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1.2 The Legendre test
The Euler-Lagrange dierential equation just introduced represents a necessary,
but not sucient, condition for the solution of the fundamental variational
problem.
The alternative functional of
I() =
Z x1
x0
f(x; Y; Y 0)dx;
may be expanded as
I() =
Z x1
x0
f(x; y + (x); y0 + 0(x))dx:
Assuming that the f function has continuous partial derivatives, the meanvalue
theorem is applicable:
f(x; y + (x); y0 + 0(x)) = f(x; y; y0)+
((x)
@f(x; y; y0)
@y
+ 0(x)
@f(x; y; y0)
@y0 ) + O(2):
By substituting we obtain
I() =
Z x1
x0
f(x; y; y0)dx+

Z x1
x0
((x)
@f(x; y; y0)
@y
+ 0(x)
@f(x; y; y0)
@y0 )dx + O(2):
With the introduction of
I1 = 
Z x1
x0
((x)
@f(x; y; y0)
@y
+ 0(x)
@f(x; y; y0)
@y0 )dx;
we can write
I() = I(0) + I1 + O(2);
where I1 is called the rst variation. The vanishing of the rst variation is a
necessary, but not sucient, condition to have an extremum. To establish a
sucient condition, assuming that the function is thrice continuously dierentiable,
we further expand as
I() = I(0) + I1 + I2 + O(3):
© 2009 by Taylor & Francis Group, LLC
8 Applied calculus of variations for engineers
Here the newly introduced second variation is
I2 =
2
2
Z x1
x0
(2(x)
@2f(x; y; y0)
@y2 +
2(x)0(x)
@2f(x; y; y0)
@y@y0 +
02(x)
@2f(x; y; y0)
@y02 )dx:
We now possess all the components to test for the existence of the extremum
(maximum or minimum). The Legendre test in [5] states that if independently
of the choice of the auxiliary (x) function
- the Euler-Lagrange equation is satised,
- the rst variation vanishes (I1 = 0), and
- the second variation does not vanish (I2 6= 0)
over the interval of integration, then the functional has an extremum. This
test manifests the necessary conditions for the existence of the extremum.
Specically, the extremum will be a maximum if the second variation is negative,
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 nally introduce the variation of the function as
y = Y (x) 􀀀 y(x) = (x);
and the variation of the derivative as
y0 = Y 0(x) 􀀀 y0(x) = 0(x):
Based on these variations, we distinguish between the following cases:
- strong extremum occurs when y is small, however, y0 is large, while
- weak extremum occurs when both y and y0 are small.
On a nal note: the above considerations did not ever state the nding or
presence of an absolute extremum; only the local extremum in the interval of
the integrand is obtained.
© 2009 by Taylor & Francis Group, LLC
The foundations of calculus of variations 9
1.3 The Euler-Lagrange dierential equation
Let us expand the derivative in the second term of the Euler-Lagrange dierential
equation as follows:
d
dx
@f
@y0 =
@2f
@x@y0 +
@2f
@y@y0 y0 +
@2f
@y02 y00:
This demonstrates that the Euler-Lagrange equation is usually of second order.
@f
@y 􀀀
@2f
@x@y0 􀀀
@2f
@y@y0 y0 􀀀
@2f
@y02 y00 = 0:
The above form is also called the extended form. Consider the case when the
multiplier of the second derivative term vanishes:
@2f
@y02 = 0:
In this case f must be a linear function of y0, in the form of
f(x; y; y0) = p(x; y) + q(x; y)y0:
For this form, the other derivatives of the equation are computed as
@f
@y
=
@p
@y
+
@q
@y
y0;
@f
@y0 = q;
@2f
@x@y0 =
@q
@x
;
and
@2f
@y@y0 =
@q
@y
:
Substituting results in the Euler-Lagrange dierential 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)
© 2009 by Taylor & Francis Group, LLC
10 Applied calculus of variations for engineers
whose total dierential is of the form
du = p(x; y)dx + q(x; y)dy = f(x; y; y0)dx:
The functional may be evaluated as
I(y) =
Z x1
x0
f(x; y; y0)dx =
Z x1
x0
du = u(x1; y1) 􀀀 u(x0; y0):
It follows from this that the necessary and sucient condition for the solution
of the Euler-Lagrange dierential equation is that the integrand of the
functional be the total dierential with respect to x of a certain function of
both x and y.
Considering furthermore, that the Euler-Lagrange dierential equation is
linear with respect to f, it also follows that a term added to f will not change
the necessity and suciency 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
dierentiations
d
dx
(y0 @f
@y0 􀀀 f) =
y0 d
dx
@f
@y0 􀀀
@f
@x 􀀀
@f
@y
y0 =
y0(
d
dx
@f
@y0 􀀀
@f
@y
) 􀀀
@f
@x
:
With the last term vanishing in this case, the dierential equation simplies to
d
dx
(y0 @f
@y0 􀀀 f) = 0:
Its consequence is the expression also known as Beltrami's formula:
y0 @f
@y0 􀀀 f = c1; (1.1)
where the right-hand side term is an integration constant. The classical problem
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
© 2009 by Taylor & Francis Group, LLC
The foundations of calculus of variations 11
and the dierential equation has the simpler
d
dx
@f
@y0 = 0
form. As above, the result is
@f
@y0 = c2
where c2 is another integration constant. The geodesic problems, also the
subject of Chapter 7, represent this type of Euler-Lagrange equations.
We can surmise that the Euler-Lagrange dierential equation's general solution
is of the form
y = y(x; c1; c2);
where the c1; c2 are constants of integration, and are solved from the boundary
conditions
y0 = y(x0; c1; c2)
and
y1 = y(x1; c1; c2):