Derivation of Euler's equation

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Derivation of Euler’s equation

Problem. Derive Euler’s equation

an equation that provides a necessary condition for a function y = y(x) to give the least value to the integral

Assume that function F(x, y, yʹ) is twice continuously differentiable with respect to the argument yʹ for all values of this argument and with respect to the arguments x and y in some region R of the x-y plane. It is assumed that we will always remain in the interior of region R.

The function

2) y = y(x)

is assumed to be continuously differentiable on the segment x₁ ≤ x ≤ x₂. Geometrically the function y = y(x) can be represented in the x-y plane as a curve m over the interval [x₁, x₂]. See Fig. 1.

Given a set M of functions of type 2) meeting certain specified conditions, our problem is to find that particular function in the set for which the integral I(y) has the least value. In the calculus of variations those functions meeting the specified conditions are usually called admissible for comparison. In general, the conditions include fixed boundary values. In our case the set of admissible functions are defined by two requirements:

1. y = f(x) is continuously differentiable on the segment [x₁, x₂].

2. At the ends of the segment y = y(x) has the values

y(x₁) = y₁, y(x₂) = y_{2 .}

Otherwise the function y(x) can be completely arbitrary. We are thus considering all possible curves over the interval [x₁, x₂] which pass through the two points A(x₁,y₁) and B(x₂,y₂) that can be represented by equation 2).

Let us now suppose that y = y(x) is that function that we are seeking, the function for which I(y) is a minimum. Consider now the family of neighboring functions y* = y + αη(x) where η(x) is any fixed function of x with η(x₁) = η(x₂) = 0 and α is any real number. We may regard

as a function of α alone with a minimum at α = 0. Thus

Using Leibniz’s Rule for differentiating an integral with respect to a parameter under an integral sign, we compute the derivative of 3) with respect to α as

For the situation where α = 0 we have

This last equation must be satisfied for every continuously differentiable function η(x) which vanishes at the end points of the segment [x₁, x₂]. In order to put equation 6) in the form we want we now transform the integral of the second term by the method of integration by parts:

The first term on the right disappears since η vanishes by hypothesis at both ends of the interval of integration. Thus 6) becomes

We now assert that the function

must be identically zero i.e. for 8) to be true it must be zero for all values of x in the interval [x₁, x₂]. Suppose this is not true. Suppose that at some point x = c of the interval it is, say, positive. Being a continuous function, it must remain positive throughout a certain neighborhood of c. Let η be chosen so as also to be positive in this neighborhood, but zero elsewhere in the interval. Then the integral could not be zero, so we have a contradiction.

We have the following lemma:

Lemma 1. If the following conditions are met a function f(x) is identically zero:

1. f(x) is continuous in the closed interval [a, b]

for an arbitrary function η(x) with continuous first derivatives

3. η(a) = η(b) = 0

We thus assert the following theorem:

Euler’s theorem. A necessary condition for a function y(x) to minimize the integral I(y) is that