12.4 Constraints, Lagrange multipliers, and isoperimetric problems

Many variational problems come with a side condition. The hanging chain minimises its potential energy — but only among curves of a fixed length. Dido’s problem asks for the maximum area enclosed by a curve of fixed perimeter. These are constrained extremisation problems, and the tool for them is the same one that handles constrained minimisation in ordinary calculus: the Lagrange multiplier.

The multiplier idea, recalled

To extremise a function $f(\mathbf x)$ subject to a constraint $g(\mathbf x) = 0$ , one does not solve the constraint and substitute; one extremises the combination $f - \lambda g$ freely, treating the number $\lambda$ as an extra unknown fixed at the end by the constraint. Geometrically, at a constrained extremum the gradient of $f$ is parallel to the gradient of $g$ — there is no way to move along the constraint surface that changes $f$ to first order — and $\lambda$ is the constant of proportionality, $\nabla f = \lambda\nabla g$ . The calculus of variations inherits this wholesale, with functionals in place of functions.

Isoperimetric problems

The characteristic constrained problem fixes the value of one functional while extremising another:

\text{extremise}\quad J[y] = \int_a^b F\,dx \qquad\text{subject to}\qquad K[y] = \int_a^b G\,dx = \ell \;(\text{fixed}).

These are called isoperimetric problems, after the original: fixed perimeter ( $K$ = length), maximal area ( $J$ ). The recipe mirrors the ordinary case exactly.

▶ Extremise F − λG freely Derivation

Perturb $y \to y + \varepsilon\eta$ as before. The catch is that $\eta$ is no longer fully arbitrary — it must preserve the constraint, $\delta K = 0$ , which couples the allowed perturbations. The multiplier untangles them. Form the combination functional

J^*[y] \;=\; J[y] - \lambda\,K[y] \;=\; \int_a^b \big(F - \lambda G\big)\,dx,

with $\lambda$ a constant to be determined. One shows (the functional version of $\nabla f = \lambda\nabla g$ ) that the constrained extremal of $J$ is an unconstrained extremal of $J^*$ for the right value of $\lambda$ . So the constrained problem is solved by the ordinary Euler–Lagrange equation applied to the modified integrand $F - \lambda G$ :

\frac{\partial (F-\lambda G)}{\partial y} - \frac{d}{dx}\frac{\partial (F-\lambda G)}{\partial y'} = 0.

Solve it; the solution carries $\lambda$ as a parameter, and the constraint $K[y] = \ell$ then pins $\lambda$ down. One extra unknown, one extra equation.

where

$J[y]$: the functional being extremised (e.g. area, energy)
$K[y] = \ell$: the constraint functional held fixed (e.g. length)
$\lambda$: Lagrange multiplier — an unknown constant, fixed at the end by the constraint

The hanging chain, properly

Now the catenary of 12.3 as it really is. A chain of fixed length $\ell$ hangs between two points; it settles to minimise its gravitational potential energy $J = \int y\,ds = \int y\sqrt{1+y'^2}\,dx$ , subject to the fixed-length constraint $K = \int \sqrt{1+y'^2}\,dx = \ell$ .

▶ The constrained chain is still a catenary Derivation

Apply the recipe to $F - \lambda G = (y-\lambda)\sqrt{1+y'^2}$ . This is exactly the surface-of-revolution integrand of 12.3 with $y$ shifted to $y - \lambda$ , and it is again independent of $x$ , so Beltrami gives

\frac{y-\lambda}{\sqrt{1+y'^2}} = a \quad\Longrightarrow\quad y = \lambda + a\cosh\!\left(\frac{x-b}{a}\right).

The same hyperbolic cosine, now with three constants $\lambda, a, b$ — fixed by the two endpoints and the length $\ell$ . The multiplier $\lambda$ enters as a vertical shift, which is physically the tension scale of the chain. The soap film (unconstrained area) and the hanging chain (constrained energy) reach the same curve by different routes — a small marvel that the multiplier method makes transparent.

Dido’s problem and the isoperimetric inequality

The problem that named the class: among all closed curves of a given perimeter $L$ , which encloses the greatest area? Applying the multiplier method to “maximise area subject to fixed perimeter” yields an extremal of constant curvature — a circle. This is the isoperimetric inequality,

A \;\le\; \frac{L^2}{4\pi},

with equality only for the circle. The same variational fact explains why soap bubbles are spherical (least area enclosing a fixed volume — the three-dimensional version) and why raindrops and stars are round: surface energy is minimised, at fixed content, by the sphere.

⏳ The history — Queen Dido's oxhide

The problem is older than the calculus that solves it. Virgil’s Aeneid tells of Dido, fleeing to the North African coast, granted as much land as she could enclose with a single oxhide. She cut the hide into a fine thread and laid it out to bound the greatest possible area — founding Carthage on the ground within. Mathematically she needed the curve of maximal area for a fixed perimeter, the isoperimetric problem, and the optimal answer (a circle, or a semicircle against the straight coastline) was proved rigorously only in the nineteenth century by Steiner and Weierstrass. The legend gives the whole family of constrained-extremum problems its name.

Pointwise constraints

A second kind of constraint holds at every point rather than as a single integral — a bead confined to a wire $g(x,y)=0$ , a system on a surface. These holonomic constraints are handled by a multiplier that is itself a function $\lambda(x)$ rather than a constant, adjoined pointwise. In mechanics this is how constraint forces enter the Lagrangian formulation: the multiplier function is the constraint force (the normal force from the wire, the tension in the rod), delivered automatically by the variational machinery. That is one of the quiet powers of the method the next lesson builds on — constraints and their forces fall out of a single scalar principle rather than being added by hand.

The final lesson collects everything into the variational formulation of physics: Hamilton’s principle, the Euler–Lagrange equations of mechanics and of fields, and Noether’s theorem.