Euler–Lagrange equation
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Under ideal conditions, the maxima and minima of a given function may be located by finding the points where its derivative vanishes (i.e., is equal to zero). By analogy, solutions of smooth variational problems may be obtained by solving the associated Euler–Lagrange equation.
Consider the functional
where and where and are constants.
The function should have at least one derivative in order to satisfy the requirements for valid application of the function; further, if the functional attains its local minimum at and is an arbitrary function that has at least one derivative and vanishes at the endpoints and , then we must have
for any number ε close to 0. Therefore, with the first variation of A must vanish,
- .
Since is a function of and ,
- .
Therefore,
where we have used the chain rule in the second line and integration by parts in the third. The last term in the third line vanishes because at the end points. Finally, according to the fundamental lemma of calculus of variations, we find that will satisfy the Euler–Lagrange equation
In general this gives a second-order ordinary differential equation which can be solved to obtain the extremal . The Euler–Lagrange equation is a necessary, but not sufficient, condition for an extremal. Sufficient conditions for an extremal are discussed in the references.
In order to illustrate this process, consider the problem of finding the shortest curve in the plane that connects two points and . The arc length is given by
with
and where , , and .
for any choice of the function . We may interpret this condition as the vanishing of all directional derivatives of in the space of differentiable functions, and this is formalized by requiring the Fréchet derivative of to vanish at . If we assume that has two continuous derivatives (or if we consider weak derivatives), then we may use integration by parts:
with the substitution
then we have
but the first term is zero since was chosen to vanish at and where the evaluation is taken. Therefore,
for any twice differentiable function that vanishes at the endpoints of the interval.
We can now apply the fundamental lemma of calculus of variations: If
for any sufficiently differentiable function within the integration range that vanishes at the endpoints of the interval, then it follows that is identically zero on its domain.
Therefore,
It follows from this equation that
and hence the extremals are straight lines.