Lesson 25
1. Why is the Laplacian operator translationally invariant in space?
2. What is the basic form of the fundamental solution to Laplace’s equation?
3. What is Poisson’s equation?
Poisson’s equation is Laplace’s equation with a non-zero right hand side:
4. What is the Green’s function approach to the solution of Laplace’s equation?
It is possible to construct solutions to Laplace’s equation using Green’s functions.
This is an extremely powerful approach, since they allow solutions to Laplace’s equation to be obtained for general boundary conditions.
Greens functions are constructed to satisfy Laplace’s equation as well as boundary conditions homogeneously i.e. irrespective of the actual boundary conditions the Green’s function is always zero at the boundary.
To construct Green’s function we make use of Green’s identities.
5. What is Green’s first identity?
0
: then -
,
,
like s coordinate
formed get trans
we such that
) , , ( to shifted
is
the of origin
the and 0
as system the
in equation
s Laplace' satisfying
have we if that means invariance
nal Translatio
2 3 2 2 2 2 2 1 2 2 0 3 3 3 0 2 2 2
0 1 1 1 0
3 0 2 0 1
3 2 2 1
3 2 2 2 2 2 1 2 2 3
2 1
x x x
x x x x x x
x x x x
x x
,x ,x x x
x ,x x
,x x
x x
equation.
s Laplace' o
solution t l
fundamenta the
of form basic the as known is
) (
) (
) (
4
1 4
ˆ 1
: as obained
, ) ( ) ˆ( the of solution The
2 3 3 2 2 2 2 1 1 2
y x y
x y
x
y x
y x y
x x
) (
2
x x
: as identity first
s Green' can write
then we ,
boundary
with
volume a
over defined
and functions smooth
two consider
we If
V
V
6. What is Green’s second identity?
7. What is Green’s third identity?
V V V V
V V
V
d ndS
d d
d d
d
y y
y n
y y
y
y
. .
.
. )
2 .(
0 )
(
: identity
second s Green' get e identity w first
s Green' from expression above
the g Subtractin
.
: get we identity, first
s Green' in and e interchang
we If
2 2
2
V V
V V
V
n dS d n
d n dS
d
y y
y
y y
identity.
third s Green' is which 1 )]
( ) ) ( ( [1 4 1
Hence
) ( ) ( ) ( )
( 4 ) ( 1 that Recalling
4 )]
( 1 ) 4 (
[ 1 4
) 1 ( -
: get we identity,
second s Green' in ng substituti Then
4 ˆ 1
i.e.
solution
l fundamenta
the be to take we
addition In
. 0
i.e.
harmonic,
is assume
we identity
third s Green' get To
2 2
2
V
V V
V V
n dS n
r φ(x) π
d r d
r dS n n
d r r
r
y x
y
y y x
y
x y y y x y
y y y
y
