Functionals

An ordinary function eats a number and hands back a number: f(3) = 9. A functional is one level up — it eats a whole function and hands back a single number. We write it with square brackets, J[y], to remind ourselves the input is a function y(x), not a point.

f : \text{number} \longmapsto \text{number}, \qquad J : \text{function} \longmapsto \text{number}.

Feed in one curve and you get back one value; feed in a different curve and you get back a different value. The calculus of variations is the study of how that value responds as you flex the whole curve.

Three functionals you already half-know

Most functionals are built from a definite integral: the integral sweeps along the curve and accumulates one running total, so the whole function collapses to a single number.

Arc length. The length of the curve y(x) from x = a to x = b sums the tiny hypotenuses \sqrt{dx^2 + dy^2} along it:

L[y] = \int_a^b \sqrt{1 + y'^2}\,dx.

Area under a curve. The signed area swept out is itself a functional of y:

A[y] = \int_a^b y\,dx.

Time of descent (the brachistochrone). A bead slides without friction down a wire bent into the shape y(x) from a high start to a lower end. Energy conservation makes its speed v = \sqrt{2gy} at drop y, and the travel time is distance over speed, accumulated:

T[y] = \int_a^b \frac{\sqrt{1 + y'^2}}{\sqrt{2gy}}\,dx.

Each of these takes a shape and returns a number — a length, an area, a time.

The central question

For an ordinary function we hunt for the input that makes the output stationary — where the derivative is zero, the language of the gradient pointing nowhere. The calculus of variations asks the very same question one level up:

Given a functional J[y] = \int_a^b L(x, y, y')\,dx over curves with fixed endpoints:

For arc length the answer is the one your intuition already shouts: the shortest curve between two points is the straight line. The machinery that proves it — and cracks the brachistochrone too — is the Euler–Lagrange equation.

Why is this harder than ordinary optimization? When you minimise an ordinary function you wiggle one number and watch the output. When you minimise a functional you must wiggle the height of the curve at every one of its infinitely many points at once and keep the total stationary against all of those wiggles simultaneously. That leap — from a finite list of variables to a continuum of them — is exactly what the next pages tame, and it is why the subject sits at the foundation of mechanics, optics and general relativity: nature, again and again, chooses the path that makes some functional stationary.

The single easiest way to trip over this topic is to slide back into ordinary-calculus habits without noticing. Keep the two ideas firmly apart:

If you catch yourself asking "what x minimises this?" for a problem that is really asking "what shape minimises this?", you've slipped back a level. The brachistochrone, arc length and the soap-film problems are never asking for a best number — they are asking for a best curve, drawn from an infinite gallery of candidate curves. That reframing — trading a handful of unknowns for a whole unknown function — is the one conceptual leap the entire subject of calculus of variations asks of you.

In June 1696, the Swiss mathematician Johann Bernoulli published a challenge to "the sharpest mathematicians of the world": a bead slides, frictionless, down a wire from a high point A to a lower point B that is not directly below it. Gravity pulls it down and along. What shape should the wire be bent into so the bead arrives at B in the least possible time?

Almost everyone's first guess is the straight line — the shortest path between the two points. It's wrong. A curve that dips more steeply at the start lets the bead pick up speed early, more than making up for the extra distance travelled. Bernoulli named the puzzle the brachistochrone problem, from the Greek for "shortest time." Newton, Leibniz, Bernoulli's own brother Jacob, and Bernoulli himself all raced to solve it — legend has it Newton, then in his fifties and running the Royal Mint, solved it overnight after receiving the challenge by post, and sent his solution back anonymously (Bernoulli reportedly recognised the author at once, remarking that he knew the lion "by his claw"). The general method that fell out of solving problems like this one — hunting for the curve, not the number, that makes some quantity stationary — is exactly the machinery built on the next page. It's fair to say this single puzzle is the problem that invented the calculus of variations as a field.

Worked example: plugging in an actual curve

A functional only becomes a genuine number once you commit to one specific curve. Let's do that twice, for the area functional A[y] = \int_0^2 y\,dx.

Try y(x) = x^2. Substitute it straight into the integral and integrate:

A[x^2] = \int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}.

Now try y(x) = 3 (a flat, constant curve at height 3) over the same interval:

A[3] = \int_0^2 3\,dx = \big[3x\big]_0^2 = 6.

Same functional, same interval — but two different curves went in, and two different numbers came out: 8/3 and 6. That's the whole mechanism in miniature: a functional is a machine that swallows a shape whole and returns a single verdict on it. Once you can picture A[y] as a dial that moves as the curve y flexes, "find the curve that makes the dial smallest" stops sounding mysterious — it's the same kind of question as "find the number that makes f(x) smallest," just asked about shapes instead of numbers.

Watch a functional change

Here is one curve between two fixed endpoints, (0,0) and (2,0): a straight chord plus a bump, y = s\,\sin(\pi x / 2). The slider s flexes the bump up and down, and the readout shows the value of the arc-length functional L[y] = \int_0^2 \sqrt{1 + y'^2}\,dx live. Slide all the way to s = 0: the curve becomes the straight chord, and the length bottoms out — the straight line minimises the functional, exactly as the theorem promises.