain a relation L(y(t)) L( t), whence Lerch cancellation law implie the olution i 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.

2 7.1 Introduction to the Laplace Method 247 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. 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 An Illutration. The idea of the Laplace method will be illutrated for the olution y(t) t of the problem y 1, y(). The method, entirely different from variation of parameter or undetermined coefficient, ue baic calculu and college algebra; ee Table 2. Table 2. Laplace method detail for the illutration y 1, y(). y (t)e t e t Multiply y 1 by e t. y (t)e t dt e t dt Integrate t to t. y (t)e t dt 1/ Ue Table 1. 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. y(t) t Apply Lerch cancellation law.

3 248 Laplace Tranform 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. Some entrie for the pecial integral table appear in Table 1 and alo in ection 7.2, Table 4. The L-notation for the direct Laplace tranform produce briefer detail, a witneed by the tranlation of Table 2 into Table 3 below. 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. 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, integrate t to t. L(y (t)) 1/ Ue Table 1. L(y(t)) y() 1/ L(y(t)) 1/ 2 Integrate by part on the left. Ue y() and divide. L(y(t)) L( t) Apply Table 1. y(t) t Invoke Lerch cancellation law. 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 cancellation law. See Theorem 2. 1 Example (Laplace method) Solve by Laplace method the initial value problem y 5 2t, y() 1. 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.

4 7.1 Introduction to the Laplace Method 249 L(y (t)) L(5 2t) Apply L acro y 5 2t. L(y (t)) Ue Table 1. L(y(t)) y() Apply the t-derivative rule, page 248. 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, page 248. y(t) 1 + 5t t 2 Invoke Lerch cancellation law. 2 Example (Laplace method) Solve by Laplace method the initial value problem y 1, y() y (). 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 t-derivative rule to y, that i, replace y by y on page 248. [L(y(t)) y()] y () L(1) Repeat the t-derivative rule, on y. 2 L(y(t)) L(1) Ue y() y (). L(y(t)) 1 Ue Table 1. Then divide. 3 L(y(t)) L(5t 2 ) y(t) 5t 2 Apply Table 1, backward. Invoke Lerch cancellation 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 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. 1 An advanced calculu background i aumed for the Laplace tranform exitence proof. Application of Laplace theory require only a calculu background.

5 25 Laplace Tranform 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, that i, how that f(t) i piecewie continuou and find α > uch that lim t f(t)/e αt. Solution: 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)). 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 follow from L(f(t)) g(t)dt M, becaue the α right ide of thi inequality ha limit zero at. The proof i complete.

6 7.1 Introduction to the Laplace Method 251 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 (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, then L(f (t)) exit for all large and L(f (t)) L(f(t)) f(). Proof: See page 276. Exercie 7.1 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!

7 252 Laplace Tranform 7.2 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 6, page 257, effectively increae the table ize manyfold, making it poible to olve typical differential equation from electrical and mechanical problem. 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 i memorized, but uually not the other. Derivation are potponed to page 27. The theory of the gamma function Γ(x) appear below on page 255. 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.

8 7.2 Laplace Integral Table Example (Laplace tranform) Let f(t) t(t 1) in 2t+e 3t. Compute L(f(t)) uing the baic 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 Table lookup. 5 Example (Invere Laplace tranform) Ue the baic Laplace table backward 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 ) Laplace table (backward). L(co 4t + 2e 3t + t t2 ) Linearity applied. f(t) co 4t + 2e 3t + t t2 Lerch cancellation 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. Solution: The detail require the ue of the Heaviide function formula { 1 a t < b, H(t a) H(t b) otherwie. The formula for f(t): 1 1 t < 2, f(t) 5 3 t < 4, otherwie { 1 1 t < 2, otherwie { 1 3 t < 4, + 5 otherwie Then f(t) f 1 (t) + 5f 2 (t) where f 1 (t) H(t 1) H(t 2) and f 2 (t) H(t 3) H(t 4). The extended 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.

9 254 Laplace Tranform 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 the Dirac impule model f(t) N n1 δ(t n). Show that 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. Show that 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. 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 >.

10 7.2 Laplace Integral Table 255 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. 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 (1) 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): (2) (3) Γ(1 + x) xγ(x) Γ(1 + n) n! for integer n 1.

11 256 Laplace Tranform Detail for relation (2) and (3): Start with e t dt 1, which give Γ(1) 1. Ue thi identity and ucceively relation (2) to obtain relation (3). To prove identity (2), 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 7.2 Laplace tranform. Compute L(f(t)) uing the baic Laplace table and the linearity propertie of the tranform. 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 ) 9. 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

12 7.3 Laplace Tranform Rule Laplace Tranform Rule In Table 6, the baic table manipulation rule are ummarized. tatement and proof of the rule appear in ection 7.7, page 275. Full The rule are applied here to everal key example. Partial fraction expanion do not appear here, but in ection 7.4, in connection with Heaviide coverup method. Table 6. 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. 11 Example (Harmonic ocillator) Solve by Laplace method the initial value problem x + x, x(), x () 1. 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.

13 258 Laplace Tranform 12 Example (-differentiation rule) Show the tep for L(t 2 e 5t ) 2 ( 5) 3. Solution: ( L(t 2 e 5t ) d ) ( d ) L(e 5t ) Apply -differentiation. d d ( 1) 2 d ( ) d 1 Baic Laplace table. d d 5 d ( ) 1 d ( 5) 2 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. L(in t H(t π)) 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 1 2 Baic Laplace table. Identity verified 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) 2a ( 2 + a 2 ) 2 (d) L(t 2 in at) 62 a a 3 ( 2 + a 2 ) 3

14 7.3 Laplace Tranform Rule 259 Solution: The detail for (a): L(t co at) (d/d)l(co at) d ( ) d 2 + a 2 The detail for (c): 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 ( ) 2 b 2 ( 2 + b 2 ) a 2 ( a)2 b 2 (( a) 2 + b 2 ) 2 Firt hifting rule. Apply -differentiation. Baic Laplace table. Calculu quotient rule. Verified (c). Left a exercie are (b) and (d).

15 26 Laplace Tranform 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)) Solution: The anwer i f(t) e t co t + e t in t. The detail: for f(t). L(f(t)) ( + 1) Signal for thi method: the denominator ha complex root. Complete the quare, denominator.

16 7.3 Laplace Tranform Rule 261 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. + 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 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. 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,

17 262 Laplace Tranform 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 Factor out e P /4, then cancel. 2 coh(p /4) ω 2 inh(p /4) 2 + ω 2 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) Compute the Laplace tranform of the half wave rectification of in ωt, denoted g(t), in which the negative cycle of in ωt have been canceled 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) Solve L(f(t)) e Solution: The anwer i f(t) e 3 t co(t 3)H(t 3). The detail: for f(t). 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.

18 7.3 Laplace Tranform Rule 263 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 () 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. Prepare for hifting rule. L(co 2t) L(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.

19 264 Laplace Tranform 7.4 Heaviide Method Thi practical method wa popularized by the Englih electrical engineer Oliver Heaviide ( ). A typical application of the method i to olve 2 ( + 1)( 2 + 1) L(f(t)) 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. More preciely, Heaviide method ytematically convert a polynomial quotient a + a a n n (1) b + b b m m into the form L(f(t)) for ome expreion f(t). 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 term of the form (2) A ( ) k where A i a real or complex contant and ( ) k divide the denominator in (1). In particular, i a root of 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 appear, where α iβ i the complex conjugate of. The correponding term in (2) turn out to be complex conjugate of one another, which can be combined in term of real number B and C a (3) A ( ) k + A ( ) k B + C (( α) 2 + β 2 ) k. 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.

20 7.4 Heaviide Method 265 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, 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 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 266. To illutrate Oliver Heaviide idea, conider the problem detail (6) ( 1)( + 1) A + B 1 + C + 1 L(A) + L(Be t ) + L(Ce t ) L(A + Be t + Ce t ) The firt line (6) ue college algebra partial fraction. The econd and third line ue the Laplace integral table and propertie of L. Heaviide myteriou method. Oliver Heaviide propoed to find in (6) 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, then evaluate on the left at the root which make the content of the box zero. The other term on the right are replaced by zero. To jutify Heaviide cover up method, multiply (6) by the denominator + 1 of partial fraction C/( + 1): (2 + 1) ( + 1) ( 1) ( + 1) A ( + 1) B ( + 1) + 1 C ( + 1) + ( + 1). Set ( + 1) in the diplay. Cancellation left and right plu annihilation of two term on the right give Heaviide precription ( 1) C. +1

21 266 Laplace Tranform The factor ( + 1) in (6) 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. Extenion to Multiple Root. An extenion of Heaviide method i poible for the cae of repeated root. The baic idea i to factor out the repeat. To illutrate, conider the partial fraction expanion detail 1 R ( + 1) 2 ( + 2) 1 ( ) ( + 1)( + 2) 1 ( ) ( + 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: 1 R ( + 1) 2 ( + 2) A B ( + 1) 2 + C + 2 A ( + 1) ( + 1) A fraction with multiple root. See equation (5). Find B and C by Heaviide cover up method. Multiply by +1. Set. 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.

22 7.5 Heaviide Step and Dirac Delta Heaviide Step and Dirac Delta The unit tep function or Heaviide func- Heaviide Function. tion i defined by H(x) { 1 for x, for x <. The mot often ued formula involving the Heaviide function i the characteritic function of the interval a t < b, given by (1) H(t a) H(t b) { 1 a t < b, t < a, t b. To illutrate, a quare wave qw(t) ( 1) floor(t) can be written in the erie form ( 1) n (H(t n) H(t n 1)). n Dirac Delta. A precie mathematical definition of the Dirac delta, denoted δ, i not poible to give here. Following it inventor P. Dirac, the definition hould be δ(t) dh(t). The latter i nonenical, becaue the unit tep doe not have a calculu derivative at t. However, dh(t) could have the meaning of a Riemann-Stieltje integrator, which retrain dh(t) to have meaning only under an integral ign. It i in thi ene that the Dirac delta δ i defined. What do we mean by the differential equation x + 16x 5δ(t t )? The equation x + 16x f(t) repreent a pring-ma ytem without damping having Hooke contant 16, ubject to external force f(t). In a mechanical context, the Dirac delta term 5δ(t t ) i an idealization of a hammer-hit at time t t > with impule 5. More preciely, the forcing term f(t) can be formally written a a Riemann- Stieltje integrator 5dH(t t ) where H i Heaviide unit tep function. The Dirac delta or derivative of the Heaviide unit tep, nonenical a it may appear, i realized in application via the two-ided or central difference quotient H(t + h) H(t h) 2h dh(t).

23 268 Laplace Tranform Therefore, the force f(t) in the idealization 5δ(t t ) i given for h > very mall by the approximation f(t) 5 H(t t + h) H(t t h). 2h The impule 2 of the approximated force over a large interval [a, b] i computed from b a h f(t)dt 5 h H(t t + h) H(t t h) dt 5, 2h due to the integrand being 1/(2h) on t t < h and otherwie. Modeling Impule. One argument for the Dirac delta idealization i that an infinity of choice exit for modeling an impule. There are in addition to the central difference quotient two other popular difference quotient, the forward quotient (H(t + h) H(t))/h and the backward quotient (H(t) H(t h))/h (h > aumed). In reality, h i unknown in any application, and the impulive force of a hammer hit i hardly contant, a i uppoed by thi naive modeling. The modeling logic often applied for the Dirac delta i that the external force f(t) i ued in the model in a limited manner, in which only the momentum p mv i important. More preciely, only the change in momentum or impule i important, b a f(t)dt p mv(b) mv(a). The precie force f(t) i replaced during the modeling by a implitic piecewie-defined force that ha exactly the ame impule p. The replacement i jutified by arguing that if only the impule i important, and not the actual detail of the force, then both model hould give imilar reult. Function or Operator? The work of phyic Nobel prize winner P. Dirac ( ) proceeded for about 2 year before the mathematical community developed a ound mathematical theory for hi impulive force repreentation. A ytematic theory wa developed in 1936 by the oviet mathematician S. Sobolev. The French mathematician L. Schwartz further developed the theory in He oberved that the idealization i not a function but an operator or linear functional, in particular, δ map or aociate to each function φ(t) it value at t, in hort, δ(φ) φ(). Thi fact wa oberved early on by Dirac and other, during the replacement of implitic force by δ. In Laplace theory, there i a natural encounter with the idea, becaue L(f(t)) routinely appear on the right of the equation after tranformation. Thi term, in the cae 2 Momentum i defined to be ma time velocity. If the force f i given by Newton law a f(t) d (mv(t)) and v(t) i velocity, then b f(t)dt mv(b) mv(a) i the dt a net momentum or impule.

24 7.5 Heaviide Step and Dirac Delta 269 of an impulive force f(t) c(h(t t h) H(t t +h))/(2h), evaluate for t > and t h > a follow: L(f(t)) t +h t h ce t c 2h (H(t t h) H(t t + h))e t dt c 2h e t dt ( e h e h ) 2h The factor eh e h i approximately 1 for h > mall, becaue of 2h L Hopital rule. The immediate concluion i that we hould replace the impulive force f by an equivalent one f uch that Well, there i no uch function f! L(f (t)) ce t. The apparent mathematical flaw in thi idea wa reolved by the work of L. Schwartz on ditribution. In hort, there i a olid foundation for introducing f, but unfortunately the mathematic involved i not elementary nor epecially acceible to thoe reader whoe background i jut calculu. Practiing engineer and cientit might be able to ignore the vat literature on ditribution, citing the example of phyicit P. Dirac, who ucceeded in applying impulive force idea without the ditribution theory developed by S. Sobolev and L. Schwartz. Thi will not be the cae for thoe who wih to read current literature on partial differential equation, becaue the work on ditribution ha forever changed the required background for that topic.

25 27 Laplace Tranform 7.6 Laplace Table Derivation Verified here are two Laplace table, the minimal Laplace Table and it extenion Table Largely, thi ection i for reading, a it i deigned to enrich lecture and to aid reader who tudy alone. Derivation of Laplace integral formula in Table 7.2-4, page 252. Proof of L(t n ) n!/ 1+n : The firt tep i to evaluate L(t n ) for n. L(1) (1)e t dt Laplace integral of f(t) 1. (1/)e t t t Evaluate the integral. 1/ Aumed > to evaluate lim t e t. The value of L(t n ) for n 1 can be obtained by -differentiation of the relation L(1) 1/, a follow. d d L(1) d d Then (1)e t dt Laplace integral for f(t) 1. d d (e t ) dt Ued d b d a F dt b df a d dt. ( t)e t dt Calculu rule (e u ) u e u. L(t) Definition of L(t). L(t) d d L(1) Rewrite lat diplay. d d (1/) Ue L(1) 1/. 1/ 2 Differentiate. Thi idea can be repeated to give L(t 2 ) d d L(t) and hence L(t2 ) 2/ 3. The pattern i L(t n ) d d L(tn 1 ) which give L(t n ) n!/ 1+n. Proof of L(e at ) 1/( a): The reult follow from L(1) 1/, a follow. L(e at ) e at e t dt Direct Laplace tranform. e ( a)t dt Ue e A e B e A+B. e St dt Subtitute S a. 1/S Apply L(1) 1/. 1/( a) Back-ubtitute S a. Proof of L(co bt) /( + b 2 ) and L(in bt) b/( + b 2 ): Ue will be made of Euler formula e iθ co θ + i in θ, uually firt introduced in trigonometry. In thi formula, θ i a real number (in radian) and i 1 i the complex unit.

26 7.6 Laplace Table Derivation 271 e ibt e t (co bt)e t + i(in bt)e t e ibt e t dt (co bt)e t dt + i (in bt)e t dt 1 ib (co bt)e t dt + i (in bt)e t dt 1 L(co bt) + il(in bt) ib Subtitute θ bt into Euler formula and multiply by e t. Integrate t to t. Ue propertie of integral. Evaluate the left ide uing L(e at ) 1/( a), a ib. Direct Laplace tranform definition. + ib 2 + b 2 L(co bt) + il(in bt) Ue complex rule 1/z z/ z 2, z A + ib, z A ib, z A2 + B 2. 2 L(co bt) + b2 b 2 L(in bt) + b2 Extract the real part. Extract the imaginary part. Derivation of Laplace integral formula in Table 7.2-5, page 252. Proof of the Heaviide formula L(H(t a)) e a /. L(H(t a)) H(t a)e t dt Direct Laplace tranform. Aume a. a (1)e t dt Becaue H(t a) for t < a. (1)e (x+a) dx Change variable t x + a. e a (1)e x dx Contant e a move outide integral. e a (1/) Apply L(1) 1/. Proof of the Dirac delta formula L(δ(t a)) e a. The definition of the delta function i a formal one, in which every occurrence of δ(t a)dt under an integrand i replaced by dh(t a). The differential ymbol dh(t a) i taken in the ene of the Riemann-Stieltje integral. Thi integral i defined in [?] for monotonic integrator α(x) a the limit b a f(x)dα(x) lim N n1 N f(x n )(α(x n ) α(x n 1 )) where x a, x N b and x < x 1 < < x N form a partition of [a, b] whoe meh approache zero a N. The tep in computing the Laplace integral of the delta function appear below. Admittedly, the proof require advanced calculu kill and a certain level of mathematical maturity. The reward i a fuller undertanding of the Dirac ymbol δ(x). L(δ(t a)) e t δ(t a)dt Laplace integral, a > aumed. e t dh(t a) Replace δ(t a)dt by dh(t a). lim M M e t dh(t a) Definition of improper integral.

27 272 Laplace Tranform e a Explained below. To explain the lat tep, apply the definition of the Riemann-Stieltje integral: M N 1 e t dh(t a) lim e tn (H(t n a) H(t n 1 a)) N n where t < t 1 < < t N M i a partition of [, M] whoe meh max 1 n N (t n t n 1 ) approache zero a N. Given a partition, if t n 1 < a t n, then H(t n a) H(t n 1 a) 1, otherwie thi factor i zero. Therefore, the um reduce to a ingle term e tn. Thi term approache e a a N, becaue t n mut approach a. e a Proof of L(floor(t/a)) (1 e a ) : The library function floor preent in computer language C and Fortran i defined by floor(x) greatet whole integer x, e.g., floor(5.2) 5 and floor( 1.9) 2. The computation of the Laplace integral of floor(t) require idea from infinite erie, a follow. F () floor(t)e t dt Laplace integral definition. n+1 n n (n)e t dt On n t < n + 1, floor(t) n. n n (e n e n ) Evaluate each integral. n ne n 1 e Common factor removed. x(1 x) n Define x e. x(1 x) d n dx Term-by-term differentiation. x(1 x) d 1 Geometric erie um. dx 1 x x Compute the derivative, implify. (1 x) e (1 e ) Subtitute x e. To evaluate the Laplace integral of floor(t/a), a change of variable i made. L(floor(t/a)) floor(t/a)e t dt Laplace integral definition. a floor(r)e ar dr Change variable t ar. af (a) Apply the formula for F (). e a (1 e a ) Simplify. Proof of L(qw(t/a)) 1 tanh(a/2): The quare wave defined by qw(x) ( 1) floor(x) i periodic of period 2 and piecewie-defined. Let P 2 qw(t)e t dt.

28 7.6 Laplace Table Derivation 273 P 1 qw(t)e t dt qw(t)e t dt Apply b a c a + b c. 1 e t dt 2 1 e t dt Ue qw(x) 1 on x < 1 and qw(x) 1 on 1 x < 2. 1 (1 e ) + 1 (e 2 e ) Evaluate each integral. 1 (1 e ) 2 Collect term. An intermediate tep i to compute the Laplace integral of qw(t): 2 L(qw(t)) dt 1 e 2 Periodic function formula, page (1 e ) e 2 Ue the computation of P above. 1 1 e 1 + e. Factor 1 e 2 (1 e )(1 + e ). 1 e /2 e /2 e /2 + e. /2 Multiply the fraction by e/2 /e /2. 1 inh(/2) coh(/2). Ue inh u (eu e u )/2, coh u (e u + e u )/2. 1 tanh(/2). Ue tanh u inh u/ coh u. To complete the computation of L(qw(t/a)), a change of variable i made: L(qw(t/a)) qw(t/a)e t dt Direct tranform. qw(r)e ar (a)dr Change variable r t/a. a tanh(a/2) See L(qw(t)) above. a 1 tanh(a/2) Proof of L(a trw(t/a)) 1 2 tanh(a/2): The triangular wave i defined by trw(t) t qw(x)dx. L(a trw(t/a)) 1 (f() + L(f (t)) Let f(t) a trw(t/a). Ue L(f (t)) L(f(t)) f(), page 251. Proof of L(t α ) 1 L(qw(t/a)) Ue f(), (a t/a qw(x)dx) qw(t/a). 1 tanh(a/2) Table entry for qw. 2 Γ(1 + α) 1+α : L(t α ) t α e t dt Direct Laplace tranform. (u/)α e u du/ Change variable u t, du dt.

29 274 Laplace Tranform 1 1+α u α e u du 1 Γ(1 + α). 1+α Where Γ(x) u x 1 e u du, by definition. The generalized factorial function Γ(x) i defined for x > and it agree with the claical factorial n! (1)(2) (n) in cae x n + 1 i an integer. In literature, α! mean Γ(1 + α). For more detail about the Gamma function, ee Abramowitz and Stegun [?], or maple documentation. π Proof of L(t 1/2 ) : L(t 1/2 Γ(1 + ( 1/2)) ) Apply the previou formula. 1 1/2 π Ue Γ(1/2) π.

30 7.7 Tranform Propertie Tranform Propertie Collected here are the major theorem and their proof for the manipulation of Laplace tranform table. Theorem 4 (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 5 (The t-derivative Rule) Let y(t) be continuou, of exponential order and let f (t) be piecewie continuou on t. Then L(y (t)) exit and L(y (t)) L(y(t)) y(). Theorem 6 (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)). Theorem 7 (The -Differentiation Rule) Let f(t) be of exponential order. Then L(tf(t)) d d L(f(t)). Theorem 8 (Firt Shifting Rule) Let f(t) be of exponential order and < a <. Then L(e at f(t)) L(f(t)) ( a). Theorem 9 (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 1 (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. Theorem 11 (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.

31 276 Laplace Tranform Proof of Theorem 4 (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 5 (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 6 (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. Proof of Theorem 7 (-differentiation): We prove the equivalent relation L(( t)f(t)) (d/d)l(f(t)). If f i of exponential order, then o i ( t)f(t), therefore L(( t)f(t)) exit. It remain to how the -derivative exit and atifie the given equality. The proof below i baed in part upon the calculu inequality (1) e x + x 1 x 2, x.

32 7.7 Tranform Propertie 277 The inequality i obtained from two application of the mean value theorem g(b) g(a) g (x)(b a), which give e x +x 1 xxe x1 with x 1 x x. In addition, the exitence of L(t 2 f(t) ) i ued to define > uch that L(t 2 f(t) ) 1 for >. Thi follow from the tranform exitence theorem for function of exponential order, where it i hown that the tranform ha limit zero at. Conider h and the Newton quotient Q(, h) (F ( + h) F ())/h for the -derivative of the Laplace integral. We have to how that lim Q(, h) L(( t)f(t)). h Thi will be accomplihed by proving for > and + h > the inequality For h, Q(, h) L(( t)f(t)) Q(, h) L(( t)f(t)) h. f(t) e t ht e t + the t h Aume h >. Due to the exponential rule e A+B e A e B, the quotient in the integrand implifie to give ( e Q(, h) L(( t)f(t)) f(t)e t ht ) + th 1 dt. h Inequality (1) applie with x ht, giving Q(, h) L(( t)f(t)) h t 2 f(t) e t dt. The right ide i h L(t 2 f(t) ), which for > i bounded by h, completing the proof for h >. If h <, then a imilar calculation i made to obtain Q(, h) L(( t)f(t)) h t 2 f(t)e t ht dt. The right ide i h L(t 2 f(t) ) evaluated at + h intead of. If + h >, then the right ide i bounded by h, completing the proof for h <. Proof of Theorem 8 (firt hifting rule): The left ide LHS of the equality can be written becaue of the exponential rule e A e B e A+B a LHS f(t)e ( a)t dt. Thi integral i L(f(t)) with replaced by a, which i preciely the meaning of the right ide RHS of the equality. Therefore, LHS RHS. Proof of Theorem 9 (econd hifting rule): The detail for (a) are LHS L(H(t a)f(t a)) H(t a)f(t a)e t dt Direct tranform. dt.

33 278 Laplace Tranform a H(t a)f(t a)e t dt Becaue a and H(x) for x <. H(x)f(x)e (x+a) dx Change variable x t a, dx dt. e a f(x)e x dx Ue H(x) 1 for x. e a L(f(t)) Direct tranform. RHS Identity (a) verified. In the detail for (b), let f(t) g(t + a), then LHS L(H(t a)g(t)) L(H(t a)f(t a)) Ue f(t a) g(t a + a) g(t). e a L(f(t)) Apply (a). e a L(g(t + a)) Becaue f(t) g(t + a). RHS Identity (b) verified. Proof of Theorem 1 (periodic function rule): LHS L(f(t)) f(t)e t dt Direct tranform. np +P f(t)e t dt Additivity of the integral. np n n P f(x + np )e x np dx Change variable t x + np. n e np P f(x)e x dx Becaue f i P -periodic and e A e B e A+B. P f(x)e x dx n rn Common factor in ummation. Define r e P. P f(x)e x dx 1 Sum the geometric erie. 1 r P dx 1 e P Subtitute r e P. RHS Periodic function identity verified. Left unmentioned here i the convergence of the infinite erie on line 3 of the proof, which follow from f of exponential order. Proof of Theorem 11 (convolution rule): The detail ue Fubini integration interchange theorem for a planar unbounded region, and therefore thi proof involve advanced calculu method that may be outide the background of the reader. Modern calculu text contain a le general verion of Fubini theorem for finite region, uually referenced a iterated integral. The unbounded planar region i written in two way: D {(r, t) : t r <, t < }, D {(r, t) : r <, r t}. Reader hould paue here and verify that D D.

34 7.7 Tranform Propertie 279 The change of variable r x + t, dr dx i applied for fixed t to obtain the identity (2) e t g(x)e x dx g(x)e x t dx t g(r t)e r dr. The left ide of the convolution identity i expanded a follow: LHS L(f(t))L(g(t)) f(t)e t dt g(x)e x dx Direct tranform. f(t) t g(r t)e r drdt Apply identity (2). D f(t)g(r t)e r drdt Fubini theorem applied. D f(t)g(r t)e r drdt Decription D and D are the ame. r f(t)g(r t)dte r dr Fubini theorem applied. Then ( ) t RHS L f(u)g(t u)du t f(u)g(t u)due t dt Direct tranform. r f(u)g(r u)due r dr Change variable name r t. r f(t)g(r t)dt e r dr Change variable name u t. LHS Convolution identity verified.

35 28 Laplace Tranform 7.8 More on the Laplace Tranform Model converion and engineering. A differential equation model for a phyical ytem can be ubjected to the Laplace tranform in order to produce an algebraic model in the tranform variable. Lerch theorem ay that both model are equivalent, that i, the olution of one model give the olution to the other model. In electrical and computer engineering it i commonplace to deal only with the Laplace algebraic model. Engineer are in fact capable of having hour-long modeling converation, during which differential equation are never referenced! Terminology for uch modeling i necearily pecialized, which give rie to new contextual meaning to the term input and output. For example, an RLC-circuit would be dicued with input F () ω 2 + ω 2, and the litener mut know that thi expreion i the Laplace tranform of the t-expreion in ωt. Hence the RLC-circuit i driven by a inuoindal input of natural frequency ω. During the modeling dicoure, it could be that the output i X() ω 2 + ω 2. Lerch equivalence ay that X() i the Laplace tranform of e t + 1 in ωt, but that i extra work, if all that i needed from the model i a tatement about the tranient and teady-tate repone to the input.