ECE 6382
Green’s Functions in Two and
Three Dimensions
D. R. Wilton
ECE Dept.
Static Potential of Point Sources
 It iswell knownthat thefreespacestaticpotential at anobservation
point r duetoapoint charge Q at r is
(r) 
Q
,

40 rr
 Ontheother hand, shouldalsosatisfyPoisson'sequation,
V
0
2
where V isthevolumechargedensity. Ques: Howcanwe
defineV soastoincorporateapoint charge, Q?
 Weusuallyavoidexplicitlyintroducingpoint chargesintoPoisson's
equationbecausethevolumechargedensity V of apoint chargeis
zeroeverywhereexcept thechargelocation, whereit isinfinite
(i.e.,afinitechargeQexistswithinavolumeof zero).
Representation of Point Sources
 But wecan mathematicallyrepresent apoint chargeasavolume
chargedensityusingdeltafunctions. Setting
V  Q(xx)(yy)(zz)
wecaneasilycheck that evaluatingthenet chargebyintegrating
over anynon-vanishingvolumeabout r yieldsthecorrecttotal charge:
z y x
   Q(xx)(yy)(zz)dxdydz  Q
z y x
(Ques:DounitsonRHSandLHS agreeintheaboveeqs.?)
 Thuswecan saythat thesolutiontoPoisson'sequation,
Q
2 (xx)(yy)(z z)
0
is
(r) 
Q
Q

40 rr
40 (xx)2 (yy)2 (zz)2
(x, y, z)
2
2
2
Superposition of Potentials
 Wefurther saythat inavolume V witha
r  r
distributedvolumesourcedensity V,every
infinitesimal volumeelement of charge,
dq V(r)dV , producesapotential
V dV 
V
dq
V(r)dV
d(r) 
.
40 rr
r
r
O
 Bythelinearityof thePoissonoperator, weconcludebysuperimposing
thecontributionsof all sourcesthat
 (r)dV
(r)  V
40 rr
V
 (r) isanintegrationof V(r) withaweightingfactor
potential at r of aunit chargeat r .
1
, the
40 rr
A Green’s Function
 Notethe roleplayedby thefactor
G(r,r) 
1
40 r r
calledtheGreens' function that allowsus to write
(r)  G(r,r)V (r)dV
V
 Aphysical interpretationof G(r,r) is that of the potential at r due
toaunit point chargeat r. Notethat it isasolutionto
2G(r,r) 
1
0
(xx)(y  y)(z z)
or inamorecoordinate-freenotation,
2G(r,r) 
1
0
(r r)
where (r r) (xx)(y  y)(z z) in rectangular coordinates,
1, rV
but moregenerally, wesimplyrequirethat (r r) dV  
in3-D

0,
r

V

V
Green’s Function Conditions
 Tocheckour claims,it sufficestoplacer at thecoordinateoriginso
spherical coordinatescanbeused:
G(r,0) 
1
40 r

1
40r
 Check that G(r,0) satisfiesthehomogeneousequationwhenr 0:
1   2 G
1 1   r2 
1 0
2
 G(r,0)  2 r



 0, r 0



2
2
2 

r r  r 
40 r r  r 
40 r
1
1
 Next,check that 2G(r,0)dV   (r)dV  , V enclosesr 0:
0 V
V
lim2G(r,0)dV limG(r,0)dV
0
0
V
2 
0
divergence
theorem

ˆ r dS
lim G(r,0)n
0
V
G(r,0)
4
1
2
ˆr
ˆ2 sindd  lim
r


,
2


0
r r
0
40 
0
lim
0
V
0
if V enclosesr 0.

V
r̂
V
Green’s Function for a General Linear Operator
 Ingeneral, aGreen'sfunctionisasolutionof thelinear operator equation
LG(r,r) (rr)
that alsosatisfiesany boundary conditions of theproblem.
(Afundamental solution satisfiestheaboveequation,but doesnot
necessarilysatisfy theboundaryconditions; toobtainaGreen's
functionweaddahomogeneoussolutiontoafundamental solution
andenforceBCs.)
 Aphysical interpretationof G(r,r) isthat of theresponseat rduetoa
unit point sourceforcingfunctionat r.
 Notethat LG(r,r) 0 except at r r where
 LG(r,r)dV  (rr)dV 1,
V
V
rin V
A Source-Weighted Superposition over the Unit
Source Response Provides a General Solution
T
hesolutiontothegeneralproblem
Lu(r)f(r), f(r) ageneralforcingfunction,
isthenfoundbyasource-w
eightedsuperpositionof the-f'nresponse:
u
u(r)G
(rr
, )f(r)dV
V
T
ocheckthatthisisasolution,notethat
Lu(r)LG
(rr
, )f(r)dVLG
(rr
, ) f(r)dV
V
 (rr)f(r)dV
V
 f(r)
V
3-D Point Source Representation in Various
Coordinate Systems
ˆ yy
ˆ zz
ˆ
rxx
(rr)  (xx)(yy)(zz)
 ()()(zz)
,  0



 
 ()(zz),
 0

2





 




(rectangular coordinates)
(cylindrical coordinateshave
acoord.singularityat 0)
(rr)()()
, r 0,  0,
r2sin
(spherical coordinateshave
(rr)()
,
r 0,  0,
acoord.singularityat r 0
2
2r sin
andat 0,)
(r)
,
r 0
2
4r
2-D Line Source Representation in Various
Coordinate Systems
ˆy
ˆ
rx
x
y
(rr) (xx)(yy)
) ( 
)
( 



,




()

,



2
(re
c
ta
n
g
u
la
rc
o
o
rd
in
a
te
s
)


0


0
(c
y
lin
d
ric
a
lc
o
o
rd
in
a
te
s
h
a
v
e
a
c
o
o
rd
.s
in
g
u
la
ritya
t
0
)
Cylindrical Coordinate Example
 If  0, thevolum
eelem
entisdVdddz
z
z  

()()(zz)
dddz 1




z  
y
x
d
dz
 d
 If  0, becom
esundefined,andthevolum
eelem
entis dV2ddz
z 2 

(
) (zz)
dddz  1



2
z 0 0
z 
(
) (zz)
 
2ddz  1
2
z 0
z
2 d 
dz
y
x
Example: A Simple Static Green’s Function with
Boundary Conditions --- Charge over a Ground
Plane
z
z
1 [C]
r
1 [C]
r  r

r
r
r


0
o
n
g
r
o
u
n
d
p
l
a
n
e
r  r
r
-1 [C]


0
o
n
g
r
o
u
n
d
p
l
a
n
e
)a
a
T
h
e
p
o
te
n
tia
lG
(,
rr
t rd
u
e
to
a
u
n
itc
h
a
rg
e
a
tr
b
o
v
e
a
g
ro
u
n
dp
la
n
e
c
a
n
b
e
fo
u
n
d
fro
m
im
a
g
e
th
e
o
ry
.Itis
g
iv
e
n
b
y
1
1
)
G
(,
rr



4

0 r
r
4

0 r
r
r

z
ˆ.
w
h
e
rer
2
z
fu
n
d
a
m
e
n
ta
ls
o
lu
tio
n
h
o
m
o
g
e
n
e
o
u
s
s
o
lu
tio
n
N
o
te
th
a
tfo
rrin
th
e
u
p
p
e
rh
a
lfs
p
a
c
e
(z
0
), G
(,
rr
)s
a
tis
fie
s


1
1
1
2


)0

G
(,
rr)   
(,
rr) (s
in
c
e


(,
rr
, z
0
)



0
4


r

r

0
 0

) 0 a
  r
 a
a
n
dG
(,
rr
t z
0(s
in
c
e r
r
r
t z
0
).
2
Static Green’s Function with Boundary
Conditions (cont.)
r  r
V dV 
z
V
r
O
r


0
o
n
g
r
o
u
n
d
p
l
a
n
e
)a
F
o
ra
n
a
r
b
itr
a
r
yc
h
a
r
g
e
d
is
tr
ib
u
tio
n
(
r
b
o
v
e
a
g
r
o
u
n
d
p
la
n
e
,w
e
V
th
u
s
h
a
v
e
1 1
1 





()
r
G
(,
r
r
)

(
r
)
d
V
,
G
(,
r
r
)




V

 r




4


r
r

r
0
V



z
a
ˆ is
w
h
e
r
er
r
2
z
th
e
r
e
fle
c
tio
n
o
fr
b
o
u
tth
e
z
0p
la
n
e
.
a
,
N
o
te
th
a
tfo
re
v
e
r
y
c
o
n
tr
ib
u
tio
n
to

()
r fr
o
m
th
e
c
h
a
r
g
e
V
tr
Vd
a
.
th
e
r
e
is
a
s
im
ila
rc
o
n
tr
ib
u
tio
n
fr
o
m
th
e
im
a
g
e
c
h
a
r
g
e

V
tr
Vd
Example: Scalar Point Source in a Rectangular Waveguide
 Weassumeaneit timedependenceand
y
a
ascalar wavefunctionthat satisfies
2(r)k2(r)f r, k 
 x,y,z

v
(r)0, x 0, a; y 0, b; z
z
b
x
 HencetheGreen'sfunctionfor theproblemsatisfies
2G(r,r)k2G(r,r)rr xx y y z z,
G(r,r)0, x 0, a; y 0, b; z withwavesoutgoingfromz z
 Assumingaseparation-of-variablesform G(r,r)  XxYyZz
andapplyingboundaryconditionsyields
d2X 2
m

k
X

0

X
x

A
sin
k
x
,
k

, m1,2,


x
x
x
2
dx
a
d2Y
n
2

k
Y

0

Y
y

B
sin
k
y
,
k

, n1,2,


y
y
y
dy2
b
d2Z 2
kz Z 0
dz2
ikzzz
 k2 k2 k2 , k2 k2 k2

, z z
x
y
x
y
Ce

 Zz   ik zz
, kz  
 
2
2
2
2
2
2
, z z
De z

i
k

k

k
,
k

k

k


x
y
x
y

Point Source in a Waveguide, cont’d
H
ence G(rr
, ) isof theform
m
x ny ikz,mnzz
 
A
s
i
n
sin
e
, zz


m
n

a
b
m1n1
G(rr
, ) 
m
x ny ikz,mnzz

B
s
i
n
sin
e
, zz


m
n

a
b
m1n1
y
a
 x,y,z
z
2
2
2
2
2
 k2 m

a

n

b
,
k

m

a

n

b








kz,mn 
2
2
2
2
2
2
i m

a

n

b

k
,
k

m

a

n

b








b
x
C
ontinuityof G(rr
, ) at zz requires(z limz)

0
Gx
( , y,z;x, y,z)Gx
( , y,z;x, y,z) (alsoderivativesw
.r.t.x, y arecontinuous!)
m
x ny  
m
x ny

A
s
i
n
s
i
n

B
s
i
n
sin




m
n
m
n
a
b
a
b
m1n1
m1n1
 
 
G(rr
, )  
Amnsin
m1n1
m
x ny ikz,mn zz
sin
e
a
b
A
m
n B
m
n


i
k
z

z


z
,m
n


e
,zz




i
k
z

z
z
,m
n
e



i
k
z

z


z
,m
n

e
,zz



Point Source in a Waveguide, cont’d
mx n y ikz,mn zz
sin
e
a
b
m1 n1
 Todetermine the constants Amn, note first that
G(r,r) 


Amn sin
z
z
  G(r,r)k G(r,r) dz   xx  y  y   z zdz
2
2
z
z
  x  x  y  y
2 2 2
 Toevaluate theLHS, note that   2  2  2 ,
x y z
2
z
2G(r,r)
G(x, y, z  ;r) G(x, y, z  ;r) 
- 
dz

lim

2


0 

z

z
z


z
Key result!
mx n y  
mx n y
 ikz,mn Amn sin
sin
ikz,mn Amn sin
sin
a
b
a
b
m1 n1
m1 n1




 2ikz,mn Amn sin
m1 n1
z
mx n y
sin
a
b
z
z
2G(r,r)
2G(r,r)
- Note 
dz  
dz   k2G(r,r)dz  0
2
2
x
y
z
z
z
Key observation!
sinceG(r,r) andhence its derivatives w.r.t. x, y are continuousat z  z
Point Source in a Waveguide, cont’d


G(r,r)  Amn sin
 Hence 
z

z
m1 n1
mx ny ikz,mn zz
sin
e
a
b
2G(rr
2G(rr
dz G(x,y,z
,
)

k
,
)




z


;r)G(x,y,z;r)
z

z
xx yy  zzdz
z
mx ny
sin
  xx  y  y
a
b
m1 n1
 Finally, todetermine Amn usetheorthogonalitypropertiesof the sin functions:


2ikz,mnAmn sin
ba
sin
00
px
mx qy ny
ab
sin
sin
sin
dxdy 
mpnq
a
a
b
b
4
px qy
sin
, usethe
a
b
orthogonalityproperties, andfinallysubstitute pm, qn, toobtain
"Project"bothsidesof theaboveontothefunctionssin
2ikz,mnAmn
ab
mx ny
sin
sin
4
a
b
 Amn 
mx ny
sin
sin
ikz,mnab
a
b
2
2
mx
mx ny ny ikz,mn zz

 G(r,r )  
sin
sin
sin
sin
e
ik
a
b
a
a
b
b
m1 n1
z,mn


2D Sources
 Atw
o-dim
ensional (noz-variation)"point"source
z
is actuallyalinesourcew
ithunitlinesourcedensity:
(rr)  (xx)(yy)
( )
 ()
, 0
or 



 (),
0

 2
or
(rr)dS (rr) dxdy 
S
1, rS

0, rS

y
x
r
1[m]
(rr) dd
ˆ
ˆ
(
R
e
m
i
n
d
e
r
:
I
n
2
D
,
rxy

xy

)
Itisoftenconvenient totreat theintegrationover Sinthexy-planeasa
volum
eintegration over,say, acircularcylinderof unit height andradius
centeredabout thepointr.
Example: Green’s Function for 2D Poisson’s
Equation
2-D Poisson's equation:
v
2
2 1     1 2
2
  ,   2  2 
  2 2

0
x y
      
2
z

Claim:
 G(r,r)  
1
20
ln is the static Green's
1[m]
functionfor the 2DPoisson's equationwith
unit line source density alongthe z-axis,

(r) 


0 

ˆ ; here wehave
ˆ  yy
ˆ  ρ
(Reminders: In2-D, r  xx
2G(r,r)  
()
20
neither a nor a z variation!)
y
x
“Proof” of Claim

G(rr
, ) isasolutionof thehomogeneousPoisson(i.e.,Laplace's)equation,
1 d  dG
1 1 d  
0




0, 0,






 d d 20  d   20
z
thesingularityat 0actuallygeneratesadeltafunctionat

0 inPoisson'sequation!
1[m]
W
emust alsoshowthat
, ) dV
G(rr
2
V
1
 
00
()
1
2ddz 
0
20
whentheintegrationdomainV istheunitheight cylinder of radius
centeredabout thepoint 0. Sincetheresult of theintegration
must beindependent of , it sufficestoconsider thelimit  0
1
()
1
2ddz 
0
20
0
00
limG(rr
, ) dV  lim
2
0
V
y
x
“Proof” of Claim (cont.)
 Toevaluate lim2G(rr
, ) dV, Vunitheight cylinder of radius, wenotethat

0
V
1
ˆ
G(rr
, )
ρ
20
1
G(rr
, ) 
ln,
20
Hencetheintegralabove is
" fluxper
unit vol."
lim2G(rr
, )dVlim 
 G(rr
, ) dV

0

0
V
divthm
.
 lim

0

V
Flux
ˆ ddz 
G(rr
, )n
VboundaryofV
dS
"
F
lu
x
p
e
r
u
n
itv
o
l."
ˆd
A
d
V 
A
n
S

V
1 2
1
1
1
ˆ
ˆ
lim
ρρ


d

d
z



0 2
0 
0

0 0
Thereforewehavefinally,
1
()
1
2ddz 

0
20
0
00
limG(rr
, ) dV  lim
2

0
V
F
lu
x
d
iv
t
h
m
.

V
Solution Is Easily Extended to 2D Sources Off
the z-Axis
 G(r,r)  
1
20
ln r  r is the two- dimensional
Line source
z
Green's function, representing a static, scalar line
y
source in unboundedhomogeneous media satisfying
x
 (r  r)
ˆ
ˆ  yy
, r  xx
2G(r,r)  
r
r
0
 The solution for the general case,
2(r)  
S
v (r)
( 0, r S)
0
y
in unboundedhomogeneous media is thus
(r) 
1
G(r,r)v (r)dS  
0 S
r  r
1
S
 ln r r v (r)dS
r
20 S
where the integrationis over the source region S
r  r
r
z
x
Example: Green’s Function for 2D Wave Equation
Claim:
H0(2) (k )
 G(r, r) 
is the outgoing - wave
4i
Green's function for the 2D wave equation
z

with unit line source density along the z - axis,
2G(r, r)  k 2G(r, r)  
 ( )
2
  (r)
and a harmonic time variation of the form eit .
(Reminders: In 2D, r  xxˆ  yyˆ  ρˆ ; here we have
1[m]
y
x
neither a  nor a z variation!)

The solution H0(2) (k ) of Bessel's equation,
1 d  dy 
2


k
y  0, which is singular at   0, actually


 d  d 
generates a delta function there!
“Proof” of Claim
 Since() 0, 0, wemust have
2G(r,r)k2G(r,r)  0, 0,
H0(2)(k)
But thisisindeedthecasesince G(r,r) 
4i
isanoutgoing solutionof the2Dwaveequation
(note n0 sincethereis no-variation))
z

1[m]
y
 Wemust next showthat
G(r,r)k G(r,r)dV
2
V
2
1
()
2ddz 1
2
00
x
 
whentheintegrationdomainV istheunit height cylinder of radius
centeredabout thepoint 0. Sincetheresult of theintegration
must beindependent of , it sufficestoconsider thelimit  0
“Proof” of Claim (cont.)
 Evaluate lim 2G(r,r)k2G(r,r)dV, Vunit height cylinder of radius
 0
V
 Since weareintegratingover aregionnear theorigin 0, wemayuse
small argument approximations totheHankel function,
2   k  
1i ln   
J (k) iN0 (k)
  2  
G(r,r)  0

,
4i
4i
Hencethefirst integral above is
G(r,r) 
"Flux per
unit vol. "
Flux
div thm.
lim 2G(r,r)dV lim  
 G(r,r) dV  lim
 0
 0
V
 0
V

 0
0
1
0 2 ρˆ ρˆ ddz 1
"
F
lu
x
p
e
r
u
n
itv
o
l."
ˆd
A
d
V 
A
n
S

V
 0
2
 0
V

0
k2 lim ln  0 0 
 0

V
1 2 
k lim G(r,r)dV  k lim 
 0
F
lu
x
d
iv
t
h
m
.
whereas thesecondis
2
ˆ dS 
G(r,r) n
VboundaryofV
1 2
lim 
1
ˆ
ρ
2
ln
0 0 2 dddz
 G(r,r)k G(r,r)dV 1,
2
V
2
rV
Extension to 2D Sources Off the z-Axis
H (k r  r )
 G(r, r) 
is the two-dimensional
4i
Green's function for outgoing wavefunctions in
unbounded homogeneous media, and satisfies
y
r
x
r
 G(r, r)  k G(r, r)   (r r)
2
Line source
z
(2)
0
2
r  r
S
 The (outgoing wave) solution for the general case,
2 (r)  k2 (r)   f (r)
y
in unbounded homogeneous media is thus
S
1
 (r)   G(r, r) f (r)dS   H0(2) (k r r ) f (r)dS
4i S
S
where the integration is over the source region S
r
r  r
r
z
x
Summary of Common 2D, 3D Greens Functions
2 -D (z  z  0) :

 2 G ( r , r )  k 2 G ( r , r )
 G ( r , r ) 
   ( r  r )
(2)
0
H
(k r  r )
y
4i
x
  2 G ( r , r )    ( r  r )

Line source
z
G ( r , r )  
r
r
r  r
1
ln r  r 
2
y
Point source
3 -D :

 2 G ( r , r )  k 2 G ( r , r )
 G ( r , r ) 

   ( r  r )
 ik r  r 
e
4 r  r 
 2 G ( r , r )    ( r  r )

r
1
G ( r , r ) 
4 r  r 
r  r
r
x
z
These Green’s functions are actually
fundamental solutions since there are
no imposed boundary conditions
Line Source Illumination of a Circular Cylinder

A line source illuminates a circular cylinder;
both are parallel to the z - axis. Hence the
y
incident field is
 (r) 

H 0(2) (k r  r )
, r   xˆ
4i
The total field  satisfies the Dirichlet boundary
inc
condition  (r)   inc (r)  sca (r)  0 at   a
on the cylinder surface.
 The scattered field is source - free and hence is an
outgoing solution to
 2 sca (r)  k 2 sca (r)  0.
In cylindrical coordinates it must have the form

 (r)   an H n(2) (k  )ein
sca
n 0

x
a

Line
source
The Addition Theorem
 Weneedanexpansionforinc(r) intermsof cylindrical wavefunctions
about x y 0. Suchanexpansionisprovidedbytheadditiontheorem
  (2)
in()

H
(
k

)
J
(
k

)
e
,  

n
n

n
H0(2)(k rr)   
 J (k)H(2)(k)ein(),  
n
n

n
y
  
x
  
where istheangular positionof thelinesourcerelativetothex-axis. For
our problem,  0.
 The additiontheoremis analogousto the Laurent expansionabout theorigin
of asimplepoleat z z:
  n n1
z z , z  z

1
 n0
  
zz

zn zn1, z  z


 n0
Solution of the Line Source Scattering Problem
 Theadditiontheoremallow
sustoeasilyapplytheD
irichletboundary
conditionat a:

1 
(2)
in
(2)
in




(
r
)


(
r
)

J
(
k
aH
)
(
k

)
e

aH
(
k
ae
)
0


n
n
n
n

a 4i
n
n
inc
sca

Jn(kaH
) n(2)(k)
an 
4iHn(2)(ka)
 Thetotalfieldisthus




( , )  



 in
Jn(kaH
) n(2)(k) (2)
1   (2)

H
(
k

)
J
(
k

)

H
(
k

)

 n
e , 
n
n
(2)
4in
Hn (ka)

Jn(kaH
) n(2)(k)  (2)
1 
in

J
(
k

)

H
(
k

)
e
 n
 n
4in
Hn(2)(ka)

, 
Interpretation as a Green’s Function
y
 Thesourceisa unit strengthlinesource


W
ecanobtaintheresult for alinesourceoff
thex-axisbysimplyreplacing by
a
Line
source


x
 HenceaGreen'sfunctionfor thecylinder scatteringproblemis
 1   (2)
 in
Jn(ka)Hn(2)(k) (2)

Hn (k)e
, 


Hn (k)Jn(k)
(2)
Hn (ka)
 4i n

G(r,r)  


Jn(ka)Hn(2)(k)  (2)
in
 1

J
(
k

)

H
(
k

)
e
, 
n
 n
(2)
 4i n
H
(
k
a
)

n


H0(2)(k rr)
1  Jn(ka)Hn(2)(k) (2)
in


H
(
k

)
e

n
4i
4i n
Hn(2)(ka)
fundamental
solution
homogeneous
solution