Isoperimetric problems — constraints (Lagrange multipliers in variational sense)

WHAT is an isoperimetric problem?

The name comes from the original problem: among all closed curves of fixed perimeter ($K=$ length $=\ell$), find the one enclosing maximum area ($J=$ area). Answer: a circle. "Iso-perimetric" = "same perimeter".

HOW to derive the rule (from first principles)

We want $\delta J=0$ for variations $\delta y$ that also keep $K$ fixed, i.e. $\delta K=0$.

Step 1 — Use a TWO-parameter family. A single $\epsilon\eta(x)$ won't work: one knob can't both vary $J$ and respect the constraint. So take

$y(x)=\bar y(x)+\epsilon_1\eta_1(x)+\epsilon_2\eta_2(x),$

with $\eta_1(a)=\eta_1(b)=\eta_2(a)=\eta_2(b)=0$ (boundary values are fixed, so the bumps vanish at the ends).

Why this step? Two parameters give us enough freedom: we'll spend one degree of freedom satisfying the constraint and still have one left to demand stationarity.

Step 2 — Reduce to ordinary calculus. Now $J$ and $K$ become ordinary functions of two numbers: $\Phi(\epsilon_1,\epsilon_2)=J[y],\qquad \Psi(\epsilon_1,\epsilon_2)=K[y].$ At $\bar y$ we sit at $\epsilon_1=\epsilon_2=0$. We want $\Phi$ stationary subject to $\Psi=\ell$. This is now a finite-dimensional constrained extremum — solved by ordinary Lagrange multipliers!

Why this step? We've converted a hard calculus-of-variations constraint into the familiar 2-variable multiplier problem we already know.

Step 3 — Apply ordinary multiplier rule. There exists $\lambda$ such that $\nabla(\Phi-\lambda\Psi)=0 \quad\text{at }(0,0),$ i.e. $\frac{\partial}{\partial\epsilon_i}\big(\Phi-\lambda\Psi\big)\Big|_{0}=0,\quad i=1,2.$

Step 4 — Compute the derivatives. Using $\dfrac{\partial f}{\partial\epsilon_i}=f_y\eta_i+f_{y'}\eta_i'$ and integrating $f_{y'}\eta_i'$ by parts (boundary terms die because $\eta_i$ vanish at ends):

$\frac{\partial\Phi}{\partial\epsilon_i}\Big|_0=\int_a^b\Big(f_y-\frac{d}{dx}f_{y'}\Big)\eta_i\,dx,$ and similarly for $\Psi$ with $g$. So the condition $\partial_{\epsilon_i}(\Phi-\lambda\Psi)=0$ reads $\int_a^b\Big[(f_y-\tfrac{d}{dx}f_{y'})-\lambda(g_y-\tfrac{d}{dx}g_{y'})\Big]\eta_i\,dx=0.$

Step 5 — Fundamental Lemma. Define $F=f-\lambda g$. The bracket is $\big(F_y-\frac{d}{dx}F_{y'}\big)$. Since $\eta_2$ is essentially arbitrary (after $\eta_1$ is used to fix the constraint), the fundamental lemma of CoV forces the bracket to vanish everywhere.

Worked Example 1 — Catenary: hanging chain of fixed length

A chain of fixed length $\ell$ hangs between two posts. It settles to minimise potential energy $\propto$ height of centre of mass, i.e. minimise $J=\int_a^b y\sqrt{1+y'^2}\,dx \quad\text{subject to length}\quad K=\int_a^b\sqrt{1+y'^2}\,dx=\ell.$

Step 1 — Augment. $f=y\sqrt{1+y'^2}$, $g=\sqrt{1+y'^2}$, so $F=f-\lambda g=(y-\lambda)\sqrt{1+y'^2}.$ Why? $F$ folds the constraint in; now we solve unconstrained.

Step 2 — Use Beltrami (since $F$ has no explicit $x$). $F-y'F_{y'}=C$. Compute $F_{y'}=(y-\lambda)\frac{y'}{\sqrt{1+y'^2}}$, so $(y-\lambda)\sqrt{1+y'^2}-(y-\lambda)\frac{y'^2}{\sqrt{1+y'^2}}=\frac{y-\lambda}{\sqrt{1+y'^2}}=C.$ Why Beltrami? It's a first integral that saves us one integration whenever $x$ is absent.

Step 3 — Solve. $y-\lambda=C\sqrt{1+y'^2}$. Separate and integrate to get $y(x)=\lambda+C\cosh\!\Big(\frac{x-x_0}{C}\Big).$ The shape is a catenary (a $\cosh$). Constants $\lambda$ (vertical shift), $C$, $x_0$ are fixed by the two endpoints and the length condition $\int\sqrt{1+y'^2}=\ell$.

Worked Example 2 — Dido's problem (maximum area, fixed perimeter)

Among curves $y(x)$ from $(-a,0)$ to $(a,0)$ of fixed arc-length $\ell$, maximise enclosed area $J=\int_{-a}^{a}y\,dx\quad\text{subject to}\quad K=\int_{-a}^{a}\sqrt{1+y'^2}\,dx=\ell.$

Step 1 — Augment. $F=y-\lambda\sqrt{1+y'^2}$. Why? $f=y$, $g=\sqrt{1+y'^2}$.

Step 2 — Beltrami ($x$ absent): $F-y'F_{y'}=C$. $F_{y'}=-\lambda\frac{y'}{\sqrt{1+y'^2}}$, so $y-\lambda\sqrt{1+y'^2}+\lambda\frac{y'^2}{\sqrt{1+y'^2}}=y-\frac{\lambda}{\sqrt{1+y'^2}}=C.$

Step 3 — Solve. $\dfrac{\lambda}{\sqrt{1+y'^2}}=y-C$. Let $u=y-C$. Then $1+y'^2=\lambda^2/u^2\Rightarrow y'^2=(\lambda^2-u^2)/u^2$, giving $\frac{u\,du}{\sqrt{\lambda^2-u^2}}=\pm dx \Rightarrow -\sqrt{\lambda^2-u^2}=\pm(x-x_0).$ Square: $(x-x_0)^2+(y-C)^2=\lambda^2$. A circular arc! Why this confirms intuition: fixed perimeter → maximum area → circle, exactly the classic isoperimetric result. The multiplier $\lambda$ turned out to be the radius.

Common Mistakes (Steel-manned)

Flashcards

Connections

• Euler–Lagrange Equation — the engine we apply to $F=f-\lambda g$.
• Beltrami Identity — first integral used in both worked examples.
• Lagrange Multipliers (finite-dimensional) — the parent idea this generalises.
• Calculus of Variations — Fundamental Lemma — justifies "bracket = 0".
• Catenary Curve / Brachistochrone Problem — sibling variational classics.
• Isoperimetric Inequality — $4\pi A\le L^2$, equality iff circle.
• Shadow Price / Envelope Theorem — interpretation of $\lambda$.

Concept Map

Hinglish (regional understanding)