Jan 30, 2002
Notes for Lecture
Notes for Lecture #6
The Green's function allows us to determine the electrostatic potential
from volume and surface integrals:
|
F(r) = |
1
4pe0
|
|
ó
õ
|
V
|
d3r¢ r(r¢) G(r,r¢) + |
1
4p
|
|
ó
õ
|
S
|
[ G(r,r¢) ÑF(r¢) - F(r¢) Ñ G(r,r¢)] · |
^
r
|
¢
|
d2 r¢. |
| (1) |
This general form can be used in 1, 2, or 3 dimensions.
Orthogonal function expansions and Green's functions
Suppose we have a ``complete'' set of orthogonal functions
{un(x)} defined in the interval x1 £ x £ x2 such
that
|
|
ó
õ
|
x2
x1
|
un(x) um(x) dx = dnm. |
| (2) |
We can show that the completeness of this functions implies that
|
|
¥
å
n = 1
|
un(x) un(x¢) = d(x-x¢). |
| (3) |
This relation allows us to use these functions to represent a
Green's function for our system. For the 1-dimensional Poisson
equation, the Green's function satisfies
|
|
¶2
¶x2
|
G(x,x¢) = -4pd(x-x¢). |
| (4) |
Therefore, if
|
|
d2
dx2
|
un(x) = -an un(x), |
| (5) |
where {un(x)} also satisfy the appropriate boundary
conditions, then we can write the Greens functions as
|
G(x,x¢) = 4 p |
å
n
|
|
un(x) un(x¢)
an
|
. |
| (6) |
For example, if un(x) = Ö[(2/a)] sin(npx/a), then
|
G(x,x¢) = |
8 p
a
|
|
å
n
|
|
sin(npx/a)sin(np x¢/a)
|
. |
| (7) |
These ideas can easily be extended to two and three dimensions.
For example if {un(x)}, {vn(x)}, and {wn(x)} denote
the complete functions in the x, y, and z directions
respectively, then the three dimensional Green's function can be
written:
|
G(x,x¢,y,y¢,z,z¢) = 4 p |
å
lmn
|
|
ul(x)ul(x¢)vm(y)vm(y¢)wn(z)wn(z¢)
al + bm + gn
|
, |
| (8) |
where
|
|
d2
dx2
|
ul(x) = -al ul(x), |
d2
dy2
|
vm(x) = -bm vm(y), and |
d2
dz2
|
wn(z) = -gn wn(z). |
| (9) |
See Eq. 3.167 in Jackson for an example.
An alternative method of finding Green's functions for second
order ordinary differential equations is based on a product of two
independent solutions of the homogeneous equation, u1(x) and
u2(x), which satisfy the boundary conditions at x1 and
x1, respectively:
|
G(x,x¢) = K u1(x < ) u2(x > ), where K º |
4p
|
, |
| (10) |
with x < meaning the smaller of x and x¢ and x >
meaning the larger of x and x¢. For example, we have
previously discussed the example of the one dimensional Poisson
equation with the boundary condition F(0) = 0 and [(dF(¥))/ dx] = 0 to have the form:
For the two and three dimensional cases, we can use this technique
in one of the dimensions in order to reduce the number of
summation terms. These ideas are discussed in Section 3.11 of
Jackson. For the two dimensional case, for example, we can
assume that the Green's function can be written in the form:
|
G(x,x¢,y,y¢) = |
å
n
|
un(x) un(x¢) gn(y,y¢). |
| (12) |
If the functions {un(x)} satisfy Eq. 5, then we must
require that G satisfy the equation:
|
Ñ2 G = |
å
n
|
un(x) un(x¢) |
é
ê
ë
|
-an + |
¶2
¶y2
|
ù
ú
û
|
gn(y,y¢) = - 4 pd(x-x¢)d(y-y¢). |
| (13) |
The y-dependence of this equation will have the required
behavior, if we choose:
|
|
é
ê
ë
|
-an + |
¶2
¶y2
|
ù
ú
û
|
gn(y,y¢) = -4 pd(y-y¢), |
| (14) |
which in turn can be expressed in terms of the two independent
solutions vn1(y) and vn2(y) of the homogeneous
equation:
|
|
é
ê
ë
|
d2
d y2
|
- an |
ù
ú
û
|
vni(y) = 0, |
| (15) |
and a constant related to the Wronskian:
|
Kn º |
4p
|
vn1 |
dvn2
dx
|
- |
dvn1
dx
|
vn2 |
|
. |
| (16) |
If these functions also satisfy the appropriate boundary
conditions, we can then construct the 2-dimensional Green's
function from
|
G(x,x¢,y,y¢) = |
å
n
|
un(x) un(x¢) Kn vn1(y < ) vn2(y > ). |
| (17) |
For example, a Green's function for a
two-dimensional rectangular system
with 0 £ x £ a and 0 £ y £ b,
which vanishes on each of the boundaries can be expanded:
|
G(x,x¢,y,y¢) = 8 |
¥
å
n = 1
|
|
|
sin |
æ
ç
è
|
|
npx
a
|
ö
÷
ø
|
sin |
æ
ç
è
|
|
npx¢
a
|
ö
÷
ø
|
sinh |
æ
ç
è
|
npy <
a
|
ö
÷
ø
|
sinh |
æ
ç
è
|
np
a
|
(b-y > ) |
ö
÷
ø
|
|
|
. |
| (18) |
As an example, we can use this result to solve the 2-dimensional Laplace
equation in the square region 0 £ x £ 1 and 0 £ y £ 1
with the boundary condition F(x,0) = F(0,y) = F(1,y) = 0 and
F(x,1) = V0. In
this case, in determining F(x,y) using Eq. (1) there is no
volume contribution (since the charge is zero) and the ``surface'' integral
becomes a line integral 0 £ x¢ £ 1 for y¢ = 1.
Using the form from Eq. (18) with a = b = 1, it can be shown that the
result takes the form:
|
F(x,y) = |
¥
å
n = 0
|
4 V0 |
sin[(2n+1)px] sinh[(2n+1)py]
|
(2n+1) psinh[(2n+1)p] |
| (19) |
File translated from TEX by TTH, version 2.20.
On 30 Jan 2002, 12:52.