Math 334 Fall 2011 Homework 10 Solutions
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1 Nov. 5, Math 334 Fall Homework Solution Baic Problem. Expre the following function uing the unit tep function. And ketch their graph. < t < a g(t = < t < t > t t < b g(t = t <t < t 3 t > Solution. a We have g =, g =, g 3 = ; t =, t =. So g(t = + ( u(t + ( u(t = +u(t u(t. ( t b We have g =t, g = t, g 3 = t 3, t =, t =. So g(t = t+(t t u(t + (t 3 t u(t. (
2 Math 334 Fall Homework Solution t 3 t t Problem. Compute the following Laplace tranform: a L{tu(t b L { cot u ( t π + (9 t + t u(t. Solution. a ecall calculating L{g(t u(t a:. Obtain f(t = g(t + a;. Compute F( = L{f. 3. Multiply it by e a to get L{g(t u(t a=e a F(. We here have g(t =t, a =. So firt get Now compute Finally we have f(t = g(t + = t+. (3 F( = L{t + = +. (4 L{t u(t =e ( +. (5 b We have to firt ue linearity: { ( L cot u t π { ( + (9 t + t u(t = L cot u t π + L{(9 t + t u(t. (6 For the firt tranform we identify: g(t = co t, a = π. So ( ( f(t = co t + π ( = co t+ π = co ( t co π 4 4 in ( t in π 4 = herefore Finally { F( = L (co ( t in ( t { ( L cot u t π = ( = e π = (co ( t in ( t. ( ( + 4. (9
3 3 Now the nd tranform: We have g(t = 9 t + t, a =. So We have So f(t = g(t+=9(t+ + (t + =9t + 3 t ( F( = L{9 t + 3 t + 39= ( ( L{(9 t + t u(t =e ( Putting thing together: { ( L cot u t π + (9 t + t u(t = Problem 3. Compute L{co (e t δ(t. Solution. ecall Here f(t = co (e t and a=. So the anwer i π e e ( (3 L{f(tδ(t a = f(ae a. (4 e co (e = (co e. (5 Intermediate Problem 4. Find the invere Laplace tranform for the following function. ( e a F( =. + b F( = e + e e 3 e 4. Solution. a Spotting e we know that the tep function i involved. We ue the formula Here a =, F( = herefore Finally b We have { e L {F = L u(t 4. Problem 5. Solve Solution. Firt write Now compute ( ( + L {e a F( = f(t a u(t a. (6. We compute f(t=l {F =L { ( ( + { = e t L = e t co t. (7 + f(t = e t co (t. ( L {F =e t co (t u(t. (9 { { { e + L e L 3 e L 4 = u(t + u(t u(t 3 ( { t/ t <6 y + y = g(t =, y( =, y 3 t 6 ( =. ( g(t=t/+(3 t/ u(t 6. ( L{y = Y y( y ( = Y. (3 L{g=L{t/+L{(3 t/u(t 6 = { + e 6 L 3 t+6 = e 6. (4 he tranformed equation i then Y + Y = e 6 + Y = ( + e 6 ( (5
4 4 Math 334 Fall Homework Solution o compute y, we compute the invere Laplace tranform of the right hand ide one by one. { L. We ue partial fraction. Write ( + ( + = / ( + = A + B + C + D + = A( + +B ( + + (C+D. (6 ( +. hu we have and = (A +C3 + (B + D +A+B (7 A+C = ( B + D = (9 A = (3 B =. (3 Conequently A = C =, B =, D =. (3 So the partial fraction repreentation i Conequently L { e 6 ( + Here a = 6, F = ( + = ( +. (33 { L = t ( + in t. (34. ecall the formula ( + L {e a F( = f(t a u(t a. (35. A we have jut computed { f = L ( + = t in t (36 We can immediately write down { [ L e 6 t = ( + 3 ] in (t 6 u(t 6. (37 he lat term i tandard: Putting everything together we have { L = in t. (3 + y = t + [ t in t 3 ]u(t in (t 6 6. (39 Problem 6. Solve y + y = δ(t π co t, y( =, y ( =. (4 Solution. Firt tranform the equation: L{y = Y y( y ( = Y ; L{δ(t π co t =e π co ( π = e π. (4 So the tranformed equation i hi give { L = in t. + ( + Y =e π. (4 Y = + + e π +. (43. In thi problem it i poible to kip the following calculation by realizing / /( + =/[ ( + ].
5 5 { L e π = f(t π u(t π with + So Summarizing, we have f(t = L { + = in t. (44 { e L π = u(t π in (t π = u(t π in (t. (45 + y = in t [ +u(t π]. (46 Advanced Problem 7. Let f atify f(t + = f(t for all t and ome fixed poitive number. Show that e t f(t dt L{f(t=. e (47 Proof. We have L{f(t = e t f(t dt = e t f(tdt + e t f(t dt + = (k+ e t f(t dt. k= k (4 Now for each integral, do change of variable: t = t k. We have (k+ e t f(t dt = e (t +k f(t + k dt = e k e t f(t dt = e k k Subtitute back: hu end the proof. L{f(t = (k+ e t f(t dt k= k = e k e t f(t dt. k= ( = e k e t f(tdt. k= ( = (e k e t f(t dt k= = e t f(t dt. e e t f(t dt. (49 Challenge Problem. Let f(t be a bounded function (not necearily continuou. Prove that it Laplace tranform F( = e t f(t dt (5 i continuou at all >. herefore uually there i no need to conider the invere tranform of function with jump. Proof. Let M > be uch that f(t M for all t. ake any >, we how that F( i continuou there. In other word, we how e t f(t dt e t f(tdt (5. hi can be replaced by continuou in it domain, but it eem the proof will become much more technical.
6 6 Math 334 Fall Homework Solution a. We ue the ε δ formality. hat i we how for any ε >, there i δ > uch that when <δ, e t f(t dt e t f(tdt < ε. (5 Now for any ε >, there i > uch that e t/ dt < ε 4 M. (53 For uch we have, for all < /, e t f(tdt e t f(tdt e t f(t dt + M e t dt + M [ ] M e t/ dt Next take δ uch that when <δ, e t f(tdt e t dt < ε. (54 e t e t < ε M for all < t <. (55 hi lead to, when < δ, e t f(tdt e t f(tdt = (e t e t f(t dt M e t e t dt < M ε M Finally let δ = min {δ, /. We ee that when < δ, e t f(t dt e t f(tdt e t f(t dt e t f(tdt < ε + ε = ε. hu end the proof. dt= ε. e t f(tdt + e t f(t dt emark. hank to Delyle, I realized that the above proof i unnecearily complicated. Following i a impler one. Let be fixed. We will how that F( F( 4 M for all uch that < /. Continuity then follow naturally. Here M i a defined at the beginning of the above proof. For uch we have (uing mean value theorem: f(a f(b = f (ξ (a b with ξ between a and b e t e t = t e ξt ( te t/. (57 he lat inequality come from the fact that if ξ lie between and, then necearily ξ > /. Now we have [ ] e t f(t dt e t f(tdt t e t/ M dt = M t e t/ dt = 4 M. (5 (56
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4e st dt. 0 e st dt. lim. f (t)e st dt. f (t) e st dt + 0. f (t) e. e (2 s)t dt + 0. e (2 s)4 1 ] = 1 = 1. te st dt + = t s e st
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