Worked examples — Isoperimetric problems — constraints (Lagrange multipliers in variational sense)

4.10.16 · D3 · Maths › Advanced Topics (Elite Level) › Isoperimetric problems — constraints (Lagrange multipliers i

Shuru karne se pehle, pieces ka ek saaf-saaf reminder, kyunki hum inhe har line pe use karte hain.

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Scenario matrix

Yeh har woh class hai jo ek isoperimetric problem tumhare saamne rakh sakti hai. Har row difficulty ki ek alag "shape" hai; last column us example ka naam deta hai jo is page pe uss par land karta hai.

#	Case class	Kya alag banaata hai	Covered by
C1	$f$ mein explicit $x$ nahi → Beltrami use karo	first integral shortcut	Ex 1, Ex 2
C2	$f$ mein $x$ hai → Beltrami fail, full E–L use karo	$\frac{d}{dx}{F}_{y^{\prime }}$ integrate karna padega	Ex 3
C3	Constraint bahut chhoti / degenerate ($\ell \le$ chord)	koi valid curve nahi; boundary case	Ex 4
C4	Limiting value: $\ell \to$ minimum → curve line mein flatten hoti hai	$\lambda \to \infty$ / arc → chord	Ex 4
C5	$\lambda$ ka sign (max vs min uska sign flip karta hai)	curve kis taraf bend karti hai	Ex 2, Ex 5
C6	Dono endpoints equal / symmetric interval	$x_0$ symmetry se pin hota hai	Ex 2, Ex 5
C7	Real-world word problem (physics units)	words ko $f,g$ mein translate karo	Ex 6
C8	Exam twist: constraint $\int y^2$ hai, arc-length nahi	$g$ depends on $y$, $y^{\prime }$ par nahi	Ex 7

Ab inhe karte hain.

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Ex 1 — Baseline sanity: fixed length par energy minimise karo (case C1)

Forecast: aage padhne se pehle shape aur uske lowest point ka andaza lagao.

Step 1 — Augment karo. $F=f-\lambda g=(y-\lambda )\sqrt{1+y^{\prime 2}}$. Yeh step kyun? $-\lambda g$ add karne se hum sab curves freely search kar sakte hain; $\lambda$ baad mein length rule ko absorb karta hai.

Step 2 — Beltrami, kyunki $F$ mein explicit $x$ nahi hai. Yeh tool kyun, full E–L kyun nahi? Jab $x$ kabhi $F$ ke andar appear nahi hota, tab Beltrami Identity $F-y^{\prime }{F}_{y^{\prime }}=C_1$ ek first integral hoti hai — yeh ek second-order ki jagah ek first-order ODE free mein de deti hai. Ise choose karne se ek poori integration bachti hai. Yahan $C_1$ hamara pehla integration constant hai. $\frac{y-\lambda }{\sqrt{1+y^{\prime 2}}}=C_1.$

Step 3 — Separate aur integrate karo. Slope isolate karne ke liye Beltrami relation rearrange karo: $\sqrt{1+y^{\prime 2}}=\frac{y-\lambda }{C_1}\Rightarrow y^{\prime 2}=(\frac{y-\lambda }{C_1}{)}^2-1.$ Nayi letter $t$ kyun introduce karein? Right side $(\text{something}{)}^2-1$ hai, aur identity ${\cosh }^{2}t-1={\sinh }^{2}t$ exactly usi shape ko perfect square mein convert karne ke liye bani hai. Toh hum substitution $\frac{y-\lambda }{C_1}=\cosh t$ lete hain — yeh ek change of variable hai jo $\sqrt{(\cdots {)}^2-1}=\sinh t$ clean bana deta hai, koi square root nahi bachta. Yahan $t$ sirf position ka ek running label hai; yeh end mein $t=\frac{x-x_0}{C_1}$ banega. Uss substitution se $y^{\prime }=\sinh t$ aur $dy=C_1\sinh tdt$, aur kyunki $y^{\prime }=dy/dx$ hai toh $dx=C_{1}dt$, isliye integrate karne par $x-x_0=C_{1}t$ milta hai. Substitution undo karne par ($t=(x-x_0)/C_1$, $y-\lambda =C_1\cosh t$): $y(x)=\lambda +C_1\cosh \left(\frac{x-x_0}{C_1}\right).$ Yeh constants kya hain? $C_1$ Beltrami se aaya (Step 2); doosri integration ne $x_0$ produce kiya, toh $x_0$ $C_2$ ka role play karta hai — horizontal shift. $\lambda$ ke saath milake yeh promised trio $C_1,C_2(=x_0),\lambda$ hai.

Step 4 — Symmetry use karo. Endpoints equal hain aur interval $[-1,1]$ symmetric hai, toh $\cosh$ ka lowest point middle mein baithta hai: $==x_0=0==$. Yeh step kyun? $\cosh$ apne vertex ke baare mein even hoti hai; equal boundary heights vertex ko symmetry axis par force karti hain. Neeche figure mein yeh pink dashed line hai: curve iske across mirror image hai.

Dekho figure mein: yellow curve $\cosh$ hai; do blue dots pinned endpoints hain equal height par; pink dot vertex hai jo exactly pink symmetry axis $x_0=0$ par baitha hai — chain ka lowest point.

Verify: $x_0=0$ ke saath, $y(-1)=y(1)=\lambda +C_1\cosh (1/C_1)$ automatically — dono endpoints ek equation se match karte hain, confirming symmetry consistent hai. Units: $y,x,\ell$ sab lengths, dimensionally clean. ✓

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Ex 2 — Dido, concrete numbers: fixed perimeter → circular arc (cases C1, C5, C6)

Forecast: chord ki length $2$ hai; string ki length $\pi$ hai. Shape aur uska area guess karo.

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

Step 2 — Beltrami (explicit $x$ nahi): $F-y^{\prime }{F}_{y^{\prime }}=C$ deta hai $y-\frac{\lambda }{\sqrt{1+y^{\prime 2}}}=C$.

Step 3 — Ek circle mein rearrange karo, ek ek algebra step karte hue. Beltrami se start karo aur square-root term isolate karo: $\frac{\lambda }{\sqrt{1+y^{\prime 2}}}=y-C.$ $u=y-C$ (ek shifted height) lete hain. Square aur invert karo: $1+y^{\prime 2}=\frac{{\lambda }^2}{u^2}\Rightarrow y^{\prime 2}=\frac{{\lambda }^2-u^2}{u^2}\Rightarrow y^{\prime }=\frac{du}{dx}=\frac{\sqrt{{\lambda }^2-u^2}}{u}.$ Ab separate kyun karein? $du/dx$ sirf $u$ ki function hai, toh hum variables split kar sakte hain: $\frac{udu}{\sqrt{{\lambda }^2-u^2}}=dx.$ Left side integrate hoke $-\sqrt{{\lambda }^2-u^2}$ deta hai (iska derivative exactly $u/\sqrt{{\lambda }^2-u^2}$ hai), toh $-\sqrt{{\lambda }^2-u^2}=x-x_0\Rightarrow (x-x_0{)}^2+u^2={\lambda }^2.$ $u=y-C$ wapas rakhne par: $(x-x_0{)}^2+(y-C{)}^2={\lambda }^2.$ Yeh radius $\lambda$ ka circle hai jiska centre $(x_0,C)$ par hai — circle wahin se aata hai. Symmetry ($C6$) centre ko $y$-axis par rakhti hai, toh $x_0=0$.

Step 4 — Endpoints fit karo. $y(\pm 1)=0$ matlab $1+C^2={\lambda }^2$. Yeh step kyun? Dono boundary points circle par hain; yeh ek equation hai $C$ aur radius $\lambda$ ko link karti hui.

Step 5 — Length fit karo. Half-angle $\theta$ (jahan $\sin \theta =1/\lambda$) subtend karne wala radius $\lambda$ ka circle chord ke upar arc length $2\lambda \theta =\pi$ deta hai. Try karo $\lambda =1$: tab $\sin \theta =1$, $\theta =\pi /2$, arc $=2\cdot 1\cdot \pi /2=\pi$. ✓ Toh $\lambda =1$, $C=0$: radius $1$ ki semicircle. $\lambda >0$ kyun (case C5)? Hum area maximise kar rahe hain, toh curve outward (upward) bow karti hai — positive radius. Minimise karne par sign flip hoga aur inward bow karti.

Step 6 — Area. Half unit disc: $J=\frac{1}{2}\pi (1{)}^2=\frac{\pi }{2}$.

Dekho figure mein: yellow arc length $\pi$ ki semicircle hai; blue segment length $2$ ka chord hai; shaded pink region enclosed area $\pi /2$ hai; pink dot origin par centre hai, aur radius arrow $\lambda =1$ dikhata hai — multiplier literally radius ke roop mein measure ho raha hai.

Verify: unit semicircle ki arc length $=\pi (1)=\pi =\ell$ ✓. Area $=\pi /2\approx 1.5708$ ✓. Multiplier $\lambda =1$ radius hai, parent ke punchline se match karta hai. ✓

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Ex 3 — Jab $x$ appear kare: Beltrami fail, full E–L use karo (case C2)

Forecast: kya answer $\cosh$ hoga, line hogi, ya polynomial?

Step 1 — Augment karo. $g=y$, toh $F=y^{\prime 2}+xy-\lambda y=y^{\prime 2}+(x-\lambda )y$.

Step 2 — Notice karo ki Beltrami yahan illegal hai. Kyun? $F$ mein explicitly $x$ hai ($xy$ term). Beltrami require karta hai $\partial F/\partial x=0$. Ise use karna ek galti hogi; hume full Euler–Lagrange Equation use karni hai.

Step 3 — E–L apply karo. $F_y=x-\lambda$, $F_{y^{\prime }}=2y^{\prime }$, toh $F_y-\frac{d}{dx}{F}_{y^{\prime }}=(x-\lambda )-2y^{\prime \prime }=0\Rightarrow y^{\prime \prime }=\frac{1}{2}(x-\lambda ).$

Step 4 — Do baar integrate karo. $y=\frac{x^3}{12}-\frac{\lambda x^2}{4}+C_{1}x+C_2.$

Step 5 — Boundary conditions. $y(0)=0\Rightarrow C_2=0$. $y(1)=0\Rightarrow \frac{1}{12}-\frac{\lambda }{4}+C_1=0\Rightarrow C_1=\frac{\lambda }{4}-\frac{1}{12}$.

Step 6 — Constraint $\lambda$ fix karta hai. Strategy: ab $y$ sirf $\lambda$ unknown ke saath likhi hai (kyunki $C_1,C_2$ pehle se $\lambda$ ke through express ho chuke hain), toh ek remaining equation — constraint — ko $\lambda$ determine karna hi hoga. ${\int }_0^{1}ydx=\ell$ compute karo: ${\int }_0^1(\frac{x^3}{12}-\frac{\lambda x^2}{4}+C_{1}x)dx=\frac{1}{48}-\frac{\lambda }{12}+\frac{C_1}{2}=\ell .$ Aise group kyun karein? Hum Step-5 ki value $C_1=\frac{\lambda }{4}-\frac{1}{12}$ substitute karte hain taaki har term ya to ek pure number ho ya $\lambda$ ka multiple — phir $\lambda$-terms ko ek side collect karne se ek move mein isolate ho jaata hai: $\underset{\text{pure numbers}}{\underbrace{\frac{1}{48}-\frac{1}{24}}}+\underset{\text{coeff. of }\lambda }{\underbrace{(-\frac{1}{12}+\frac{1}{8})}}\lambda =-\frac{1}{48}+\frac{\lambda }{24}=\ell .$ Hence $\frac{\lambda }{24}=\ell +\frac{1}{48}$, giving $\lambda =24\ell +\frac{1}{2}$.

Verify: $\ell =0$ lo. Tab $\lambda =\frac{1}{2}$, $C_1=\frac{1}{8}-\frac{1}{12}=\frac{1}{24}$, aur $y=\frac{x^3}{12}-\frac{x^2}{8}+\frac{x}{24}$. Check karo $y(1)=\frac{1}{12}-\frac{1}{8}+\frac{1}{24}=0$ ✓ aur ${\int }_0^{1}ydx=0$ ✓ (VERIFY mein kiya). ✓

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Ex 4 — Degenerate & limiting budget (cases C3, C4)

Forecast: $\ell \downarrow 2$ hone par shape guess karo, aur kya $\ell =1.5$ solve bhi ho sakta hai.

Step 1 — $\ell$ ki lower bound seedha chord hai. Kyun? $(-1,0)$ se $(1,0)$ join karne wali sabse chhoti curve seedha segment hai, length $2$. Koi bhi admissible curve isse chhoti nahi ho sakti.

Step 2 — Case C3, $\ell =1.5$. Kyunki $1.5<2$, koi curve exist nahi karti. Problem infeasible hai — constraint surface $K[y]=\ell$ in endpoints ke liye empty hai. Yeh kyun matter karta hai: multiplier dhundhne se pehle hamesha check karo $\ell \ge$ chord. Yahan jo "solution" compute karo woh spurious hoga.

Step 3 — Limiting case C4, $\ell \to 2^{+}$. Arc almost straight hona chahiye, toh radius $\lambda \to \infty$ aur arc chord par collapse hota hai. Area $J\to 0$. Kyun? Ek fixed chord par bahut bade radius ka circular arc almost flat hota hai; flat curve almost no area enclose karti hai.

Step 4 — Small-bulge area quantify karo. Maximum height (sagitta) $h$ ke saath length $L=2$ ke chord par ek circular segment ke liye, enclosed area exactly $A={\int }_{-1}^1\left(\text{arc height}\right)dx.$ Ek shallow parabola-like bulge $y\approx h(1-x^2)$ (low arc ki leading shape over $[-1,1]$) ke liye, yeh integral $A=h{\int }_{-1}^1(1-x^2)dx=h\cdot \frac{4}{3}=\frac{2}{3}Lh$ hai jahan $L=2$. Toh $A\approx \frac{2}{3}Lh\to 0$ jaise $h\to 0$. Yeh approximation kyun? Flat limit ke paas exact circular arc is parabola se leading order tak indistinguishable hai, aur parabola ek line mein integrate hoti hai — $A\to 0$ linearly in sagitta dekhne ke liye kaafi hai.

Step 5 — Range ke dono ends. $\ell =2$ exactly par, $\theta \to 0$, $\lambda \to \infty$, $J\to 0$. $\ell =\pi$ par (Ex 2) humne $J=\pi /2$ paya, semicircle. Toh area monotonically badhta hai $0$ se jaise $\ell$ $2$ se $\pi$ ki taraf climb karta hai.

Verify: $\ell =\pi$ par, $J=\pi /2\approx 1.571$; semicircle single-valued $y(x)$ arc ke compatible sabse bade $\ell$ ka use karti hai (half-angle $\pi /2$). $2$ aur $\pi$ ke beech ka koi bhi $\ell$ strictly $0$ aur $\pi /2$ ke beech area deta hai. Feasibility test $\ell \ge 2$ confirmed. ✓

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Ex 5 — $\lambda$ ka sign: maximise vs minimise (cases C5, C6)

Forecast: arc kis taraf bow karta hai ab?

Step 1 — Same augmented $F$, opposite optimisation. E–L / Beltrami algebra identical hai: hume phir bhi circle $(x{)}^2+(y-C{)}^2={\lambda }^2$ milti hai.

Step 2 — Sign flip hota hai. Kyun? Signed area minimise karna chahta hai ki $y$ jitna ho sake utna negative ho, toh arc downward bow karta hai. Centre ab chord ke upar hai; radius $\lambda$ effectively maximisation case ke relative opposite sign carry karta hai.

Step 3 — Solution. Minimiser lower semicircle hai: $y(x)=-\sqrt{1-x^2}$, area $J=-\pi /2$.

Verify: arc length $=\pi$ ✓ (same circle, mirrored). Signed area $=-\pi /2\approx -1.571$ ✓. $\lambda$ ka sign exactly bulge ka sign hai — parent ke "$\lambda$ = radius" reading se match karta hai (orientation tak). ✓

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Ex 6 — Real-world word problem with units (case C7)

Forecast: guess karo ki $6\text{ m}$ gap par $10\text{ m}$ ki rope $1\text{ m}$ se zyada sagte gi ya kam.

Step 1 — Words ko $f,g$ mein translate karo, aur shape yaad karo. Hanging rope potential energy $J=\int y\sqrt{1+y^{\prime 2}}dx$ minimize karti hai subject to length $K=\int \sqrt{1+y^{\prime 2}}dx=\ell$. Yeh exactly Ex 1 ka setup hai, jiska solution humne wahan derive kiya tha: $y(x)=\lambda +C\cosh \left(\frac{x-x_0}{C}\right),\text{(from Ex 1, Step 3)}$ jahan $C$ Beltrami constant hai. Hum yahan woh result reuse karte hain, re-derive nahi karte.

Step 2 — Symmetry $x_0$ pin karta hai. $x=\pm 3$ par pins symmetric hain, toh $x_0=0$ (case C6, jaise Ex 1 mein).

Step 3 — Catenary ka length formula. Yaad ki gayi shape differentiate karo: $y^{\prime }=\sinh \left(\frac{x}{C}\right)$ (kyunki $x_0=0$). Tab arc-length integrand hai $\sqrt{1+y^{\prime 2}}=\sqrt{1+{\sinh }^2\frac{x}{C}}=\cosh \frac{x}{C},$ identity $1+{\sinh }^2={\cosh }^2$ use karte hue. Toh $\ell ={\int }_{-3}^3\cosh \frac{x}{C}dx=2C\sinh \frac{3}{C}.$ Yeh step kyun? $\sqrt{1+y^{\prime 2}}$ ko known derivative $y^{\prime }=\sinh (x/C)$ ke terms mein wapas likhne se messy root ek plain $\cosh$ mein collapse ho jaati hai, jiska integral $C\sinh$ hai — arc length ko ek clean closed form mein badal deta hai.

Step 4 — $C=2$ plug karo. $\ell =2\cdot 2\cdot \sinh (1.5)=4\sinh (1.5)\approx 4(2.1293)=8.517\text{ m}$. Yeh $10\text{ m}$ nahi hai — toh $C=2$, $10\text{ m}$ ki rope ke saath inconsistent hai. Sahi $C$ solve karta hai $2C\sinh (3/C)=10$, jo deta hai $C\approx 1.62\text{ m}$. Check kyun karein? Handed-down constant par kabhi trust mat karo; length constraint hi $C$ fix karta hai.

Step 5 — Sag. True $C\approx 1.62$ ke saath: sag $=y(3)-y(0)=C(\cosh \frac{3}{C}-1)=1.62(\cosh (1.852)-1)\approx 1.62(3.199-1)\approx 3.56\text{ m}$.

Verify: $2(1.62)\sinh (3/1.62)=3.24\sinh (1.852)\approx 3.24(3.086)\approx 10.0\text{ m}$ ✓ (length recover ho gayi). Units sab metres ✓. Lesson: false $C=2$ ne wrong length di — constraint use correct karta hai. ✓

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Ex 7 — Exam twist: constraint $y$ par depend karta hai, $y^{\prime }$ par nahi (case C8)

Forecast: konse familiar functions ka ${\int }_0^{\pi }(\cdot {)}^2$ easily normalise hota hai?

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

Step 2 — Full E–L ($x$ absent, lekin safe rahte hain): $F_y=-2\lambda y$, $F_{y^{\prime }}=2y^{\prime }$, toh $-2\lambda y-\frac{d}{dx}(2y^{\prime })=0\Rightarrow y^{\prime \prime }+\lambda y=0.$ Yeh step kyun? Multiplier ne ek constrained variational problem ko eigenvalue equation $y^{\prime \prime }=-\lambda y$ mein badal diya — constraint ki "price" $\lambda$ hi eigenvalue hai.

Step 3 — $\lambda$ ke sign par split karo (edge case). Solution ki character puri tarah $\lambda$ ke sign par depend karti hai:

• Agar $\lambda <0$, likho $\lambda =-k^2$: tab $y^{\prime \prime }=k^{2}y$ ke solutions $y=A{e}^{kx}+B{e}^{-kx}$ hain (hyperbolic). Aisi combination dono $x=0$ aur $x=\pi$ par zero hone ka ek hi tarika hai $A=B=0$, yaani $y\equiv 0$ — jo constraint $\int y^2=1$ tod deta hai. Toh $\lambda <0$ yahan impossible hai.
• Agar $\lambda =0$, tab $y^{\prime \prime }=0$, $y=ax+b$; dono endpoints zero force karta hai $y\equiv 0$ phir se — impossible.
• Agar $\lambda >0$, likho $\lambda =k^2$: tab $y^{\prime \prime }+k^{2}y=0$ oscillating solutions $y=A\sin (kx)+B\cos (kx)$ deta hai, jo dono ends par zero ho sakta hai. Yahi ek viable sign hai.

$\lambda$ positive kyun hona chahiye: sirf oscillatory (trigonometric) solutions do alag points par zero hit kar sakti hain non-trivially rahte hue; uske liye $\lambda >0$ chahiye.

Step 4 — Boundary conditions. $y(0)=0$ se humein $B=0$ chahiye; $y(\pi )=0$ se $\sin (k\pi )=0$ chahiye, toh $k=n$ integer. Isliye $y=A\sin (nx)$ with $\lambda =n^2$, $n=1,2,3,\dots$

Step 5 — Constraint se normalise karo. ${\int }_0^{\pi }{A}^2{\sin }^2(nx)dx=A^2\cdot \frac{\pi }{2}=1\Rightarrow A=\sqrt{2/\pi }$.

Step 6 — Minimiser choose karo. Sabse chhota $J={\int }_0^{\pi }{y}^{\prime 2}$ sabse chhote $\lambda =1$ se aata hai: $y=\sqrt{2/\pi }\sin x$, aur phir $J=\lambda \int y^2=\lambda =1$. $\lambda$ = $J$ ki value kyun? $y^{\prime \prime }+\lambda y=0$ ko $y$ se multiply karo aur parts se integrate karo: $\int y^{\prime 2}=\lambda \int y^2=\lambda$. Multiplier optimum ke equal hai — shadow price literally sach sabit hoti hai.

Verify: $A^2\pi /2=(2/\pi )(\pi /2)=1$ ✓. Minimum $J=1$ at $n=1$ ✓. Higher modes $n=2,3$ larger $J=4,9$ dete hain — stationary but not minimal ✓ (VERIFY mein check kiya). ✓

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Recall Quick self-test

Beltrami kab legal hai? Sirf jab augmented F = f - lambda g mein explicit x na ho; warna full Euler–Lagrange equation use karo. Ek standard isoperimetric solve mein kitne unknowns aur kitni conditions hain? Teen unknowns C1, C2, lambda; teen conditions — do boundary values plus integral constraint. $\lambda$ numerically kya matlab rakhta hai? Yeh marginal value dJ*/d-ell hai: per unit extra budget optimum J kitna improve hota hai (Dido mein radius, Ex 7 mein eigenvalue). Agar ell chord length se chhoti ho toh kya hota hai? Koi admissible curve exist nahi karti — problem infeasible hai (case C3).

Back to the parent topic. Related: Catenary Curve, Isoperimetric Inequality, Brachistochrone Problem, Lagrange Multipliers (finite-dimensional), Calculus of Variations — Fundamental Lemma.