Ewald s Method Revisited: Rapidly Convergent Series Representations of Certain Green s Functions. Vassilis G. Papanicolaou 1

Transcription

1 wald s Mthod Rvisitd: Rapidly Convrgnt Sris Rprsntations of Crtain Grn s Functions Vassilis G. Papanicolaou 1 Suggstd Running Had: wald s Mthod Rvisitd Complt Mailing Addrss of Contact Author for offic purposs: Vassilis G. Papanicolaou Dpartmnt of Mathmatics and Statistics Wichita Stat Univrsity Wichita, KS tl FAX mail: papanico@cs.twsu.du 1 Dpartmnt of Mathmatics and Statistics, Wichita Stat Univrsity, Wichita, KS

2 ABSTRACT W propos a justification of wald s mthod of obtaining rapidly convrgnt sris for th Grn s function of th 3-dimnsional Hlmholtz quation. Our point of viw nabls us to xtnd th mthod to Grn s functions for th Hlmholtz quation in crtain domains of R d, with quit gnral boundary conditions. Ky words. Hlmholtz quation, Grn s function, wald s mthod. 2

3 1 Introduction In 1921 P. P. wald publishd a snsational papr s [1]. H prsntd a mthod that was usd to transform th slowly convrgnt sris of th Grn s function of th 3-d Hlmholtz quation, with Floqut.g. priodic boundary conditions in th x and y dirctions, to a sris that convrgs vry rapidly. wald s mthod givs spctacular computational rsults and hnc it has bn usd xtnsivly by physicist, nginrs, and numrical analysts as an xampl, s [2]. Howvr, it is not transparnt at all. Th radr is usually lft with th imprssion that it works by coincidnc. Th mthod is basd on th formula ikr R = 1 π xp R 2 z 2 + K2 4z 2 dz, which somhow coms out of th blu. In this not w propos a point of viw that shds som undrstanding to th undrlying structur of wald s approach. This nabls us to asily xtnd his mthod to any dimnsion and a varity of boundary conditions. 2 A Rapidly Convrgnt Sris for th Grn s Function Lt L b a slf-adjoint oprator on th spac L 2 D, whr D is a domain in R d, such that inf σl = λ, whr σl is th spctrum of L. If λ is a complx numbr with R {λ} < λ, thn L λ 1 = λt tl dt. In trms of intgral krnls, th abov quation can b writtn as Gx, y; λ = λt pt, x, ydt, 1 whr Gx, y; λ is th Grn s function and pt, x, y th hat krnl associatd to L thus, th Grn s function is th Laplac transform of th hat krnl, viwd as a function of t. Th intgral in 1 convrgs if R {λ} < λ and divrgs if R {λ} > λ if R {λ} = λ, convrgnc is possibl. On th othr hand, Gx, y; λ is analytic in λ, on C \ σl. If L =, acting on R d hnc σl = [, ; L is th fr d- dimnsional ngativ Laplacian oprator, thn its hat krnl is 3

4 1 pt, x, y = 4πt d/2 xp x y 2. 4t Lt Γ d x, y; λ b th corrsponding Grn s function, i.. th intgral krnl of L λ 1. Thn 1 givs 1 x y 2 Γ d x, y; λ = xp λt dt df = Γ d/2 d x y ; λ 4πt 4t 2 Γ d x, x; λ =, if d 2. By dforming th contour of th abov intgral, w can gt Γ d x, y; λ, for R {λ}, but on can do vn bttr, sinc Γ d x y ; λ can b computd xplicitly. If d = 1, it is asy to s by solving th associatd ordinary diffrntial quation for Γ 1 that λ x y Γ 1 x, y; λ = Γ 1 x y ; λ = 2 λ. Furthrmor, from 2 on obsrvs that d dλ Γ dr; λ = 1 4π Γ d 2R; λ. From this quation on can comput Γ d x, y; λ for all odd dimnsions d: λ x y Γ 3 x, y; λ = 4π x y, tc. If d is vn, Γ d x, y; λ is not an lmntary function. On can find th radial solutions of th quation to conclud that for all d Γ d x, y; λ = u = λu + δx, x R d, 1 λ 2π d/2 λd/4 1/2 x y 1 d/2 K d/2 1 x y, whr K ν is th modifid Bssl function of th 2nd kind, of ordr ν. Finally, w notic that Γ d R; λ dcays xponntially, as R. This is asy to stablish if d is odd, whras, if d is vn, on can, for xampl, us 2 to gt th stimat Γ d R; λ < Γ d+1 R; λ + Γ d 1 R; λ 4

5 altrnativly, on can do asymptotic analysis, as x y, to th intgral in 2. Now considr th domain D in R d D =, b 1, b r R d r r is som intgr btwn 1 and d, and lt L α = act on D with α- Floqut-Bloch boundary conditions. This mans that thr ar numbrs α 1,..., α r such that, for u to b in th domain of L α, w must hav ux 1,..., b j,..., x d = iα jb j ux 1,...,,..., x d, ux 1,..., b j,..., x d = iα jb j ux 1,...,,..., x d, 3 for all j = 1,..., r. Without loss of gnrality w assum that π < α j π, j = 1,..., r. b j b j Th mthod of imags givs that th Grn s function of L α is Gx, y; λ = m Z iα mb Γ d x + mb, y; λ, r whr α = α 1,..., α r,,..., and mb = m 1 b 1,..., m r b r,,..., ar considrd in R d. Using 2 in th abov quation, w gt Gx, y; λ = m Z iα mb 1 x y + mb 2 xp λt dt, d/2 r 4πt 4t 4 whr, for λ < or, mor gnrally, R {λ} <, th sris convrgs absolutly, providd x y x and y ar in D. W now show how to obtain a sris rprsntation of Gx, y; λ, x y, that convrgs vry rapidly, for all λ C \ N, whr N is a discrt countabl st. First w writ following wald s approach Gx, y; λ = G 1 x, y; λ + G 2 x, y; λ, whr G 1 x, y; λ = m Z r iα mb 1 4πt d/2 xp λt x y + mb 2 dt, 4t 5 5

6 G 2 x, y; λ = m Z r iα mb 1 4πt d/2 xp λt x y + mb 2 dt, 4t 6 and is a positiv numbr. Obsrv that G 1 x, y; λ and G 2 x, y; λ ar th intgral krnls of th oprators tlα λ dt and tlα λ dt = L α λ 1 Lα λ, rspctivly. Th sris for G 1 x, y; λ, givn in 5, is alrady rapidly convrgnt, for all λ C as long as x y, sinc its gnral trm dcays in m lik C mb 2 xp mb 2 /4. In th cas r = d, G 2 x, y; λ, bing th intgral krnl of L α λ 1 L α λ, has th ignfunction xpansion G 2 x, y; λ = 1 D m Z d [ α 2 +4π 2 m/b 2 +4πα m/b λ] i[2πm/b+α] x y α 2 + 4π 2 m/b 2, + 4πα m/b λ 7 whr D = b 1 b 2 b d is th volum of D and m/b = m 1 /b 1,..., m d /b d. Formula 7 is valid for all λ α 2 + 4π 2 m/b 2 + 4πα m/b, i.. all λ / σl α, at which G 2 x, y; λ has simpl pols. Th trms of th sris abov dcay in m lik C m/b 2 xp 4π 2 m/b 2, for any fixd λ C \ σl α. Now, lt us discuss th cas r < d. For th sris in 6 w first assum that R {λ} <. In this cas w can intrchang summation and intgration and gt λt xr y r 2 /4t G 2 x, y; λ = 4πt d/2 m Z xr yr+mb 2 /4t iα mb dt, r 8 whr, for x = x 1,..., x d, w hav introducd x r = x 1,..., x r,,..., and x r =,...,, x r+1,..., x d. Nxt following th spirit of wald s approach w invok a lmma. 6

7 Lmma 1. If s > and z, γ C, thn n Z s2z+n2 iγn = π s iγz γ2 /4s2 n Z π2 n2 /s2 πγn/s2 +2πizn. Proof. Poisson Summation Formula s,.g. [3] applid to th function fx = Ax2 +Bx, A >, B C. in fact, th formula is also a thta function idntity. whr Th lmma implis immdiatly m Z xr yr+mb 2 /4t iα mb = r 4πt r/2 iα xr y r α 2 t b 1 b r m Z [4π 2 m/b 2 +4πα m/b]t+2πix r y r m/b, r Thus 8 bcoms G 2 x, y; λ = i[2πm/b+α] xr yr m Z b r 1 b r m/b = m 1 /b 1,..., m r /b r,,...,. [ α 2 +4π 2 m/b 2 +4πα m/b λ]t x r y r 2 /4t dt 4πt. d r/2 9 So far w hav assumd R {λ} <. Howvr, th trms of th sris abov dcay in m lik C m/b 2 xp 4π 2 m/b 2, for any fixd λ C. Thus, in th tail of th sris, w can tak λ to b any complx numbr. Thr sms to b a problm though with th first fw trms of th sris, whr w may hav R {λ} α 2 +4π 2 m/b 2 +4πα m/b so that th corrsponding intgrals divrg. Th following lmma shows how to ovrcom this problm. Lmma 2. Lt a, c, b positiv constants, and n a positiv intgr. For R {λ} > w st fλ = a λt c2 /4t 7 dt 4πt. n/2

8 Thn th only singularity of fλ on C is a branch point at λ = a. Proof. Dfin f λ = a λt c2 /4t dt 4πt. n/2 Notic that f λ is ntir in λ and 1 fλ + f λ = 2π a n/2 λn/4 1/2 c 1 n/2 K n/2 1 c a λ, whr K ν is th modifid Bssl function of th 2nd kind, of ordr ν. Thus, th xact typ of th branch point of f λ at λ = a is known. Thus, if r < d, th lmma implis that th only singularitis of G 2 x, y; λ, viwd as a function of λ, ar branch points at λ = α 2 + 4π 2 m/b 2 + 4πα m/b. Ths ar also th singularitis of Gx, y; λ. Notic that, if d r > 2, thn Gx, y; λ stays finit at ths valus of λ. 3 Th Main Rsult W summariz th main rsult stablishd in th prvious sction s 5 and 9: Thorm. Considr th domain D in R d D =, b 1, b r R d r r is som intgr btwn 1 and d, and lt L α = act on D with α- Floqut-Bloch boundary conditions s 3. Thn, for any >, th Grn s function Gx, y; λ of L α has th following rapidly convrgnt sris rprsntation Gx, y; λ = G 1 x, y; λ + G 2 x, y; λ, G 1 x, y; λ = m Z iα mb 1 x y + mb 2 xp λt dt, d/2 r 4πt 4t G 2 x, y; λ = i[2πm/b+α] xr yr m Z b r 1 b r [ α 2 +4π 2 m/b 2 +4πα m/b λ]t x r y r 2 /4t whr α = α 1,..., α r,,...,, mb = m 1 b 1,..., m r b r,,...,, m/b = m 1 /b 1,..., m r /b r,,...,, x r = x 1,..., x r,,...,, and x r =,...,, x r+1,..., x d. 8 dt 4πt, d r/2

9 Rmarks. a If n = 1, thn fλ = a λt c2 /4t It follows diffrntiat with rspct to that whr dt = 1 a λs2 c 2 /4s 2 ds. 4πt π a λ fλ = c 4 a λ rfc c a λ + 2 a λ + c 4 a λ rfc c a λ 2, rfcz = 1 rfz = 1 2/ π z xp s 2 ds = 2/ π xp s 2 ds a wll known ntir function. Thus, if r = d 1, thn w hav a mor xplicit rprsntation of th intgrals in th sris for G 2 x, y; λ, in trms of th spcial function rfcz. b Lmma 2 implis that th rprsntation of G 2 x, y; λ, is valid for all λ α 2 + 4π 2 m/b 2 + 4πα m/b, vn though σl α = [λ,, whr λ = α 2. c If w want to balanc th dcays of th trms of th sris for G 1 x, y; λ and for G 2 x, y; λ, a rasonabl choic is = 1 [ ] b b b 2 1/2 r 4π b b b 2 r d Th cas of Dirichlt or Numann boundary conditions can b tratd similarly, sinc th corrsponding Grn s function can b writtn as a sum altrnating sum in th Dirichlt cas of fr Grn s functions mthod of imags so w again hav a formula vry similar to 4. Lt us tak d = 3. If D =, b 1, b 2, b 3 and α =,,, w hav Gx, y; λ = 1 λ x y+mb 4π x y + mb, whr G 1 x, y; λ = m Z 3 m Z 3 Gx, y; λ = G 1 x, y; λ + G 2 x, y; λ, 1 4πt 3/2 xp λt x y + mb 2 dt, 4t z 9

10 and, by 7, G 2 x, y; λ = 1 b 1 b 2 b 3 4π 2 m/b 2 λ m Z 3 4π 2 m/b 2 λ 2iπm x y. f Again, lt d = 3. Obsrv that by substituting t = s 2 xp λt c2 dt 4t t = 2 c 2 /4s 2 +λ/s 2 ds. 3/2 Thus, by a abov w gt xp 1/ λt c2 dt 4t t 3/2 = π c [ ] c c ic λ rfc 2 + i λ + ic λ rfc 2 i λ. 1 For d = 3, th formula for G 1 x, y; λ bcoms G 1 x, y; λ = m Z iα mb 1 xp λt 3/2 3 4πt x y + mb 2 dt, 4t and thrfor, by 1, th intgrals in th sris trms can b computd xplicitly, in trms of th function rfcz. g If D =, b 1, b 2 R hnc d = 3 and r = 2, and α = α 1, α 2,, w obtain th cas of th papr of Kirk Jordan t al. s [2]. Notic that this is th only cas whr th trms of both G 1 x, y; λ and G 2 x, y; λ can b computd in trms of th function rfcz. h In th cas d > 1 w hav that Gx, y; λ, as x y. All th singular bhavior of Gx, y; λ is capturd in th trm of th sris for G 1 x, y; λ that corrsponds to m =. APPNDIX Th mastr formula in wald s work [1], ikr R = 1 xp R 2 z 2 + K2 dz π 4z 2 th formula is in th sns of analytic continuation with rspct to K; strictly spaking it is valid in th rgion R{K 2 } ; altrnativly w can 1

11 tak K ral in th formula abov and think of th intgral as a contour intgral whr th contour is th ntir x-axis xcpt for a small intrval containing which can b rplacd by a small smicircl avoiding s also [2], can b sn as a variant of 2 for d = 3. An altrnativ, mor dirct way to justify this formula is by applying th corollary of th wllknown lmma, givn blow, to th function fx = x2. Lmma 3. If f L 1 R, thn f x 1 dx = f x dx. x Proof. By using th substitution u = 1/x w obtain f x 1 dx = f u 1 du x u u 2 and f x 1 dx = x f u 1 du u u. 2 Adding up and rplacing th dummy variabl u by x w gt f x 1 dx = f x 1 dx x x x = 1 f x x dx. 2 2 x 2 Now, th substitution u = x 1/x givs f x x dx = f x x dx = f u du x 2 x 2 and th proof is compltd. Corollary. If f L 1 R, a >, and b, thn f ax b dx = 1 f x dx x a thus th first intgral dos not dpnd on b. Proof. Th cas b = is obvious. If b >, th substitution u = ab 1/2 x givs f ax b 1/2 b [ ab dx = f u 1 ] du. x a u 11

12 Thus, by th prvious lmma f ax b dx = x 1/2 b f u ab du a and th proof is finishd by substituting u = ab 1/2 x in th intgral of th right hand sid. 12

13 Acknowldgmnts. Th author wants to thank Profssors Mansoor Haidr and Stphanos Vnakids for finding a srious misprint in Rmark a and Formula 1 of th original manuscript. W also want to thank Profssor Ptr Kuchmnt for his ncouragmnt and suggstions. 13

14 Rfrncs [1]P. P. wald, Di Brchnung Optischr und lctrostatischr Gittrpotntial, Annaln dr Physik, 64, ; Translatd by A. Cornll, Atomics Intrnational Library, [2]K.. Jordan, G. R. Richtr, and P. Shng, An fficint Numrical valuation of th Grn s Function for th Hlmholtz Oprator on Priodic Structurs, Journal of Computational Physics, 63, [3]Y. Katznlson, An Introduction to Harmonic Analysis, Dovr Publications, Inc., Nw York,