Green's Functions for the Wave, Helmholtz and Poisson Equations

Revista Brasileira de Ensino de F´ısica, v. 35, n. 1, 1304 (2013) www.sbﬁsica.org.br

Green’s functions for the wave, Helmholtz and Poisson equations in a two-dimensional boundless domain (Fun¸c˜oesde Green para as equa¸c˜oesda onda, Helmholtz e Poisson num dom´ıniobidimensional sem fronteiras)

Roberto Toscano Couto1

Departamento de Matem´atica Aplicada, Universidade Federal Fluminense, Niter´oi,RJ, Brasil Recebido em 11/4/2012; Aceito em 23/6/2012; Publicado em 18/2/2013

In this work, Green’s functions for the two-dimensional wave, Helmholtz and Poisson equations are calculated in the entire plane domain by means of the two-dimensional Fourier transform. New procedures are provided for the evaluation of the improper double integrals related to the inverse Fourier transforms that furnish these Green’s functions. The integrals are calculated by using contour integration in the complex plane. The method consists basically in applying the correct prescription for circumventing the real poles of the integrand as well as in using well-known integral representations of some Bessel functions. Keywords: Green’s function, Helmholtz equation, two dimensions.

Neste trabalho, as fun¸cõesde Green para as equa¸cõesbidimensionais da onda, Helmholtz e Poisson sãocal- culadas na totalidade do dom´ınioplano por meio da transformada de Fourier bidimensional. Sãoapresentados novos modos de se efetuarem as integrais duplas imprópriasrelacionadas àstransformadas de Fourier inversas que fornecem essas fun¸cõesde Green. As integrais sãocalculadas a partir de integrais no plano complexo. O método consiste basicamente em determinar o caminho de integra¸cãoque se desvia corretamente dos polos reais do integrando bem como em usar representa¸cõesintegrais bem conhecidas de algumas fun¸cõesde Bessel. Palavras-chave: fun¸cãode Green, equa¸cãode Helmholtz, duas dimensões.

1. Introduction sider the physical conditions to determine the correct prescription. For this reason, the wave and Helmholtz Green’s functions for the wave, Helmholtz and Poisson equations solved in this work refer to concrete situa- equations in the absence of boundaries have well known tions. expressions in one, two and three dimensions. A stan- Sections 2, 3 and 4 are devoted to the wave, dard method to derive them is based on the Fourier Helmholtz and Poisson equations, respectively. Section transform. Nevertheless, its derivation in two dimen- 5 concludes the body of the paper with final comments. sions (the most difficult one), unlike in one and three, is hardly found in the literature, this being particularly 2. Wave equation true for the Helmholtz equation.2 This work aims at providing new ways of performing the improper dou- For the reasons given in the Introduction, in order to ble integrals related to the inverse Fourier transforms calculate Green’s function for the wave equation, let that furnish those Green’s functions in the bidimen- us consider a concrete problem, that of a vibrating, sional case. stretched, boundless membrane It is assumed that the reader is acquainted with the usual prescriptions for circumventing the real poles of ∇2 − −2 − −1 a function being integrated over the real axis: the iϵ- z(r, t) c z[tt = T f(r, t]), prescription [2–5] and that which leads to the Cauchy r in R2, t in R . (1) principal value. In this respect, Ref. [6] is closely fol- lowed. As described in this reference, in physical appli- In this, z(r, t) is the membrane vertical displacement at cations, the improper integrals that arise are often not the point r of the xy-plane at the time t, f(r, t)√ is the well defined mathematically, being necessary to con- external vertical force per unit area, and c ≡ T/σ, 1E-mail: [email protected]ff.br.

Copyright by the Sociedade Brasileira de F´ısica.Printed in Brazil. 2For this equation, the author is aware of the procedure sketched in Ref. [1], p. 173-176 1304-2 Toscano Couto

where T is the stretching force per unit length (uni- state - horizontally and at rest - in accordance with the form and isotropic) and σ is the mass per unit area. causality principle. We admit that the membrane has always been (since To solve the problem deﬁned by Eqs. (2) and (3), t → −∞) lying at rest on the xy-plane until the source we Fourier transform (2), both in the spatial and time of vibrations f starts generating waves. variables - according to the deﬁnitions The associated Green’s function G(r, t | r′, t′) is the ′ ′ solution of (1) with a unit, point, instantaneous source Fr {G(r, t | r , t )} ≡ ′ ′ ∫ at the point r at the time t 1 d 2r eik·rG(r, t | r′, t′) ≡ G¯(k, t | r′, t′) (4) ∇2 | ′ ′ − −2 − −1 − ′ − ′ 2π R2 G(r, t r , t ) c[ Gtt = T δ(r r ) δ(t t]) r and r′ in R2, t and t′ in R . (2) F {G¯(k, t | r′, t′)} ≡ We consider here the causal Green’s function, for which t ∫ 1 ∞ G(r, t | r′, t′) = 0 if t < t′ , (3) √ dt eiωtG¯(k, t | r′, t′) ≡G¯˜(k, ω | r′, t′), 2π −∞ meaning that, before the instantaneous action of the unit point source, the membrane is found in its original and solve the resulting equation to obtain G¯˜(k, ω | r′, t′)

⌋

2 ′ ′ −2 −1 ′ ′ Fr ∇ G(r, t | r , t ) − c Gtt = −T δ(r − r ) δ(t − t ) =⇒

′ F −k2G¯(k, t | r′, t′) − c−2d 2G/dt¯ 2 = [−T −1eik·r /(2π)] δ(t − t′) =⇒t

′ ′ √ −k2G¯˜(k, ω | r′, t′) + (ω/c)2G¯˜ = [−T −1 eik·r /(2π)] eiωt / 2π =⇒

′ ′ −c2 T −1 eik·r eiωt / (2π)3/2 G¯˜(k, ω | r′, t′) = . ω2 − k2c2 We then calculate the inverse Fourier transforms, approaching both poles from above (which is equivalent F −1{ ¯˜} ¯ → + t G = G ﬁrst [carried out below, in Eq. (5)]. This to calculate the limit as ϵ 0 of the Fourier integral is an integral of the type discussed in Ref. [6], whose with both poles shifted of −iϵ), thus leaving both poles evaluation as part of a contour integral in the ω-plane outside the contour and leading to the expected null ′ imposes the need of prescribing the way to circumvent result. For t > t , we adopt this same prescription in- the two real poles at ω = ±kc. dicating how the path along the real axis circumvents For t < t′, in closing the contour with a semicircle the poles, but, because of Jordan’s lemma, we close the CR of radius R → ∞, we obtain a vanishing integral contour as in Fig. 1(b), with CR in the lower half- along CR if we let it lying in the upper half-plane (as plane, thus enclosing both poles, at which the residues Fig. 1(a) shows), according to Jordan’s lemma. There- now contribute to the result. fore, fulﬁllment of Eq. (3) necessitates the path of in- In the notation of Ref. [6], such a way of calculating 3 tegration along the real axis to be that in Fig. 1(a), the improper integral leads to its D−−-value

− ∫ ∞ − − ′ − 2 1 ′ iω(t t ) ′ ′ −1 ˜ ′ ′ c T ik·r e G¯(k, t | r , t ) = F {G¯(k, ω | r , t )} = e D−− dω = (5) t (2π)2 ω2 − k2c2    −∞  ′  0 (t < t )  2 −1   c T ik·r′  c i k·r′ sin kcτ  e · 2πi · ′  = e U(τ) ,  (2π)2  Res(−kc) + Res(kc) (t > t )  2πT k  | {z } | {z } eikcτ /(−2kc) e−ikcτ /(2kc)

where τ ≡ t − t′ and U(τ) is the unit step function (equal to 0 for τ < 0 and to 1 for τ > 0).

3 “ D−−” means that, in applying the iϵ-prescription method, both the left and right poles are shifted of −iϵ. Cauchy’s P-value as ′ well as the D++-, D+−- and D−+-value of the improper integral are not physically acceptable, because they do not vanish for t < t and are thus inconsistent with the causality principle. Green’s functions for the wave, Helmholtz and Poisson equations in a two-dimensional boundless domain 1304-3

- plane (a) t" t ! kc kc

CR (b) t# t ! kc kc

Figure 1 - The contours used to evaluate the integral in Eq. (5) for (a) t < t′ and (b) t > t′.

F −1 Now evaluating r of the result above, we obtain The brackets in this equation enclose an integral representation of the Bessel function J (kρ), which also ad- G(r, t | r′, t′) = F −1{G¯(k, t | r′, t′)} = 0 ∫ r mits another well-known integral representation; that U ′ 4 c (τ) 2 −ik·(r−r ) sin kcτ is, 2 d k e . (6) (2π) T R2 k ∫2π ∫∞ k 1 ikρ cos φ 2 sin kρu y dφ e = J0(kρ) = du √ . (8) k 2π π u2 − 1 0 1 k Therefore, with the substitution of this latter integral r " r representation of J0(kρ) for the former one in Eq. (7), we can continue the calculation as follows kx ∫∞ [ ∫∞ ] c U(τ) 2 sin kρu G(r, t | r′, t′) = dk sin kcτ du√ 2 − Figure 2 - The axes of the Cartesian coordinates kx and ky of the 2πT π u 1 vector k used to evaluate the double integrals in Eqs. (6), (14), 0 1 ∫∞ [ ∫∞ ] (19) and (25). c U(τ) du 2 = √ dk sin kρu sin kcτ . (9) This double integral becomes easier to perform if we 2πT u2 − 1 π express the area element of the k-plane in the polar co- 1 0 ordinates, d 2k = k dk dφ (instead of the Cartesian ones, 2 Now recognizing that the last pair of brackets en- d k = dkx dky), and choose the kx-axis in the opposite direction of the vector ρ ≡ r − r′, as shown in Fig. 2. closes an integral representation of the delta function − In addition, noticing that k · (r − r′) = kρ cos(π − φ), δ(ρu cτ)[8,9] and then changing the variable of inte- − we can write gration from u to ξ = ρu cτ, we can write ∫2π∫∞ ∫∞ c U(τ) c U(τ) du δ(ρu − cτ) G(r, t | r′, t′) = eikρ cos φ sin kcτ dk dφ G(r, t | r′, t′) = √ = (2π)2T 2πT u2 − 1 0 0 1 ∫∞ [ ∫2π ] ∫∞ c U(τ) 1 c U(τ) dξ δ(ξ) ikρ cos φ √ . = dk sin kcτ dφ e . (7) ( )2 2πT 2π 2πT ρ ξ+cτ − 1 0 0 ρ−cτ ρ This integral is easily evaluated by using the sifting property of the delta function;5 we obtain { c U(τ) √ 0 if ρ − cτ > 0 (i.e. τ − ρ/c < 0) G(r, t | r′, t′) = × ( ) cτ 2 − − − 2πT ρ 1/ ρ 1 if ρ cτ < 0 (i.e. τ ρ/c > 0) { U(τ) 0 if τ − ρ/c < 0 U(τ) U(τ − ρ/c) = √ × = √ . 2 − 2 1 if τ − ρ/c > 0 2 − 2 2πT τ (ρ/c) | {z } 2πT τ (ρ/c) U(τ−ρ/c)

4The ﬁrst integral representation, by bisecting the range of integration and making the changing of variable φ → 2π − φ in the latter part, becomes the Eq. (2) in Ref. [7], §2.3, with n = 0 and z = kρ. The second one is also found in this reference, §6.13, Eq. (3), with ν = 0 and x =∫ kρ. 5 b That is, a dx δ(x) f(x) is equal to f(0) if a < 0 < b and is zero otherwise. 1304-4 Toscano Couto

But, in the product U(τ) U(τ − ρ/c), we can drop the ﬁrst unit step function without altering the result (this is readily seen by plotting them both). We then obtain the ﬁnal result

1 U(τ − ρ/c) G(r, t | r′, t′) = G(ρ, τ) = √ [ ρ ≡ | r − r′| , τ ≡ t − t′ ] . 2πT τ 2 − (ρ/c)2

3. Helmholt equation 3.1. First method

In this section, we calculate Green’s function for the Helmholtz equation in an unbounded two-dimensional To evaluate Eq. (14), we again choose the kx-axis in domain. Being more specific, we solve Eq. (13), which the opposite direction of the vector ρ, as in Fig. 2, arises in the following concrete problem. but we now proceed with the Cartesian coordinates of Suppose that, in the membrane problem of the pre- k = kx ex + ky ey vious section, the source term is given by − iω0t ≥ ∫ ∫ f(r, t) = ϕ(r) e (t t0), (10) −1 ∞ ∞ ′ T dky Γ (r | r ) = dk eikxρ . that is, the external vertical force per unit area varies 2 x 2 2 − 2 (2π) −∞ −∞ kx + ky k0 harmonically with time with frequency ω0 at all points | {z } of the membrane, being ϕ(r) (a non-negative function I(kx) everywhere) its maximum magnitude at the point r. (15) In addition, we admit that we are only interested in The integral denoted by I(kx) above can be calcu- the stationary solution zst(r, t) that will prevail after lated as part of a contour integral in the ky-plane. It a very long time has elapsed as a consequence of the does not matter if we close the contour with a semicircle forced harmonic oscillation. of radius R→∞ through the upper or lower half-plane; It is well-known that, when this steady-state motion let us close it through the upper half-plane (Figs. 3 and is reached, all points of the membrane will be vibrat- 4). However, to investigate the poles of the integrand, ing harmonically with the same frequency ω0, but with we need to consider two separate cases: |kx| > k0 and amplitudes described by some function ζ(r), that is |kx| < k0. −iω0t zst(r, t) = ζ(r) e (t ≫ t0). (11) For |kx| > k0, by writing the denominator of the Therefore, the desired stationary solution becomes de- integrand in the form termined as soon as ζ(r) is calculated. By substituting Eqs. (10) and (11) in Eq. (1), we verify that ζ(r) is the ( √ )( √ ) solution of the two-dimensional Helmholtz equation ( ) 2 2 − 2 2 − 2 − 2 − 2 ky + kx k0 = ky + i kx k0 ky i kx k0 , ∇2ζ + k2ζ(r) = −T −1ϕ(r)[k ≡ ω /c], (12) 0 ∫ 0 0 2 ′ | ′ ′ whose solution is given by ζ(r) = R2 d r Γ (r r ) ϕ(r ), ′ √ where Γ (r | r ) is the Green’s function for the above 2 − 2 we verify that only the pole i kx k0 is inside the equation, the solution of contour (Fig. 3); therefore ∇2 2 | ′ − −1 − ′ ′ R2 Γ + k0Γ (r r ) = T δ(r r )[ r and r in ]. (13) Since Eq. (13) is Eq. (12) with ϕ(r) = δ(r − r′), we see that Γ (r | r′) describes the amplitude of the stationary motion when the external vertical force is concentrated ′ k - plane at r and oscillates with unit amplitude. y Let us solve Eq. (13). We first take the same Fourier transform Fr defined in Eq. (4), obtaining − 2 ¯ | ′ 2 ¯ | ′ − −1 ik·r′ ⇒ k Γ (k r ) + k0Γ (k r ) = T e /(2π) = ′ 2 2 T −1 eik·r i k k Γ¯ (k | r′) = , x 0 2 − 2 2π k k0 and then calculate the inverse Fourier transform ∫ −1 −ik·ρ 2 2 ′ T e ′ i k k 0 Γ (r | r ) = d2k [ ρ ≡ r − r ]. (14) x 2 2 − 2 (2π) R2 k k0 We present two methods of calculating this integral in Figure 3 - The closed contour used to evaluate the integral I(kx) defined in Eq. (15) for |k | > k . the next two subsections. x 0 Green’s functions for the wave, Helmholtz and Poisson equations in a two-dimensional boundless domain 1304-5

and Eq. (17) for I(k ), we split the interval of integra- Im ky x tion in two∫ (for |kx| = |k0|, convergence occurs when P ∞ 2 I(kx) = −∞ dky/ky = 0 , which makes no contribu- tion) (a) D ! [ ∫ ∫ ] −1 ′ T | ikxρ Γ (r r ) = 2 dkx + dkx e I(kx) 2 2 (2π) k k | | | | 0 x kx >k0 kx k0 |kx|

Γ (r | r′) = (d) D [ ] −1 ∫ ∫ T ∞ cos kxρ k0 cos kxρ 2π dkx √ ± 2πi dkx √ . (2π)2 k0 2 − 2 0 2 − 2 kx k0 k0 kx Figure 4 - Four ways of circumventing the real poles in the evalua- | | tion of the integral I(kx) defined in Eq. (15) for kx < k0. These The first integral, with the change of variable given are the possible iϵ-prescriptions; for each one, the corresponding by k = k u, becomes a known integral representation D-value (D−+, etc) is indicated. x 0 of the Neumann function of order zero6 ( √ ) ∫ ∞ ∫ ∞ 1 cos kxρ cos k0ρu π 2 − 2 √ dkx √ = du√ = − N0(k0ρ). I(kx) = 2πi Res i kx k0 = 2πi 2 2 2 2 − 2 k k − k 1 u − 1 2 2i kx k0 0 x 0 π = √ [ |kx| > k0 ] . (16) 2 − 2 The second integral, with the change of variable kx = kx k0 7 k0 cos ϑ, also becomes a known integral representation | | For kx < k0 , that denominator in the form ∫ ∫ ( ) k0 cos k ρ π/2 π 2 2 2 x k − k − k = dkx √ = dϑ cos(k0ρ cos ϑ) = J0(k0ρ). y 0 x 2 − 2 2 ( √ )( √ ) 0 k0 kx 0 2 − 2 − 2 − 2 ky + k0 kx ky k0 kx We thus get clearly√ shows the existence of the two real poles 2 −1 [ ] ± k − k2 . Since the residues at them are given by ′ T π π ( 0 √ x ) ( √ ) Γ (r | r ) = − N (k ρ) ± i J (k ρ) = ± 2 − 2 ± 2 − 2 0 0 0 0 Res k0 kx = 1/ 2 k0 kx , we can deduce 2π 2 2 that ± i T −1 √ [± i N0(k0ρ) + J0(k0ρ)] , / 4 ± 2 − 2 | | I(kx) = πi k0 kx [ kx < k0 ] (17) which is equal to (i T −1/4) H(1)(k ρ) for the plus sign − 0 0 are possible values for I(kx). In fact, the + and signs − −1 (2) and to ( i T /4) H0 (k0ρ) for the minus sign. correspond, in the notation of Ref. [6], to the D−+- and D However, the stationary membrane motion +−-value of I(kx), obtained with the first and second ′ − Γ (r | r ) e iω0t, according to the radiation condition prescriptions shown in Fig. 4. The D - and D−−-value ++ (see Ref. [10], p. 471), must be a wave moving away (obtained with the third and fourth prescriptions) as from the unit point source (i.e., from the harmonic well as the Cauchy P-value of I(k ) all vanish. We x force of unit amplitude at r′). This imposes the choice proceed with both signs in Eq. (17), because only at of the first Hankel function.8 The final result is then the end, by physically analyzing the two corresponding solutions, we will be able to take the correct sign. i Γ (r | r′) = Γ (ρ) = H(1)(k ρ)[ ρ ≡ | r − r′| ] . Let us now perform the integral with respect to kx 4T 0 0 in Eq. (15). In view of the distinct results in Eq. (16) (18) 6 Ref. [7], §6.21, Eq. (15), where this function is denoted by Y0 instead of N0. 7 Ref. [7], §2.2, Eq. (9), with n = 0 and z = k0ρ. 8A good explanation of the representation of outgoing and incoming cylindrical waves by means of the Hankel functions is given in Ref. [10], Sec. 9.12, p. 470. Notice, however, that, in this reference, the outward-traveling wave is formed with the second Hankel (2) iω0t −iω0t function H0 , because the time factor used there is e , instead of the e used here. 1304-6 Toscano Couto

3.2. Second method The integral in Eq. (14) can also be calculated in polar coordinates. The ﬁrst steps are similar to those performed in going from Eq. (6) to Eq. (9)

∫ ∫2π∫∞ T −1 e−ik·ρ T −1 k dk dφ Γ (r | r′) = d2k = eikρ cos φ 2 2 − 2 2 2 − 2 (2π) R2 k k0 (2π) k k0 0 0 ∫∞ [ ∫2π ] ∫∞ [ ∫∞ ] T −1 dk k 1 T −1 dk k 2 sin kρu = dφ eikρ cos φ = du √ 2 − 2 2 − 2 2 2π k k0 2π 2π k k0 π u − 1 0 | 0 {z } 0 | 1 {z }

J0(kρ) J0(kρ) ∫∞ [ ∫∞ ] ∫∞ T −1 du k sin kρu T −1 du S(u) = √ dk = √ . (19) 2 2 2 − 2 2 2 π u − 1 k k0 π u − 1 1 | 0 {z } 1 ≡ S(u)

To evaluate the integral S(u) deﬁned above, we write it in a more appropriate form ∫∞ ( ) [ ∫ ] 1 k eikρu − e−ikρu E+(u) − E−(u) ∞ k e±ikρu S(u) = dk = E±(u) ≡ dk . 2 − 2 2 − 2 2 2i (k k0) 4i −∞ k k0 −∞

+ − The residues of the integrands in E (u) and E (u) (denoted by Res+ and Res−, respecively) at the poles ±k0 are ±ik0ρu ∓ik0ρu Res+(±k0) = e /2 and Res−(±k0) = e /2 . Therefore, referring to the contours shown in Fig. 5 (which are closed with inﬁnite semicircles that, according to Jordan’s lemma, do not contribute to the integrals), we calculate the four possible D-values of S(u) as follows [ ] + − D− S(u) = D− E (u) − D− E (u) /(4i) = [2πiRes (k ) − (−2πi)Res−(−k )] /(4i) + + [ + ] + 0 0 = (π/2) eik0ρu/2 + eik0ρu/2 = (π/2) eik0ρu (20) [ ] + − D −S(u) = D −E (u) − D −E (u) /(4i) = [2πiRes (−k ) − (−2πi)Res−(k )] /(4i) + + [ + ] + 0 0 − − − = (π/2) e ik0ρu/2 + e ik0ρu/2 = (π/2) e ik0ρu (21) 0 [ z }| { ] D S(u) = D E+(u) − D E−(u) /(4i) = 2πi [Res (−k ) + Res (k )] /(4i) ++ ++ [ ++ ] + 0 + 0 −ik0ρu ik0ρu = (π/2) e /2 + e /2 = (π/2) cos(k0ρu) (22) 0 [ z }| { ] + − D−−S(u) = D−−E (u) −D−−E (u) /(4i) = −(−2πi) [2πiRes−(−k ) + Res−(k )] /(4i) [ ] 0 0 ik0ρu −ik0ρu = (π/2) e /2 + e /2 = (π/2) cos(k0ρu) (23) In order to determine the P-value of S(u), let us ﬁrst calculate the P-values of E+(u) and E−(u) employing the contours in Figs. 5(c) and 5(h), respectively. Denoting the semicircles of radius r → 0 used to circumvent the poles at ±k0 by (±k0, r) and the semicircle of radius R → ∞ centered at the origin by (0,R), we can write [ ] I ∫ ∫ ∫ 0 ± k e ik0ρu dk PE±(u) = lim − − − 7 → 2 − 2 r 0 k k0 R→∞ ± C (−k0,r) (+k0,r) (0,R) ± − − ± − − ± − = 2πi [Res±( k0) + Res±(k0)] [( πi)Res±( k0) ( π] i)Res±(k0) 0 ∓ik0ρu ±ik0ρu = ±πi [Res±(−k0) + Res±(k0)] = ±πi e /2 + e /2 = ±πi cos(k0ρu) , from which [ ] + − PS(u) = PE (u) − PE (u) = (π/2) cos(k0ρu). (24) Green’s functions for the wave, Helmholtz and Poisson equations in a two-dimensional boundless domain 1304-7

Among the several results for S(u) calculated above, and then calculate the inverse Fourier transform:

given by Eqs. (20) to (24), it is its D− -value, given ∫ + −ik·ρ by Eq. (20), which is to be taken and substituted in | ′ − 1 e 2 ≡ − ′ Ω(r r ) = 2 d k [ ρ r r ]. (25) Eq. (19), because, by doing so, we obtain the correct 2π R2 k result, that given by Eq. (18) To evaluate this integral, we act as in the case of the − ∫ ∞ 1 ik0ρu wave equation, that is, we choose the kx-axis in the op- ′ T du (√π/2) e i (1) Γ (r | r ) = = H (k0ρ). posite direction of the vector ρ, as in Fig. 2, and adopt π2 u2 − 1 4T 0 1 the polar coordinates k and φ of k. Next, we recognize Here, in the last step, we used the integral repre- the integral with respect to φ as the ﬁrst integral rep- sentation of the ﬁrst∫ Hankel function of order zero resentation of J0(kρ) given in Eq. (8). The result is the (1) ∞ √ (−2i/π) du eiux/ u2−1 H0 (x) = 1 , given in Ref. [7], following function of ρ only §6.13, Eq. (1). ∫2π∫∞ 1 eikρ cos φ Ω(r | r′) = − k dk dφ = 4. Poisson equation 2π k2 0 0 To calculate Green’s function for the Poisson equation ∫∞ ∫2π ∫∞ dk 1 J (kρ) in an unbounded two-dimensional domain, that is, the − eikρ cos φ = − dk 0 = Ω(ρ). k 2π k solution of 0 0 0 [ ] | {z } 2 ′ ′ ′ 2 ∇ Ω(r | r ) = 2π δ(r − r ) r and r in R , J0(kρ)

we again take the Fourier transform deﬁned in Eq. (4), The integral with respect to k becomes straightfor- obtaining ward if we diﬀerentiate it with respect to ρ and then use the chain rule to change to a derivative with respect ′ ′ −k2Ω¯ (k | r′) = eik·r =⇒ Ω¯ (k | r′) = −eik·r/k2, to k

k-plane r k 0 k 0 r E ! D ! R D ! E R r (a) k (e) 0 k 0 r

! (b) D! E D! E (f)

! (c) D!! E D!! E (g)

! (d) D E D E (h)

Figure 5 - The contours in the k-plane associated to the four possible D-values of the integrals E+(u) (at the left) and E−(u) (at the right). 1304-8 Toscano Couto

means of a one-dimensional Fourier transform (in the x ∫ variable). ∞ ∂ J (kρ) Ω ′(ρ) = − dk 0 = For the Helmholtz equation, two procedures for 0 ∂ρ k evaluating the inverse Fourier transform which fur- ∫ ∞ [ ]∞ ∂ J (kρ) J (kρ) 1 nishes the Green’s functions were presented. They − dk 0 = − 0 = , diﬀer on the coordinates used to carry on the dou- 0 ∂k ρ ρ k=0 ρ ble integrals. It seems that the calculation employing since J0(x) equals 1 at x = 0 and goes to 0 as x → ∞ . the Cartesian coordinates is somewhat less cumbersome Now, integrating with respect to ρ, we obtain the ﬁnal than that based on the polar coordinates. result

Ω(r | r′) = Ω(ρ) = ln ρ + constant [ ρ ≡ | r − r′| ] , References [1] B. Davies, Integral Transforms and Their Application, also obtained in Ref. [1], p. 169-170 (by a much more Texts in Applied Mathematics (Springer-Verlag, New complicated method). The arbitrary additive constant York, 2002), 3rd ed. is physically irrelevant. In fact, Green’s function for the [2] R. Shankar, Principles of Quantum Mechanics Poisson equation can be interpreted, for instance, as the (Plenum Press, New York, 1994), 2nd ed., sec. 19.4, electrostatic potential at r due to electrical charge con- p. 661 and p. 541. centrated at r′, and only potential differences are rele- vant. However, unlike in the three-dimensional case, in [3] F.W. Byron Jr, and R.W. Fuller, Mathematics fo Clas- sical and Quantum Physics (Dover Publications, New which such constant is found to be zero by choosing the York, 1992), sec. 7.4, p. 415. zero potential at infinite distances away from the point charge at r′, this cannot be done in the two-dimensional [4] E. Butkov, Mathematical Physics (Addison-Wesley case, because the potential diverges (logarithmically) as Publishing Company, Reading, 1973), sec. 7.7, p. 281. the distance from the line charge at r′ increases. [5] E. Merzbacher, Quantum Mechanics (John Wiley & Sons, New York, 1970), 2nd ed., sec. 11.3, p. 225. [6] R. Toscano Couto, Revista Brasileira de Ensino de 5. Final comments F´ısica 29, 3 (2007). As said in the Introduction, the Green’s functions con- [7] G.N. Watson, A Treatise on the Theory of Bessel Func- sidered here have well known expressions, which are tions (Cambridge University Press, London, 1944), 2nd obtained in a number of ways (e.g., by descenting from ed. the easier three-dimensional case). Therefore, we did [8] G.B. Arfken and H.J. Weber, Mathematical Methods not aimed at presenting new results but new meth- for Physicists (Harcourt Academic Press, San Diego, ods. In this respect, concerning the footnote in the first 2001), 5th ed., Prob. 1.15.24. page, we should say that, for the Helmholtz equation, [9] R. Toscano Couto, TEMA Tend. Mat. Apl. Comput. the procedure adopted here differs considerably from 11, 57 (2010). that in Ref. [1], where that equation is converted into [10] F.B. Hildebrand, Advanced Calculus for Applications an ordinary differential equation (in the y variable) by (Prentice-Hall, Englewood Cliffs, 1976), 2nd ed.