Revista Brasileira de Ensino de Fsica, v. 35, n. 1, 1304 (2013)www.sbsica.org.br

Green's functions for the wave, Helmholtz and Poisson equations

in a two-dimensional boundless domain(Func~oes de Green para as equac~oes da onda, Helmholtz e Poisson num domnio bidimensional sem fronteiras)

Roberto Toscano Couto1

Departamento de Matematica Aplicada, Universidade Federal Fluminense, Niteroi, RJ, BrasilRecebido 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 calculatedin the entire plane domain by means of the two-dimensional Fourier transform. New procedures are providedfor the evaluation of the improper double integrals related to the inverse Fourier transforms that furnish theseGreen's functions. The integrals are calculated by using contour integration in the complex plane. The methodconsists basically in applying the correct prescription for circumventing the real poles of the integrand as well asin using well-known integral representations of some Bessel functions.Keywords: Green's function, Helmholtz equation, two dimensions.

Neste trabalho, as func~oes de Green para as equac~oes bidimensionais da onda, Helmholtz e Poisson s~ao cal-culadas na totalidade do domnio plano por meio da transformada de Fourier bidimensional. S~ao apresentadosnovos modos de se efetuarem as integrais duplas improprias relacionadas as transformadas de Fourier inversasque fornecem essas func~oes de Green. As integrais s~ao calculadas a partir de integrais no plano complexo. Ometodo consiste basicamente em determinar o caminho de integrac~ao que se desvia corretamente dos polos reaisdo integrando bem como em usar representac~oes integrais bem conhecidas de algumas func~oes de Bessel.Palavras-chave: func~ao de Green, equac~ao de Helmholtz, duas dimens~oes.

1. Introduction

Green's functions for the wave, Helmholtz and Poissonequations in the absence of boundaries have well knownexpressions in one, two and three dimensions. A stan-dard method to derive them is based on the Fouriertransform. Nevertheless, its derivation in two dimen-sions (the most dicult one), unlike in one and three,is hardly found in the literature, this being particularlytrue for the Helmholtz equation.2 This work aims atproviding new ways of performing the improper dou-ble integrals related to the inverse Fourier transformsthat furnish those Green's functions in the bidimen-sional case.

It is assumed that the reader is acquainted with theusual prescriptions for circumventing the real poles ofa function being integrated over the real axis: the i-prescription [2{5] and that which leads to the Cauchyprincipal value. In this respect, Ref. [6] is closely fol-lowed. As described in this reference, in physical appli-cations, the improper integrals that arise are often notwell dened mathematically, being necessary to con-

sider the physical conditions to determine the correctprescription. For this reason, the wave and Helmholtzequations solved in this work refer to concrete situa-tions.

Sections 2, 3 and 4 are devoted to the wave,Helmholtz and Poisson equations, respectively. Section5 concludes the body of the paper with nal comments.

2. Wave equation

For the reasons given in the Introduction, in order tocalculate Green's function for the wave equation, letus consider a concrete problem, that of a vibrating,stretched, boundless membrane

r2z(r; t) c2ztt = T1f(r; t);r in R2; t in R

: (1)

In this, z(r; t) is the membrane vertical displacement atthe point r of the xy-plane at the time t, f(r; t) is theexternal vertical force per unit area, and c pT=,

1E-mail: toscano@im.u.br.

Copyright by the Sociedade Brasileira de Fsica. 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-form and isotropic) and is the mass per unit area.We admit that the membrane has always been (sincet! 1) lying at rest on the xy-plane until the sourceof vibrations f starts generating waves.

The associated Green's function G(r; t j r0; t0) is thesolution of (1) with a unit, point, instantaneous sourceat the point r0 at the time t0

r2G(r; t j r0; t0) c2Gtt = T1 (r r0) (t t0)r and r0 in R2; t and t0 in R

: (2)

We consider here the causal Green's function, for which

G(r; t j r0; t0) = 0 if t < t0 ; (3)meaning that, before the instantaneous action of theunit point source, the membrane is found in its original

state - horizontally and at rest - in accordance with thecausality principle.

To solve the problem dened by Eqs. (2) and (3),we Fourier transform (2), both in the spatial and timevariables - according to the denitions

Fr fG(r; t j r0; t0)g 1

2

ZR2d 2r eikrG(r; t j r0; t0) G(k; t j r0; t0) (4)

Ftf G(k; t j r0; t0)g 1p2

Z 11

dt ei!t G(k; t j r0; t0) ~G(k; ! j r0; t0);

and solve the resulting equation to obtain ~G(k; ! j r0; t0)c

r2G(r; t j r0; t0) c2Gtt = T1 (r r0) (t t0) Fr=)

k2 G(k; t j r0; t0) c2d 2 G=dt2 = [T1eikr0=(2)] (t t0) Ft=)

k2 ~G(k; ! j r0; t0) + (!=c)2 ~G = [T1 eikr0=(2)] ei!t0=p2 =)

~G(k; ! j r0; t0) = c2 T1 eikr

0ei!t

0= (2)3=2

!2 k2c2 :

We then calculate the inverse Fourier transforms,

F1t f ~Gg = G rst [carried out below, in Eq. (5)]. Thisis an integral of the type discussed in Ref. [6], whoseevaluation as part of a contour integral in the !-planeimposes the need of prescribing the way to circumventthe two real poles at ! = kc.

For t < t0, in closing the contour with a semicircleCR of radius R ! 1, we obtain a vanishing integralalong CR if we let it lying in the upper half-plane (asFig. 1(a) shows), according to Jordan's lemma. There-fore, fulllment of Eq. (3) necessitates the path of in-tegration along the real axis to be that in Fig. 1(a),

approaching both poles from above (which is equivalentto calculate the limit as ! 0+ of the Fourier integralwith both poles shifted of i), thus leaving both polesoutside the contour and leading to the expected nullresult. For t > t0, we adopt this same prescription in-dicating how the path along the real axis circumventsthe poles, but, because of Jordan's lemma, we close thecontour as in Fig. 1(b), with CR in the lower half-plane, thus enclosing both poles, at which the residuesnow contribute to the result.

In the notation of Ref. [6], such a way of calculatingthe improper integral leads to its D-value3

G(k; t j r0; t0) = F1t f ~G(k; ! j r0; t0)g =c2T1(2)2

eikr0D

Z 11

ei!(tt0)

!2 k2c2 d! = (5)26664c2T1(2)2 eikr0 2i 8>>>>>:

0 (t < t0)

Res(kc)| {z }eikc=(2kc)

+ Res(kc)| {z }eikc=(2kc)

(t > t0)

37775 = c2T eikr0 sin kck U() ;where t t0 and U() is the unit step function (equal to 0 for < 0 and to 1 for > 0).

3\D" means that, in applying the i-prescription method, both the left and right poles are shifted of i. Cauchy's P-value aswell as the D++-, D+- and D+-value of the improper integral are not physically acceptable, because they do not vanish for t < t0and 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

kc

(a) t t !"

kc

RC

kc kc

RC

(b) t t !#

Figure 1 - The contours used to evaluate the integral in Eq. (5) for (a) t < t0 and (b) t > t0.

Now evaluating F1r of the result above, we obtain

G(r; t j r0; t0) = F1r f G(k; t j r0; t0)g =c U()(2)2T

ZR2d2k eik(rr

0) sin kc

k: (6)

k

yk

! "r r

xk

k

Figure 2 - The axes of the Cartesian coordinates kx and ky of thevector k used to evaluate the double integrals in Eqs. (6), (14),(19) and (25).

This double integral becomes easier to perform if weexpress the area element of the k-plane in the polar co-ordinates, d 2k = k dk d' (instead of the Cartesian ones,d 2k = dkx dky), and choose the kx-axis in the oppositedirection of the vector r r0, as shown in Fig. 2.In addition, noticing that k (r r0) = k cos( '),we can write

G(r; t j r0; t0) = c U()(2)2T

2Z0

1Z0

eik cos' sin kc dk d'

=c U()2T

1Z0

dk sin kc

"1

2

2Z0

d' eik cos'

#: (7)

The brackets in this equation enclose an integral repre-sentation of the Bessel function J0(k), which also ad-mits another well-known integral representation; thatis,4

1

2

2Z0

d' eik cos' = J0(k) =2

1Z1

dusin kupu2 1 : (8)

Therefore, with the substitution of this latter integralrepresentation of J0(k) for the former one in Eq. (7),we can continue the calculation as follows

G(r; t j r0; t0) = c U()2T

1Z0

dk sin kc

"2

1Z1

dusin kupu2 1

#

=c U()2T

1Z1

dupu2 1

"2

1Z0

dk sin ku sin kc

#: (9)

Now recognizing that the last pair of brackets en-closes an integral representation of the delta function(u c) [8,9] and then changing the variable of inte-gration from u to = u c , we can write

G(r; t j r0; t0) = c U()2T

1Z1

du (u c)pu2 1 =

cU()2T

1Zc

d ()q+c

2 1 :This integral is easily evaluated by using the sifting property of the delta function;5 we obtain

G(r; t j r0; t0) = cU()2T

(

0 if c > 0 (i:e: =c < 0)1=q

c

2 1 if c < 0 (i:e: =c > 0)=

U()2T

p2 (=c)2

0 if =c < 01 if =c > 0| {z }

U(=c)

=U()U( =c)2T

p2 (=c)2 :

4The rst integral representation, by bisecting the range of integration and making the changing of variable '! 2' in the latterpart, becomes the Eq. (2) in Ref. [7], x2.3, with n = 0 and z = k. The second one is also found in this reference, x6.13, Eq. (3), with = 0 and x = k.

5That is,R ba dx (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()U( =c), we can drop the rst unit step function without altering the result (this isreadily seen by plotting them both). We then obtain the nal result

G(r; t j r0; t0) = G(; ) = 12T

U( =c)p2 (=c)2 [ j r r

0j ; t t0 ] :

3. Helmholt equation

In this section, we calculate Green's function for theHelmholtz equation in an unbounded two-dimensionaldomain. Being more specic, we solve Eq. (13), whicharises in the following concrete problem.

Suppose that, in the membrane problem of the pre-vious section, the source term is given by

f(r; t) = (r) ei!0t (t t0); (10)that is, the external vertical force per unit area variesharmonically with time with frequency !0 at all pointsof the membrane, being (r) (a non-negative functioneverywhere) its maximum magnitude at the point r.In addition, we admit that we are only interested inthe stationary solution zst(r; t) that will prevail aftera very long time has elapsed as a consequence of theforced harmonic oscillation.

It is well-known that, when this steady-state motionis reached, all points of the membrane will be vibrat-ing harmonically with the same frequency !0, but withamplitudes described by some function (r), that is

zst(r; t) = (r) ei!0t (t t0): (11)

Therefore, the desired stationary solution becomes de-termined as soon as (r) is calculated. By substitutingEqs. (10) and (11) in Eq. (1), we verify that (r) is thesolution of the two-dimensional Helmholtz equation

r2 + k20(r) = T1(r) [k0 !0=c]; (12)whose solution is given by (r) =

RR2 d

2r0 (r j r0)(r0),where (r j r0) is the Green's function for the aboveequation, the solution of

r2 + k20 (r j r0) = T1(r r0) [ r and r0 in R2 ]:(13)

Since Eq. (13) is Eq. (12) with (r) = (r r0), we seethat (r j r0) describes the amplitude of the stationarymotion when the external vertical force is concentratedat r0 and oscillates with unit amplitude.

Let us solve Eq. (13). We rst take the same Fouriertransform Fr dened in Eq. (4), obtaining

k2 (k j r0) + k20 (k j r0) = T1eikr0=(2) =)

(k j r0) = T1

2

eikr0

k2 k20;

and then calculate the inverse Fourier transform

(r j r0) = T1

(2)2

ZR2d2k

eik

k2 k20[ r r0 ]: (14)

We present two methods of calculating this integral inthe next two subsections.

3.1. First method

To evaluate Eq. (14), we again choose the kx-axis inthe opposite direction of the vector , as in Fig. 2,but we now proceed with the Cartesian coordinates ofk = kx ex + ky ey

(r j r0) = T1

(2)2

Z 11

dkx eikx

Z 11

dkyk2x + k

2y k20| {z }

I(kx)

:

(15)

The integral denoted by I(kx) above can be calcu-lated as part of a contour integral in the ky-plane. Itdoes not matter if we close the contour with a semicircleof radius R!1 through the upper or lower half-plane;let us close it through the upper half-plane (Figs. 3 and4). However, to investigate the poles of the integrand,we need to consider two separate cases: jkxj > k0 andjkxj < k0.

For jkxj > k0, by writing the denominator of theintegrand in the form

k2y+k2x k20

=ky+i

qk2x k20

ky i

qk2x k20

;

we verify that only the pole ipk2x k20 is inside the

contour (Fig. 3); therefore

-yk plane

2 20i xk k

2 20i xk k

Figure 3 - The closed contour used to evaluate the integral I(kx)dened in Eq. (15) for jkxj > k0.

Green's functions for the wave, Helmholtz and Poisson equations in a two-dimensional boundless domain 1304-5

Im yk

D(a) !

2 20 xk k

Re yk

D(b) !

D(c) !!

(d) D

2 20 xk k

!

! !

!

! !

!

!

Figure 4 - Four ways of circumventing the real poles in the evalua-tion of the integral I(kx) dened in Eq. (15) for jkxj < k0. Theseare the possible i-prescriptions; for each one, the correspondingD-value (D+, etc) is indicated.

I(kx) = 2i Resiqk2x k20

= 2i

1

2ipk2x k20

=p

k2x k20[ jkxj > k0 ] : (16)

For jkxj < k0 , that denominator in the formk2y

k20 k2x

=

ky +pk20 k2x

ky

pk20 k2x

clearly shows the existence of the two real polespk20 k2x . Since the residues at them are given by

Respk20 k2x = 1=2pk20 k2x , we can deduce

that

I(kx) = i.q

k20 k2x [ jkxj < k0 ] (17)

are possible values for I(kx). In fact, the + and signscorrespond, in the notation of Ref. [6], to the D+- andD+-value of I(kx), obtained with the rst and secondprescriptions shown in Fig. 4. The D++- and D-value(obtained with the third and fourth prescriptions) aswell as the Cauchy P-value of I(kx) all vanish. Weproceed with both signs in Eq. (17), because only atthe end, by physically analyzing the two correspondingsolutions, we will be able to take the correct sign.

Let us now perform the integral with respect to kxin Eq. (15). In view of the distinct results in Eq. (16)

and Eq. (17) for I(kx), we split the interval of integra-tion in two (for jkxj = jk0j, convergence occurs whenI(kx) = P

R 11 dky=k

2y = 0 , which makes no contribu-

tion)

(r j r0) = T1

(2)2

" Zjkxj>k0

dkx +

Zjkxj

1304-6 Toscano Couto

3.2. Second method

The integral in Eq. (14) can also be calculated in polar coordinates. The rst steps are similar to those performedin going from Eq. (6) to Eq. (9)

(r j r0) = T1

(2)2

ZR2d2k

eik

k2 k20=

T1

(2)2

2Z0

1Z0

eik cos'k dk d'

k2 k20

=T1

2

1Z0

dk k

k2 k20

"1

2

2Z0

d' eik cos'

#| {z }

J0(k)

=T1

2

1Z0

dk k

k2 k20

"2

1Z1

dusin kupu2 1

#| {z }

J0(k)

=T1

2

1Z1

dupu2 1

" 1Z0

dkk sin ku

k2 k20

#| {z }

S(u)

=T1

2

1Z1

du S(u)pu2 1 : (19)

To evaluate the integral S(u) dened above, we write it in a more appropriate form

S(u) =1

2

1Z1

dkkeiku eiku2i (k2 k20)

=E+(u) E(u)

4i

E(u)

Z 11

dkk eiku

k2 k20

:

The residues of the integrands in E+(u) and E(u) (denoted by Res+ and Res, respecively) at the poles k0are

Res+(k0) = eik0u=2 and Res(k0) = eik0u=2 :Therefore, referring to the contours shown in Fig. 5 (which are closed with innite semicircles that, according toJordan'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) = [2iRes+(k0) (2i)Res(k0)] =(4i)

= (=2)eik0u=2 + eik0u=2

= (=2) eik0u (20)

D+S(u) =D+E+(u)D+E(u) =(4i) = [2iRes+(k0) (2i)Res(k0)] =(4i)

= (=2)eik0u=2 + eik0u=2

= (=2) eik0u (21)

D++S(u) =D++E+(u)

0z }| {D++E(u)

=(4i) = 2i [Res+(k0) + Res+(k0)] =(4i)

= (=2)eik0u=2 + eik0u=2

= (=2) cos(k0u) (22)

DS(u) = 0z }| {DE+(u)DE(u)=(4i) = (2i) [2iRes(k0) + Res(k0)] =(4i)

= (=2)eik0u=2 + eik0u=2

= (=2) cos(k0u) (23)

In order to determine the P-value of S(u), let us rst calculate the P-values of E+(u) and E(u) employingthe contours in Figs. 5(c) and 5(h), respectively. Denoting the semicircles of radius r ! 0 used to circumvent thepoles at k0 by (k0; r) and the semicircle of radius R!1 centered at the origin by (0; R), we can write

PE(u) = limr!0R!1

" IC

Z

(k0;r)

Z

(+k0;r)

Z

(0;R)

70#k eik0u dkk2 k20

= 2i [Res(k0) + Res(k0)] (i)Res(k0) (i)Res(k0) 0= i [Res(k0) + Res(k0)] = i

eik0u=2 + eik0u=2

= i cos(k0u) ;

from whichPS(u) = PE+(u) PE(u) = (=2) cos(k0u): (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,given by Eqs. (20) to (24), it is its D+-value, givenby Eq. (20), which is to be taken and substituted inEq. (19), because, by doing so, we obtain the correctresult, that given by Eq. (18)

(r j r0) = T1

2

Z 11

du (=2) eik0upu2 1 =

i

4TH(1)0 (k0):

Here, in the last step, we used the integral repre-sentation of the rst Hankel function of order zeroH(1)0 (x) = (2i=)

R11du eiux=

pu21 , given in Ref. [7],

x6.13, Eq. (1).

4. Poisson equation

To calculate Green's function for the Poisson equationin an unbounded two-dimensional domain, that is, thesolution of

r2(r j r0) = 2 (r r0) r and r0 in R2 ;we again take the Fourier transform dened in Eq. (4),obtaining

k2 (k j r0) = eikr0 =) (k j r0) = eikr0=k2;

and then calculate the inverse Fourier transform:

(r j r0) = 12

ZR2

eik

k2d2k [ r r0 ]: (25)

To evaluate this integral, we act as in the case of thewave equation, that is, we choose the kx-axis in the op-posite direction of the vector , as in Fig. 2, and adoptthe polar coordinates k and ' of k. Next, we recognizethe integral with respect to ' as the rst integral rep-resentation of J0(k) given in Eq. (8). The result is thefollowing function of only

(r j r0) = 12

2Z0

1Z0

eik cos'

k2k dk d' =

1Z0

dk

k

1

2

2Z0

eik cos'

| {z }J0(k)

= 1Z0

dkJ0(k)

k= ():

The integral with respect to k becomes straightfor-ward if we dierentiate it with respect to and thenuse the chain rule to change to a derivative with respectto k

D E !

!

D E !

D E!

!

D E!!

!

D E

!

R

r

r

D E!

D E!!

D E

r

r

R

0k

0k

0k

0k

(a)

(b)

(c)

(d)

-planek

(e)

(f)

(g)

(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 theright).

1304-8 Toscano Couto

0() = Z 10

dk@

@

J0(k)

k=

Z 10

dk@

@k

J0(k)

=

J0(k)

1k=0

=1

;

since J0(x) equals 1 at x = 0 and goes to 0 as x!1 .Now, integrating with respect to , we obtain the nalresult

(r j r0) = () = ln + constant [ j r r0j ] ;

also obtained in Ref. [1], p. 169-170 (by a much morecomplicated method). The arbitrary additive constantis physically irrelevant. In fact, Green's function for thePoisson equation can be interpreted, for instance, as theelectrostatic potential at r due to electrical charge con-centrated at r0, and only potential dierences are rele-vant. However, unlike in the three-dimensional case, inwhich such constant is found to be zero by choosing thezero potential at innite distances away from the pointcharge at r0, this cannot be done in the two-dimensionalcase, because the potential diverges (logarithmically) asthe distance from the line charge at r0 increases.

5. Final comments

As said in the Introduction, the Green's functions con-sidered here have well known expressions, which areobtained in a number of ways (e.g., by descenting fromthe easier three-dimensional case). Therefore, we didnot aimed at presenting new results but new meth-ods. In this respect, concerning the footnote in the rstpage, we should say that, for the Helmholtz equation,the procedure adopted here diers considerably fromthat in Ref. [1], where that equation is converted intoan ordinary dierential equation (in the y variable) by

means of a one-dimensional Fourier transform (in the xvariable).

For the Helmholtz equation, two procedures forevaluating the inverse Fourier transform which fur-nishes the Green's functions were presented. Theydier on the coordinates used to carry on the dou-ble integrals. It seems that the calculation employingthe Cartesian coordinates is somewhat less cumbersomethan that based on the polar coordinates.

References

[1] B. Davies, Integral Transforms and Their Application,Texts in Applied Mathematics (Springer-Verlag, NewYork, 2002), 3rd ed.

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[3] F.W. Byron Jr, and R.W. Fuller, Mathematics fo Clas-sical and Quantum Physics (Dover Publications, NewYork, 1992), sec. 7.4, p. 415.

[4] E. Butkov, Mathematical Physics (Addison-WesleyPublishing Company, Reading, 1973), sec. 7.7, p. 281.

[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 deFsica 29, 3 (2007).

[7] G.N. Watson, A Treatise on the Theory of Bessel Func-tions (Cambridge University Press, London, 1944), 2nded.

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