Proof of the Euler Equation (first order condition) for the
Calculus of Variations
Assume that you have the following functional that is to be maximized
b
V   f ( x, dx / dt, t )dt
a
We can use the notation dx / dt  x to simplify our exposition. We also assume,
as in all calculus of variations problems, that x(a) = xa and x(b) = xb where xa and
xb are fixed and known constants.
Suppose also that we have the integral constraint
b
Bo   g ( x, x, t )dt
a
for some fixed constant Bo.
Our goal is to find the first order condition (necessary condition) for a maximum.
That is, we are looking for a condition that will occur when the best x(t) has been
chosen and maximizes V subject to the constraint. Remember that we are
choosing a whole function x(t) and not simply one variable as in usual calculus.
The solution is to introduce a new function which we may call  (t ) , where
 (t ) is ANY differentiable function with  ( a )  0 and  (b)  0 . We then add
the  (t ) to an assumed optimal x(t) which we can call x*(t) and therefore our
functional becomes
b
V * ( )   f ( x*  (t ), x*  (t ), t )dt
a
while our constraint becomes
b
Bo   g ( x*  (t ), x*  (t ), t )dt .
a
Now suppose we choose the value of  to maximize V ( ) subject to our
constraint. This is just a simple calculus problem of choosing the maximizing
value for  . What is more, we know the answer is that the optimal  = 0, since
x*(t) is the optimal x(t) by assumption.
Therefore, we form the Lagrangian
b
L(  ) =  f ( x*  (t ), x*  (t ), t )dt
a
b
+ λ{ Bo   g ( x*   , x*   (t ), t )dt }
a
and differentiate with respect to  .
After differentiating with respect to  and setting equal to zero, we get
b f
b g
f
g
{  (  (t )  (t ))dt}    (  (t )  (t ))dt}  0
x
x
a x
a x
and rearranging terms, we can write
b f
b f
g
g
(


)

(
t
)
dt

(   )(t )dt  0
 x

x
x
a
a x
The second integral can be rewritten, using integration by parts, and therefore
the above becomes
b f
g
g
d f
 {( x   x )  dt ( x   x )} (t )dt  0
a
Now, since  (t ) can be any differentiable function such that  (a)   (b)  0 , it
follows that
(
g
g
f
d f
λ ) ( λ ) 0.
x
 x dt  x
 x
which is the first order condition known as the Euler Equation.
Example #1: A Straight Line Minimizes the Distance between Two
Points
Suppose that you have two point and you want to draw a curve that minimizes
the distance between the two points. This curve is obviously a straight line.
We can show this simple fact using the calculus of variations as a simple example
of how to use the Euler Equation above.
The graph above shows that the length of the curve connecting the two points is
exactly equal to
b
b
L   1  ( x) 2 dt   f ( x, x, t )dt
a
a
If we seek to minimize this, then we have a very clear calculus of variations
problem with no side constraint. That is, g  0.
The Euler Equation becomes
(
g
g
f
d f
f d f
 λ )  (  λ )  (  ( ))  0
x
 x dt  x
 x
 x dt  x
Therefore, the first order condition is
d
(
dt

x
1  (x ) 2
)0
which implies
( x) 2  C(1  ( x) 2 )
and thus renaming the constants,
x  C
1
or
x(t )  C  C t
2
1
To determine the constants C1 and C2 we use x(a) = xa and x(b) = xb. This leads
to
x(t )  (
ax  bx
x x
a )  ( a b )t
b
a b
a b
This is just the equation of a straight line. A straight line minimizes the distance
between two points.
Example #2: Maximizing the Area Enclosed by a Rope
Suppose that you drove two stakes in the ground and attached a rope to these
two stakes. Now imagine that the two stakes are on a horizontal axis. The
issue we are concerned with is -- how can we lay the rope to maximize the area
between the rope and the horizontal axis? Here is a graph to help visualize the
problem.
To prove this use Green's theorem coupled with a line integral defining the
region to be maximized. This kind of problem is called an isoperimetric problem.