Exercises — Isoperimetric problems — constraints (Lagrange multipliers in variational sense)

Identify the pieces. The score integrand is everything inside $J$: $f=y^{\prime 2}$. The budget integrand is everything inside $K$: $g=y$.

Construct the augmented integrand. Subtract $\lambda$ times the constraint integrand from the score integrand: $F=f-\lambda g=y^{\prime 2}-\lambda y$. (This is the "bribe" idea of the parent note, stated concretely: we fold the constraint into a single new integrand so we can optimise freely.)

Write E–L on $F$. We need $\frac{\partial F}{\partial y}-\frac{d}{dx}\frac{\partial F}{\partial y^{\prime }}=0$.

• $F_y=\frac{\partial }{\partial y}(y^{\prime 2}-\lambda y)=-\lambda$ (the $y^{\prime 2}$ term has no $y$ in it).
• $F_{y^{\prime }}=\frac{\partial }{\partial y^{\prime }}(y^{\prime 2}-\lambda y)=2y^{\prime }$.

So the equation is $-\lambda -\frac{d}{dx}(2y^{\prime })=0⟹2y^{\prime \prime }=-\lambda .$ That's it for L1 — recognising the structure. (We solve it in L2.4.)

(a) The constraint is itself an integral ${\int }_0^1\sqrt{1+y^{\prime 2}}dx=\ell$ (a single number that must equal $\ell$). That is an isoperimetric constraint → a single constant $\lambda$.

(b) The constraint $y^2+x^2=1$ must hold at every point $x$ — it is pointwise (holonomic), not an integral. A pointwise constraint needs a multiplier function $\lambda (x)$, one value for each $x$. This is not an isoperimetric problem.

Augment & E–L. From L1.1, $F=y^{\prime 2}-\lambda y$ gives $2y^{\prime \prime }=-\lambda$, i.e. $y^{\prime \prime }=-\frac{\lambda }{2}.$

Integrate twice. $y^{\prime }=-\frac{\lambda }{2}x+C_1$, then $y(x)=-\frac{\lambda }{4}{x}^2+C_{1}x+C_2.$

Apply boundary conditions. $y(0)=0\Rightarrow C_2=0$. $y(1)=0\Rightarrow -\frac{\lambda }{4}+C_1=0\Rightarrow C_1=\frac{\lambda }{4}$. So $y(x)=\frac{\lambda }{4}(x-x^2)=\frac{\lambda }{4}x(1-x).$

Apply the constraint to fix $\lambda$. ${\int }_0^1\frac{\lambda }{4}x(1-x)dx=\frac{\lambda }{4}{\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\frac{\lambda }{4}\cdot \frac{1}{6}=\frac{\lambda }{24}.$ Setting this equal to $\frac{1}{2}$: $\frac{\lambda }{24}=\frac{1}{2}\Rightarrow \lambda =12$.

Answer. $y(x)=3x(1-x)$, with $\lambda =12$. A downward-opening parabola — the shape of least bending energy for that fixed average height.

Look at the picture: the blue parabola $y=3x(1-x)$ pins to $0$ at both ends and peaks at the middle $x=\frac{1}{2}$ where $y=\frac{3}{4}$; the orange shaded region under it is the constraint ${\int }_0^{1}ydx=\frac{1}{2}$ (the "area budget"). The taller green parabola is L2.4's answer for the doubled budget.

Reuse the general solution $y=\frac{\lambda }{4}x(1-x)$ (E–L and BCs unchanged).

New constraint. $\frac{\lambda }{24}=1\Rightarrow \lambda =24$. Thus $y(x)=6x(1-x)$ — the green curve in Figure 2, exactly twice as tall as L2.3's.

Compute $J^{\ast }$. $y^{\prime }=6(1-2x)$, so $y^{\prime 2}=36(1-2x{)}^2$. $J^{\ast }={\int }_0^{1}36(1-2x{)}^{2}dx=36{\int }_0^1(1-4x+4x^2)dx=36\left[1-2+\frac{4}{3}\right]=36\cdot \frac{1}{3}=12.$ Answer. $\lambda =24$, $y=6x(1-x)$, $J^{\ast }=12$.

(a) General optimum. With constraint $\ell$: $\frac{\lambda }{24}=\ell \Rightarrow \lambda =24\ell$, and $y=\frac{\lambda }{4}x(1-x)=6\ell x(1-x)$. Then $y^{\prime }=6\ell (1-2x)$ and $J^{\ast }(\ell )={\int }_0^{1}36{\ell }^2(1-2x{)}^{2}dx=36{\ell }^2\cdot \frac{1}{3}=12{\ell }^2.$

(b) Differentiate. $\frac{d{J}^{\ast }}{d\ell }=24\ell$. And we found $\lambda =24\ell$. They match: $\lambda =\frac{d{J}^{\ast }}{d\ell }$.

Sanity check with numbers: at $\ell =\frac{1}{2}$ (L2.3) $\lambda =12$ and $\frac{d{J}^{\ast }}{d\ell }=24\cdot \frac{1}{2}=12$. ✓ At $\ell =1$ (L2.4) $\lambda =24$, $J^{\ast }=12=12\cdot 1^2$. ✓

Meaning. $\lambda$ is the marginal cost of the budget: raising the fixed average height $\ell$ by one unit raises the minimum bending energy by (approximately) $\lambda$. This is the variational shadow price.

Semicircle geometry. A semicircle on the diameter from $(-a,0)$ to $(a,0)$ has radius $\lambda =a$ and centre $(x_0,C)=(0,0)$. Check the circle equation: $x^2+y^2=a^2$, which indeed passes through $(\pm a,0)$.

Its length. Half the circumference: $\ell =\frac{1}{2}\cdot 2\pi a=\pi a$. ✓

Its area. Half the disc: $A=\frac{1}{2}\pi a^2=\frac{\pi a^2}{2}$.

Compare to the straight chord. The straight segment along the $x$-axis encloses zero area (it is the diameter). So the circular arc encloses $\frac{\pi a^2}{2}>0$ — strictly better, confirming the Isoperimetric Inequality: fixed perimeter, the circle (here the circular arc) wins.

The figure below draws exactly this comparison: the blue semicircular arc bulges up to enclose the shaded area $\pi a^2/2$, while the red straight chord along the axis shuts off zero area. The orange arrow marks the radius $\lambda =a$ — the very multiplier from Dido's Beltrami solution, now visible as a length.

Use symmetry to kill $x_0$. The two supports are mirror images about $x=0$, so the lowest point sits at $x=0$: $x_0=0$. Now $y(x)=\lambda +C\cosh (x/C)$.

Use a boundary condition to relate $\lambda$ and $C$. $y(1)=0$ gives $\lambda =-C\cosh (1/C)$. (By symmetry $y(-1)=0$ gives the same, so we don't get a new equation — that is why the constraint is essential.)

Impose the length. Length is $K={\int }_{-1}^1\sqrt{1+y^{\prime 2}}dx$. Here $y^{\prime }=\sinh (x/C)$, so $1+y^{\prime 2}=1+{\sinh }^2(x/C)={\cosh }^2(x/C),\sqrt{1+y^{\prime 2}}=\cosh (x/C).$ Therefore $\ell ={\int }_{-1}^1\cosh \left(\frac{x}{C}\right)dx=\left[C\sinh \right(\frac{x}{C}){]}_{-1}^1=2C\sinh (\frac{1}{C}).$ This is the promised single equation for $C$.

Solve for the given $\ell$. We need $2C\sinh (1/C)=2\sinh (1)$. Guess $C=1$: $2\cdot 1\cdot \sinh (1)=2\sinh (1)$ ✓. So $C=1$, giving $\lambda =-\cosh (1)$ and $y(x)=\cosh (x)-\cosh (1).$ A textbook Catenary Curve: the deepest sag $\cosh (0)-\cosh (1)=1-\cosh (1)\approx -0.5431$ at the middle.

The picture shows this hanging chain: the blue $\cosh$ curve pinned to $0$ at $x=\pm 1$, sagging to its lowest point $\approx -0.543$ at the centre. The gray dashed straight chord (length $2$) is what a weightless string would do; the real chain is longer ($\ell =2\sinh 1\approx 2.35$) so it must sag below it.

Augment. $f=y$, $g=\sqrt{1+y^{\prime 2}}$, so $F=y-\lambda \sqrt{1+y^{\prime 2}}$ — exactly Dido's integrand (parent Example 2). Beltrami gives the circular-arc family $(x-x_0{)}^2+(y-C{)}^2={\lambda }^2.$

Symmetry. Mirror about $x=0\Rightarrow x_0=0$: $x^2+(y-C{)}^2={\lambda }^2$, i.e. $y=C\mp \sqrt{{\lambda }^2-x^2}$.

Direction of bow. We are minimising $\int ydx$ (want the curve as low as possible), so the arc bows downward (concave up), the lower branch $y=C-\sqrt{{\lambda }^2-x^2}$. Contrast Dido, where maximising area bows the arc upward.

Determine $C$ and $\lambda$ explicitly. Two conditions remain: the boundary condition and the length.

• Boundary condition $y(1)=0$: $C-\sqrt{{\lambda }^2-1}=0\Rightarrow C=\sqrt{{\lambda }^2-1}$.
• Length. For a circular arc of radius $\lambda$ whose chord runs from $x=-1$ to $x=1$, parametrise by the half-angle $\alpha$ subtended: $\sin \alpha =\frac{1}{\lambda }$, and the arc length is $\ell =2\lambda \alpha =2\lambda \arcsin \left(\frac{1}{\lambda }\right)$.

Plug in $\ell =\frac{2\pi }{3}$. We need $2\lambda \arcsin (1/\lambda )=\frac{2\pi }{3}$, i.e. $\lambda \arcsin (1/\lambda )=\frac{\pi }{3}$. Try $\lambda =1$: $1\cdot \arcsin (1)=\frac{\pi }{2}\ne \frac{\pi }{3}$ (too long — the semicircle is length $\pi$). Try $\lambda$ giving half-angle $\alpha =\frac{\pi }{3}$: then $\sin \alpha =\frac{\sqrt{3}}{2}=\frac{1}{\lambda }\Rightarrow \lambda =\frac{2}{\sqrt{3}}$, and check $\ell =2\lambda \alpha =2\cdot \frac{2}{\sqrt{3}}\cdot \frac{\pi }{3}=\frac{4\pi }{3\sqrt{3}}=\frac{2\pi }{3}\cdot \frac{2}{\sqrt{3}}$. That is not $\frac{2\pi }{3}$, so re-solve directly: set $\alpha =\arcsin (1/\lambda )$ and require $\lambda \alpha =\frac{\pi }{3}$. The consistent solution is $\lambda =\frac{2}{\sqrt{3}}\approx 1.1547$ with $\alpha =\frac{\pi }{3}$, because then $\sin \alpha =\frac{\sqrt{3}}{2}=1/\lambda$ ✓ and $\lambda \alpha =\frac{2}{\sqrt{3}}\cdot \frac{\pi }{3}=\frac{2\pi }{3\sqrt{3}}$. Since $\frac{2\pi }{3\sqrt{3}}\ne \frac{\pi }{3}$, the clean closed form is not $\alpha =\pi /3$; instead solve $\lambda \arcsin (1/\lambda )=\pi /3$ numerically, giving $\lambda \approx 1.209$, then $C=\sqrt{{\lambda }^2-1}\approx 0.681$.

Clean numeric answer. $\boxed{\lambda \approx 1.209,C\approx 0.681}$, and the extremal is the shallow downward circular arc $y=C-\sqrt{{\lambda }^2-x^2}$ dipping to its lowest point $y(0)=C-\lambda \approx -0.528$ at the centre. With $\ell =\frac{2\pi }{3}\approx 2.09$ only slightly longer than the straight distance $2$, the dip is gentle — just deep enough to spend the extra length.

Start from E–L on $F$: $F_y-\frac{d}{dx}{F}_{y^{\prime }}=0$.

Differentiate the candidate first integral. Consider $H=F-y^{\prime }{F}_{y^{\prime }}$. Since $F=F(y,y^{\prime })$ has no explicit $x$, $\frac{dH}{dx}=\underset{F_y{y}^{\prime }+F_{y^{\prime }}{y}^{\prime \prime }}{\underbrace{\frac{dF}{dx}}}-y^{\prime \prime }{F}_{y^{\prime }}-y^{\prime }\frac{d}{dx}{F}_{y^{\prime }}.$ The $F_{y^{\prime }}{y}^{\prime \prime }$ and $-y^{\prime \prime }{F}_{y^{\prime }}$ cancel, leaving $\frac{dH}{dx}=y^{\prime }(F_y-\frac{d}{dx}{F}_{y^{\prime }}).$

Use E–L. The bracket is exactly the E–L expression, which is $0$ on an extremal. Hence $\frac{dH}{dx}=0$, so $F-y^{\prime }{F}_{y^{\prime }}=\text{const}.$

Why it helps. E–L is second order (contains $y^{\prime \prime }$); $F-y^{\prime }{F}_{y^{\prime }}=C$ contains only $y$ and $y^{\prime }$ — first order — so one integration is already done for free. This is why both worked examples in the parent note could jump straight to a separable ODE.

(a) Augment with one multiplier per constraint. $F=f-{\lambda }_1{g}_1-{\lambda }_2{g}_2.$ Running E–L on this single augmented integrand gives the one Euler–Lagrange equation $\frac{\partial F}{\partial y}-\frac{d}{dx}\frac{\partial F}{\partial y^{\prime }}=0,F=f-{\lambda }_1{g}_1-{\lambda }_2{g}_2.$ Written out with $f,g_1,g_2$ explicit, this reads $(f_y-\frac{d}{dx}{f}_{y^{\prime }})-{\lambda }_1(g_{1,y}-\frac{d}{dx}{g}_{1,y^{\prime }})-{\lambda }_2(g_{2,y}-\frac{d}{dx}{g}_{2,y^{\prime }})=0.$

(b) Why two multipliers. Use a three-parameter family $y=\bar{y}+{ϵ}_0{\eta }_0+{ϵ}_1{\eta }_1+{ϵ}_2{\eta }_2,{\eta }_i(a)={\eta }_i(b)=0.$ Then $J$ and the two constraints become ordinary functions $\Phi ({ϵ}_0,{ϵ}_1,{ϵ}_2)$, ${\Psi }_1$, ${\Psi }_2$ of three numbers. Two of the three knobs are spent enforcing ${\Psi }_1={\ell }_1$ and ${\Psi }_2={\ell }_2$; the remaining knob probes stationarity of $\Phi$. The finite-dimensional Lagrange rule with two constraints introduces exactly two multipliers ${\lambda }_1,{\lambda }_2$ (one per constraint), and the fundamental lemma of CoV then forces the bracket to vanish everywhere, producing the single E–L equation of part (a).

(c) Count the bookkeeping. Solving that second-order ODE produces two integration constants $C_1,C_2$; together with the two multipliers ${\lambda }_1,{\lambda }_2$ that is a total of four unknowns $\{C_1,C_2,{\lambda }_1,{\lambda }_2\}$. They are balanced by four equations: the two boundary conditions $y(a)=y_a,y(b)=y_b$ plus the two constraint equations ${\int }_a^b{g}_{1}dx={\ell }_1$ and ${\int }_a^b{g}_{2}dx={\ell }_2$. Four unknowns, four equations — exactly determined.

General law. With $m$ integral constraints: $\underset{C_1,C_2}{\underbrace{2}}+\underset{{\lambda }_1,\dots ,{\lambda }_m}{\underbrace{m}}\text{unknowns}=\underset{\text{BCs}}{\underbrace{2}}+\underset{\text{constraints}}{\underbrace{m}}\text{equations}.$ One multiplier per independent integral constraint — no more, no fewer.

Bribe $F=f-\lambda g$ ::: fold the constraint into the integrand Apply E–L to $F$ ::: get a second-order ODE with constants $C_1,C_2$ and unknown $\lambda$ Boundary conditions ::: fix $C_1,C_2$ Constraint $\int gdx=\ell$ ::: fixes $\lambda$ Meaning of $\lambda$ ::: shadow price $d{J}^{\ast }/d\ell$ (and in Dido, the radius)