Numerical methods of calculus of variations 99
This function assumes the role of enforcing the boundary condition and the
following set of simpler, low order basis functions are adequate to present the
solution:
b
1
(x, y)=ω(x, y),
b
2
(x, y)=ω(x, y)x,
b
3
(x, y)=ω(x, y)y,
b
4
(x, y)=ω(x, y)x
2
,
b
5
(x, y)=ω(x, y)xy,
b
6
(x, y)=ω(x, y)y
2
,
and so on, following the same pattern. It is clear that all these basis func-
tions vanish on the boundary by the virtue of ω(x, y), even though the power
function components do not.
The question is how to generate ω(x, y) for various shapes of domains. For
a centrally located circle with radius r, the equation
x
2
+ y
2
= r
2
implies very intuitively the form of
ω(x, y)=r
2
− x
2
− y
2
.
Obviously the function is zero everywhere on the circle