δJ

x2

= δy(x) dx. (1.2)

δy(x)

x1

Here δy(x) ≡ ·(x), and the quantity

δJ ‚f d ‚f

≡ ’ (1.3)

δy(x) ‚y dx ‚y

1.2. FUNCTIONALS 3

is called the functional (or Fr´chet) derivative of J with respect to y(x). We

e

can think of it as a kind of generalization of the notion of a partial derivative

‚J/‚yi , with the discrete subscript “i” on y being replaced by a continuous

label, “x”. Thus

‚J δJ

x2

δyi ’

δJ = dx δy(x). (1.4)

‚yi δy(x)

x1

i

The condition for the functional to be stationary under variations y ’

y + δy is

δJ ‚f d ‚f

’

= = 0, (1.5)

δy(x) ‚y dx ‚y

and this is usually called the Euler-Lagrange equation.

If the functional depends on more than one function y, then stationarity

under all possible variations requires one equation

δJ ‚f d ‚f

’

= =0 (1.6)

δyi (x) ‚yi dx ‚yi

for each function yi (x).

If the function depends on higher derivatives, y , y (3), etc., then we have

to integrate by parts more times, and we end up with

d2 d3

δJ ‚f d ‚f ‚f ‚f

’ ’3 + ···.

= +2 (1.7)

‚y (3)

δy(x) ‚y dx ‚y dx ‚y dx

1.2.2 Examples

Now we apply our new derivative to solve some simple problems.

Soap ¬lm supported by a pair of coaxial rings.

y(x)

x1 x

x2

4 CHAPTER 1. CALCULUS OF VARIATIONS

Here we wish to minimize the free energy of the ¬lm, which is equal to twice

(once for each liquid-air interface) the surface tension σ of the soap solution

times the area of the ¬lm. We therefore need to minimize

x2

y 1 + y 2 dx.

J[y] = 4σπ (1.8)

x1

with y(x1 ) = y1 and y(x2 ) = y2 . We form the partial derivatives

‚f ‚f 4πσyy

= 4πσ 1 + y 2 , = (1.9)

‚y ‚y 1+y2

and thus write down the Euler-Lagrange equation

«

d yy

1+y 2’ = 0. (1.10)

dx 2

1+y

Performing the indicated derivative with respect to x gives

(y )2 y(y )2 y

yy

2

1+y ’ ’ + = 0. (1.11)

(1 + y 2 )3/2

2 2

1+y 1+y

Collecting terms, this is

1 yy

’ = 0. (1.12)

(1 + y 2 )3/2

2

1+y

This di¬erential equation looks a tri¬‚e intimidating. To simplify, we multiply

by y to get

y yy y

’

0=

(1 + y 2 )3/2

1+y2

«

d y

= . (1.13)

dx 1+y2

The solution to the minimization problem therefore reduces to solving

y

= κ, (1.14)

2

1+y

1.2. FUNCTIONALS 5

where κ is an as yet undetermined integration constant. Fortunately this

non-linear, ¬rst order, di¬erential equation is elementary. We write it as

y2

dy

’1

= (1.15)

κ2

dx

and separate variables

dy

dx = . (1.16)

y2

’1

κ2

We now make the natural substitution y = κ cosh t, whence

dx = κ dt. (1.17)

Thus we ¬nd that x + a = κt, leading to

x+a

y = κ cosh . (1.18)

κ

We select κ and a to ¬t the endpoints y(x1 ) = y1 and y(x2 ) = y2 .

Heavy Chain over Pulleys. We cannot yet consider the form of a hanging

chain of ¬xed length, but we can solve a simpler problem of a heavy cable

draped over a pair of pulleys located at x = ±L, y = h, and with the excess

cable resting on a horizontal surface.