Feb 5, 2001
Notes for Lecture
Notes for Lecture #9
Finite element method
The finite element approach is based on an expansion of the
unknown electrostatic potential in terms of known grid-based
functions of fixed shape. In two dimensions, using the indices
{i,j} to reference the grid, we can denote the shape
functions as {fij(x,y)}. The finite element expansion
of the potential in two dimensions can take the form:
|
4 pe0 F(x,y) = |
å
ij
|
yij fij(x,y), |
| (1) |
where yij represents the amplitude associated with the
shape function fij(x,y). The amplitude values can be
determined for a given solution of the Poisson equation:
|
- Ñ2 ( 4 pe0 F(x,y) ) = 4 pr(x,y), |
| (2) |
by solving a linear algebra problem of the form
where
|
Mkl,ij º |
ó
õ
|
dx |
ó
õ
|
dy Ñfkl(x,y) ·Ñ fij(x,y) and Gkl º |
ó
õ
|
dx |
ó
õ
|
dy fkl(x,y) 4 pr(x,y) . |
| (4) |
In obtaining this result, we have assumed that the boundary values
vanish. In order for this result to be useful, we need to be
able evaluate the integrals for Mkl,ij and for Gkl. In
the latter case, we need to know the form of the charge density.
The form of Mkl,ij only depends upon the form of the shape
functions. If we take these functions to be:
where
and Yj(y) has a similar expression in the variable
y. Then
|
Mkl,ij º |
ó
õ
|
dx |
ó
õ
|
dy |
é
ê
ë
|
|
dXk(x)
dx
|
|
dXi(x)
dx
|
Yl(y)Yj(y)+ Xk(x)Xi(x) |
dYl(y)
dy
|
|
dYj(y)
dy
|
|
ù
ú
û
|
. |
| (7) |
There are four types of non-trivial contributions to these values:
|
|
ó
õ
|
xi+h
xi-h
|
(Xi(x) )2 dx = h |
ó
õ
|
1
-1
|
(1 - |u|)2 du = |
2h
3
|
, |
| (8) |
|
|
ó
õ
|
xi+h
xi-h
|
(Xi(x) Xi+1(x)) dx = h |
ó
õ
|
1
0
|
(1 - u) u du = |
h
6
|
, |
| (9) |
|
|
ó
õ
|
xi+h
xi-h
|
|
æ
ç
è
|
d Xi(x)
dx
|
ö
÷
ø
|
2
|
dx = |
1
h
|
|
ó
õ
|
1
-1
|
du = |
2
h
|
, |
| (10) |
and
|
|
ó
õ
|
xi+h
xi-h
|
|
æ
ç
è
|
|
d Xi(x)
dx
|
|
d Xi+1(x)
dx
|
ö
÷
ø
|
dx = - |
1
h
|
|
ó
õ
|
1
0
|
du = |
-1
h
|
. |
| (11) |
These basic ingredients lead to the following distinct values for
the matrix:
|
Mkl,ij = |
ì
ï
ï
ï
ï
ï
í
ï
ï
ï
ï
ï
î
|
|
| |
| |
for k-i = ±1 and/or l-j = ±1 |
|
| |
|
. |
| (12) |
For problems in which the boundary values are 0, Eq. 3
then can be used to find all of the interior amplitudes
yij.
In order to use this technique to solve the boundary value problem
discussed in Lecture Notes #5, we have to make one modification.
The boundary value of F(x,a) = V0 is not consistent with the
derivation of Eq. (4), however, since we are only interested
in the region 0 £ y £ a, we can extend our numerical
analysis to the region 0 £ y £ a+h and require F(x,a+h) = 0 in addition to F(x,a) = V0. Using the same indexing as
in Lecture Notes #5, this means that y1 = y2 = y3 = V0.
The finite element approach for this problem thus can be put into
the matrix form for analysis by Maple:
| | | | | |
| |
| | | | | |
| |
| | | | |
|
) |
|
|
| |
|
| |
|
| |
|
| |
|
|
) = |
|
|
| |
|
| |
|
| |
|
| |
|
|
) V0. |
(13) |