h

’L +L

The potential energy of the system is

L

y 1 + (y )2 dx + const.

P.E. = mgy = ρg (1.19)

’L

6 CHAPTER 1. CALCULUS OF VARIATIONS

Here the constant refers to the unchanging potential energy of the vertically

hanging cable and the cable on the horizontal surface. Notice that the tension

in the cable is being tacitly determined by the weight of the vertical segments.

The Euler-Lagrange equations coincide with those of the soap ¬lm, so

(x + a)

y = κ cosh (1.20)

κ

where we have to ¬nd κ and a. We have

h = κ cosh(’L + a)/κ,

= κ cosh(L + a)/κ, (1.21)

so a = 0 and h = κ cosh L/κ. Setting t = L/κ this reduces to

h

t = cosh t. (1.22)

L

By considering the intersection of the line y = ht/L with y = cosh t we see

that if h/L is too small there is no solution (the weight of the suspended

cable is too big for the tension supplied by the dangling ends) and once h/L

is large enough there will be two possible solutions.

y

y= cosh t

y= ht/L

t=L/κ

Intersection of y = ht/L with y = cosh t.

Further investigation will show that only one of these is stable.

Example: The Brachistochrone. This problem was posed as a challenge by

Johann Bernoulli in 1696. He asked what shape should a wire with endpoints

1.2. FUNCTIONALS 7

(0, 0) and (a, b) take in order that a frictionless bead will slide from rest down

the wire in the shortest possible time (βρ±χισ„ o‚: shortest, χρoνo‚: time).

x

g

(a,b)

y

When presented with an ostensibly anonymous solution, Johann made his

famous remark: Tanquam ex unguem leonem 1 , ” meaning that he recognized

that the author was Isaac Newton.

Johann gave a solution himself, but that of his brother Jacob Bernoulli

was superior and Johann tried to pass it o¬ as his. This was not atypical.

Johann later misrepresented the publication date of his book on hydraulics

to make it seem that he had priority in this ¬eld over his own son, Daniel

Bernoulli.

We begin our solution of the problem by observing that the total energy

1 1

E = m(x2 + y 2) ’ mgy = mx2 (1 + y 2) ’ mgy,

™ ™ ™ (1.23)

2 2

of the bead will be constant. From the initial condition we see that this

constant is zero. We therefore wish to minimize

1+y2

1

T a a

T= dt = dx = dx (1.24)

x

™ 2gy

0 0 0

so as ¬nd y(x), given that y(0) = 0 and y(a) = b. The Euler-Lagrange

equation is

1

yy + (1 + y 2 ) = 0. (1.25)

2

Again this looks intimidating, but we can use the same trick of multiplying

through by y to get

1 1d

yy + (1 + y 2 ) = y(1 + y 2) = 0.

y (1.26)

2 2 dx

1

I recognize the lion by his clawmark.

8 CHAPTER 1. CALCULUS OF VARIATIONS

Thus

2c = y(1 + y 2 ). (1.27)

This has a parametric solution

x = c(θ ’ sin θ),

y = c(1 ’ cos θ), (1.28)

(as you should verify) and the solution is a cycloid.

x

(0,0)

(x,y)

θ

θ

(a,b)

y

A wheel rolls on the x axis. The dot, which is ¬xed to the rim of the wheel,

traces out a cycloid.

The parameter c is determined by requiring that the curve does in fact pass

through the point (a, b).

1.2.3 First Integral

How did we know that we could simplify both the soap-¬lm problem and

the brachistochrone by multiplying the Euler equation by y ? The answer

is that there is a general principle, closely related to energy conservation in

mechanics, that tells us when and how we can make such a simpli¬cation. It

works when the f is of the form f (y, y ), i.e. has no explicit dependence on

x. In this case the last term in

df ‚f ‚f ‚f

=y +y + (1.29)

dx ‚y ‚y ‚x

is absent, and we have

d ‚f ‚f ‚f ‚f d ‚f

f ’y ’y ’y

=y +y

dx ‚y ‚y ‚y ‚y dx ‚y

1.3. LAGRANGIAN MECHANICS 9

‚f d ‚f