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

bα

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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