8 Applied calculus of variations for engineers
Here the newly introduced second variation is
δI
2
=
2
2
x
1
x
0
(η
2
(x)
2
f(x, y, y
)
∂y
2
+
2η(x)η
(x)
2
f(x, y, y
)
∂y∂y
+
η
2
(x)
2
f(x, y, y
)
∂y
2
)dx.
We now possess all the components to test for the existence of the extremum
(maximum or minimum). The Legendre test in [7] states that if indepen-
dently of the choice of the auxiliary η(x) function
- the Euler-Lagrange equation is satisfied,
- the first variation vanishes (δI
1
=0),and
- the second variation does not vanish (δI
2
=0)
over the interval of integration, then the functional has an extremum. This
test manifests the necessary conditions for the existence of the extremum.
Specifically, the extremum will be a maximum if the second variation is neg-
ative, and conversely a minimum if it is positive. Certain similarities to the
extremum evaluation of regular functions by the teaching of classical calculus
are obvious.
We finally introduce the variation of the function as
δy = Y (x) y(x)=η(x),
and the variation of the derivative as
δy
= Y
(x) y
(x)=η
(x).
Based on these variations, we distinguish between the following cases:
- strong extremum occurs when δy is small, however, δy
is large, while
- weak extremum occurs when both δy and δy
are small.
On a final note: the above considerations did not ever state the finding or
presence of an absolute extremum; only the local extremum in the interval of
the integrand is obtained.
The foundations of calculus of variations 9
1.3 The Euler-Lagrange differential equation
Let us expand the derivative in the second term of the Euler-Lagrange differ-
ential equation as follows:
d
dx
∂f
∂y
=
2
f
∂x∂y
+
2
f
∂y∂y
y
+
2
f
∂y
2
y

.
This demonstrates that the Euler-Lagrange equation is usually of second or-
der.
∂f
∂y
2
f
∂x∂y
2
f
∂y∂y
y
2
f
∂y
2
y

=0.
The above form is also called the extended form. Consider the case when the
multiplier of the second derivative term vanishes:
2
f
∂y
2
=0.
In this case f must be a linear function of y
,intheformof
f(x, y, y
)=p(x, y)+q(x, y)y
.
For this form, the other derivatives of the equation are computed as
∂f
∂y
=
∂p
∂y
+
∂q
∂y
y
,
∂f
∂y
= q,
2
f
∂x∂y
=
∂q
∂x
,
and
2
f
∂y∂y
=
∂q
∂y
.
Substituting results in the Euler-Lagrange differential equation of the form
∂p
∂y
∂q
∂x
=0,
or
∂p
∂y
=
∂q
∂x
.
In order to have a solution, this must be an identity, in which case there must
be a function of two variables
u(x, y)
10 Applied calculus of variations for engineers
whose total differential is of the form
du = p(x, y)dx + q(x, y)dy = f(x, y, y
)dx.
The functional may be evaluated as
I(y)=
x
1
x
0
f(x, y, y
)dx =
x
1
x
0
du = u(x
1
,y
1
) u(x
0
,y
0
).
It follows from this that the necessary and sufficient condition for the solu-
tion of the Euler-Lagrange differential equation is that the integrand of the
functional be the total differential with respect to x of a certain function of
both x and y.
Considering furthermore, that the Euler-Lagrange differential equation is
linear with respect to f , it also follows that a term added to f will not change
the necessity and sufficiency of that condition.
Another special case may be worthy of consideration. Let us assume that
the integrand does not explicitly contain the x term. Then by executing the
differentiations
d
dx
(y
∂f
∂y
f)=
y
d
dx
∂f
∂y
∂f
∂x
∂f
∂y
y
=
y
(
d
dx
∂f
∂y
∂f
∂y
)
∂f
∂x
.
With the last term vanishing in this case, the differential equation simplifies to
d
dx
(y
∂f
∂y
f)=0.
Its consequence is the expression also known as Beltrami’s formula:
y
∂f
∂y
f = c
1
, (1.1)
where the right-hand side term is an integration constant. The classical prob-
lem of the brachistochrone, discussed in the next section, belongs to this class.
Finally, it is also often the case that the integrand does not contain the y
term explicitly. Then
∂f
∂y
=0
The foundations of calculus of variations 11
and the differential equation has the simpler
d
dx
∂f
∂y
=0
form. As above, the result is
∂f
∂y
= c
2
where c
2
is another integration constant. The geodesic problems, also the
subject of Chapter 8, represent this type of Euler-Lagrange equations.
We can surmise that the Euler-Lagrange differential equation’s general so-
lution is of the form
y = y(x, c
1
,c
2
),
where the c
1
,c
2
are constants of integration, and are solved from the bound-
ary conditions
y
0
= y(x
0
,c
1
,c
2
)
and
y
1
= y(x
1
,c
1
,c
2
).
1.4 Application: minimal path problems
This section deals with several classical problems to illustrate the methodol-
ogy. The problem of finding the minimal path between two points in space
will be addressed in different senses.
The first problem is simple geometry, the shortest geometric distance be-
tween the points. The second one is the well-known classical problem of the
brachistochrone, originally posed and solved by Bernoulli. This is the path
of the shortest time required to move from one point to the other under the
force of gravity.
The third problem considers a minimal path in an optical sense and leads
to Snell’s law of reflection in optics. The fourth example finds the path of
minimal kinetic energy of a particle moving under the force of gravity.
All four problems will be presented in two-dimensional space, although they
may also be posed and solved in three dimensions with some more algebraic
difficulty but without any additional instructional benefit.

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