Visual walkthrough — Isoperimetric problems — constraints (Lagrange multipliers in variational sense)

Step 1 — What are we actually optimising over?

WHAT. In ordinary calculus you slide a point $x$ along a line to find the lowest spot of a curve $h(x)$. In the calculus of variations you slide a whole curve $y(x)$ around to find the "best" curve — the one that makes a number called a functional as small (or large) as possible.

A functional is a machine: feed it an entire function, it spits out one number. We write

$J[y]={\int }_a^{b}f(x,y,y^{\prime })dx.$

• $y$ — the whole curve you are choosing (the input to the machine).
• $y^{\prime }$ — its slope $\frac{dy}{dx}$ at each point (how steep the curve is there).
• $f(x,y,y^{\prime })$ — a little rule that scores the curve locally, at each $x$.
• ${\int }_a^b\dots dx$ — adds up all those local scores from $x=a$ to $x=b$ into one total.
• $J[y]$ — the single output number. The square brackets $[\cdot ]$ are a reminder: the input is a function, not a number.

WHY. Before touching constraints we must see the arena clearly: we are searching a space of curves, and each curve is a single point in that abstract space.

PICTURE. Look at the figure. Left: ordinary calculus slides a dot on a hill. Right: variations slide an entire curve; a dial on the side reads the value $J[y]$.

Step 2 — What does a constraint look like in curve-space?

WHAT. We now add a second machine, a second functional,

$K[y]={\int }_a^{b}g(x,y,y^{\prime })dx=\ell .$

• $g$ — a second local scoring rule (for the catenary it will measure arc-length).
• $\ell$ — a fixed number we are not allowed to change (the budget: e.g. the length of the chain).
• $K[y]=\ell$ — the rule "the total must equal exactly $\ell$."

WHY. We may not choose any curve — only curves whose $K$-value is exactly $\ell$. In the space of all curves, this equation carves out a constraint surface: the set of curves that spend exactly the budget $\ell$.

PICTURE. Curve-space drawn as a plane. The grey shading is all curves; the cyan surface is the sliver where $K[y]=\ell$. We are only ever allowed to walk on the cyan surface. Free Euler–Lagrange would happily walk off it — that is the whole problem.

Step 3 — One knob is not enough

WHAT. To move a curve we nudge it. The standard nudge uses one bump function $\eta (x)$ and one small number $ϵ$:

$y(x)=\bar{y}(x)+ϵ\eta (x).$

• $\bar{y}$ — the candidate curve we are testing (the "centre").
• $\eta (x)$ — a chosen wiggle, forced to be $0$ at both ends ($\eta (a)=\eta (b)=0$) so the endpoints stay pinned.
• $ϵ$ — a dial: turn it and the wiggle grows or shrinks. At $ϵ=0$ we are back at $\bar{y}$.

WHY. With one dial $ϵ$, both $J$ and $K$ become functions of that single number. But we have two demands: keep $K$ fixed and make $J$ stationary. One dial cannot serve two masters — turning it to keep $K=\ell$ generally leaves $J$ still changing. We are stuck.

PICTURE. A single slider $ϵ$ drags the curve. The value $K$ (amber) drifts away from $\ell$ as soon as we move — we cannot both hold the budget and hunt for the best $J$.

Step 4 — Two knobs: one to obey the budget, one to hunt

WHAT. Use two independent bumps and two dials:

$y(x)=\bar{y}(x)+{ϵ}_1{\eta }_1(x)+{ϵ}_2{\eta }_2(x),$

with ${\eta }_1,{\eta }_2$ both vanishing at the ends. Now the two machines collapse into two ordinary two-variable functions:

$\Phi ({ϵ}_1,{ϵ}_2)=J[y],\Psi ({ϵ}_1,{ϵ}_2)=K[y].$

• $\Phi$ — the value of $J$ as an ordinary function of the two dial-settings.
• $\Psi$ — the value of $K$, likewise. At $({ϵ}_1,{ϵ}_2)=(0,0)$ we sit at $\bar{y}$.

WHY. With two dials we have exactly enough freedom: spend one degree of freedom keeping $\Psi =\ell$, and still have one left over to demand $\Phi$ is stationary. Crucially, we have now shrunk an infinite-dimensional curve problem into an ordinary two-variable constrained-extremum problem — the kind finite-dimensional Lagrange multipliers already solve.

PICTURE. The $({ϵ}_1,{ϵ}_2)$ plane. The amber curve is $\Psi =\ell$ (the budget, obeyed). Along it, contour lines of $\Phi$ (cyan) show where $J$ is bigger or smaller. We must find the point on the amber curve where $\Phi$ stops changing.

Step 5 — The multiplier appears (gradients line up)

WHAT. From ordinary Lagrange multipliers: a point on the constraint curve $\Psi =\ell$ makes $\Phi$ stationary exactly when the two gradients are parallel:

$\nabla \Phi =\lambda \nabla \Psi ,\text{i.e.}\nabla (\Phi -\lambda \Psi )=0.$

• $\nabla \Phi$ — arrow pointing in the direction $\Phi$ increases fastest (in the $ϵ$-plane).
• $\nabla \Psi$ — arrow pointing where the budget $K$ increases fastest; it is perpendicular to the amber constraint curve.
• $\lambda$ — one single number, the Lagrange multiplier: the ratio of the two gradients' lengths.

WHY. If $\nabla \Phi$ had any component along the amber curve, we could slide that way and improve $\Phi$ while staying on budget — not stationary yet. Stationarity means $\nabla \Phi$ points purely across the curve, i.e. parallel to $\nabla \Psi$. That parallel condition is the birth of $\lambda$.

PICTURE. At the winning point the cyan arrow $\nabla \Phi$ and amber arrow $\nabla \Psi$ point the same way (one is $\lambda \times$ the other). At a losing point they are misaligned — a red slide-arrow along the curve still improves $\Phi$.

Step 6 — Turn the two conditions back into an integral

WHAT. Write out $\nabla (\Phi -\lambda \Psi )=0$ component by component. Because $\Phi ={\int }_a^{b}fdx$ depends on ${ϵ}_i$ only through $y=\bar{y}+{ϵ}_1{\eta }_1+{ϵ}_2{\eta }_2$, we differentiate under the integral sign:

$\frac{\partial \Phi }{\partial {ϵ}_i}={\int }_a^b(f_y{\eta }_i+f_{y^{\prime }}{\eta }_i^{\prime })dx.$

Now integrate the $f_{y^{\prime }}{\eta }_i^{\prime }$ piece by parts (the boundary terms die because ${\eta }_i(a)={\eta }_i(b)=0$). Introduce right here the augmented integrand

$F=f-\lambda g,$

so that combining ${\partial }_{{ϵ}_i}\Phi -\lambda {\partial }_{{ϵ}_i}\Psi =0$ collapses the two integrands $f$ and $g$ into the single $F$, giving for $i=1,2$:

${\int }_a^b\left[\underset{F_y-\frac{d}{dx}{F}_{y^{\prime }}}{\underbrace{(f_y-\frac{d}{dx}{f}_{y^{\prime }})-\lambda (g_y-\frac{d}{dx}{g}_{y^{\prime }})}}\right]{\eta }_{i}dx=0.$

• $\frac{\partial \Phi }{\partial {ϵ}_i}=\int (\dots )dx$ — the derivative of the whole integral, not of $f$ alone; that is what "differentiate under the integral sign" means (justified by the box above).
• $f_y$ — how the local score changes if the curve rises a bit; $f_{y^{\prime }}$ — how it changes if the slope changes.
• $\frac{d}{dx}{f}_{y^{\prime }}$ — the Euler–Lagrange "bookkeeping" term left over from integration by parts.
• $F=f-\lambda g$ — the augmented integrand; the whole bracket is exactly $F_y-\frac{d}{dx}{F}_{y^{\prime }}$.
• ${\eta }_i$ — our bump function, still riding along inside the integral.

WHY. Integration by parts is the tool that removes ${\eta }_i^{\prime }$ and leaves only ${\eta }_i$ multiplying everything — because only then can we invoke the lemma in Step 7. We use it precisely to trade a derivative-of-the-wiggle for a plain wiggle. And we name $F$ now, the instant the bracket appears, so nothing is used before it is defined. (The by-parts step also needs $f_{y^{\prime }}$ to be $C^1$ in $x$ so that $\frac{d}{dx}{f}_{y^{\prime }}$ exists — guaranteed by the $C^2$ assumption above.)

PICTURE. The by-parts move drawn as a see-saw: the derivative hops off ${\eta }_i^{\prime }$ and lands on $f_{y^{\prime }}$, while the endpoint terms (shown in faded amber) vanish because the wiggle is pinned to zero at both posts.

Step 7 — The Fundamental Lemma snaps the bracket to zero

WHAT. We first make precise how one bump function is consumed by the budget. Staying on the constraint means $\Psi ({ϵ}_1,{ϵ}_2)=\ell$. Since $\Psi (0,0)=\ell$ already, Taylor-expand $\Psi$ about the origin:

$\Psi ({ϵ}_1,{ϵ}_2)-\ell =\frac{\partial \Psi }{\partial {ϵ}_1}{ϵ}_1+\frac{\partial \Psi }{\partial {ϵ}_2}{ϵ}_2+O({ϵ}^2).$

Setting $\Psi =\ell$ and keeping only first order (the $O({ϵ}^2)$ terms are negligible for the infinitesimal variations that define stationarity) gives the first-order necessary condition for staying on budget:

$\frac{\partial \Psi }{\partial {ϵ}_1}{ϵ}_1+\frac{\partial \Psi }{\partial {ϵ}_2}{ϵ}_2=0.$

This single equation ties ${ϵ}_1$ to ${ϵ}_2$: pick ${ϵ}_2$ freely and ${ϵ}_1$ is forced. In bump-language we need $\partial \Psi /\partial {ϵ}_1={\int }_a^b(g_y-\frac{d}{dx}{g}_{y^{\prime }}){\eta }_{1}dx\ne 0$ so that ${\eta }_1$ can spend itself enforcing the budget.

Now apply the lemma to the free direction ${\eta }_2$. The Fundamental Lemma of the Calculus of Variations says: if ${\int }_a^{b}B(x)\eta (x)dx=0$ for every smooth $\eta$ vanishing at the ends, then $B(x)=0$ everywhere. With $B=F_y-\frac{d}{dx}{F}_{y^{\prime }}$ (the bracket named in Step 6), it must vanish:

$\boxed{F_y-\frac{d}{dx}{F}_{y^{\prime }}=0,F=f-\lambda g}$

• $F=f-\lambda g$ — the augmented integrand introduced in Step 6: maximise-target minus $\lambda$ times budget-cost.
• The boxed equation is just the ordinary Euler–Lagrange equation, run on $F$ instead of $f$.
• $\lambda$ is exactly the number that made the two directions' contributions cancel, i.e. the multiplier from the linearised constraint above.

WHY. The linearisation shows precisely why we needed two bumps: one to satisfy $\Psi =\ell$ (via $\partial \Psi /\partial {ϵ}_1$), one to remain free for the lemma. And an integral that is zero against every free wiggle can only mean the multiplied function is zero — otherwise pick a bump concentrated where it is positive and the integral would be positive.

PICTURE. Suppose the bracket $B(x)$ were positive on some interval (amber hump). Choose a wiggle $\eta$ (cyan) that is a bump exactly there and zero elsewhere. Then $\int B\eta dx>0$ — a contradiction. So $B$ must be flat zero.

Step 8 — Counting unknowns (why nothing is left dangling)

WHAT. The boxed ODE is second order, so its general solution carries two constants $C_1,C_2$, plus the still-unknown $\lambda$. That is three unknowns. We have exactly three conditions:

$y(a)=y_a,y(b)=y_b,{\int }_a^{b}gdx=\ell .$

• The two boundary conditions pin the endpoints.
• The constraint equation ${\int }_a^{b}gdx=\ell$ is what finally determines $\lambda$ — the price of the budget.

WHY. A common fear is that $\lambda$ is a loose end. It is not: the budget equation is one extra equation, and it is exactly the one that fixes $\lambda$. Books that "set $\lambda$ to anything" quietly break this count — see the parent's Mistake box.

PICTURE. A ledger: three unknown columns ($C_1$, $C_2$, $\lambda$) balanced against three condition rows (two posts + one budget). Every unknown has a home.

Step 9 — Edge and degenerate cases

WHAT & WHY. The rule quietly assumes a few things; here is what happens when they fail.

• $\lambda =0$ (constraint is slack / non-binding). Then $F=f$ and we recover the free Euler–Lagrange solution. This happens when the unconstrained optimum already satisfies $\int g=\ell$ by luck — the budget wasn't fighting us. By the shadow-price reading $\lambda =\frac{d{J}^{\ast }}{d\ell }$, a zero price means "one more unit of budget buys no improvement."
• The abnormal case ($\nabla \Psi =0$, i.e. $G\equiv 0$). If $\bar{y}$ is itself a stationary point of $K$ (its constraint-gradient vanishes identically, so no ${\eta }_1$ from Step 7 exists), the gradients cannot be made parallel by any $\lambda$; the tidy $\nabla \Phi =\lambda \nabla \Psi$ rule needs a more general $({\mu }_0\nabla \Phi =\lambda \nabla \Psi )$ form. Rare, but real.
• Pointwise vs integral constraint. If the constraint is holonomic, $g(x,y)=0$ for every $x$ (not just its integral), one constant $\lambda$ is far too few — you need a whole function $\lambda (x)$, one price per point. Isoperimetric (integral) constraints get one constant $\lambda$; this is the single most common confusion.

PICTURE. Three mini-panels: (a) the constraint contour passes right through the free optimum → $\lambda =0$; (b) $\nabla \Psi =0$, arrows can't align (abnormal); (c) integral budget (one price) vs pointwise budget (a price at every $x$).

The one-picture summary

Everything above, compressed: we ride the amber budget-surface $K[y]=\ell$; among curves on it we seek where the cyan value-contours $J[y]$ stop changing; that happens where the gradients align, which births $\lambda$; feeding $F=f-\lambda g$ into Euler–Lagrange and closing with the budget equation delivers the extremal — a catenary for the hanging chain, a circular arc for Dido's fixed-perimeter problem.

Quick self-checks: