Isoperimetric problems — constraints (Lagrange multipliers in variational sense)

4.10.16 · Maths › Advanced Topics (Elite Level)

Isoperimetric problem KYA hota hai?

Yeh naam original problem se aaya hai: fixed perimeter ($K=$ length $=\ell$) wali saari closed curves mein se, woh curve dhundho jo maximum area ($J=$ area) enclose kare. Answer: a circle. "Iso-perimetric" = "same perimeter".

Rule derive kaise karein (first principles se)

Hum chahte hain $\delta J=0$ un variations $\delta y$ ke liye jo saath mein $K$ ko bhi fixed rakhen, yaani $\delta K=0$.

Step 1 — TWO-parameter family use karo. Ek akela $ϵ\eta (x)$ kaam nahi karega: ek knob se $J$ ko vary bhi nahi kar sakte aur constraint ka bhi dhyan nahi rakh sakte. Toh lo

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

jahan ${\eta }_1(a)={\eta }_1(b)={\eta }_2(a)={\eta }_2(b)=0$ (boundary values fixed hain, isliye bumps endpoints par vanish ho jaate hain).

Yeh step kyun? Do parameters hamen kaafi freedom dete hain: ek degree of freedom hum constraint satisfy karne mein spend karenge aur phir bhi ek bachega jisse stationarity demand kar sakein.

Step 2 — Ordinary calculus par reduce karo. Ab $J$ aur $K$ do numbers ke ordinary functions ban jaate hain: $\Phi ({ϵ}_1,{ϵ}_2)=J[y],\Psi ({ϵ}_1,{ϵ}_2)=K[y].$ $\bar{y}$ par hum ${ϵ}_1={ϵ}_2=0$ par baithe hain. Hum chahte hain $\Phi$ stationary ho subject to $\Psi =\ell$. Yeh ab ek finite-dimensional constrained extremum hai — jo ordinary Lagrange multipliers se solve hota hai!

Yeh step kyun? Humne calculus of variations ki ek mushkil constraint ko us jaane-pehchaane 2-variable multiplier problem mein convert kar diya jo hum pehle se jaante hain.

Step 3 — Ordinary multiplier rule apply karo. Ek aisa $\lambda$ exist karta hai ki $\nabla (\Phi -\lambda \Psi )=0\text{at }(0,0),$ yaani $\frac{\partial }{\partial {ϵ}_i}(\Phi -\lambda \Psi ){|}_0=0,i=1,2.$

Step 4 — Derivatives compute karo. $\frac{\partial f}{\partial {ϵ}_i}=f_y{\eta }_i+f_{y^{\prime }}{\eta }_i^{\prime }$ use karte hue aur $f_{y^{\prime }}{\eta }_i^{\prime }$ ko parts se integrate karte hue (boundary terms zero ho jaate hain kyunki ${\eta }_i$ ends par vanish karte hain):

$\frac{\partial \Phi }{\partial {ϵ}_i}{|}_0={\int }_a^b(f_y-\frac{d}{dx}{f}_{y^{\prime }}){\eta }_{i}dx,$ aur similarly $\Psi$ ke liye $g$ ke saath. Toh condition ${\partial }_{{ϵ}_i}(\Phi -\lambda \Psi )=0$ yeh ban jaati hai: ${\int }_a^b[(f_y-\frac{d}{dx}{f}_{y^{\prime }})-\lambda (g_y-\frac{d}{dx}{g}_{y^{\prime }})]{\eta }_{i}dx=0.$

Step 5 — Fundamental Lemma. $F=f-\lambda g$ define karo. Bracket $(F_y-\frac{d}{dx}{F}_{y^{\prime }})$ hai. Kyunki ${\eta }_2$ essentially arbitrary hai (jab ${\eta }_1$ constraint fix karne mein use ho jaata hai), CoV ka fundamental lemma bracket ko har jagah zero hone par majboor karta hai.

Worked Example 1 — Catenary: fixed length ki hanging chain

Ek chain jo fixed length $\ell$ ki hai, do posts ke beech latki hai. Woh settle hoti hai potential energy $\propto$ centre of mass ki height ko minimise karke, yaani minimise karo $J={\int }_a^{b}y\sqrt{1+y^{\prime 2}}dx\text{subject to length}K={\int }_a^b\sqrt{1+y^{\prime 2}}dx=\ell .$

Step 1 — Augment karo. $f=y\sqrt{1+y^{\prime 2}}$, $g=\sqrt{1+y^{\prime 2}}$, toh $F=f-\lambda g=(y-\lambda )\sqrt{1+y^{\prime 2}}.$ Kyun? $F$ constraint ko fold kar leta hai; ab hum unconstrained solve karte hain.

Step 2 — Beltrami use karo (kyunki $F$ mein explicit $x$ nahi hai). $F-y^{\prime }{F}_{y^{\prime }}=C$. Compute karo $F_{y^{\prime }}=(y-\lambda )\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}$, toh $(y-\lambda )\sqrt{1+y^{\prime 2}}-(y-\lambda )\frac{y^{\prime 2}}{\sqrt{1+y^{\prime 2}}}=\frac{y-\lambda }{\sqrt{1+y^{\prime 2}}}=C.$ Beltrami kyun? Yeh ek first integral hai jo ek integration bachata hai jab bhi $x$ absent ho.

Step 3 — Solve karo. $y-\lambda =C\sqrt{1+y^{\prime 2}}$. Separate karke integrate karo: $y(x)=\lambda +C\cosh \left(\frac{x-x_0}{C}\right).$ Shape ek catenary ($\cosh$) hai. Constants $\lambda$ (vertical shift), $C$, $x_0$ do endpoints aur length condition $\int \sqrt{1+y^{\prime 2}}=\ell$ se fix hote hain.

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

$(-a,0)$ se $(a,0)$ tak ki curves $y(x)$ mein se jo fixed arc-length $\ell$ ki hain, enclosed area maximise karo: $J={\int }_{-a}^{a}ydx\text{subject to}K={\int }_{-a}^a\sqrt{1+y^{\prime 2}}dx=\ell .$

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

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

Step 3 — Solve karo. $\frac{\lambda }{\sqrt{1+y^{\prime 2}}}=y-C$. Maano $u=y-C$. Toh $1+y^{\prime 2}={\lambda }^2/u^2\Rightarrow y^{\prime 2}=({\lambda }^2-u^2)/u^2$, jisse milta hai $\frac{udu}{\sqrt{{\lambda }^2-u^2}}=\pm dx\Rightarrow -\sqrt{{\lambda }^2-u^2}=\pm (x-x_0).$ Square karo: $(x-x_0{)}^2+(y-C{)}^2={\lambda }^2$. Ek circular arc! Yeh intuition confirm kyun karta hai: fixed perimeter → maximum area → circle, bilkul classic isoperimetric result. Multiplier $\lambda$ radius nikla.

Common Mistakes (Steel-manned)

Flashcards

Connections

• Euler–Lagrange Equation — woh engine jise hum $F=f-\lambda g$ par apply karte hain.
• Beltrami Identity — dono worked examples mein use hua first integral.
• Lagrange Multipliers (finite-dimensional) — parent idea jise yeh generalize karta hai.
• Calculus of Variations — Fundamental Lemma — "bracket = 0" justify karta hai.
• Catenary Curve / Brachistochrone Problem — sibling variational classics.
• Isoperimetric Inequality — $4\pi A\le L^2$, equality iff circle.
• Shadow Price / Envelope Theorem — $\lambda$ ki interpretation.

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