Lift the multiplier trick to functions
In finite dimensions the way to optimise under a constraint is
Lagrange
multipliers: to extremise f subject to
g = c you introduce a number \lambda and
extremise f - \lambda g freely. The calculus of variations inherits
the trick wholesale — the only change is that the "variables" are now the values of a function
at its continuum of points.
Step 1 — combine into one Lagrangian. Fold the constraint into the integrand
with a single multiplier \lambda:
L^\ast = L + \lambda G.
Step 2 — vary freely. Apply the ordinary
Euler–Lagrange
equation to the combined Lagrangian L^\ast, as if there
were no constraint at all:
\frac{\partial L^\ast}{\partial y} - \frac{d}{dx}\frac{\partial L^\ast}{\partial y'} = 0.
Step 3 — pin \lambda with the constraint. The
solution carries the unknown \lambda; the requirement
\int G\,dx = C is the extra equation that fixes its value.
To extremise \int L\,dx subject to
\int G\,dx = C:
-
introduce a Lagrange multiplier \lambda and
extremise the augmented functional \int (L + \lambda G)\,dx
freely;
-
the extremal solves the Euler–Lagrange equation for
L^\ast = L + \lambda G;
-
the constraint \int G\,dx = C then determines
\lambda — exactly the finite-dimensional recipe, lifted to
functions.
It's tempting to skip the multiplier: just solve the plain, unconstrained
Euler–Lagrange equation for L alone, then hope the constraint
happens to hold — or fudge a leftover constant of integration until it does. That does not
work, because the constraint changes the shape of the extremal, not merely its
constants.
Take the hanging chain: minimising \int y\,ds with
no length constraint at all has an obvious answer — pile the whole chain onto
the floor, as low as possible. That is nothing like the graceful catenary
y = a\cosh(x/a). The catenary only appears once the fixed-length
requirement is woven directly into the functional as
L + \lambda G and the Euler–Lagrange equation is solved for
that. Skip the multiplier and you are solving a different problem entirely.
Two famous answers
Dido's problem. Of all closed curves of a fixed perimeter, which encloses the
greatest area? Here L builds the area and
G builds the perimeter; extremising
L + \lambda G forces the boundary to have constant curvature. The
answer is the circle — the same shape that, legend says, Queen Dido enclosed
with an oxhide to found Carthage.
The hanging chain. A flexible chain of fixed length hangs between two posts and
settles into the shape that minimises its gravitational potential energy
\int y\,ds, subject to a fixed length
\int ds = \ell. Both integrands are functions of
y and y' only, so after folding in
\lambda the Beltrami
identity applies, and the curve that comes out is the catenary:
y = a\cosh\!\left( \frac{x}{a} \right).
It looks like a parabola but isn't — it is the hyperbolic cosine, and the constant
a (set by the chain's length and the multiplier) controls how sharply
it sags.
Galileo guessed a hanging chain was a parabola; he was close, but wrong. The true curve,
named the catenary from the Latin catena ("chain"), was found in 1691 by
Huygens, Leibniz and Johann Bernoulli, responding to a challenge from Jacob Bernoulli. The
difference matters: turn a catenary upside down and it becomes the ideal self-supporting arch,
carrying its own weight in pure compression — which is exactly why Gaudí hung chains as
model "arches" and why the St. Louis Gateway Arch is a flattened catenary, not a parabola.