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

å
ij 
Mkl,ij yij = Gkl,
(3)
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:
fij(x,y) º Xi(x)Yj(y),
(5)
where
Xi(x) º ì
ï
ï
í
ï
ï
î
æ
ç
è
1 - |x-xi|
h
ö
÷
ø
for xi-h £ x £ xi+h
0
otherwise
,
(6)
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
8
3
for k = i and l = j
- 1
3
for k-i = ±1 and/or l-j = ±1
0
otherwise
.
(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:
















(
8/3
-1/3
-1/3
-1/3
0
0
-2/3
8/3
-2/3
-1/3
0
0
-1/3
-1/3
8/3
-1/3
-1/3
-1/3
-2/3
-1/3
-2/3
8/3
-2/3
-1/3
0
0
-1/3
-1/3
8/3
-1/3
0
0
-2/3
-1/3
-2/3
8/3\end array
)
y5
y6
y8
y9
y11
y12 \end array
) =
1
1
0
0
0
0\end array
) V0. (13)