D. R. Wilton ECE Dept. ECE 6382 Green’s Functions in Two and Three Dimensions.
Transcript of D. R. Wilton ECE Dept. ECE 6382 Green’s Functions in Two and Three Dimensions.
Static Potential of Point SourcesStatic Potential of Point Sources
0
2
0
( ) ,4
V
Q
Q
r r
rr r
It is well known that the free space static potential at an observation
point due to a point charge at is
On the other hand, should also satisfy Poisson's equation,
V
V Q
where is the volume charge density. Ques : How can we
define so as to incorporate a point charge, ?
We usually avoid explicitly introducing point charges into Poisson's
equation be
caus V
Q
e the volume charge density of a point charge is
except the charge location, where it is
(i.e., a finite charge exists within a volume of zero ).
zero everywhere infinite
Representation of Point Sources Representation of Point Sources
( ) ( ) ( )V Q x x y y z z
But we mathematically represent a point charge as a volume
charge density using delta functions. Setting
we can easily check that evaluating the net charge by integrating
can
( ) ( ) )
( )
(yz x
z y x
Q x x y y z z dxdydz Q
r
over any non- vanishing volume about yields the correct total charge :
Ques : Do units on RHS and LHS agree in the above eqs.?
Thus we can
2
0
2 2 20 0
( ) ( ) ( )
( )4 4 ( ) ( ) ( )
Qx x y y z z
Q Q
x x y y z z
rr r
say that the solution to Poisson's equation,
is
22
2( , , )x y z
Superposition of PotentialsSuperposition of Potentials
0
( )
( )( ) .
4
V
V
V
dq
V
dq dV
dVd
r
rr
r r
We further say that in a volume with a
distributed volume source density , every
infinitesimal volume element of charge,
, produces a potential
By th
0
0
( )( )
4
1( ) ( )
4
V
V
V
dV
r
rr r
r rr r
e linearity of the Poisson operator, we conclude by
that
is an integration of with a weighting factor , t
superimposing
the contributions of all sources
r r
he
potential at of a unit charge at .
rr
r r
V dV
V
O
A Green’s FunctionA Green’s Function
0
1( , )
4
( ) ( , ) ( )
( , )
V
V
G
G dV
G
r rr r
r r r r
r r r
Note the role played by the factor
called the that allows us to write
A physical interpretation of is that of the
Greens' function
potential at du
2
0
2
0
.
1( , ) ( ) ( ) ( )
1( , ) ( )
( ) ( ) ( ) ( )
G x x y y z z
G
x x y y z z
r
r r
r r r r
r r
Note that it is a solution to
or in a more coordinate - free notation,
where in rect
e
to a unit point charge at
1,( )
0,V
VdV
V
rr r
r
angular coordinates,
but more generally, we simply require that in 3 -D
Green’s Function ConditionsGreen’s Function Conditions
0 0
2
1 1( , )
4 4
( , ) 0
(
Gr
G r
G
r
r 0r
r 0
To check our claims, it suffices to place at the coordinate origin so
spherical coordinates can be used :
Check that satisfies the homogeneous equation when :
2
22 2
0
1 1 1, )
4
G rr
r r r r r
r 02r
20
2
0 0
2
0 0 0
2
0
1 00, 0
4
1 1( , ) ( ) ,
ˆlim ( , ) lim ( , ) lim ( , )
( , )ˆ ˆlim sin
V V
rV V V
r
rr
G dV dV V
G dV G dV G dS
Gd d
r
r 0 r r 0
r 0 r 0 r 0 n
r 0r r
divergencetheorem
Next, check that encloses :
2
00 0
4lim
4 2
0 2
0
1,
.V
r 0 if encloses
r̂V
V
Green’s Function for a General Linear OperatorGreen’s Function for a General Linear Operator
( , ) ( )G
r r r r
In general, a Green's function is a solution of the linear operator equation
that of the problem.
(A satisfies the above e
L
also satisfies any boundary conditions
fundamental solution
quation, but does
necessarily satisfy the boundary conditions; to obtain a Green's
function we add a homogeneous solution to a fundamental solution
and enforce BCs.)
A physical interpreta
not
( , )
.
( , ) 0
( , ) ( ) 1,V V
G
G
G dV dV V
r r r
r
r r r r
r r r r r
tion of is that of
Note that at where
in
L
L
the response at due to a
unit point source forcing function at
except
A Source-Weighted Superposition over the Unit A Source-Weighted Superposition over the Unit Source Response Provides a General Solution Source Response Provides a General Solution
( ) ( ) , ( )
( ) ( , ) ( )V
u f f
u G f dV
r r r
r r r r
The solution to the general problem
a general forcing function,
is then found by a source - weighted superposition of the f 'n response :
To check that this is
-
a so
Lu
u
( ) ( , ) ( ) ( , ) ( )
( ) ( )
( )
V V
V
u G f dV G f dV
f dV
f
r r r r r r r
r r r
r
lution, note that
Lu L L
3-D Point Source Representation in Various 3-D Point Source Representation in Various Coordinate Systems Coordinate Systems
ˆ ˆ ˆ
( ) ( ) ( ) ( )
( ) ( ) ( ), 0
( ) ( ) 0, 0
2
( ) ( ) (
x y z
x x y y z z
z z
z z
r r
r x y z
r r (rectangular coordinates)
(cylindrical coordinates have
a coord. singularity at )
2
2
2
), 0, 0,
sin( ) ( )
, 0, 0, 02 sin
0,( ), 0
4
rr
r rr r
rr
rr
(spherical coordinates have
a coord. singularity at
and at )
2-D Line Source Representation in Various 2-D Line Source Representation in Various Coordinate Systems Coordinate Systems
ˆ ˆ
( ) ( ) ( )
( ) ( ), 0
( ) 0, 0
2
x y
x x y y
r x y
r r (rectangular coordinates)
(cylindrical coordinates have
a coord. singularity at )
Cylindrical Coordinate ExampleCylindrical Coordinate Example
0
( ) ( ) ( )
dV d d dz
z z
If , the volume element is
2
0 0
1
0 2
( ) ( )1
2
( ) ( )
2
z
z
z
z
d d dz
dV d dz
z zd d dz
z z
If , becomes undefined, and the volume element is
20
1z
z
d dz
d
d dz
xy
z
dz
2 d z
xy
Example: A Simple Static Green’s Function with Example: A Simple Static Green’s Function with Boundary Conditions --- Charge over a Ground Boundary Conditions --- Charge over a Ground
PlanePlane
0 0
( , )
1 1( , )
4 4fundamental solution homogeneous soluti
The potential at due to a unit charge at
can be found from image theory. It is given by
G
G
r r r r
r rr r r r
above a ground plane
2 2
0 0 0
ˆ2 .
0 ( , )
1 1 1( , ) ( , ) ( , ) 0, 0
4
( , ) 0 0
onwhere
Note that for in the upper half space ( ), satisfies
(since )
at (since
z
z G
G z
G z
r r z
r r r
r r r r r rr r
r r r r r
and
0at ).z r
1 [C]r r
z
0 on ground plane
1 [C]r r
z
0 on ground plane
r
r r
-1 [C]
r r
Static Green’s Function with Boundary Static Green’s Function with Boundary Conditions (cont.)Conditions (cont.)
0
( )
1 1 1( ) ( , ) ( ) , ( , )
4
ˆ2 0
For an charge distribution above a ground plane, we
thus have
where is the reflection of about the plane.
V
V
V
G dV G
z z
r
r r r r r rr r r r
r r z r
arbitrary
( )
.
Note that for every contribution to from the charge at ,
at V
V
dV
dV
r r
r there is a similar contribution from the image charge
z
0 on ground plane
rr
r r
V dV
V
O
Example: Scalar Point Source in a Rectangular WaveguideExample: Scalar Point Source in a Rectangular Waveguide
2 2
2 2
( ) ( ) ,
( ) 0, 0, ; 0, ;
( , ) ( , )
i te
k f kv
x a y b z
G k G
r r r
r
r r r r r
We assume an time dependence and
a scalar wavefunction that satisfies
Hence the Green's function for the problem satisfies
2
22
,
( , ) 0, 0, ; 0, ;
( , )
0 sin ,x x
x x y y z z
G x a y b z z z
G X x Y y Z z
d Xk X X x A k x k
dx
r
r r
r r
with waves outgoing from
Assuming a separation - of - variables form
and applying boundary conditions yie
lds
22
2
2 2 2 2 2 22
22
2 2 2 2 2 2
, 1,2,
0 sin , , 1,2,
,,0 ,
, ,
z
z
x
y y y
ik z zx y x y
z zik z z
x y x y
mm
a
d Y nk Y Y y B k y k n
dy b
k k k k k kCe z zd Zk Z Z z k
dz De z z i k k k k k k
x , y ,z
y
z
x
a
b
Point Source in a Waveguide, cont’dPoint Source in a Waveguide, cont’d
,
,
1 1
1 1
2 2 2 22 2
,2 2 2 22 2
( , )
sin sin ,
( , )
sin sin ,
,
,
z mn
z mn
ik z z
mnm n
ik z z
mnm n
z mn
G
m x n yA e z z
a bG
m x n yB e z z
a b
k m a n b k m a n bk
i m a n b k k m a n b
r r
r r
Hence is of the form
0
1 1 1
( , ) lim
( , , ; , , ) ( , , ; , , ) ,
sin sin sin sinmn mnn m n
G z z z z
G x y z x y z G x y z x y z x y
m x n y m x n yA B
a b a b
r rContinuity of at requires ( )
(also derivatives w.r.t. are continuous!)
,
1
1 1
( , ) sin sin z mn
mn mnm
ik z zmn
m n
A B
m x n yG A e
a b
r r
x , y ,z
y
z
x
a
b
,
,
,
,
,
z mn
z mn
z mn
ik z zik z z
ik z z
e z ze
e z z
Point Source in a Waveguide, cont’dPoint Source in a Waveguide, cont’d
,
1 1
2 2
( , ) sin sin
( , ) ( , )
To determine the constants , note first that
To evaluate the LHS, note that
z mnik z zmn
m n
mn
z z
z z
m x n yG A e
a b
A
G k G dz x x y y z z dz
x x y y
r r
r r r r
2 2 22
2 2 2
2
2 0
, ,1 1 1 1
,1 1
,
( , ) ( , , ; ) ( , , ; )lim
sin sin sin sin
2 sin sin
z
z
z mn mn z mn mnm n m n
z mn mnm n
x y z
G G x y z G x y zdz
z z z
m x n y m x n yik A ik A
a b a b
m x n yik A
a b
r r r r
-
2 22
2 2
( , ) ( , )( , ) 0
( , ) ,
Note
since and hence w.r.t. are continuous at
z z z
z z z
G Gdz dz k G dz
x y
G x y z z
r r r r
r r
r r
-
its derivatives
Key result!
Key observation!
Point Source in a Waveguide, cont’dPoint Source in a Waveguide, cont’d
,
1 1
,1 1
2 2 ( , , ; ) ( , , ; )( , ) ( , )
( , ) sin sin
2 sin sin
z mnik z zmn
m n
z mn mnm n
z
z
G x y z G x y zG k G dzz z x x y y
m x n yG A e
a b
m x n yik A x x y y
a b
r rr r r r
r r
Hence
0 0
sin
sin sin sin sin4
"
mn
b a
mp nq
z
z
z z dz
A
p x m x q y n y abdxdy
a a b b
Finally, to determine use the orthogonality properties of the functions :
Project" both sides of the above onto the
,,
,
sin sin ,
, ,
22 sin sin sin sin
4
2( , ) sin sin sin
z mn mn mnz mn
z mn
p x q y
a bp m q n
ab m x n y m x n yik A A
a b ik ab a b
m x m x nG
ik ab a a
r r
functions use the
orthogonality properties, and finally substitute to obtain
,
1 1
sin z mnik z z
m n
y n ye
b b
2D Sources 2D Sources
( ) ( ) ( )
( ) ( ), 0
( ), 0
2
( ) ( )S
x x y y
dS dxdy
r r
r r r r
or
o
A two - dimensional (no - variation) "point" source
is actually a with unit line source density :
z
line source
( )
1,
0,
d d
S
S
S xy
r r
r
r
r
It is often convenient to treat the integration over in the -plane as a
integration over, say, a circular cylinder of and radius
volume unit height
.r centered about the point
ˆ ˆx y r x y(Reminder : In 2D, )
1[m]x
z
y
r
Example: Green’s Function for 2D Poisson’s Example: Green’s Function for 2D Poisson’s EquationEquation
2 2 22 2
2 2 2 20
0
1 1,
1( , ) ln
2
:
v
x y
G
r r
2 - D Poisson's equation
is the Green's
function for the 2D Poisson's equation with
unit line source density alon
Claim:
static
2
0 0
( ) ( )( , )
2
ˆˆ ˆ
G
x y
z
rr r
r x y ρ
g the - axis,
(Reminders : In 2 - D, ; here we have
neither a nor a variation!)
z
1[m]
y
x
z
““Proof” of Claim Proof” of Claim
0
( , )
1 1 1
2
G
d dG d
d d d
r r is a solution of the homogeneous Poisson (i.e., Laplace's) equation,
0
2
00, 0
2
0
0
( )( , )
2V
G dV
r r
,
the singularity at actually generates a delta function at
in Poisson's equation!
We must also show that
0
2
1
00 0
1
0.
d dz
V
when the integration domain is the unit height cylinder of radius
centered about the point Since the result of the integration
must be independent of , it suffices to c
1
2
0 00 00 0
0
( ) 1lim ( , ) lim 2
2V
G dV d dz
r r
onsider the limi
t
1[m]
y
x
z
““Proof” of Claim (cont.)Proof” of Claim (cont.)
2
0
0 0
2
0 0
lim ( , )
1 1ˆ( , ) ln , ( , )
2 2
lim ( , ) lim ( , )
V
V
G dV V
G G
G dV G
r r
r r r r ρ
r r r r
" f
To evaluate , unit height cylinder of radius , we note that
Hence th
e integral above is
0
00
ˆlim ( , )
1 1lim
2
V
V V dS
dV
G d dz
r r n
lux per unit vol. "
Flux
div thm.
boundary of
ˆ ˆ ρ ρ
1 2
00 0
12
0 00 00 0
1
( ) 1lim ( , ) lim 2
2V
d dz
G dV d dz
r r
Therefore we have finally,
ˆV V
dV dS
A A n
" Flux per Flux unit vol. "
div thm.
Solution Is Easily Extended to 2D Sources Off Solution Is Easily Extended to 2D Sources Off the the zz-Axis -Axis
0
2
0
1( , ) ln
2
( )ˆ ˆ( , ) ,
G
G x y
r r r r
r rr r r x y
is the two - dimensional
Green's function, representing a static, scalar line
source in unbounded homogeneous medi
a satisfying
The solution
for
2
0
0 0
( )( ) ( 0, )
1 1( ) ( , ) ( ) ln ( )
2
v
v v
S S
S
G dS dS
S
rr r
r r r r r r r
the case,
in unbounded homogeneous media is thus
where the integration is over the source region
general
x
r r
y
rr
S
z
x
z
y
rr
Line source
r r
S
Example: Green’s Function for 2D Wave EquationExample: Green’s Function for 2D Wave Equation
(2)0
2 2
( )( , )
4
( )( , ) ( , ) ( )
2
H kG
i
G k G
r r
r r r r r
is the
Green's function for the 2D wave equation
with unit line source density along the - axis,
and a harmonic tim
outgoing - wave
z
Claim:
(2)0
2
.
ˆˆ ˆ
( )
10
i te
x y
z
H k
d dyk y
d d
r x y ρ
e variation of the form
(Reminders : In 2D, ; here we have
neither a nor a variation!)
The solution of Bessel's equation,
, which is singular at 0, actually
generates a delta function there!
1[m]
y
x
z
““Proof” of Claim Proof” of Claim
2 2
(2)0
( ) 0, 0 ,
( , ) ( , ) 0, 0,
( )( , )
4
0
G k G
H kG
i
n
r r r r
r r
Since we must have
But this is indeed the case since
is an outgoing solution of the 2D wave equation
(note since there is no - variatio
1
2 2
0 0
( )( , ) ( , ) 2 1
2
0.
V
G k G dV d dz
V
r r r r
n))
We must next show that
when the integration domain is the unit height cylinder of radius
centered about the point Since the result of t
0 he integration
must be independent of , it suffices to consider the limit
1[m]
y
x
z
““Proof” of Claim (cont.)Proof” of Claim (cont.)
2 2
0lim ( , ) ( , )
0V
G k G dV V
r r r r
Evaluate , unit height cylinder of radius
Since we are integrating over a region near the origin , we may use
small argument approximations to the Hankel functi
0 0
2
0 0
21 ln
( ) ( ) 12ˆ( , ) , ( , )
4 4 2
lim ( , ) lim ( , ) limV V
ki
J k iN kG G
i i
G dV G dV
r r r r ρ
r r r r
" Flux per unit vol. "
div thm.
on,
Hence the first integral above is
0
1 2
00 0
1 202 2
0 00 0 0
2 2
00
ˆ( , )
1ˆ ˆlim 1
2
lnlim ( , ) lim
2
lim ln 0 ( , )
V V
V
G dS
d dz
k G dV k d d dz
k G
r r n
ρ ρ
r r
r r
Flux
boundary of
whereas the second i
s
2 ( , ) 1,V
k G dV V r r r
ˆV V
dV dS
A A n
" Flux per Flux unit vol. "
div thm.
Extension to 2D Sources Off the Extension to 2D Sources Off the zz-Axis -Axis
(2)0
2 2
( )( , )
4
( , ) ( , ) ( )
is the two - dimensional
Green's function for outgoing wavefunctions in
unbounded homogeneous media, and satisfies
The (outgoing wave) solution
f
H kG
i
G k G
r rr r
r r r r r r
2 2
(2)0
( ) ( ) ( )
1( ) ( , ) ( ) ( ) ( )
4
or the case,
in unbounded homogeneous media is thus
where the integration is over the source region S S
k f
G f dS H k f dSi
S
r r r
r r r r r r r
general
x
r r
y
rr
S
z
x
z
y
rr
Line source
r r
S
Summary of Common 2D, 3D Greens FunctionsSummary of Common 2D, 3D Greens Functions
2 2
(2)0
2
2 2
2
0
( , ) ( , ) ( )
( )( , )
4
( , ) ( )
1( , ) ln
2
( , ) ( , ) ( )
( , )4
( ,
( )
) ( )
1( , )
4
ik
z z
G k G
H kG
i
G
G
G k G
eG
G
G
r r
r r r r r r
r rr r
r r r r
r r r r
r r r r r r
r rr r
r r r r
r rr r
2-D :
3 -D :
x
z
y
rr
Line source
r r
x
r r
y
r
r
z
Point source
These Green’s functions are actually fundamental solutions since there areno imposed boundary conditions
Line Source Illumination of a Circular Cylinder Line Source Illumination of a Circular Cylinder
x
y
a
Line source
(2)0inc ( )
ˆ( ) ,4
z
H k
i
r r
r r x
A line source illuminates a circular cylinder;
both are parallel to the -axis. Hence the
incident field is
The field satisfies the Dirichlet boundary
condition
totalinc sca
2 sca 2 sca
( ) ( ) ( ) 0
( ) ( ) 0.
a
k
r r r
r r
at
on the cylinder surface.
The scattered field is source - free and hence is an
outgoing solution to
In cylindrical coordinates it must have th r
e fo
sca (2)
0
( ) ( ) inn n
n
a H k e
r
m
The Addition TheoremThe Addition Theoreminc
(2) ( )
(2)0
( )
0.
( ) ( ) ,
( )
(
inn n
n
n
x y
H k J k e
H k
J k
r
r r
We need an expansion for in terms of cylindrical wavefunctions
about Such an expansion is provided by the addition theorem
(2) ( )) ( ) ,
0.
inn
n
H k e
x
where is the angular position of the line source relative to the -axis. For
our problem,
The addition theorem is analogous to the Laurent expansion abo
1
0
1
0
,1
,
n n
n
n n
n
z z
z z z z
z zz z z z
ut the origin
of a simple pole at :
x
y
Solution of the Line Source Scattering Problem Solution of the Line Source Scattering Problem
inc sca (2) (2)
(2)
(2)
:
1( ) ( ) ( ) ( ) ( ) 0
4
( ) ( )
4 (
in inn n n na
n n
n nn
n
a
J ka H k e a H ka ei
J ka H ka
i H k
r r
The addition theorem allows us to easily apply the Dirichlet boundary
condition at
(2)(2) (2)
(2)
(2)(2)
(2)
)
( ) ( )1( ) ( ) ( ) ,
4 ( )( , )
( ) ( )1( ) ( ) ,
4 ( )
inn nn n n
n n
inn nn n
n n
a
J ka H kH k J k H k e
i H ka
J ka H kJ k H k e
i H ka
The total field is thus
Interpretation as a Green’s FunctionInterpretation as a Green’s Function
The source is a unit strength line source
We can obtain the result for a line source
the x -axis by simply replacing by
Hence a Green's function for the cylinder scattering problem is
off
(2)(2) (2)
(2)
(2)(2)
(2)
(2)0
( ) ( )1( ) ( ) ( ) ,
4 ( )( , )
( ) ( )1( ) ( ) ,
4 ( )
( )
4fundamentalsolution
inn nn n n
n n
inn nn n
n n
J ka H kH k J k H k e
i H kaG
J ka H kJ k H k e
i H ka
H k
i
r r
r r
(2)
(2)(2)
( ) ( )1( )
4 ( )homogeneous solution
inn nn
n n
J ka H kH k e
i H ka
Line source
x
y
a