THE EULER EQUATION

2. THE EULER EQUATION

Before we do the general problem, let us first do the problem of a geodesic on a plane; we shall show that a straight line gives the shortest distance between two points. (The reason for doing this is to clarify the theory; you will not do problems this way.) Our problem is to find y = y(x) which will make

I=

1+y ′2 dx

as small as possible. The y(x) which does this is called an extremal. Now we want some way to represent algebraically all the curves passing through the given end- points, but differing from the (as yet unknown) extremal by small amounts. (We

Section 2 The Euler Equation 475

assume that all the curves have continuous second derivatives so that we can carry out needed differentiations later.) These curves are called varied curves; there are infinitely many of them as close as we like to the ex- tremal. We construct a function representing these varied curves in the following way (Fig- ure 2.1). Let η(x) represent a function of x

which is zero at x 1 and x 2 , and has a contin- uous second derivative in the interval x 1 to

x 2 , but is otherwise completely arbitrary. We define the function Y (x) by the equation

(2.1) Y (x) = y(x) + ǫη(x), Figure 2.1 where y(x) is the desired extremal and ǫ is a

parameter. Because of the arbitrariness of η(x), Y (x) represents any (single-valued) curve (with continuous second derivative) you want to draw through (x 1 ,y 1 ) and (x 2 ,y 2 ). Out of all these curves Y (x) we want to pick the one curve that makes

a minimum. Now I is a function of the parameter ǫ; when ǫ = 0, Y = y(x), the desired extremal. Our problem then is to make I(ǫ) take its minimum value when ǫ = 0. In other words, we want

Differentiating (2.2) under the integral sign with respect to the parameter ǫ, we get

Differentiating (2.1) with respect to x, we get (2.5)

Y ′ (x) = y ′ (x) + ǫη ′ (x).

Then from (2.5) we have

We see from (2.1) that putting ǫ = 0 means putting Y (x) = y(x). Then substituting (2.6) into (2.4) and putting dI/dǫ equal to zero when ǫ = 0, we get

x 2 y ′ (x)η ′ (x)

We can integrate this by parts (since we assumed that η and y have continuous second derivatives). Let

u=y ′ / 1+y ′2 , dv = η ′ (x)dx.

476 Calculus of Variations Chapter 9

Then

du = dx, v = η(x),

1+y ′2 The first term is zero because η(x) = 0 at the endpoints. In the second term, recall

1+y ′2

x 1 x 1 dx

that η(x) is an arbitrary function. This means that

for otherwise we could select some function η(x) so that the integral would not be zero. Notice carefully here that we are not saying that when an integral is zero, the integrand is also zero; this is not true (as, for example 2π

0 x sin x dx = 0 shows). 2 What we are saying is that the only way x 1 f (x)η(x) dx can always be zero for

every η(x) is for f (x) to be zero. You can prove this by contradiction in the following way. If f (x) is not zero, then, since η(x) is arbitrary, choose η to be positive where

f is positive and negative where f is negative. Then f η is positive, so its integral is not zero, in contradiction to the statement that Integrating (2.8) with respect to x, we get

y ′ = const.

1+y ′2

or y ′ = const. Thus the slope of y(x) is constant, so y(x) is a straight line as we expected.

Now we could go through this process with every calculus of variations problem. It is much simpler to do the general problem once for all and find a differential equation which we can use to solve later problems. The problem is to find the y which will make stationary the integral

where F is a given function. The y(x) which makes I stationary is called an extremal whether I is a maximum or minimum or neither. The method is the one we have just used with the straight line. We consider a set of varied curves

Y (x) = y(x) + ǫη(x)

just as before. Then we have

and we want (d/dǫ)I(ǫ) = 0 when ǫ = 0. Remembering that Y and Y ′ are functions of ǫ, and differentiating under the integral sign with respect to ǫ, we get

x dI 2 dY

∂F dY ′

dx.

dǫ

x 1 ∂Y dǫ

∂Y dǫ

Section 2 The Euler Equation 477

Substituting (2.1) and (2.5) into (2.11), we have

We want dI/dǫ = 0 at ǫ = 0; recall that ǫ = 0 means Y = y. Then (2.12) gives

Assuming that y ′′ is continuous, we can integrate the second term by parts just as in the straight-line problem:

η(x) dx. The integrated term is zero as before because η(x) is zero at x 1 and x 2 . Then we

η(x) dx = 0.

As before, since η(x) is arbitrary, we must have

Euler equation

dx ∂y ′

∂y

This is the Euler (or Euler-Lagrange) equation. Any problem in the calculus of variations, then, is solved by setting up the integral which is to be stationary, writing what the function F is, substituting it into the Euler equation, and solving the resulting differential equation.

Example. Let’s find the geodesics in a plane again, this time using the Euler equation as you will do in problems. We are to minimize

1+y ′2 dx,

so we have F = 1+y ′2 . Then

and the Euler equation gives

as we had in (2.8).

478 Calculus of Variations Chapter 9

PROBLEMS, SECTION 2

Write and solve the Euler equations to make the following integrals stationary. In solving the Euler equations, the integrals in Chapter 5, Section 1, may be useful.

7. R x 2 x 1 e x p1 + y ′2 dx Hint: In the last integration, let u = e x and see Chapter 5, Problem 1.6. Z x 2 p