A challenge thrown across Europe
In 1696 Johann Bernoulli posed a public dare: between two points at different heights, find the shape of wire down which a bead slides — under gravity, without friction — in the least time. Not the straight line, not the obvious arc, but the genuinely fastest curve. This is the brachistochrone (Greek for "shortest time"). The trick is that the quantity you minimise is not a number but a whole curve, and the descent time is a functional — a machine that eats a function y(x) and returns a number. That single shift, from optimising over numbers to optimising over functions, is the seed of the entire field.
How do you even write the time down? Recall from Volume I that arc length over a tiny step is ds = sqrt(1 + (dy/dx)^2) dx, built from the derivative y' = dy/dx. Energy conservation gives the speed v = sqrt(2 g y) after dropping a height y. Time is distance over speed, so the total descent time is the definite integral T[y] = integral of sqrt(1 + y'^2) / sqrt(2 g y) dx. We now want the y(x) that makes this integral as small as possible — exactly the kind of problem the Euler–Lagrange machinery was built for.
The Beltrami shortcut
All three classic problems share a lucky feature: the integrand F(y, y') contains no explicit x. When that happens you do not need the full second-order Euler–Lagrange equation. Instead the Beltrami identity gives a first integral for free: F - y' (partial F / partial y') = constant. It is the variational echo of energy conservation — when the rules of the game do not depend on where you are along x, a conserved quantity drops out, much as a time-independent law conserves energy. This is your first whisper of Noether's theorem.
Full equation: d/dx ( dF/dy' ) - dF/dy = 0 If F has no explicit x (F = F(y, y')): Beltrami: F - y' * (dF/dy') = C (a constant) This turns a 2nd-order ODE into a 1st-order one.
Solving the brachistochrone
Feed F = sqrt(1 + y'^2) / sqrt(y) into Beltrami (drop the constant 2g, it only rescales). After tidying, the conserved quantity says y (1 + y'^2) = constant. Solving this first-order relation, the curve that emerges is a cycloid — the path traced by a point on the rim of a rolling wheel. Parametrically x = r(theta - sin theta), y = r(1 - cos theta). The answer is not a parabola or a circular arc; it dives steeply at the start to build speed, then flattens out. Nature's fastest slide is a rolling-wheel curve.
- Write the descent time as the functional T[y] = integral of sqrt(1 + y'^2) / sqrt(2 g y) dx.
- Notice F has no explicit x, so apply the Beltrami identity instead of the full equation.
- Simplify the first integral to y (1 + y'^2) = constant, a separable first-order ODE.
- Substitute y = r(1 - cos theta); the integral resolves into the cycloid x = r(theta - sin theta).
The hanging chain and the shortest path
Next, the catenary: hang a flexible chain from two nails and ask what shape it settles into. Physically it minimises gravitational potential energy, which means minimising the height-weighted length integral of y sqrt(1 + y'^2) dx — but only among curves of a fixed total length. That length constraint makes it an isoperimetric problem, handled with a Lagrange multiplier for functionals. Apply Beltrami again and the solution is the hyperbolic cosine: y = a cosh(x/a). The catenary is famously NOT a parabola, though the two look deceptively similar near the bottom — a classic misconception worth retiring.
Now the geodesic: the shortest path between two points on a surface. On a flat plane the length functional integral of sqrt(1 + y'^2) dx is minimised by a straight line — fall through to Euler–Lagrange and you get y'' = 0, exactly as expected. On a curved surface the same idea, run through the geodesic equation, bends the answer: on a sphere the shortest routes are great circles, which is why long flights arc poleward. Geodesics are where the calculus of variations hands the baton to differential geometry and, eventually, to general relativity.
Soap films and honest caveats
Dip two coaxial rings into soapy water and the film between them is a minimal surface — it minimises area because surface tension pulls it taut. Spinning the area functional through Euler–Lagrange (with x absent again, so Beltrami applies) yields the catenoid, the surface you get by revolving a catenary y = a cosh(x/a). The same hyperbolic cosine governs both the minimal surface of a soap film and the hanging chain — a quiet, beautiful coincidence that is really the same variational structure wearing two costumes.
One more honesty: a stationary solution need not exist or be unique. Stretch the two rings of the soap-film problem too far apart and the catenoid solution simply fails — the film snaps to two flat disks instead. The Euler–Lagrange condition is necessary, not a magic existence guarantee, and these classic problems are precious precisely because they show both the method's power and the places where you must check the answer against physical sense.