Isoperimetric problems

In one sentence

An isoperimetric problem extremises one functional subject to another being fixed, and the method is a Lagrange multiplier applied to functionals rather than to functions.

Why the exam asks it

Because the extension from ordinary Lagrange multipliers is exact and short, and because the classical answer — the circle — is a fact with a clean statement that can be tested against plausible alternatives.

The idea

Extremise $J[y] = \displaystyle\int_a^b F(x,y,y')\,dx$ subject to $K[y] = \displaystyle\int_a^b G(x,y,y')\,dx = \ell$, with $\ell$ fixed.

Form $H = F + \lambda G$ and solve the Euler–Lagrange equation for $H$. The constant $\lambda$ is an unknown, determined at the end by imposing the constraint.

That is the whole method. The count works out: the Euler–Lagrange equation gives two constants, the boundary conditions fix them, and $\lambda$ is fixed by the constraint — three conditions, three unknowns.

The multiplier is a genuine unknown, not a parameter to choose. Solving for $y$ in terms of $\lambda$ and then imposing $K[y]=\ell$ is the order of operations, and reversing it is the common error.

The classical problem

Among closed curves of given perimeter $L$, which encloses the greatest area? The answer is the circle, with area $L^2/(4\pi)$.

Comparison makes the size of the gap visible. A square of the same perimeter encloses $L^2/16$, and since $4\pi \approx 12.57 < 16$, the circle wins. The isoperimetric inequality

$$4\pi A \le L^2$$

holds for every closed curve, with equality only for the circle. The ratio $4\pi A/L^2$ is $1$ for a circle and strictly less for anything else, which is the cleanest way to state it.

Dido's problem is the variant with a straight boundary available: maximise area against a straight coastline with a curve of given length. The answer is a semicircle, and recognising which variant is being asked is usually the question.

Degenerate cases

If the constraint functional's own Euler–Lagrange equation is satisfied by the extremal — that is, the constraint is itself stationary — the multiplier method degenerates and needs the more general form $\mu F + \lambda G$ with $(\mu,\lambda)\ne(0,0)$. Rare in exam problems, worth recognising as the reason the general statement carries $\mu$.

Several constraints

With constraints $K_1 = \ell_1, \dots, K_m = \ell_m$, use $F + \sum\lambda_i G_i$ — one multiplier per constraint, exactly as in finite dimensions.

Note the contrast with holonomic constraints of the form $g(x,y)=0$, which do not involve an integral. Those are handled by a multiplier function $\lambda(x)$ rather than a constant, and mixing the two up is a standard trap.

Where intuition breaks

"Choose $\lambda$ to make the algebra convenient." $\lambda$ is determined by the constraint. Solve for $y$ with $\lambda$ carried through, then impose $K[y]=\ell$.

"The square is optimal because it is symmetric." $4\pi < 16$, so the circle encloses more with the same perimeter. Symmetry alone does not decide; the isoperimetric ratio does.

"The isoperimetric inequality is an approximation." It is exact, with equality precisely for circles. The ratio $4\pi A/L^2$ characterises them.

"A constraint $\int G = \ell$ and a constraint $g(x,y)=0$ are handled the same way." The first takes a constant multiplier, the second a function $\lambda(x)$. Different problems.

"More constraints need one multiplier." One per constraint.

"The multiplier method gives a maximum." Stationary, as always. Whether it is a maximum needs a separate argument, and for the classical problem the answer is known rather than derived from the second variation.

The exam's angle

1. Write $H = F+\lambda G$ immediately and apply the same Euler–Lagrange machinery.
2. Check which variable is absent from $H$ — the shortcuts still apply, and usually do.
3. Carry $\lambda$ through and impose the constraint last.
4. Recognise the classical answers: circle for the closed isoperimetric problem, semicircle for Dido, catenary for a hanging chain of fixed length.
5. Distinguish integral constraints from pointwise ones — constant multiplier against multiplier function.

The night before

• Extremise $J$ subject to $K=\ell$: solve Euler–Lagrange for $F+\lambda G$, then fix $\lambda$ from the constraint.
• Isoperimetric inequality $4\pi A \le L^2$, equality only for the circle. Circle $L^2/(4\pi)$ against square $L^2/16$, and $4\pi<16$.
• Dido: straight boundary available $\Rightarrow$ semicircle.
• Hanging chain of fixed length $\Rightarrow$ catenary.
• One multiplier per integral constraint; a multiplier function for a pointwise constraint $g(x,y)=0$.
• The result is stationary; optimality is a separate argument.