1
The foundations of calculus of variations
The problem of the calculus of variations evolves from the analysis of func-
tions. In the analysis of functions the focus is on the relation between two
sets of numbers, the independent (x) and the dependent (y) set. The func-
tion f creates a one-to-one correspondence between these two sets, denoted as
y = f (x).
The generalization of this concept is based on allowing the two sets not to be
restricted to being real numbers and to be functions themselves. The relation-
ship between these sets is now called a functional. The topic of the calculus
of variations is to find extrema of functionals, most commonly formulated in
the form of an integral.
1.1 The fundamental problem and lemma of calculus of
variations
The fundamental problem of the calculus of variations is to find the extremum
(maximum or minimum) of the functional
I(y)=
x
1
x
0
f(x, y, y
)dx,
where the solution satisfies the boundary conditions
y(x
0
)=y
0
and
y(x
1
)=y
1
.
This problem may be generalized to the cases when higher derivatives or multi-
ple functions are given and will be discussed in Chapters 3 and 4, respectively.
These problems may also be extended with constraints, the topic of Chapter 2.
A solution process may be arrived at with the following logic. Let us as-
sume that there exists such a solution y(x) for the above problem that satisfies
3
4 Applied calculus of variations for engineers
the boundary conditions and produces the extremum of the functional. Fur-
thermore, we assume that it is twice differentiable. In order to prove that
this function results in an extremum, we need to prove that any alternative
function does not attain the extremum.
We introduce an alternative solution function of the form
Y (x)=y(x)+η(x),
where η(x) is an arbitrary auxiliary function of x, that is also twice differen-
tiable and vanishes at the boundary:
η(x
0
)=η(x
1
)=0.
In consequence the following is also true:
Y (x
0
)=y(x
0
)=y
0
and
Y (x
1
)=y(x
1
)=y
1
.
A typical relationship between these functions is shown in Figure 1.1 where
the function is represented by the solid line and the alternative function by
the dotted line. The dashed line represents the arbitrary auxiliary function.
Since the alternative function Y (x) also satisfies the boundary conditions
of the functional, we may substitute into the variational problem:
I()=
x
1
x
0
f(x, Y, Y
)dx.
where
Y
(x)=y
(x)+η
(x).
The new functional in terms of is identical with the original in the case when
= 0 and has its extremum when
∂I()
∂
|
=0
=0.
Executing the derivation and taking the derivative into the integral, since the
limits are fixed, with the chain rule we obtain
∂I()
∂
=
x
1
x
0
(
∂f
∂Y
dY
d
+
∂f
∂Y
dY
d
)dx.
Clearly
dY
d
= η(x),
The foundations of calculus of variations 5
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 0.2 0.4 0.6 0.8 1
y(x)
eta(x)
Y(x)
FIGURE 1.1 Alternative solutions example
and
dY
d
= η
(x),
resulting in
∂I()
∂
=
x
1
x
0
(
∂f
∂Y
η(x)+
∂f
∂Y
η
(x))dx.
Integrating the second term by parts yields
x
1
x
0
(
∂f
∂Y
η
(x))dx =
∂f
∂Y
η(x)|
x
1
x
0
−
x
1
x
0
(
d
dx
∂f
∂Y
)η(x)dx.
Due to the boundary conditions, the first term vanishes. With substitution
and factoring the auxiliary function, the problem becomes
∂I()
∂
=
x
1
x
0
(
∂f
∂Y
−
d
dx
∂f
∂Y
)η(x)dx.
The extremum is achieved when = 0 as stated above, hence
∂I()
∂
|
=0
=
x
1
x
0
(
∂f
∂y
−
d
dx
∂f
∂y
)η(x)dx.
6 Applied calculus of variations for engineers
Let us now consider the following integral:
x
1
x
0
η(x)F (x)dx,
where x
0
≤ x ≤ x
1
and F (x) is continuous, while η(x) is continuously differ-
entiable, satisfying
η(x
0
)=η(x
1
)=0.
The fundamental lemma of calculus of variations states that if for all such η(x)
x
1
x
0
η(x)F (x)dx =0,
then
F (x)=0
in the whole interval.
The following proof by contradiction is from [18]. Let us assume that there
exists at least one such location x
0
≤ ζ ≤ x
1
where F (x) is not zero, for
example,
F (ζ) > 0.
By the condition of continuity of F (x) there must be a neighborhood of
ζ − h ≤ ζ ≤ ζ + h
where F (x) > 0. In this case, however, the integral becomes
x
1
x
0
η(x)F (x)dx > 0,
for the right choice of η(x), which contradicts the original assumption. Hence
the statement of the lemma must be true.
Applying the lemma to this case results in the Euler-Lagrange differen-
tial equation specifying the extremum
∂f
∂y
−
d
dx
∂f
∂y
=0.
The foundations of calculus of variations 7
1.2 The Legendre test
The Euler-Lagrange differential equation just introduced represents a neces-
sary, but not sufficient, condition for the solution of the fundamental varia-
tional problem.
The alternative functional of
I()=
x
1
x
0
f(x, Y, Y
)dx,
may be expanded as
I()=
x
1
x
0
f(x, y + η(x),y
+ η
(x))dx.
Assuming that the f function has continuous partial derivatives, the mean-
value theorem is applicable:
f(x, y + η(x),y
+ η
(x)) = f(x, y, y
)+
(η(x)
∂f(x, y, y
)
∂y
+ η
(x)
∂f(x, y, y
)
∂y
)+O(
2
).
By substituting we obtain
I()=
x
1
x
0
f(x, y, y
)dx+
x
1
x
0
(η(x)
∂f(x, y, y
)
∂y
+ η
(x)
∂f(x, y, y
)
∂y
)dx + O(
2
).
With the introduction of
δI
1
=
x
1
x
0
(η(x)
∂f(x, y, y
)
∂y
+ η
(x)
∂f(x, y, y
)
∂y
)dx,
we can write
I()=I(0) + δI
1
+ O(
2
),
where δI
1
is called the first variation. The vanishing of the first variation is a
necessary, but not sufficient, condition to have an extremum. To establish a
sufficient condition, assuming that the function is thrice continuously differ-
entiable, we further expand as
I()=I(0) + δI
1
+ δI
2
+ O(
3
).
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