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

4.10.16 · D2 · Maths › Advanced Topics (Elite Level) › Isoperimetric problems — constraints (Lagrange multipliers i

Step 1 — Hum actually optimise kya kar rahe hain?

KYA. Ordinary calculus mein tum ek point $x$ ko ek line par slide karte ho taaki curve $h(x)$ ka sabse nichal spot dhundh sako. Calculus of variations mein tum ek poori curve $y(x)$ ko idhar-udhar move karte ho taaki "best" curve dhundh sako — woh curve jo functional naam ki ek number ko jitna ho sake chota (ya bada) banaye.

Ek functional ek machine hai: usse ek poora function do, woh ek number ugalti hai. Hum likhte hain

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

• $y$ — woh poori curve jo tum choose kar rahe ho (machine ka input).
• $y^{\prime }$ — har point par uski slope $\frac{dy}{dx}$ (wahan curve kitni steep hai).
• $f(x,y,y^{\prime })$ — ek chhota sa rule jo curve ko locally, har $x$ par, score karta hai.
• ${\int }_a^b\dots dx$ — un sab local scores ko $x=a$ se $x=b$ tak ek total mein jodta hai.
• $J[y]$ — woh single output number. Square brackets $[\cdot ]$ ek reminder hain: input ek function hai, number nahi.

KYUN. Constraints ko touch karne se pehle hume arena clearly dekhna zaroori hai: hum curves ki ek space mein search kar rahe hain, aur har curve us abstract space mein ek akela point hai.

PICTURE. Figure dekho. Left: ordinary calculus ek dot ko ek hill par slide karta hai. Right: variations ek poori curve ko slide karte hain; side par ek dial value $J[y]$ read karta hai.

Step 2 — Curve-space mein ek constraint kaisi dikhti hai?

KYA. Ab hum ek doosri machine, ek doosra functional, add karte hain:

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

• $g$ — ek doosra local scoring rule (catenary mein yeh arc-length measure karega).
• $\ell$ — ek fixed number jise hum badal nahi sakte (budget: jaise chain ki length).
• $K[y]=\ell$ — rule yeh hai ki "total exactly $\ell$ hona chahiye."

KYUN. Hum koi bhi curve choose nahi kar sakte — sirf woh curves jinka $K$-value exactly $\ell$ ho. Saari curves ki space mein, yeh equation ek constraint surface kaat ke nikalti hai: un curves ka set jo exactly $\ell$ ka budget kharach karti hain.

PICTURE. Curve-space ko ek plane ki tarah draw kiya gaya hai. Grey shading saari curves hai; cyan surface woh patli si patti hai jahan $K[y]=\ell$ hai. Hume sirf cyan surface par chalna allowed hai. Free Euler–Lagrange khushi-khushi us surface se baahar chala jaayega — yahi toh poora problem hai.

Step 3 — Ek knob kaafi nahi hai

KYA. Kisi curve ko move karne ke liye hum usse nudge karte hain. Standard nudge ek bump function $\eta (x)$ aur ek chhote number $ϵ$ ka use karta hai:

$y(x)=\bar{y}(x)+ϵ\eta (x).$

• $\bar{y}$ — woh candidate curve jise hum test kar rahe hain ("centre").
• $\eta (x)$ — ek chosen wiggle, jo dono ends par $0$ hone ke liye forced hai ($\eta (a)=\eta (b)=0$) taaki endpoints pinned rehein.
• $ϵ$ — ek dial: ise ghumao aur wiggle badhti ya ghatti hai. $ϵ=0$ par hum $\bar{y}$ par waapas hain.

KYUN. Ek dial $ϵ$ se, $J$ aur $K$ dono us single number ke functions ban jaate hain. Lekin hamare paas do demands hain: $K$ ko fixed rakho aur $J$ ko stationary banao. Ek dial do maalik ki sewa nahi kar sakta — use $K=\ell$ rakhe rakhne ke liye ghumao toh generally $J$ phir bhi change karta rehta hai. Hum phans gaye hain.

PICTURE. Ek akela slider $ϵ$ curve ko kheenchta hai. Value $K$ (amber) $\ell$ se door drift kar jaati hai jaise hi hum move karte hain — hum ek saath budget nahi pakad sakte aur best $J$ bhi nahi dhundh sakte.

Step 4 — Do knobs: ek budget obey karne ke liye, ek dhundhne ke liye

KYA. Do independent bumps aur do dials use karo:

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

jahan ${\eta }_1,{\eta }_2$ dono ends par vanish karte hain. Ab dono machines do ordinary two-variable functions mein collapse ho jaati hain:

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

• $\Phi$ — $J$ ki value, do dial-settings ka ek ordinary function ke roop mein.
• $\Psi$ — $K$ ki value, usi tarah. $({ϵ}_1,{ϵ}_2)=(0,0)$ par hum $\bar{y}$ par baithte hain.

KYUN. Do dials se hamare paas exactly itni freedom hai: ek degree of freedom kharach karo $\Psi =\ell$ rakhe rakhne mein, aur ek abhi bhi bacha rahega $\Phi$ ko stationary demand karne ke liye. Khaas baat yeh hai ki ab hum ek infinite-dimensional curve problem ko ek ordinary two-variable constrained-extremum problem mein shrink kar chuke hain — woh waali problem jo finite-dimensional Lagrange multipliers pehle se solve kar lete hain.

PICTURE. $({ϵ}_1,{ϵ}_2)$ plane. Amber curve $\Psi =\ell$ hai (budget, maana gaya). Us par $\Phi$ ki contour lines (cyan) dikhati hain kahan $J$ bada ya chota hai. Hume amber curve par woh point dhundhhna hai jahan $\Phi$ change karna band kar de.

Step 5 — Multiplier appear hota hai (gradients line up ho jaate hain)

KYA. Ordinary Lagrange multipliers se: constraint curve $\Psi =\ell$ par ek point $\Phi$ ko stationary banata hai exactly tab jab do gradients parallel hoon:

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

• $\nabla \Phi$ — woh arrow jo direction mein point karta hai jidhar $\Phi$ sabse fast badhta hai ($ϵ$-plane mein).
• $\nabla \Psi$ — woh arrow jo batata hai kahan budget $K$ sabse fast badhti hai; yeh amber constraint curve ke perpendicular hai.
• $\lambda$ — ek akela number, Lagrange multiplier: dono gradients ki lengths ka ratio.

KYUN. Agar $\nabla \Phi$ ka koi bhi component amber curve ke saath (along) hota, toh hum us direction mein slide kar ke $\Phi$ ko improve kar sakte the budget mein rehte hue — abhi stationary nahi. Stationarity ka matlab hai $\nabla \Phi$ poori tarah curve ke aaar-paar (across) point kare, yaani $\nabla \Psi$ ke parallel ho. Yahi parallel condition $\lambda$ ka janm hai.

PICTURE. Winning point par cyan arrow $\nabla \Phi$ aur amber arrow $\nabla \Psi$ ek hi direction mein point karte hain (ek doosre ka $\lambda \times$ hai). Losing point par woh misaligned hain — curve ke saath ek red slide-arrow abhi bhi $\Phi$ ko improve kar sakta hai.

Step 6 — Do conditions ko waapas ek integral mein badlo

KYA. $\nabla (\Phi -\lambda \Psi )=0$ ko component by component likho. Kyunki $\Phi ={\int }_a^{b}fdx$, ${ϵ}_i$ par sirf $y=\bar{y}+{ϵ}_1{\eta }_1+{ϵ}_2{\eta }_2$ ke through depend karta hai, hum integral sign ke under differentiate karte hain:

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

Ab $f_{y^{\prime }}{\eta }_i^{\prime }$ waale piece ko parts se integrate karo (boundary terms mar jaate hain kyunki ${\eta }_i(a)={\eta }_i(b)=0$). Yahan abhi augmented integrand introduce karo:

$F=f-\lambda g,$

taaki ${\partial }_{{ϵ}_i}\Phi -\lambda {\partial }_{{ϵ}_i}\Psi =0$ combine karne par do integrands $f$ aur $g$ ek single $F$ mein collapse ho jaayein, aur $i=1,2$ ke liye yeh mile:

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

• $\frac{\partial \Phi }{\partial {ϵ}_i}=\int (\dots )dx$ — poore integral ka derivative, sirf $f$ ka nahi; yahi "integrate sign ke under differentiate" ka matlab hai (upar ke box se justified).
• $f_y$ — local score kitna change hota hai agar curve thodi si upar jaaye; $f_{y^{\prime }}$ — kitna change hota hai agar slope change ho.
• $\frac{d}{dx}{f}_{y^{\prime }}$ — Euler–Lagrange ka "bookkeeping" term jo integration by parts ke baad bachta hai.
• $F=f-\lambda g$ — augmented integrand; poora bracket exactly $F_y-\frac{d}{dx}{F}_{y^{\prime }}$ hai.
• ${\eta }_i$ — haara bump function, abhi bhi integral ke andar saath chala aa raha hai.

KYUN. Integration by parts woh tool hai jo ${\eta }_i^{\prime }$ ko remove karta hai aur sirf ${\eta }_i$ ko sabkuch multiply karte hua chodta hai — kyunki tabhi hum Step 7 ka lemma invoke kar sakte hain. Hum ise precisely isliye use karte hain taaki wiggle-ke-derivative ko ek plain wiggle se trade kar sakein. Aur hum $F$ ko abhi naam dete hain, jis instant bracket appear hota hai, taaki koi cheez define hone se pehle use na ho. (By-parts step ko bhi $f_{y^{\prime }}$ ka $x$ mein $C^1$ hona chahiye taaki $\frac{d}{dx}{f}_{y^{\prime }}$ exist kare — upar wale $C^2$ assumption se guaranteed.)

PICTURE. By-parts move ko ek see-saw ki tarah draw kiya gaya hai: derivative ${\eta }_i^{\prime }$ se uchhal ke $f_{y^{\prime }}$ par land karta hai, jabki endpoint terms (faded amber mein dikhaye gaye) vanish ho jaate hain kyunki wiggle dono posts par zero par pinned hai.

Step 7 — Fundamental Lemma bracket ko zero par snap kar deta hai

KYA. Hum pehle precisely karte hain ki budget dwara ek bump function kaise consume hota hai. Constraint par rehne ka matlab hai $\Psi ({ϵ}_1,{ϵ}_2)=\ell$. Kyunki $\Psi (0,0)=\ell$ pehle se hai, $\Psi$ ko origin ke baare mein Taylor-expand karo:

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

$\Psi =\ell$ set karo aur sirf first order rakho ($O({ϵ}^2)$ terms un infinitesimal variations ke liye negligible hain jo stationarity define karte hain) toh budget par rehne ki first-order necessary condition milti hai:

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

Yeh single equation ${ϵ}_1$ ko ${ϵ}_2$ se baandhti hai: ${ϵ}_2$ freely choose karo aur ${ϵ}_1$ forced ho jaata hai. Bump-language mein hume chahiye $\partial \Psi /\partial {ϵ}_1={\int }_a^b(g_y-\frac{d}{dx}{g}_{y^{\prime }}){\eta }_{1}dx\ne 0$ taaki ${\eta }_1$ budget enforce karne mein khud ko kharach kar sake.

Ab lemma ko free direction ${\eta }_2$ par apply karo. Calculus of Variations ka Fundamental Lemma kehta hai: agar ${\int }_a^{b}B(x)\eta (x)dx=0$ har smooth $\eta$ ke liye jo ends par vanish kare, toh $B(x)=0$ har jagah hai. $B=F_y-\frac{d}{dx}{F}_{y^{\prime }}$ (Step 6 mein naam diya gaya bracket) ke saath, yeh vanish hona chahiye:

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

• $F=f-\lambda g$ — Step 6 mein introduce kiya gaya augmented integrand: maximise-target minus $\lambda$ times budget-cost.
• Boxed equation bas ordinary Euler–Lagrange equation hai, $f$ ki jagah $F$ par chalaya gaya.
• $\lambda$ exactly woh number hai jisne do directions ke contributions ko cancel karaya, yaani upar linearised constraint se multiplier.

KYUN. Linearisation precisely yeh dikhata hai ki hume do bumps kyun chahiye the: ek $\Psi =\ell$ satisfy karne ke liye ($\partial \Psi /\partial {ϵ}_1$ ke zariye), ek lemma ke liye free rehne ke liye. Aur ek integral jo har free wiggle ke against zero ho, uska matlab sirf yeh ho sakta hai ki multiplied function zero hai — warna wahan ek bump choose karo jahan woh positive hai aur integral positive ho jaayega.

PICTURE. Maano bracket $B(x)$ kisi interval par positive ho (amber hump). Wahan exactly ek bump $\eta$ (cyan) choose karo jo wahan bump ho aur baaki jagah zero. Tab $\int B\eta dx>0$ — ek contradiction. Toh $B$ bilkul zero hona chahiye.

Step 8 — Unknowns count karna (kyun kuch dangling nahi rehta)

KYA. Boxed ODE second order hai, toh uska general solution do constants $C_1,C_2$ laata hai, plus abhi-bhi-unknown $\lambda$. Yeh teen unknowns hain. Hamare paas exactly teen conditions hain:

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

• Do boundary conditions endpoints ko pin karte hain.
• Constraint equation ${\int }_a^{b}gdx=\ell$ woh cheez hai jo finally $\lambda$ determine karti hai — budget ki price.

KYUN. Ek common dar yeh hai ki $\lambda$ ek loose end hai. Aisa nahi hai: budget equation ek extra equation hai, aur yahi exactly woh hai jo $\lambda$ fix karta hai. Woh books jo "$\lambda$ ko kuch bhi set karo" kehti hain woh quietly is count ko tod rahi hain — parent note ka Mistake box dekho.

PICTURE. Ek ledger: teen unknown columns ($C_1$, $C_2$, $\lambda$) teen condition rows ke against balanced (do posts + ek budget). Har unknown ka ghar hai.

Step 9 — Edge aur degenerate cases

KYA & KYUN. Rule kuch cheezein quietly assume karta hai; yahan yeh hai ki jab woh fail hoti hain toh kya hota hai.

• $\lambda =0$ (constraint slack / non-binding hai). Tab $F=f$ aur hum free Euler–Lagrange solution recover karte hain. Yeh tab hota hai jab unconstrained optimum pehle se $\int g=\ell$ satisfy karta ho kismat se — budget hum se lad nahi raha tha. Shadow-price reading $\lambda =\frac{d{J}^{\ast }}{d\ell }$ se, zero price ka matlab hai "budget ki ek aur unit se koi improvement nahi khareedi ja sakti."
• Abnormal case ($\nabla \Psi =0$, yaani $G\equiv 0$). Agar $\bar{y}$ khud $K$ ka stationary point ho (uski constraint-gradient identically vanish karti ho, toh Step 7 se koi ${\eta }_1$ exist nahi karta), toh gradients ko kisi $\lambda$ se parallel nahi banaya ja sakta; tidy $\nabla \Phi =\lambda \nabla \Psi$ rule ko ek zyada general $({\mu }_0\nabla \Phi =\lambda \nabla \Psi )$ form chahiye. Rare hai, par real bhi hai.
• Pointwise vs integral constraint. Agar constraint holonomic hai, $g(x,y)=0$ har $x$ ke liye (sirf uska integral nahi), toh ek constant $\lambda$ far too kam hai — tumhe ek poora function $\lambda (x)$ chahiye, har point par ek price. Isoperimetric (integral) constraints ko ek constant $\lambda$ milta hai; yeh sabse common confusion hai.

PICTURE. Teen mini-panels: (a) constraint contour seedha free optimum se guzarti hai → $\lambda =0$; (b) $\nabla \Psi =0$, arrows align nahi ho sakte (abnormal); (c) integral budget (ek price) vs pointwise budget (har $x$ par ek price).

Ek-picture summary

Upar sab kuch compress karke: hum amber budget-surface $K[y]=\ell$ par savar hote hain; us par curves mein se woh dhundhhte hain jahan cyan value-contours $J[y]$ change karna band kar dein; yeh wahan hota hai jahan gradients align hoon, jo $\lambda$ ko janm deta hai; $F=f-\lambda g$ ko Euler–Lagrange mein daal do aur budget equation se close karo, extremal deliver ho jaata hai — hanging chain ke liye ek catenary, Dido's fixed-perimeter problem ke liye ek circular arc.

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