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 and non-zero in the
interior of the domain. It is also non-zero on the outside of the domain, but
that is irrelevant in connection with our problem.
One can also consider a domain consisting of overlapping circular regions,
some of which represent voids in the domain. Figure 7.2 shows a domain of
two circles with equations
x
2
+ y
2
= r
2
and
(x r/2)
2
+ y
2
=(r/2)
2
.
Reordering the latter yields
x
2
xr + y
2
=0,
andinturnresultsin
ω(x, y)=(r
2
x
2
y
2
)(x
2
rx + y
2
).
Clearly on the boundary of the larger circle the left term is zero and on the
boundary of the smaller circle the right term is zero. Hence the product
100 Applied calculus of variations for engineers
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
FIGURE 7.2 Domain with overlapping circular regions
function vanishes on the perimeter of both circles, which constitutes the now
nontrivial boundary.
Let us now consider the boundary of a rectangle of width 2w and height
2h, also centrally located around the origin. The equations of the sides
x = ±w,
and
y = ±h,
imply the very simple form of
ω(x, y)=(w
2
x
2
)(h
2
y
2
).
The verification is very simple,
ω(x, y)=0, (x, y)=(±w, ±h).
The construction technique clearly shows signs of difficulties to come with
very generic, and especially three-dimensional domains. In fact such difficul-
ties limited the practical usefulness of this otherwise innovative method until
more recent work enabled the automatic creation of the ω functions for generic
Numerical methods of calculus of variations 101
two- or three-dimensional domains with the help of spline functions, a topic
that will be discussed in Chapter 11 at length.
We shall now demonstrate the correctness of such a solution. For this we
consider the solution of a specific Poisson’s equation:
2
u
∂x
2
+
2
u
∂y
2
= 2,
with
u(x, y)=0, (x, y) ∂D,
where we designate the domain to be the rectangle whose ω function was
specified above. We will search for Kantorovich’s solution in the form of
u(x, y)=(w
2
x
2
)(h
2
y
2
)(α
1
+ α
2
x + α
3
y + ...).
The method, as all direct methods, is approximate, so we may truncate the
sequence of power function terms at a certain power. It is sufficient for the
demonstration to use only the first term.
We will apply the method in connection with Galerkin’s method of the last
section. Therefore the extremum is sought from
+w
w
+h
h
(
2
u
∂x
2
+
2
u
∂y
2
+2)ω(x, y)dydx =0.
Executing the posted differentiations and substituting results in
+w
w
+h
h
2α
1
(w
2
x
2
)(h
2
y
2
)
2
2α
1
(w
2
x
2
)
2
(h
2
y
2
)+
2(w
2
x
2
)(h
2
y
2
)dydx =0.
Since we only have a single coefficient, the system of equations developed ear-
lier boils down to a single scalar equation of
1
= f,
with
b =
+w
w
+h
h
((w
2
x
2
)(h
2
y
2
)
2
+(w
2
x
2
)
2
(h
2
y
2
))dydx,
and
f =
+w
w
+h
h
(w
2
x
2
)(h
2
y
2
)dydx.
After the (tedious) evaluation of the integrals, the value of
α
1
=
5
4(w
2
+ h
2
)
102 Applied calculus of variations for engineers
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
FIGURE 7.3 Solution of Poisson’s equation
emerges. In turn, the approximate Kantorovich-Galerkin solution is
u(x, y)=
5
4
(w
2
x
2
)(h
2
y
2
)
w
2
+ h
2
.
The solution is depicted graphically in Figure 7.3 using
w = h =1.
The figure demonstrates that the solution function satisfies the zero boundary
condition on the circumference of the square. To increase accuracy, additional
terms of the power series may be used. The method also enables the exploita-
tion of the symmetry of the domain. For example, if the above domain would
exhibit the same height as width,
s = w = h,
the solution may be sought in the form of
u(x, y)=(s
2
x
2
)(s
2
y
2
)(α
1
+ α
23
(x + y)),
where α
23
denotes the single constant accompanying both the second and
third terms.

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