7

Numerical methods of calculus of variations

In the last chapter we focused on analytical solutions. Application problems

in engineering practice, however, may not be easily solved by such techniques,

if solvable at all. Hence, before we embark on applications, it seems prudent to

discuss solution techniques that are amenable for practical problems. These

methods produce approximate solutions and are, as such, called numerical

methods.

It was mentioned in the introduction that the solution of the Euler-Lagrange

diﬀerential equation resulting from a certain variational problem may not be

easy. This gave rise to the idea of directly solving the variational problem.

The classical method is the Euler method.

The most inﬂuential method is that of Ritz. The methods of Galerkin and

Kantorovich, both described in [13], could be considered extensions of Ritz’s.

They are the most well-known by engineers and used in the industry. Finally,

the boundary integral method is also useful for certain kind of engineering

problems.

7.1 Euler’s method

Euler proposed a numerical solution for the variational problem of

I(y)=

x

n

x

0

f(x, y, y

)dx = extremum

with the boundary conditions

y(x

0

)=y

0

; y(x

n

)=y

n

,

by subdividing the interval of the independent variable as

x

i

= x

0

+ i

x

n

− x

0

n

; i =1, 2,...,n.

89

90 Applied calculus of variations for engineers

Introducing

h =

x

n

− x

0

n

,

the functional may be approximated as

I(y

i

)=

x

1

x

0

f(x

i

,y

i

,y

i

)=h

n−1

i=1

f(x

0

+ ih, y

i

,

y

i+1

− y

i

h

)dx = extremum.

Here the approximated solution values y

i

are the unknowns and the extremum

may be found by diﬀerentiation:

∂I

∂y

i

=0.

The process is rather simple and follows from Euler’s other work in the nu-

merical solution of diﬀerential equations. For illustration, we consider the

following problem:

I(y)=

1

0

(2xy + y

2

+ y

2

)dx = extremum,

with the boundary conditions

y(0) = y(1) = 0.

Let us subdivide the interval into n = 5 equidistant segments with

h =0.2,

and

x

i

=0.2i.

The approximate functional with the appropriate substitutions becomes

I(y

i

)=0.2

4

i=1

(0.4iy

i

+ y

2

i

+(5(y

i+1

− y

i

))

2

).

The computed partial derivatives are

∂I

∂y

1

=0.2(0.4+2y

1

−

2(y

2

− y

1

)

0.04

)=0,

∂I

∂y

2

=0.2(0.8+2y

2

−

2(y

3

− y

2

)

0.04

+

2(y

2

− y

1

)

0.04

)=0,

∂I

∂y

3

=0.2(1.2+2y

3

−

2(y

4

− y

3

)

0.04

+

2(y

3

− y

2

)

0.04

)=0,

and

∂I

∂y

4

=0.2(1.6+2y

4

+

2y

4

0.04

+

2(y

4

− y

3

)

0.04

)=0.

Numerical methods of calculus of variations 91

TA BL E 7 . 1

Accuracy of Euler’s

method

i x

i

y

i

y(x

i

)

1 0.2 -0.0286 -0.0287

2 0.4 -0.0503 -0.0505

3 0.6 -0.0580 -0.0583

4 0.8 -0.0442 -0.0444

This system of four equations yields the values of the approximate solution.

The analytic solution of this problem is

y(x)=−x + e

e

x

− e

−x

e

2

− 1

.

The comparison of the Euler solution (y

i

) and the analytic solution (y(x

i

)) at

the four discrete points is shown in Table 7.1.

The boundary solutions of y(0) and y(1) = 0 are not shown since they are

in full agreement by deﬁnition.

7.2 Ritz method

Let us consider the variational problem of

I(y)=

x

1

x

0

f(x, y, y

)dx = extremum,

under the boundary conditions

y(x

0

)=y

0

; y(x

1

)=y

1

.

The Ritz method is based on an approximation of the unknown solution

function with a linear combination of certain basis functions. Finite element

or spline-based approximations are the most commonly used and will be the

subject of detailed discussion in Chapters 9 and 11. Let the unknown function

be approximated with

y(x)=α

0

b

0

(x)+α

1

b

1

(x)+...+ α

n

b

n

(x),

where the basis functions are also required to satisfy the boundary conditions

and the coeﬃcients are as yet unknown. Substituting the approximate solu-

tion into the variational problem results in

92 Applied calculus of variations for engineers

I(

y)=

x

1

x

0

f(x, y, y

)dx = extremum.

In order to reach an extremum of the functional, it is necessary that the

derivatives with respect to the unknown coeﬃcients vanish:

∂I(

y)

∂α

i

=0;i =0, 1,...,n.

It is not intuitively clear that the approximated function approaches the

extremum of the original variational problem, but it has been proven, for

example in [13]. Let us just demonstrate the process with a small analytic

example. Consider the variational problem of

I(y)=

1

0

y

2

(x)dx = extremum,

with the boundary conditions

y(0) = y(1) = 0,

and constraint of

1

0

y

2

(x)dx =1.

Since this is a constrained problem, we apply the Lagrange multiplier tech-

nique and rewrite the variational problem as

I(y)=

1

0

(y

2

(x) − λy

2

)dx = extremum.

Let us use, for example, the basis functions of

b

0

(x)=x(x − 1)

and

b

1

(x)=x

2

(x − 1).

It is trivial to verify that these also obey the boundary conditions. The ap-

proximated solution function is

y = α

0

x(x − 1) + α

1

x

2

(x − 1).

The functional of the constrained, approximated variational problem is

I(

y)=

1

0

(y

2

− λy

2

)dx.

Evaluating the integral yields

I(

y)=

1

3

(α

2

0

+ α

0

α

1

+

2

5

α

2

1

) − λ(

1

30

α

2

0

+

1

30

α

0

α

1

+

1

105

α

2

1

).

Numerical methods of calculus of variations 93

The extremum requires the satisfaction of

∂I

∂α

0

= α

0

(

2

3

−

λ

15

)+α

1

(

1

3

−

λ

30

)=0

and

∂I

∂α

1

= α

0

(

1

3

−

λ

30

)+α

1

(

4

15

−

2λ

105

)=0.

A nontrivial solution of this system of equations is obtained by setting its

determinant to zero, resulting in the following quadratic equation

λ

2

− 52λ + 420

6300

=0.

Its solutions are

λ

1

= 10; λ

2

=42.

Using the ﬁrst value and substituting into the second condition yields

α

1

=0

with arbitrary α

0

. Hence

y(x)=α

0

x(x − 1).

The condition

1

0

y

2

dx =

1

0

α

2

0

x

2

(x − 1)

2

dx =1

results in

α

0

= ±

√

30.

The approximate solution of the variational problem is

y(x)=±

√

30x(x − 1).

It is very important to point out that the solution obtained as a function

of the chosen basis functions is not the analytic solution of the variational

problem. For this particular example the corresponding Euler-Lagrange dif-

ferential equation is

y

+ λy =0

whose analytic solution, based on Section 5.3, is

y = ±

√

2sin(πx).

Figure 7.1 compares the analytic and the approximate solutions and plots

the error of the latter.

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