APPM 4360/5360 Homework Assignment #7 Solutions Spring 2016

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1 APPM 436/536 Homework Assignment #7 Solutions Spring 6 Problem # ( points: Evlute the following rel integrl by residue integrtion: x 3 sinkx x 4 4, k rel, 4 > Solution: Since the integrnd is even function, I x 3 sinkx x 4 4 i ImJ J, x 3 e i kx J x 4 4 For k >, closing the contour in the upper hlf-plne nd using Jordn lemm, we find J n:imz n > Res(f (z; z n, where f (z z 3 e i kz /(z 4 4 Its only sp in the upper hlf-plne re two simple poles t z e i/4 nd z e 3i/4, so J z3 e i kz 4z 3 z3 e i kz 4z 3 zz zz i ( e i k(i / e i k( i / i e k/ cos(k / i e k/ cos(k / Also I ( k I (k I sign(ke k / cos(k / Problem # ( points: Show tht cosh x coshx sec (, < Use rectngulr contour with corners t±r nd ±R i Solution: Let R 3 4 be the rectngulr contour nd n, n,,4 re its sides in counterclockwise order, [,R] Then cosh x I coshx cosh x f (x coshx Inside R, the integrnd f (z is nlytic except points where coshz, ie z i / Thus f (zd z Res(f (z;i / R cosh z sinhz cos(/ zi / On the other hnd, f (zd z becuse < Besides, 3 f (zd z R 4 f (zd z, cosh (x i cosh(x i cosh x coshi sinh x sinhi R coshx coshi cos R cosh x I cos coshx R f (zd z cos(/ I I cos I cos (/, so I sec(/ nd cosh x coshx I sec(/ Problem #3 ( points: onsider rectngulr contour with corners t b± i R nd b ±ir Use this contour to show tht bi R b i R where <b<, Im < e z sinz d z (e, Solution: Let R 3 4 be the clockwise rectngulr contour nd n, n,,4 re its sides in clockwise order, [b i R,b i R] Then I bi R b i R e z d z f (zd z, sinz where f (z e z sinz Inside R, the integrnd f (z is nlytic except points where sin z, ie only z Thus f (zd z Res(f (z; R e z cosz i e z

2 On the other hnd, f (zd z becuse Im < Besides, 4 f (zd z, (bi y e f (zd z 3 R sin(b i y i d y nd I R e (bi y e sin(b i ycos i d y e I (e (e I i e e Problem #4 ( points: onsider rectngulr contour R with corners t (±R, nd (±R, Show tht R R e z d z e x e (xi J R, R where J R e (Ri y i d y e (i y i d y Show J R, whereupon we hve e (xi e x, nd consequently, deduce tht e x cos x e Solution: Let R 3 4 be the counterclockwise rectngulr contour nd n, n,,4 re its sides in counterclockwise order, [,R] Then e z d z 3 e z d z 4 e z d z e z d z R R e x, e (Ri y i d y, e (xi e (i y i d y R e (xi, e (i y i d y, nd, by uchy theorem, R e z d z, which shows the first point We hve J R e (Ri y i d y e (i y i d y ( e e y d y e y d y, which proves tht e (xi e x Finlly, e (xi e e x e i x e e x cos x, the lst equlity becuse the integrl of imginry (odd prt is zero Problem #5 ( points: Use sector contour with rdius R s in figure 46, centered t the origin with ngle θ 5, to find, for >, x sin 5 Solution: ontour x R L ; on x, z x, x R; on R, z Re iθ, θ 5 ; on L, z r e /5, r R Then I x 5 5 f (zd z L f (zd z R /5 R x f (zd z, e /5 dr r 5 5 e /5 I, Rdθ R 5 5, nd, since the only sp inside is z e i/5, f (zd z Res(f (z; e i/5 5 4 e 4i/5 so I ( e /5 I 5 4 e 4i/5, 5 4 e 4i/5 ( e /5 5 4 (i sin(/5 5 4 sin(/5 Problem #6 ( points: ( Use principl vlue integrls to show tht cos k x cosmx x ( k m, k,m rel Hint: note tht the function f (z(e i kz e i mz /z hs simple pole t the origin

3 (b Let k, m to deduce tht Solution: sin x x ( Let k > nd m> onsider f (zd z where [, ǫ] ǫ [ǫ,r] R, ǫ {ǫe iθ, θ } nd R {Re iθ, θ Then f (zd z Res(f (z; i k z i mz z z z (k m nd, since R f (zd z by Jordn lemm, I (k,m 4 pv cos k x cosmx x cos k x cosmx x f (zd z (k m Since I (±k,±m I (k,m, we get I (k,m ( k m (b For k, m, I (k,m, but I (, cos x x so the c follows sin x x, Problem #7 ( points: Show tht sin x x(x ( e Solution: It is convenient to write the integrl s follows: I sin x x(x e i x i x(x sin x x(x e i x i x(x hnging the integrtion vrible x x in the second integrl on the right we find e i x x(x e i x x(x, so e i x I i x(x onsider the lst integrl tken over the closed contour in complex plne, [,R] R, R {Re i t, t }: e i z i z(z d z R e i x i x(x e i z R i z(z d z When we tke the it R, the first integrl on the right-hnd side becomes I while the second tends to zero since e i z R i z(z d z (e i Rei t Ri e i t d t i Re i t (R e i t (e sint d t (R R e i z I i z(z d z f (zd z Res(f (z;i e i i ( e Problem #8 ( points: Use rectngulr contour with corners t ±R nd ±R i /k, with n pproprite indenttion, to show tht x sinhkx 4k k for k,k rel Solution: Let R 3 4 be the rectngulr contour nd n, n,,4 re its sides in counterclockwise order, [,R] Since sinhkx on the contour t x i /k, we must use indenttion round this point Let k > Ie consider insted contour R,ǫ 3 4 ǫ, where 3 [R i /k,ǫi/k] [ ǫi/k, i/k] nd ǫ {z i /k ǫe iθ sinθ< (trced from right to left nd let ǫ Then, by uchy theorem, R,ǫ f (zd z onsider we lso hve I x sinhkx f (zd z, f (zd z 3 3

4 ( ǫi/k i/k z Ri/k ǫi/k sinhkz d z ( ǫ x i /k R ǫ sinhk(x i /k ( ǫ x i /k R ǫ sinhkx coshi ( R ǫ x sinhkx I, ǫ where we used tht sinhkx is odd in going to the lst line Besides, f (zd z f (zd z, 4 becuse there sinh k z sinhk(±r i y, y /k, so sinhk(±r i y± we get I f (zd z, ǫ ǫ where ǫ f (zd z i /k ǫe iθ sinhk(i /k ǫe iθ ǫi eiθ dθ i /k ǫe iθ cosh i sinhkǫe iθ ǫi eiθ dθ ǫ I /k, nd i k ǫe iθ ǫi eiθ dθ k x sinhkx I 4k Since sinh( k x sinh k x, in generl, the nswer is 4k k Problem #9 ( points: Use the keyhole contour of Figure 436 to show tht, on the principl brnch of x k, x k I x sink k, <k <, > Solution: Let be the whole closed contour, ± be the upper/lower sides of the keyhole Then, in the it s R nd ǫ, z k z d z zk z d zi, (ze k z d z e (k I, nd ǫ zk R z d z R k Rdθ, R z k z d z ǫ k e i (k θ i ǫei θdθ ǫei θ I ( e (k z k d z Res(f (z; z ( k k e i(k, nd it follows tht I sin(k k sink k Problem # ( points: Use the technique described in problem bove, using f (z(log z d z, to estblish tht log x I ( x log, > Solution: We use log z J ( z d z log x x (log x x 4i I ( ( x 4i I ( 4 ( R nd ǫ s usul From residues, J ( log z z log z zi z z i (log i / i I ( J ( i (log 3i / log i i log i i log Problem # ( points: By using lrge semicirculr contour, enclosing the left hlf plne with suitble keyhole contour, show tht i i e zt z d z t for, t > This is the inverse Lplce trnsform of the function / z (The Lplce trnsform nd its inverse re discussed in Section 45 4

5 Solution: Let I be the integrl to find, the whole closed contour As prt of it, we use the keyhole contour round the brnch cut on the negtive rel hlf-xis Let ± be the upper/lower sides of the cut, respectively Then e zt d z i ± ± z i e xt x e t u du i t As integrl over lrge semicircle vnishes in the it (due to e zt e (Rei θ t, cosθ<, R, it is Jordn lemm gin, nd integrl over little circle round z lso vnishes s it rdius ǫ, we get I i e zt d z, t z since the (brnch of the integrnd is nlytic inside I t Problem # ( points: onsider the integrl I R R d z (z, >, / where R is circle of rdius R centered t the origin enclosing the points z ± Tke the principl vlue of the squre root ( Evlute the residue of the integrnd t infinity nd show tht I R (b Evlute the integrl by defining the contour round the brnch points nd long the brnch cuts between z nd z to find tht I R x Use this rel-vlued integrl to obtin the sme result s in prt ( Solution: Mking intervl [, ] the brnch cut, we get nlytic brnches of (z / outside the cut ( About z, (z / ( z z / ( z z so ( z z z z 3, ( I R Res (z ; / (b On the other hnd, ( I R (e i x x / (e i x e x / 4 x x / rcsin u Problem #3 ( points: Show f (z d z N, f (z f du u where N is the number of points z where f (z f ( constnt inside ; f (z nd f (z re nlytic inside nd on ; nd f (z f on the boundry of Solution: The points z where f (z f ( constnt inside re the zeros of g (z f (z f inside nd it hs no poles there So the formul follows by Argument Principle Problem #4 ( points: Use the Argument Principle to show tht f (z z 7 hs two zeroes in the first qudrnt Solution: Since f (z is entire, by Argument Principle, f (z f (z d z N, where N is the number of zeros of f (z inside Let be the contour in Fig 443 in the first qudrnt with corner t z Then on [, R] the rgument rg f (z does not chnge, on the qurter-circle, rgf (z 7 7, nd on the intervl of imginry xis [i R,] we hve f (z(i y 7 i y 7, so tnrgf (z Im f (z Ref (z y 7 <, 5

6 it chnges from to for lrge R, nd therefore rgf (z chnges from 7/ to 4 two zeros indeed N t ot l rgf (z 4, Problem #5 ( points: ( Show tht e z (4z hs exctly two roots for z < Hint: in Rouche s Theorem use f (z 4z nd g (ze z, so tht when is the unit circle f (z 4 nd g (z e z e z (b Show tht the improved estimte g (z e cn be deduced from e z z e w d w nd tht this llows us to estblish tht e z (z, hs exctly one root for z < Solution: ( Since on z, g (z e z e z e <4, we hve f (z > g (z there So, by Rouche s theorem, f (z nd f (z g (z hve equl number of zeros inside, nd this number is for f (z, so lso for f (z g (ze z (4z (b g (z z e w d w e z e on z Since e < z on the unit circle, Rouche theorem yields the second c Problem #6 ( points: Obtin the inverse Fourier trnsform of the following function: (k w, w > Note the dulity between direct nd inverse Fourier trnsforms Solution: We close the contour by lrge semicircle in the upper hlf-plne for x>, then (using Jordn lemm F (x e i kx (k w dk e i xz (z w d z i Imz n > Res(f (z; z n i Res(f (z;i w d e i xz i d z (z i w zi w ( i xe i xz i (z i w ei xz (z i w 3 zi w i e xw (xw e xw ( xw (i w 3 4w 3 Since F ( x F (x, in generl, F (x ( x w e x w 4w 3 Problem #7 ( points: Show tht the Fourier trnsform of the Gussin" f (xexp( (x x /, x, rel, is lso Gussin: Solution: e (k/ e i kx ˆF (k e (k/ e i kx ˆF (k f (xe i kx exp( (x x / e i kx exp( (x x i k / / e (k/ e i kx, the lst step is forml but cn be justified exp( (x x i k / / i k / exp( z / d z i k / onsider rectngulr closed contour with corners t±r nd±r i k / Then exp( z / d z by nlyticity Then the integrls over the upper nd lower sides re equl up to the opposite sign, nd the integrls over left nd right sides vnish in the it s R (there z±r i y, y [,k / is finite the result Problem #8 ( points: Obtin the Fourier trnsform of the following function, nd thereby show tht the Fourier trnsforms of hyperbolic secnt functions re lso relted to hyperbolic secnt functions sech[(x x ]e iωx,, x,ω rel 6

7 Solution: It is convenient to use rectngulr contour with corners t±r nd±r i / nd then let R Let R 3 4 be the rectngulr contour nd n, n,,4 re its sides in counterclockwise order, [,R] Then I sech[(x x ]e iωx sech[(x x ]e iωx f (x Inside R, the integrnd f (z is nlytic except points where cosh (z x, ie z x i /( Thus R f (zd z Res(f (z; x i /( e iωz sinh[(z x ] On the other hnd, zx i/( f (zd z e ω/iωx 4 f (zd z, becuse sech[(z x ] there Besides, f (zd z 3 so R e ω/ sech[(x x i /]e iω(xi/ R sech[(x x ]e iωx Ie ω/ I (e ω/ e ω/iωx, I e iωx cosh[ω/(] Extr-redit Problem #9 (3 points: #436 ψ (z is nlytic in the upper hlf-plne H, ψ (z is nlytic in the lower hlf-plne H, ψ (x ψ (x f (x, ψ ± (z z ; Solution: ψ ± (x ǫ f (ζ ζ (x± i ǫ dζ ( Introduce opertors Then (b Let P ± ǫ P ψ (x ǫ ǫ dζ ζ (x± i ǫ ψ (ζ ζ (x i ǫ dζ ψ (ζ ζ (x i ǫ dζ ǫ ψ (x i ǫψ (x, where we closed the contour in H And, by nlyticity of ψ (ζ in H, using the sme contour we obtin P ψ (x Similrly, closing the contour in H (then it is oriented clockwise we get P ψ (x ψ (x nd P ψ (x P ± re indeed projection opertors Then ψ ± (x ǫ ψ (x ψ (x x 4 (ζ 4 (ζ (x± i ǫ dζ The roots of the eqution ζ 4 re ζ e i/4, ζ e 3i/4, ζ 3 e 3i/4, ζ 4 e i/4, the first two re in H nd the lst two re in H Denote the integrted functions bove f ± (ζ, respectively Then ψ (x ( ǫ Res(f ; x i ǫres(f ;ζ Res(f ;ζ x 4 4ζ 3 (ζ x 4ζ 3 (ζ x Since ζ 4 ζ4, we get ψ (x x 4 ζ 4(ζ x ζ 4(ζ x Similrly we obtin ψ (x x 4 ζ 3 4(ζ 3 x ζ 4 4(ζ 4 x, where the opposite signs ccount for the opposite orienttion of the contour in H hecking the obtined formuls, we tke their difference, ψ (x ψ (x 7

8 x 4 ( 4 ζ x ζ ζ x ζ ζ 3 x ζ 3 ζ 4 x ζ 4 x 4 x 3 j ζ j x k j ζ j ζ k x l k j ζ j ζ k ζ l 4 j ζ j 4(x ζ (x ζ (x ζ 3 (x ζ 4 x 4 x 4 x 4, s should be (in the lst line we used ζ 4 (ζ ζ (ζ ζ (ζ ζ 3 (ζ ζ 4 nd so 4 ζ j, 4 ζ j nd the expression in the numertor of the second frction equls ((x 4 x 4 finlly (since ζ 3 ζ, ζ 4 ζ (c Show tht ψ (z z 4 ζ 4(z ζ ζ 4(z ζ, ψ (z ζ ζ z 4 4(z ζ 4(z ζ ψ ± (x We strt with f (ζ ζ x ψ ± (x ǫ f (x dζ± f (x H(f (x± i f (ζ ζ (x± i ǫ dζ For ψ, let z ζ iǫ, then ψ (x ǫ ǫ ( x ǫ iǫ iǫ f (z i ǫ z x d z f (z i ǫ xǫ ǫ z x d z, where the first two integrtions go over intervls of rel line nd the third over smll semicircle ǫ {x ǫe i t, t } The it of the sum of the first two integrls is the integrl in the sense of principl vlue (or Hilbert trnsform by definition, so ǫ ψ (x f (z z x d z f (x ǫe i t i ǫ ǫe i t i ǫe i t d t f (z z x d z f (x The formul for ψ is proved quite similrly, the minus sign rising from clockwise orienttion of the semicircle in this cse 8