Øving 2 - Laplacetransform II - LF

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1 Øving - Laplaceranform II - LF Jon Vegard Venå, Tai Terje Huu Nguyen Obligaorike oppgaver See he Lecure noe Malab: clear %e oving.m hvi du lurer paa noe her [-pi:.:pi]; fheaviide; plo,f axi[-pi pi - ] Pyhon: impor numpy a np impor maplolib.pyplo a pl N #-aken np.linpace-np.pi,np.pi,numn # de andre argumene er funkjonverdien for x. f np.heaviide, #lage plo av funkjonen pl.plo,f #korrek uni pl.axi[-np.pi,np.pi,-,] # navn paa akene pl.xlabelr'$x$' pl.ylabelr'$y$' #vie plo pl.how 3 a Taking he Laplace ranform of y + 4y + 5y δ yield Y y y + 4 Y y + 5Y e. Wih he iniial condiion y and y 3 we have Y 3 + 4Y + 5Y e uch ha Y e e +

2 Taking he invere Laplace ranform uing -hifing, -hifing and a andard able of Laplace ranform link y 3e in + e in u b Taking he Laplace ranform of y + 5y + 6y δ π + u π co δ π u π co π yield Y y y + 5 Y y + 6Y e π/ e π +. Wih he iniial condiion y and y we have uch ha Y + 5Y + 6Y e π/ e π + Y e π/ e π Uing parial fracion expanion we ge and A + B + + A + + B A + B + 3A + B { A + B 3A + B A, B C + + D + 3 A + B C + D + 3C + D B + C + D3 + A + 5B + 3C + D + 5A + 6B + C + D + 6A + 3C + D B + C + D A + 5B + 3C + D 5A + 6B + C + D 6A + 3C + D A, B, C 5, D 3. Hence, Y e π/ e π/ e π e π

3 Taking he invere Laplace ranform uing -hifing and a able of Laplace ranform y e π/ e 3 π/ u π in π + co π 4e π + 3e 3 π u π e π/ e 3 π/ u π + in + co + 4e π 3e 3 π u π c We have L {y } d d Y y Y Y and uing y L {y } d Y y y Y Y +. d Taking he Laplace ranform of he ODE herefore yield Y Y + + Y + Y + Y Y + Y. From earlier coure we know ha we can olve hi fir order ODE uing inegraing facor which i in hi cae. Muliplicaion by he inegraing facor yield Y + Y d d Y Y + C Y + C for ome conan C. Taking he invere ranform y + C. Since y we have C. The oluion of he iniial value problem i herefore y +. d Noe ha we can wrie he inegral equaion in he form y y g wih g. Uing he convoluion heorem and he infamou able we ge Y Y 3 Y,

4 and by parial fracion expanion we ge Hence, A + B + C + A A + B + C + B C A A + B + C B C A A, B, C. Y Taking he invere Laplace ranform look here yield y + e + e + coh. + B + C + B C Anbefale oppgaver See he Lecure noe a We wrie + +. Hence: L + b We wrie ++ L + + Tha i L ++ L L + + in Hence, uing alo -hifing: L + e. L + e e. c We wrie: + + : F G, where we recognize ha F and + G are Laplace ranform of ome known funcion: F Lco and G Lin. Thu by he convoluion formula: L + co in. 3

5 By definiion of he convoluion, we ge: Now, we have: and co in in coτ in τ dτ coτ in coτ inτ co dτ co τ dτ co coτ + in dτ co inτ in + τ co 4 coτ inτ dτ inτ dτ coτ 4 in in + in + co co co. 4 in in in co 5 co co co co co co in 6 and hence i follow ha only he erm in i lef. Tha i: L + in. 7 d We wrie 3 5. Now recall ha, if h 3 5 m m, hen Lh m m! m+ m N {}. We ake m 4 o ha m + 5. Then we ee ha Lh 4 4! Conidering hi and -hifing, we conclude ha: when L e a We fir recall he -hifing formula. Le θ a be he funcion defined o be given by θ a u a. Tha i he uni-ep funcion a a. We conider Lθ a f for ome general funcion f whoe Laplace ranform exi. Then by definiion: Lθ a f a e θ a f d e u af d e f d. 9 The la equaliy follow becaue u a for < a and i equal o when > a. Thi i by definiion. Now, le u change variable. We define x : a. Then x run from x o x. Moreover dx d and x + a. Subiuing hi, we find: Lθ a f e x+a fx + a dx e a e x fx + a dx.

6 Now, le u define f a o be he funcion uch ha f a x fx + a. The inegral on he righ-hand ide i hen by definiion Lf a. Hence we conclude ha: Lθ a f e a Lf a. A perhap more familiar way of wriing hi, o ha he Laplace ranform of f appear on he righ-hand ide i hen: Uing he noaion in he book, hi i: Lθ a f a e a Lf. Lu af a e a Lf. 3 We now conider he Laplace ranform of f from he exercie. Tha i, of f, where f θ θ π co. According o he formula we derived above, we have: Lθ co e Lco + Similarly, from he formula above, we have wih a π: Thu we finally have: Lco +. 4 Lθ π co e π Lco + π e π L co e π +. 5 Lf + + e π + + e π. 6 + To draw he graph of f, we conider how u u π behave like for differen value of. Noe ha for < boh erm u and u π are zero, o he whole expreion i. When < < π, u while u π. Hence he whole expreion i equal o. Finally for > π, u and u π. So he whole expreion i again zero. Thu he only place i i non-zero, in which cae i i equal o, i when, π. Thu he graph of f i he graph of co for, π and zero elewhere. b We ue he -hifing formula. Le h. We ge: We have h a + a + a + a. Hence: Lθ a h e a Lh a. 7 and hence: Lh a L + a + a 3 + a + a 8 Lθ a h e a 3 + a + a. 9

7 A for he graph of θ a h, we ue he definiion of θ a and h. I follow ha θ a h for < a and i equal o h for > a. c Perhap he imple way o olve hi exercie i o look a he graph of f given below wih a π and ee ha i alernae beween and in inerval of lengh a. Thu we can pli he inegral in he Laplace ranform o ge Lf ai+ i e d i i ai i i e ai+ e ai i e ai e a i i[ e ] ai+ ai e a e a i. i Now, ince boh and a are poiive, e a <, o we can ue he formula for a geomeric erie o obain e a i e a i e a e a + e a e a anha/ Noe: When we olve hi exercie we aume ha we can change he order of inegraion and ummaion. In general, for infinie um, hi canno be done. In our cae hi i poible, ince e i u ia d i + N e N i u ia i i u ia d in i + N in i + Na in e i u ia d i e u ia d e i u ia d i e u ia d.

8 A i u ia i bounded by, we ge he bound i e u ia d i e u ia d Na in Na in Na in which converge o zero when N ince e become very mall. e d, 3 4 a We ake he Laplace ranform on boh ide and ue he iniial condiion Y y y +Y e π }{{}}{{}. We collec he erm uing he flye-bye rule Y e π + }{{} :F e π + We noe ha he invere Laplace ranform of F i f co, and uing -hifing we ge y co πu π. b We ake he Laplace ranform on boh ide and ue ha Lf g LfLg. We e Ly : Y and recall Lh where h. Then we ge afer aking Laplace ranform on boh ide of he given equaion: Y Y. 4 Tha i: Y Y. 5 Tha i, for >, we obain: Hence y inh. Y. 6 5 The RLC-circui i governed by he following inegro-differenial equaion where in our cae v Li + Ri + C { 34e < < 4 oherwie iτ dτ v 7 34e u 4 34e 34e 4 e 4 u 4.

9 The Laplace ranform of hi expreion i given by uing -hifing and a andard able of Laplace ranform V L {v} e 4 e e Equaing hi reul wih he Laplace ranform of he lef hand ide of 7, we obain uing Laplace ranform of derivaive and Laplace ranform of inegral Thu, 34 e 4 4 L + { Li + Ri + C L I i + RI + C I + 4I + I I Uing parial fracion expanion we ge 34 e A + + B + D Hence, I } iτ dτ I I. A + D + 4A + B + D + A + B A + D 4A + B + D 34 A + B A, B 4, D. 4 + e e 4 e 4. The invere Laplace ranform of hi funcion yield he final reul uing -hifing, -hifing and andard able of Laplace ranform i e + e co 4 + 9e in 4 u 4e 4 e 4 + e 4 co[4 4] + 9e 4 in[4 4] e + e co in 4 u 4 e + e co[4 4] + 9 in[4 4]. Nø a We fir recall he definiion of he Gamma funcion Γ i i alo aed in he exercie: Γx + z x e z dz. 8

10 Le h n be he funcion uch ha h n n. Then by definiion of he Laplace ranform of h n, we have: Lh n n e d. 9 We ue a change of variable. Se x :. Then, for >, x and d dx. Alo x run from o. Subiuing hi, we ge ha: Lh n x n e x dx n+ b We fir how Γx + xγx. By definiion we have: Γx + We ue inegraion by par, which hen yield: x n e x dx Γn + n+. 3 y x e y dy. 3 Γx + y x e y + x y x e y dy xγx where he la equaliy follow from he definiion of Γx and he fac ha lim y y x e y. Obervaion: noe ha Γ e y dy and hence in he cae n N, i follow by hi formula and by inducion ha Γn + n!. In order o calculae Γ k+ noe ha Γ k+ Γ k +. From hi and he formula we ju proved Γx + xγx, i follow by inducion ha Γk + / Γk + / + Γ/ k j j + /, when k. Here k j g j for ome quaniie g,..., g k, imply mean g g g k ; ha i, he produc of all hee quaniie. Γ3/ Γ + / Γ/ Γ5/ Γ + / Γ + 3/ 3 Γ/ Γ7/ Γ3 + / Γ + 5/ 5 3 Γ/ Γ9/ Γ4 + / Γ + 7/ Γ/ For he cae k, hi reduce immediaely o Γ/. Hence we will need o know how o calculae Γ/ in order for hi o be ueful. We do hi in d below. c By definiion, we have: Γ/ x / e x dx. 3 We ue a change of variable. Se u : x. Then u run from o and dx u du. Subiuing hi, we herefore have: Γ/ e u du. 33

11 d We conider he inegral e x dx. Thi i he claical Gau inegral. I ha imporan applicaion in many area, for inance in aiic where i appear in relaion wih e x +y dx dy. he normal diribuion. Denoe he inegral by I. Then we have I We now change coordinae o polar coordinae, eing x + y r and dx dy r dr dθ. Noe ha he inegraion area i over he whole fir quadran in R. Tha i θ, π/ and r,. Thu we find: I π/ e r r dr dθ π e u du π 4 34 and hence I du dr. π. In he econd equaliy we have ued he ubiuion u : r giving e We fir noe ha by combining c and d we have Γ/ π. We expand he exponenial, uing he ideniy: Taking z p we hu ge: e z n z n n!. 35 e p n p n. 36 n! n Thi i a power erie wih infinie radiu of convergence i converge for all p, and o from calculu we know ha we can inegrae erm-wie over all of R for hoe familiar wih he erm: we can do hi becaue he erie converge uniformly a we know from he heory of power erie; c.f Abel lemma. Hence inegraing erm-wie we find: e p dp n p n dp n! n n n n n! p n dp n n!n + n+/ 37 / 3/ 3 + 5/ 7/ + 5! 7 3! Now we wan o ake he Laplace ranform of he above erie, and alo here move he inegral inide he um:

12 L n n n!n + n+/ n n n!n + L n+/. 38 From a we know he Laplace ranform of n+/, i.e., L n+/ Γn++/ n++/, >, o we have n n n!n + L n+/ n n Γn + + / n!n + n++/. 39 We now ue b wih x n + /, o n n Γn + / + n!n + n+/+ n Γ3/ 3/ n n + /Γn + / n!n + n+3/. 4 Γ5/ 3 5/ + Γ7/ 5! 7/ + Γ9/ 7 3! 9/ + A la we ue he formula for Γk + ha we found in b and wha we noed in he beginning: ha Γ/ π, o ge n n n! Γ n+ n Γ/ n j j / n+3/ n! n+3/ 4 n π n n j j / n! n+3/ n π 3/ 5/ / / + π 3/ We inroduce now he generaliaion of wha i known a binomial formula, i.e., r x + y r x r k y k, k k which hold for x > y and r i a real. The o-called Pochhammer ymbol r k i defined by r rr r k + :. k k! Thi hen ell u ha x + y r x r + rx r y + rr x r y +! rr r x r 3 y !

13 Hence, wih x, y /, >, and r / we find Going back o 4, we find + / / + n n n! Thu, he Laplace ranform i Γ n+ π n+3/ + / / 3/ π + / / π +. L e p dp π An alernaive mehod uing direcly he definiion of he Laplace ranform and changing he order of inegraion in muliple inegral, i a follow. From he definiion we have: L e x dx e e x dx d. 44 π π The idea i now o change order of inegraion. Noe ha x run from o while run from o. The curve x i he righ par of he curve x i.e. he par where x. Hence in he x-plane we are inegraing over he infinie region above he curve x wih x. Thu we ee ha hi i he region where run from x o and x run from o. Hence: L e x dx e x e d dx π π π π x x e x e dx e x + dx. 45 We now ue a change of variable u : x +. Then u run from o and du + dx. Subiuing hi, we herefore finally have L e x dx π π + e u du + 46 where he la equaliy follow from he Gau inegral I e u du π from d.

14 Finally, we can alo proceed a follow. Le f : π e x dx. Then by he fundamenal heorem of calculu, we have f π e π e. Noe ha f. Hence by he formula for he Laplace ranform of he derivaive, we ge: Tha i Lf Lf. Now, we have: Lf π Lf Lf. 47 e e d π e + d. 48 We ue a change of variable. Pu u : +. Then du o. Subiuing hi, we herefore have: + d and u run from Lf e u du π + π π where he econd equaliy follow from he Gau inegral I e x dx Hence we ge: π from d. Lf Lf +. 5