Problem 1 (4 5 points)

Transcription

1 ACM95/b Problem Set 3 Solution /7/4 Problem (4 5 point) The following (trivial once you 'get it') problem i deigned to help thoe of you who had trouble with Problem Set ' problem 7c. It will alo help you to find aymptotic epanion of Laplace integral and their relative. a) Evaluate a ep H- tl t () for a= -4, -3, 3 µ -3, 9 µ -3,., and. Alo give your anwer in fied decimal form to 7 digit [i.e. number like.34567], and eplain any trend you notice. b) Now conider the integral a I HaL = ep H-tL t () for >>. What range of value make I(a)=I()(+e) with e <. (i.e. give I() to % accuracy)? Your anwer may depend on. c) Ue reaoning motivated by part (b) to find the implet function of which approimate -t I HL = Å + t t to % accuracy for >>. Jutify your error etimate. (3) d) Ue reaoning motivated by part (b) to find the implet function of which approimate ê5 -t I HL = ÅÅÅÅ!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ÅÅ ÅÅÅ t t + t 3 ê 4 + Co t to % accuracy for >>. Jutify your error etimate. Solution to Problem (4) a) J HaL = aep H- tl t = - - a ÅÅ Å a J HaL µ µ Table For a>. there i no difference between J(a) and J() to even decimal digit. (5) b)

2 I HaL = aep H-tL t = ÅÅ - -a Å Since >>> we have a imple epreion for I() I HL = ep H-tL t = ÅÅÅÅÅ Thi give: (6) (7) I HaL - I HL ÅÅ Å = -a I HL (8) We require thi to be le than.. Thi give u an inequality involving and a: -a <. (9) Since the eponential i a monotone function the inequality i preerved when we compute the natural log of both ide -a< ln H.L Further implification give: () a >-ln H.Lê = ê () So for >> we have found a uitable approimation provided that a>.46 c) -t I HL = Å + t t We want to find a function f() uch that: () I HL - f HL ÅÅ Å <. (3) f HL Notice that for large value of, the integrand i nearly ero ecept in a very mall interval near t=. Thi i made clear by the following plot of the integrand for = t Figure The mallne of the integrand away from t= eem to ugget that the integral might be approimated well by Taylor epanding êh + t L near t= and keeping only the firt term:

3 f HL = -t t = ÅÅÅÅÅ The error in uch an approimation i on the order of the integral of the net term in the Taylor erie: (4 Error HL = O i j t -t t y = O J Å k { N 3 (5) Thi tatement i made more precie by what' known a Waton' Lemma (ee Bender & Orag). For our purpoe we'll take thi to be a good meaure of the error. We find: I HL - f HL ê 3 ÅÅ Å ~ f HL ê = << =. (6) Thi i below the deired % threhold. For a more precie treatment, ee the appendi. Below i a plot of the percent error a a function of : d) % Error Figure ê5 -t I HL = ÅÅÅÅ!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ÅÅ ÅÅÅ t t + t 3 ê 4 + Co t Again the integrand i very mall for t> a we ee from thi plot when =. (7) t Figure 3 Becaue of thi, only a mall error i made by replacing I() with the integral over a emi-infinite interval:

4 -t J HL = ÅÅÅÅ!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ÅÅÅ t t + t 3 ê 4 + Co t The notion of "mall error" i made precie by Waton' Lemma, however it i eay to ee that J HL - I HL = ê5 -t ÅÅÅÅ!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ÅÅ ÅÅÅ t i y j ÅÅÅÅ t + t 3!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ÅÅ ÅÅÅ ê 4 + Co t k t + t 3 ê 4 + Co t { t=ê5 ê5 -t t =.856 -ê5 `.7536 µ - Now we need to approimate J(). Proceeding a in part (c). We ue the firt term in the Taylor erie: ÅÅÅÅ!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ÅÅ ÅÅÅÅ = - t +... t + t 3 ê 4 + Co t and we find: So we find: f HL = -t t = ÅÅÅÅÅ Error HL = O i j -t -t t y k { = Å - J HL - f HL ê ÅÅ Å ~ f HL ê = ÅÅÅÅÅ ` ÅÅÅÅ =. Combining thi with the previou error bound and uing the triangle inequality give: (8 (9) () () () I HL - f HL I HL - J HL + J HL - f HL ÅÅ Å = Å f HL f HL I HL - J HL J HL - f HL ÅÅ Å + ÅÅ Å f HL f HL ~.856 -ê5 + ÅÅÅÅÅ `.7536 µ - +. ~. (3) Therefor f()=/ give u an approimation to I() accurate to %. Below i a plot of % Error a a function of. % Error Figure 4 Comparing to the figure from part (c) how that the approimation i not a good, becaue the Taylor erie for the integrand in (c) had a econd term that wa O(t ) while the Taylor erie for the integrand in (d) had a econd term that wa O(t). Problem (5 point)

5 Prove that for an analytic function f(t) with Laplace tranform F() lim F HL = lim f HtL Ø tø + Solution to Problem (4) F HL = -t f HtL t (5) For very large poitive Re(), the integrand will be very mall for all t>. Motivated by the lat problem, we upect that a good approimation to thi integral will come from uing the firt term in the Taylor erie for f(t): -t f HL t = f HL ÅÅÅÅÅ To ee if thi i a good approimation we check the error: F HL - ÅÅÅÅÅ f HL = -t f HtL t - -t f HL t = -t Hf HtL - f HLL t -t f HtL - f HL t Since f i analytic, it i differentiable, o, by the mean value theorem we have for ome cœ[,t]: f HtL - f HL = tf' HcL So our bound become: (6) (7) (8) F HL - ÅÅÅÅÅ f HL f' HcL t -t t = f' HcL ÅÅÅ (9) Hence a Ø we have: F HL - f HL So we have hown that: f' HcL ÅÅÅ Ø lim F HL = lim f HtL Ø tø + Problem 3 (6 4 point) (3) (3) Fun with Dirac a) Show that Å d HL =-d HL (Hint: ue the Gauian d equence, d n =n/ p ep(-n ); integration by part may alo be helpful) (3) b) Show that for f'() continuou at = - d' HL f HL =-f' HL c) Show that if i the olution of f( )=, - f HL d Hf HLL = ÅÅÅ d H - L (33) (34)

6 d)what happen in (c) if f() ha more than one ero? Ue your reult to find for - L H - LD e) Conider the three dimenional delta function in carteian coordinate: (35) d Hr - r L ªd H - L d Hy - y L d H - L (36) Introduce a new coordinate ytem with the three new coordinate, h, and by =X(,h,), y=y(,h,), =Z(,h,). Show that the equation of part (c) generalie to HX, Y, ZL d H - L d Hy - y L d H - L =d H - L d Hh-h L d H - L H, h, L - (37) where (X,Y,Z)/ (,h,) i the Jacobian determinant of partial derivative of X,Y,Z with repect to,h, and, h, and are the olution of = X(, h, ), y =Y(, h, ), =Z(, h, ) f) In particular, how that in cylindrical coordinate r, f, : d H - L d Hy - y L d H - L = ÅÅÅÅÅ r d Hr - r L d Hf -f L d H - L in pherical coordinate r, q, f d H - L d Hy - y L d H - L = ÅÅ ÅÅÅÅÅÅ r Sin q d Hr - r L d Hq -q L d Hf-f L and if m= Co q i ubtituted in pherical coordinate d H - L d Hy - y L d H - L = ÅÅÅÅÅÅ r d Hr - r L d Hm -m L d Hf-f L Solution to Problem 3 (38) (39) (4) a) Å d HL =-d HL (4) The delta function i defined by it' action on mooth tet function, i.e. the delta function i a "function" which atifie f HL = - d HL f HL for all infinitely differentiable (mooth) function f() which are ero outide of ome interval (-a,a) (i.e. have compact upport). Let u conider (4) - d n ' HL f HL Method : Integrate by part - d n ' HL f HL = lim Hf HL d n HLL - lim Hf HL d n HLL - Hf HL + f' HLL d n HL = Ø Ø f HL d n HL - - f' HL d n HL Since the Gauian delta equence ha d a it' limit, we have: (43) (44)

7 i lim j- nø k i lim j- nø k So we conclude: - - f HL d n HL y =-f HL { f' HL d n HL y =- f' HL = { (45 lim nø - d n ' HL f HL =-f HL = lim nø - -d n HL f HL (46) Hence d n ' HL and -d n HL have the ame limit o that we mut have: d' HL =-d HL Method : Approimate the integral directly - d n ' HL f HL = - i n j ÅÅÅÅÅ k p -n y - n3 ' f HL = ÅÅ { p f HL -n - By the ame reaoning a in problem, we replace term in the integrand with their taylor erie epanion: (47) (48) - n 3 Å p f HL -n - n3 ~ Å - p f HL -n - The error i given approimately by: =-f HL (49) Error = O i - n 3 j Å k p ÅÅÅÅÅ - 4 f'' HL -n y = O J- ÅÅÅÅÅ 3 f'' HLN { 4n And thi vanihe a nø. Likewie we find: (5) - So we conclude that: -d n HL f HL =- ÅÅÅÅÅ n p f HL -n - nf HL ~- ÅÅÅ p - -n =- f HL (5) lim nø - and hence d'()=-d(). d n ' HL f HL =-f HL = lim nø - -d n HL f HL (5) b) d' HL f HL = lim d n ' HL f HL - nø - Method : Integrate by part lim d n ' HL f HL = nø - i lim jlim Hf HL d n HLL - lim nø k Hf HL d n HLL - Ø Ø- - We conclude: - d' HL f HL = lim nø - d n ' HL f HL =-f' HL d n HL f' HL y =-lim d n HL f' HL =-f' HL { nø - (53) (54) (55)

8 Method : Approimate the integral directly Ue the defiition of the delta equence and keep two term in the Taylor erie: - n3 d n ' HL f HL = Å f - p HL -n - n3 ~ Å - p Hf HL + f' HLL -n - The error i given by: Error = O i - n 3 j Å k p J ÅÅÅÅÅ f ''' HLN -n y -f ''' HL = O J ÅÅ N { 4n Thi vanihe a nø. We conclude: =-f' HL (56) (57) c) - d' HL f HL = lim d n ' HL f HL =-f' HL nø - (58) Let g() be a tet function and uppoe that f() ha only one ero, located at. Furthermore, uppoe that the function f() i invertible near, i.e. by the invere function theorem we require f'( ). We wih to calculate: - d Hf HLL g HL Since d() i ero outide of any neighborhood of, we could replace the integral with one about where the function f() ha an invere. Since f'( ) the invere of f eit in ome neighborhood of which we'll call ( -a, +b). Thee two obervation let u write: (59) - +b d Hf HLL g HL = d Hf HLL g HL -a (6) and then change variable according to y=f(): f H +bl d HyL g Hf- HyLL ÅÅ ÅÅÅÅÅÅ (6) f H -al f' Hf - HyLL y Since f' in thi interval, it i alway of one ign. If it i poitive, then the range of integration i poitively oriented, and by the definition of the delta function, thi lat integral i: g Hf - HLL ÅÅ ÅÅÅÅÅÅ f' Hf - HLL If f'< then the range of integration i negatively oriented and the integral i - g Hf- HLL ÅÅ ÅÅÅÅÅÅ f' Hf - HLL So in general, thi integral i g Hf - HLL ÅÅ Å f' Hf - HLL Recalling that f( )= give: (6) (63) (64) - d Hf HLL g HL = g Hf- HLL ÅÅ Å = f' Hf - HLL J Å f - N = g H L (65)

9 Alo notice that: d H - L f - Å - g HL = J Å f - N g H L = So we have hown that when f i a differentiable function with eactly one ero at with f' then: (66) d) d Hf HLL = Å f - d H - L (67) Suppoe f ha countably many ero at the point i. Then ince d()= for we may write the integral - d Hf HLL g HL a countably many integral about each of the i. Further, we require that f'( i ) o that f i locally invertible near each of the i. Let the neighborhood where the local invere i valid be ( i - a i, i + b i ). Putting thee fact together give: (68) i +b i d Hf HLL g HL = d Hf HLL g HL - i= i -a i and allow u to evaluate each of the integral by the ame method a the previou problem. We find: Thi how: - d Hf HLL g HL = J Å f - N g H i L = i i= (69) (7) d Hf HLL = i= Å f - d H - i L (7) A a pecial cae, conider the function: f HL = H - L H - L Uing the formula jut derived we have: (7) d HH - L H - LL = J Å - H - L H - d H - i L = d H - L +d H - L Å LN= i - i= (73) e) In part (c) we howed that for an invertible -D function f(y) with a ingle ero at y the following i true: - f d Hf HyLL = Å d Hy - y (74) y L We now prove a imilar reult for higher dimenion. In what follow ' and y' are vector, and f i a vector valued function. Let u calculate: d Hf HyLL g HyL y (75)

10 where the integral i taken over all of n-dimenional pace. Suppoe that f i invertible (i.e. ha non-ero jacobian) at the point y and that y i the only ero of f(y). Since d()= for thi integral may be confined to the region around = where f i invertible. Call thi region R. d Hf HyLL g HyL y = R d Hf HyLL g HyL y If under the invere mapping, R i mapped to S, then the formula for thi new integral come from any advanced calculu tet (e.g. Apotol, vol). Recall from calculu, that under the mapping =h(u) the following formula hold (76) k HL = k Hh HuLL J HuL u (77) Where J(u) i the jacobian of the function h(u). Letting y= f - HL Thi formula give: R d Hf HyLL g HyL y = S d HL g Hf - HLL J HL (78) Where J() i the Jacobian of the invere function f - HL. From the definition of the delta function, thi lat integral i g Hf - HLL J HL = g Hy L J HL = g Hy L i k j ÅÅÅÅÅÅ f- y { = (79) Alo from calculu, the Jacobian of f - HL i divided by the Jacobian of f(y). So we have: i d Hf HyLL g HyL y = g Hy L k j Å f y y { y=y - (8) We have thu hown that: d Hf HyLL =d Hy - y L f Å y - If the function f i that which map yø and y Ø, then we may write: (8) d H - L =d Hy - y L f Å y - (8) In coordinate form in 3-D thi i: Hf,f,f 3 L d H - L d H - L d H 3-3 L =d Hy - y L d Hy - y L d Hy 3 - y 3 L ÅÅ ÅÅ ÅÅÅÅÅÅ Hy,y,y 3 L A quick change to Greek variable give the deired anwer. - (83) f) If coordinate are being changed according to = X H, h, L y = Y H, h, L = Z H, h, L then the Jacobian appearing above i: (84)

11 X X h X Y Y h Y Z ÅÅÅÅÅÅ Z ÅÅÅÅÅÅ h Z ÅÅÅÅÅÅ For cylindrical coordinate the change i given by: (85 = X Hr, f, L = r Co f y = Y Hr, f, L = r Sin f = Z Hr, f, L = (86) So the jacobian i X r X f X Y r Y f Y Z ÅÅÅÅÅÅ r Z ÅÅÅÅÅÅ f Z ÅÅÅÅÅÅ Co f Sin f Cof Sin f = -r Sin f r Co f = -r Sin f r Co f = r (87) The formula from part (e) then give the deired anwer. d H - L d Hy - y L d H - L = ÅÅÅÅÅ r d Hr - r L d Hf -f L d H - L in pherical coordinate r, q, f the change i given by (88) = X Hr, q, fl = r Co f Sin q y = Y Hr, q, fl = r Sin f Sin q = Z Hr, q, fl = r Co q (89) So the jacobian i X r X q X f Y r Y q Y f Z ÅÅÅÅÅÅ r Z ÅÅÅÅÅÅ q Z ÅÅÅÅÅÅ f = Co f Sin q Sin q Sin f Co q r Co f Co q r Sin f Co q -r Sin q -r Sin f Sin q r Sin q Co f = r Sin q (9) d H - L d Hy - y L d H - L = ÅÅ ÅÅÅÅÅÅ r Sin q d Hr - r L d Hq -q L d Hf-f L and if m= Co q i ubtituted in pherical coordinate (9) = X Hr, m, fl = r Co f H -m L ê y = Y Hr, m, fl = r Sin f H -m L ê = Z Hr, m, fl = r m (9) X r X m X f Y r Y m Y f Z ÅÅÅÅÅÅ r Co f H -m L ê Sin f H -m L ê m Z ÅÅÅÅÅÅ m = -mr Co f H -m L -ê -r m H -m L -ê Sin f = r Z ÅÅÅÅÅÅ f -r Sin f H -m L ê r H -m L ê Co f r (93) Applying the formula from part (e) give:

12 d H - L d Hy - y L d H - L = ÅÅÅÅÅÅ r d Hr - r L d Hm -m L d Hf-f L Problem 4 ( point) (94 Laplace tranform on dicontinuou function a) (4 point) Conider the function f HtL = t < t t > t f t = t Compute the Laplace tranform F() of f(t). Then by eplicit integration of the Mellin inverion formula along the Bromwich contour (being careful about principle value when needed), how that the invere Laplace tranform of F() i (95) 8L - F< HtL = t < t t > t ê t = t and thu need not agree with f(t) at t=t. (96) b) ( point) I thi conitent with Lerch' theorem (a tated in cla: if f HtL and f HtL have the ame Laplace tranform, then f and f differ by a null function, i.e. f - f = NHtL, where Ÿ t NHtL t= for all t >.)? c) (4 point) Now conider y HtL = f HtL t < t g HtL t > t By conidering the continuou (ecept poibly at the iolated point t ) function y HtL - H Ht - t L Hg Ht + L - f Ht - LL - 8LHyL<D Ht L? Solution to Problem 4 (97) (98) f HtL = t < t t > t f t = t Compute the tranform from the definition: (99) t F HL = -t f HtL t = -t t + -t t = -t t () Compute the invere from the inverion formula: 8L - F< HtL = ÅÅÅÅÅ c+â pâ t -t c-â = ÅÅÅÅÅ pâ c-â c+â Ht-t L ÅÅÅÅÅ () cae (i): t > t Let c be any real number and integrate along the following contour:

13 R -R c Figure 5 Call thi contour G R and call the emicircular part C R. By the reidue theorem we have: -R pâ= pâre i ÅÅÅÅÅ k j Ht-t L, y = { G R Ht-t L ÅÅÅÅÅ Integrating counterclockwie along each piece of the contour eparately give: G R Ht-t L ÅÅÅÅÅ = c-âr c+âr Ht-t L ÅÅÅÅÅ ÂR + c+âr Each of thee integral may be computed or approimated: Similarly ÂR Ht-t L ÅÅÅÅÅ c+âr = c Ht-t L ÂR Ht-t L y ÅÅÅ Å y +ÂR y = c Ht-t L y ÅÅÅ Å y +ÂR y -ÂR Ht-t L ÅÅÅÅÅ c-âr Ø Now eamine the emicircular part: c Ht-t L y ÅÅÅ Å y +ÂR Ht-t L ÅÅÅÅÅ + C R c Ht-t L y y = ÅÅ!!!!!!!!!!!! y + R Ht-t L ÅÅÅÅÅ + -ÂR L c y c ÅÅÅÅÅ Ht-t Ø R c-âr Ht-t L ÅÅÅÅÅ () (3) (4) (5) C R Ht-t L ÅÅÅÅÅ = pê 3 pê Ht-t L R Âq ÅÅ ÅÅÅ Å ÂR Âq q R Âq pê 3 pê 3 pê Ht-t L R Âq q = Ht-t L R Co q q pê Since Co q i non-poitive in thi region, and t-t >, thi integral vanihe by Jordan' Lemma. (6) So we have hown for t> t : pâ= lim RØ c-âr So for t>t we have c+âr Ht-t L ÅÅÅÅÅ (7) 8L - F< HtL = ÅÅÅÅÅ pâ c-â c+â Ht-t L ÅÅÅÅÅ = (8)

14 cae (ii) t < t Now ue the contour drawn below: R c R Figure 6 By the Cauchy-Gourat theorem, the integral around thi contour i ero. -R = G R Ht-t L ÅÅÅÅÅ = c-âr Bounding the emicircular contour: c+âr Ht-t L ÅÅÅÅÅ + C R Ht-t L ÅÅÅÅÅ (9) C R Ht-t L ÅÅÅÅÅ = pê -pê Ht-t LHc+R Âq L ÅÅ ÅÅ ÅÅÅÅÅ Â R Âq q c + R Âq -pê R Ht-t L c Ht-t L R Âq ÅÅ ÅÅÅÅ pê c + R Âq q R L c -pê ÅÅ Ht-t ÅÅÅ R - c Ht-t L R Co q q pê Since Co q i poitive in thi region and t - t < Jordan' lemma tell u that thi integral vanihe. () So we have hown for t < t : 8L - F< HtL = ÅÅÅÅÅ pâ c-âr c+âr Ht-t L ÅÅÅÅÅ = () cae (iii) t = t 8L - F< Ht L = ÅÅÅÅÅ c+âr pâ ÅÅÅÅÅ c-âr () The integrand i analytic along the contour provided c>. So we may evaluate thi integral uing anti-derivative provided that we define a branch for the log whoe cut doen't cro the contour. We define our log by -p < arg HL <p So we have: c+âr c-âr ÅÅÅÅÅ c +ÂR = Log Hc +ÂRL - Log Hc -ÂRL = ln ÅÅÅÅÅ +ÂHarg Hc +ÂRL-arg Hc -ÂRLL c -ÂR (3) (4)

15 Obviouly ln ÅÅÅÅÅ c +ÂR Ø ln - = c -ÂR Alo, for our branch, we have: arg Hc +ÂRL Øpê arg Hc -ÂRL Ø-pê So that we have: (5) (6) 8L - F< Ht L = ÅÅÅÅÅ c+âr pâ ÅÅÅÅÅ c-âr = ÅÅÅÅÅ pâ Â J ÅÅÅÅÅ p - Å -p N = ÅÅÅÅÅ (7) b) Define the following two function: f HtL = f HtL = Their difference i: t < t t > t f t = t t < t t > t ê t = t (8) (9) t t N HtL = f - ê t = t Thi function N(t) i indeed a null function ince: () t N HL = for all () So even though f f pointwie, the integral of their difference i alway ero ince the function differ only on a et of meaure (in thi cae, a point). So our previou reult are conitent with Lerch' theorem. c) y HtL = f HtL t < t g HtL t > t () h HtL =y HtL - H Ht - t L Hg Ht + L - f Ht - LL (3) Notice that: lim h HtL = f Ht tøt - - L lim h HtL = f Ht - L tøt + (4) So that h(t) i a continuou function. Tranforming h(t) give: L HhL = L HyL - Hg Ht + L - f Ht - LL -t t = L HyL - Hg Ht + L - f Ht - LL -t t (5)

16 Since h i continuou, we may invert both ide to get: h HtL = L HyLD - Hg Ht + L - f Ht - LL L - A -t E (6) In part (a) we computed the invere tranform on the right: Thu we have: L - A -t E = t < t t > t ê t = t (7) L HyLD HtL = h HtL = f HtL for t < t L HyLD Ht L = h Ht L + g Ht + L - f Ht - L ÅÅÅÅ = g Ht + L + f Ht - L ÅÅÅÅ L HyLD HtL = h HtL + g Ht + L - f Ht - L = g HtL for t > t (8) Problem 5 ( 7 point) Ue the hifting theorem to ait you in olving the following initial value problem (you may in addition ue Laplace tranform table -e.g. the one handed out in cla): a) b) 4 y'' - 4 y' + 37 y = y HL = 3 y' HL = 3 ê (9) y''+ y = r HtL = t < t < t > y HL = y' HL = Solution to Problem 5 (3) a) 4 y'' - 4 y' + 37 y = y HL = 3 y' HL = 3 ê Applying the Laplace tranform give: 4 H Y - y HL - y' HLL - 4 HY-y HLL + 37 Y = Simplifying: (3) (3) - 6 Y = ÅÅ ÅÅ ÅÅÅ = 3 - ÅÅÅÅ ÅÅ ÅÅ H - ÅÅÅÅ L + 3 (33) Knowing one tranform and a hifting theorem

17 L HCo k L = ÅÅ ÅÅÅÅÅ + k 8L H a f HLL< HL = 8L Hf HLL< H - al (34 ) allow u to write: y HL = 3 ÅÅÅÅ Co 3 (35) b) y''+ y = r HtL = t < t < t > y HL = y' HL = Applying the Laplace tranform give: Y - y HL - y' HL + Y = t -t t = - - J + ÅÅÅÅÅ N Simplifying: (36) (37) Y = ÅÅÅÅÅ H + L - ÅÅÅ - + J + ÅÅÅÅÅ N (38) We ue partial fraction to rewrite thi: Y = - ÅÅÅ J + ÅÅÅÅÅ - ÅÅÅ + - ÅÅÅ + N We then ue the following 4 tranform and a hifting theorem: (39) L HSin L = ÅÅÅ + L HCo L = ÅÅÅ + L HL = ÅÅÅÅÅ L HL = L Hf H - al H H - all = -a L Hf HLL (4) Thee give: y HL = - Sin - H - Sin H - L - Co H - LL H H - L (4) Thi i a continuou function with continuou firt derivative. The econd derivative i of coure dicontinuou. Thee fact are reflected in the plot below.

18 yhl y HL y HL.5.5 Problem 6 ( point) Figure 7 Solve the differential equation y''+ y'-y = y HL = (4) y' HL = by taking the Laplace tranform of both ide (auming the tranform Y() of y() eit). Solve the reulting firt-order differential equation for Y(). Be careful to chooe the contant of integration o that Y() behave a Ø in a manner conitent with Laplace tranform. Invert Y() to find y() and check that y() atifie the IVP. Solution to Problem 6 y''+ y'-y = y HL = y' HL = The Laplace tranform of n f() may be computed a follow. Firt Laplace tranform f(): (43) F HL = L 8f HL< HL = - f HL (44) Then, auming that f i of eponential order, thi integral may be differentiated with repect to and the derivative paed through the integral: F HnL HL = H-L n - f HL Finally we have: (45) L 8 n f HL< HL = H-L n n ÅÅÅÅ L 8f HL< HL n (46) We may now tranform the equation: = L 8y'' + y' - y< = Y - y HL - y' HL - HY - y HLL - Y Simplifying give: (47) Y' + J ÅÅÅÅÅ - N Y =- ÅÅÅÅÅ (48) Thi firt order ODE for Y i eaily olved uing integration factor or the general formula ued in previou week.

19 Y = A ÅÅÅÅ + (49 ) The Laplace tranform of any piecewie continuou function of eponential order can not grow eponentially. Proving thi i imple. Suppoe f i uch a function, then f HL a M For ome contant M and a. Then we have: L HfL HL = - f HL - f HL If i real, then thi bound i: L HfL HL - f HL M -H-aL = ÅÅÅÅÅÅ M - a So the Laplace tranform of f Ø a real Ø. For thi reaon, we et A= to get: Inverting: Y = (5) (5) (5) (53) y HL = (54) See appendi for another way to how A=. Problem 7 (5 4 point) Laplace Tranform [L{f}]()= Ÿ eph-tl f HtL t from Taylor erie: a) Show that for n=,,3... L Ht n-ê p ÿ 3 ÿ 5... H n - L L = Å Å n n+ê b) Find the power erie epanion about = for the error function (55) erf HL = ÅÅÅÅÅ (56) p ep H-u L u c) Let = è!! in your erie from (b). Take the Laplace tranform term by term of the reulting erie, to how that for, L Ierf I MM = ÅÅ ÅÅÅ!!!!!! + d) What happen in part (c) for <? Could thi equation alo be true for <? Think about thi for a while before checking your anwer by continuing with part (e). e) Show that erf I M ÅÅ = Å - (58)!!! p Uing the integral definition of the Laplace tranform given at the beginning of thi problem, compute the Laplace tranform of the right hand ide of thi equation. Uing the epreion for the Laplace tranform of a derivative, find the Laplace tranform of erf( ), and compare to your reult in (c). Check your anwer to part (d) and dicu (57)

20 Solution to Problem 7 a) From the formula for the derivative of a Laplace tranform derived in problem 6 we may write L Ht n-ê L = t n-ê -t t = H-L n- n- ÅÅ t n- ê -t t Let t=y, integrate once by part, and ue the known epreion for the Gauian integral t ê -t t = y -y y = ÅÅÅÅÅ p -y y = Å 3ê Thi i valid for all Re()>. So we have: (59) (6) L Ht n-ê L = H-L n- p n- ÅÅÅÅÅ ÅÅ n- -3ê = H-L n- p ÅÅÅÅÅ J Å -3 N J Å -5 -H n - L N... J ÅÅ ÅÅÅÅÅÅ N -n-ê = H-L n- p H-L n- ÅÅÅÅÅ ÅÅÅÅÅÅ H3 ÿ 5 ÿ... ÿ H n - LL -n-ê p ÿ 3 ÿ 5... H n - L = Å Å n- n n+ê (6) b) erf HL = ÅÅÅÅÅ p -u u Differentiate thi: erf ' HL = ÅÅÅÅÅ p - Uing the known erie for the eponential, we write: (6) (63) - = n= H- L n ÅÅÅÅÅ n! = n= H-L n Å n n! We find a power erie for erf() by integration (which i permitted ince thi power erie converge uniformly for all ): (64) erf HL = erf ' HL = ÅÅÅÅÅ p H-L n Å n = ÅÅÅÅÅ n! n= p H-L n Å n! n = ÅÅÅÅÅ n= p n= H-L n ÅÅ Å n! H n + L n+ (65) c) Plug = è!! into our previou um: erf I M = ÅÅÅÅÅ p H-L n ÅÅ Å n! H n + L I M n+ = ÅÅÅÅÅ n= p H-L n ÅÅ Å n! H n + L n+ê n= Now Laplace tranform thi term by term (66)

21 L Ierf I MM = ÅÅÅÅÅ p H-L n n! H n + L L Hn+ê L = n= n= H-L n Å n! ÿ 3 ÿ 5... H n - L ÅÅÅÅ n n+3ê = -3ê n= ÿ 3 ÿ 5... H n - L ÅÅÅÅ n! J Å - n N (67 ) Oberve the following: f HyL = H - yl -ê f HnL ÿ 3 ÿ.. ÿ H n - L HyL = ÿ ÿ.. ÿ So that we have for y <: -H n+lê H - yl (68) H - yl -ê ÿ 3 ÿ.. ÿ H n - L = n! n= J y ÅÅÅÅÅ Nn (69) So for > L Ierf I MM = -3ê ÿ 3 ÿ 5... H n - L ÅÅÅÅ n! n= J Å - n N = -3ê J + ÅÅÅÅÅ -ê N = ÅÅÅ!!!!!! + (7) d) Recall that we have: L Ierf I MM = -3ê ÿ 3 ÿ 5... H n - L ÅÅÅÅ n! n= The um converge provided that J Å - n N (7) ÿ3ÿ5... H n+l H - Hn+L! ÅÅ ÅÅÅÅÅÅ ÅÅÅÅÅÅ lim Ln+ Å ÅÅÅÅ < nø ÿ3ÿ5... H n-l ÅÅÅÅÅ H ÅÅÅÅÅÅ - n! Ln Simplifying thi condition give: (7) ÅÅÅÅÅ (73) < Since we know that power erie diverge outide of their radiu of convergence, the erie we found diverge when <. A i often the cae in comple analyi, formulae containing erie which diverge can often be etended to include value outide the radiu. For eample, the erie f HL = n= n diverge for >. The function f() can be computed for < (74) f HL = for < (75) - If we epect that f() i analytic (ecept perhap at certain pole or branch point), then we might be able to analytically continue f() to the whole comple plane (ecept =) to define a new function, the analytic continuation of f():

22 F HL = - for We might imilarly epect that (76 ) L Ierf I MM HL i analytic in a larger domain that jut >. So we might be able to define it' analytic continuation to the whole plane (ecept the pole at =, the branch point at =-, and ome branch cut) (77) F HL = ÅÅÅ!!!!!! + In part (e) we will ue another method to cover a different part of the plane. (78) e) Recall that in part (b) we howed: erf HL ÅÅ = ÅÅÅÅÅ p By the chain rule we have: - erf I M ÅÅ Å = i erf I M y i y j ÅÅÅÅÅÅ k j Å { k = i j ÅÅÅÅÅ { k p -y i y j Å { k = Å -!!! { p We now Laplace tranform the right hand ide: L i - y j Å!!! k p = - Å!!! { p - = ÅÅÅÅÅ p -ê -H+L Make the change of variable =y ÅÅÅÅÅ p -H+L y y From the known Gauian integral, thi i defined for Re()>-: (79) (8) (8) (8) - L i y j Å!!! k p = ÅÅ!!!!!! ÅÅÅÅÅÅ { + (83) We now compute the tranform of the derivative: L i erf I M y j ÅÅ k { = L Ierf I MM - erf HL = L Ierf I MM (84) For and Re()>- thi give: L Ierf I MM = ÅÅÅÅÅ L i erf I M y j ÅÅ Å = ÅÅÅÅÅ L i - y j Å!!! k { k p HL = ÅÅÅ {!!!!!! + (85) Which i the ame a that found in (c). In part (c) we howed thi to be true for > and in (d) it' true for Re()>- and. Thee two region are illutrated below.

23 = and ReHL= Figure 8 - You can ee how the region A={: Re()>- and } contain ome of the region B={: >} but that the combination of thee two region (A B) cover the entire comple plane ecept =- and =. So by tranforming a function in two different way we have been able to define the tranform for all,-. Appendi: Waton' Lemma and problem A mentioned above, there i a lemma which make approimating integral eaier. The baic idea i a follow. For an integral of the form b I HL = f HtL -t t a with b>a and where: (86) f HtL = Ht - al a a n Ht - al bn n= i a erie for f near t=a, the integral I() may be approimated by keeping only the firt n term of the erie and integrating over [a,): (87) n I HL - k= a k a -t Ht - al bk+a t << ÅÅ ÅÅÅÅÅ H - al +bn+a In problem c Waton' Lemma may be applied with a=, a=, and b= to find: I HL - ÅÅÅÅ ÅÅ ÅÅÅ ÅÅÅÅ << ÅÅÅÅÅ ê ê = In problem d with a=, a=, and b= Waton' Lemma may be applied to find: (88) (89) I HL - ÅÅÅÅ ÅÅ ÅÅÅ (9) ÅÅÅÅ << ÅÅÅÅÅ ê ê = Evidently, in both problem, we hould keep more term to be ure that our approimation i accurate enough. For eample, in problem d we keep two term and Waton' lemma tell u that

24 I HL - ÅÅÅÅ + ÅÅÅÅÅ ÅÅÅ ÅÅÅÅ ÅÅÅÅÅ << ÅÅ ê 3 ÅÅÅÅ ÅÅÅÅÅ - - = ÅÅÅÅ - << - =. Without uing Waton' Lemma, it i till ometime eay to find your own bound. For eample, conider c. Let u try to bound the error. (9 ) I HL - ÅÅÅÅÅ = So we have: i j -t y Å k + t - -t t { = -t -t ÅÅ Å t + t = I HL - ÅÅÅÅ ÅÅ ÅÅÅ ÅÅÅÅ Compare thi to uing Waton' lemma where we found the lightly better bound I HL - ÅÅÅÅ ÅÅ ÅÅÅ ÅÅÅÅ << t -t ÅÅÅÅÅ + t t Appendi: Showing that A= in problem 6 Y = A ÅÅÅÅ + Another way of howing that A= give the correct anwer i to write: y HL = L - HYL = L - J A ÅÅÅÅ + N = Ag HL + For ome non-trivial g(). Now, you may check that g() mut atify: -t t = ÅÅÅÅÅ (9) (93) (94) (95) (96) g''+ g'-g = g HL = (97) g' HL = Thi linear equation ha analytic coefficient and by the Cauchy-Kowalevki theorem (F. John, pg74) mut have a unique olution which i analytic near =. Since every analytic function ha a unique power erie repreentation we have: But we find: g HL = n= g HnL HL ÅÅÅÅÅ n n! (98) g HL = g' HL = g'' HL = Hg - g'l = = (99) g''' HL = HHg - g'l'l = = etc. So that any derivative of g at = vanihe. Thi mean that the power erie for g ha all ero cofficient. Hence g()= in ome neighborhood of. By analytic continuation we mut have g()= for all. But thi i the ame reult a you get by etting A=.