USING THE EULER EQUATION

3. USING THE EULER EQUATION

Other Variables We have used x and y as our variables. But the mathematics is just the same if we use some other letters, for example polar coordinates r and θ. To minimize (make stationary) the integral

θ ′ = dθ/dr, we solve the Euler equation

To minimize

Notice that the first derivative in the Euler equation [d/dx in (2.16), d/dr in (3.1), d/dt in (3.2)] is with respect to the integration variable in the integral. The partial derivatives are with respect to the other variable and its derivative [y and y ′ in (2.16), θ and θ ′ in (3.1), x and ˙x in (3.2)].

Example 1. Find the path followed by a light ray if the index of refraction ( in polar coordinates) is proportional to r −2 . We want to make stationary

n ds 2 or r −2 ds = r −2 dr 2 +r 2 dθ =

r −2 1+r 2 θ ′2 dr.

Section 3 Using the Euler Equation 479

The Euler equation is then (3.1) with F = r −2 1+r 2 θ ′2 . Since ∂F/∂θ = 0, we have

d r −2 r 2 θ ′

= const. = K. dr

Solve for θ ′ and integrate (see Chapter 5, Problem 1.5): θ ′2 =K 2 (1 + r 2 θ ′2 ) so θ ′2 (1 − K 2 r 2 )=K 2 ,

θ = arc sin Kr + const.

First Integrals of the Euler Equation In some problems the integrand F in I [see equation (1.1)] does not contain y (that is, F does not contain the dependent variable). Then ∂F/∂y = 0 and the Euler equation becomes

This happened in the example and most of your problems in Section 2. Because ∂F/∂y was zero, we were able to integrate the Euler equation once; the equation ∂F/∂y ′ = const. is for this reason called a first integral of the Euler equation.

There is another less obvious case in which we can easily find a first integral of the Euler equation. Let us show this by an example (the soap film problem mentioned in Section 1).

Example 2. Our problem is this: Given two points P 1 and P 2 (not too far apart), we are going to draw a curve joining P 1 and P 2 and revolve it about the x axis to form a surface of revolution. We want the equation of the curve so that the surface area will be a minimum. That is, we want to minimize I =

ds = 1+y ′2 dx. Instead, let us write ds = 1+x ′2 dy, where x ′ = dx/dy. Then I=

1+x ′2 dy. Recall from (3.1) and (3.2) and the discussion following them how to write the Euler equation in various sets of variables. Here y is the

variable of integration, F = y 1+x ′2 , and the Euler equation is

Since ∂F/∂x = 0, (3.3) becomes

d yx ′ = 0.

dy

1+x ′2

480 Calculus of Variations Chapter 9

This is the simplified equation we wanted. We integrate once, solve for x ′ and integrate again (see Chapter 5, Problem 1.3):

x=c 1 cosh −1

+c 2 ,

c 1 x−c 2

y=c 1 cosh

c 1 Figure 3.1 The graph of this equation is called a catenary; it is shown in Figure 3.1 for the

special case c 1 = 1, c 2 = 0, y = cosh x = 1 2 (e x +e −x ). The catenary does not always give the solution to the soap film problem. If the given points (or the hoops in Figure 1.1) are too far apart, the soap film may break into two parts (circular films on the hoops). For further discussion see Courant and Robbins, Chapter 7, Section 11, and Arfken and Weber, Chapter 17. For another problem involving a catenary, see Problem 6.4.

Observe that the method used in this example will simplify any problem in which I =

′ ) dx does not have the independent variable x in the integrand. We change to y as the integration variable making the substitutions

in I. Then the integrand is a function of y and x ′ , so the Euler equation [now (3.3)] simplifies since ∂F/∂x = 0. (See also Problem 8.1.)

Example 3. Find a first integral of the Euler equation to make stationary the integral

Since x is missing in the integrand, we change to y as the integration variable; then by (3.4)

We see that ∂F/∂x = 0; from (3.3) the Euler equation is

dy ∂x

dy

y x ′2 +1

Section 3 Using the Euler Equation 481

The first integral of the Euler equation is, then,

√ = 8(x +y ). Using cylindrical coordinates, we have z = 8r ,z=r

Example 4. Find the geodesics on the cone z 2 2 2

8, dz = dr

8, so

ds 2 = dr 2 +r 2 dθ 2 + dz 2 = dr 2 +r 2 dθ 2 + 8 dr 2 = 9 dr 2 +r 2 dθ 2 . We want to minimize

9+r 2 θ ′2 dr. Note that we use r as the integration variable since the integrand contains r but

not θ. Then ∂F/∂θ = 0, and we can immediately write a first integral of the Euler equation:

d ∂F

= const. = K. dr ∂θ

We solve for θ ′ and integrate again.

From computer or tables (or see Chapter 5, Problem 1.6):

θ + α = 3 arc cos (α = const. of integration)

PROBLEMS, SECTION 3

Change the independent variable to simplify the Euler equation, and then find a first integral of it.

Write and solve the Euler equations to make the following integrals stationary. Change the independent variable, if needed, to make the Euler equation simpler.

5. 1+y 2 y ′2 dx

yy ′2

6. ′ dx

x 1 x 1 1 + yy

482 Calculus of Variations Chapter 9 Z x 2

2 +r ′ dθ, r = dr/dθ

9. θ ′2 + sin 2 θ dφ, θ − ′ p

= dθ/dφ ′ 10. s 1 s 2 +s ′2 dt, s = ds/dt

Use Fermat’s principle to find the path followed by a light ray if the index of refraction is proportional to the given function.

15. Find the geodesics on a plane using polar coordinates. 16. Show that the geodesics on a circular cylinder (with elements parallel to the z axis)

are helices az + bθ = c, where a, b, c are constants depending on the given endpoints. (Hint: Use cylindrical coordinates.) Note that the equation az + bθ = c includes the circles z = const. (for b = 0), straight lines θ = const. (for a = 0), and the special helices az + bθ = 0.

17. Find the geodesics on the cone x 2 +y 2 =z 2 . Hint: Use cylindrical coordinates. 18. Find the geodesics on a sphere. Hints: Use spherical coordinates with constant

r = a. Choose your integration variable so that you can write a first integral of the Euler equation. For the second integration, make the change of variable w = cot θ. To recognize your result as a great circle, find, in terms of spherical coordinates θ and φ, the equation of intersection of the sphere with a plane through the origin.