Visual walkthrough — Calculus of variations — functionals, functional derivative

4.10.12 · D2 · Maths › Advanced Topics (Elite Level) › Calculus of variations — functionals, functional derivative

Hum ek hi sawaal ka jawaab denge: point $A$ se point $B$ tak jaane wali saari curves mein se, kaun si ek total "cost" ko minimum karti hai — aur hum usse kaise pehchaante hain?

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Step 1 — Hum kya choose kar rahe hain: ek poori curve, koi point nahi

KYA. Ordinary "minimum dhundho" problems mein tum ek single dot ko left-right slide karte ho jab tak valley ke bottom tak na pahuncho. Yahan hum jo cheez change kar sakte hain woh ek poori curve $y(x)$ hai — ek bendy wire jo do nails pe pin ki gayi hai, left nail $A$ pe aur right nail $B$ pe.

KYUN. Length, travel-time, energy — yeh sab wire ki poori shape pe depend karte hain, kisi ek number pe nahi. Isliye unknown khud ek shape hai.

PICTURE. Figure dekho. Horizontal axis $x$ hai (zameen ke saath position). Vertical axis $y$ hai (wire ki height). Teal curve ek candidate wire hai. Do orange dots fixed endpoints hain — yeh kabhi nahi hilte.

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Step 2 — Curve ko "wiggle" kaise karein: the variation

KYA. Yeh test karne ke liye ki koi curve best hai ya nahi, hum use thoda nudge karte hain. Hum ek chhota bump $\eta (x)$ add karte hain jise ek tiny knob $\epsilon$ se scale kiya jaata hai: $\tilde{y}(x)=y(x)+\epsilon \eta (x).$

KYUN. Ordinary calculus mein tum minimum check karte ho $x$ ko har direction mein thoda sa nudge karke. Ek curve ko nudge karne ke liye infinitely many "directions" hote hain — bump ki har shape ke liye ek. Toh $\eta$ hamara "kis direction mein wiggle karna hai" hai, aur $\epsilon$ hai "kitni tezi se."

PICTURE. Teal curve candidate $y$ hai. Plum dashed curve $\eta$ hai, yani bump. Orange curve wiggled wire $y+\epsilon \eta$ hai. Sabse zaroori baat: bump dono nails pe flat hai — wahan zero hona chahiye taaki endpoints pin rahe.

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Step 3 — Problem ko ek ordinary variable tak collapse karo

KYA. Bump shape $\eta$ ko freeze karo aur sirf dial $\epsilon$ ko move karne do. Tab cost ek ordinary one-variable function ban jaati hai: $\Phi (\epsilon )=J[y+\epsilon \eta ].$

KYUN. Yahi saara trick hai. Hum directly "space of all curves" pe calculus karna nahi jaante, isliye hum iske andar se ek 1-D beam chalate hain: is beam ke saath cost sirf ek normal function $\Phi (\epsilon )$ hai, aur hum pehle se jaante hain unhe kaise minimise karein. Agar $y$ sach mein best curve hai, toh isse kisi bhi direction mein hat'ne se cost sirf badh sakti hai, isliye $\epsilon =0$ $\Phi$ ka bottom hai.

PICTURE. Parabola-jaisi teal curve $\Phi (\epsilon )$ hai. Uska lowest point exactly $\epsilon =0$ pe baitha hai (orange dot). Wahan tangent line flat hai — slope zero.

Yeh single equation, khol ke dekho, wohi hai Euler–Lagrange equation.

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Step 4 — Integral ke andar differentiate karo (slices pe chain rule)

KYA. Dial ghuma ke dekho ki har slice-cost kaise badalti hai. $\Phi (\epsilon )={\int }_a^{b}L(x,y+\epsilon \eta ,y^{\prime }+\epsilon {\eta }^{\prime })dx$ ko $\epsilon$ ke saath differentiate karo: ${\Phi }^{\prime }(\epsilon )={\int }_a^b(\frac{\partial L}{\partial y}\eta +\frac{\partial L}{\partial y^{\prime }}{\eta }^{\prime })dx.$

KYUN. $L$ ke andar do slots hain jo $\epsilon$ pe depend karte hain: height slot ($y+\epsilon \eta$) aur slope slot ($y^{\prime }+\epsilon {\eta }^{\prime }$). Chain rule kehta hai: rate of change $=$ (height ke baare mein sensitivity)$\times$(height kitni tezi se badalti hai) $+$ (slope ke baare mein sensitivity)$\times$(slope kitni tezi se badalti hai). Dial ghhumane se height $\eta$ se upar push hoti hai aur slope ${\eta }^{\prime }$ se — isliye $\eta$ ke saath $\partial L/\partial y$ hai aur ${\eta }^{\prime }$ ke saath $\partial L/\partial y^{\prime }$.

PICTURE. Ek sample point pe, do arrows: orange arrow dikhata hai ki slice cost height badhne pe kaise react karti hai ($\partial L/\partial y$, $\eta$ se weighted); teal arrow dikhata hai ki slope tilt karne pe kaise react karti hai ($\partial L/\partial y^{\prime }$, ${\eta }^{\prime }$ se weighted).

Ek mushkil hai: do terms mein alag alag wiggles hain — ek mein $\eta$ hai, doosre mein ${\eta }^{\prime }$. Hum abhi tak common $\eta$ factor out nahi kar sakte. Step 5 yeh theek karta hai.

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Step 5 — Integration by parts: derivative ko wiggle se hatao

KYA. Awkward ${\eta }^{\prime }$ term ko rewrite karo taaki usme bhi plain $\eta$ ho: ${\int }_a^b\frac{\partial L}{\partial y^{\prime }}{\eta }^{\prime }dx=\underset{\text{boundary term}}{\underbrace{[\frac{\partial L}{\partial y^{\prime }}\eta {]}_a^b}}-{\int }_a^b\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)\eta dx.$

KYUN. Integration by parts woh tool hai jo ek factor se doosre pe derivative swap karta hai. Yahi exactly woh surgery hai jo hume chahiye: yeh ${\eta }^{\prime }$ se prime utha ke $\partial L/\partial y^{\prime }$ pe daal deta hai, ek plain $\eta$ chhod deta hai jise hum factor out kar sakte hain. Hum yeh tool precisely isliye choose karte hain kyunki $\eta$ woh object hai jise hum isolate karna chahte hain.

PICTURE. Ek "before / after" strip. Before: derivative $\eta$ pe baitha hai (plum bump pe ek chhota prime tag dikhaya gaya hai). After: derivative $\partial L/\partial y^{\prime }$ pe kood gaya hai (tag teal factor ke paar jaata hai), aur ek boundary term dono nails pe pop up hota hai.

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Step 6 — Ab ek single common factor $\eta$

KYA. Bacha hua pieces ek saath jodo: ${\Phi }^{\prime }(0)={\int }_a^b\left[\underset{\frac{\delta J}{\delta y}}{\underbrace{\frac{\partial L}{\partial y}-\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)}}\right]\eta (x)dx=0.$

KYUN. Step 5 ke baad dono terms mein $\eta$ hai, isliye hum ise factor out karte hain. Bracket functional derivative $\frac{\delta J}{\delta y}$ hai — ek "gradient" jiska har point $x$ pe ek component hai. Equation kehti hai: chahe hum koi bhi wiggle $\eta$ choose karein, yeh weighted sum zero hai.

PICTURE. Bracket ko $x$ ke saath ek burnt-orange gradient field ke roop mein draw kiya gaya hai: har point pe ek arrow jo curve ko bolta hai "mujhe yahan neeche push karo, wahan upar." Wiggle $\eta$ plum curve hai. Unka product, integrate kiya gaya, shaded area hai — aur yeh har possible $\eta$ ke liye total zero hona chahiye.

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Step 7 — Fundamental Lemma: "for all $\eta$" se "bracket zero hai" tak

KYA. Agar ${\int }_a^{b}f(x)\eta (x)dx=0$ har smooth $\eta$ ke liye jo ends pe vanish kare, tab $f(x)\equiv 0$ har jagah.

KYUN. Maano bracket $f$ kisi chhoti stretch pe positive ho. Hum koi bhi $\eta$ choose karne ke liye free hain. Toh aisa $\eta$ choose karo jo ek chhota positive lump ho exactly us stretch pe aur baaki jagah zero ho. Tab $f\eta$ wahan positive hai aur baaki jagah zero, isliye integral strictly positive hai — "yeh har $\eta$ ke liye zero hai" se contradiction. Wohi argument kisi bhi negative stretch ko bhi khatam kar deta hai. Toh $f$ sirf zero hi ho sakta hai.

PICTURE. Ek plum bump $\eta$ us jagah park ki gayi hai jahan bracket $f$ (burnt orange) positive hai; shaded product area clearly positive hai — contradiction, drawn.

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Step 8 — Edge case: kya hoga agar hum endpoints nail nahi karte?

KYA. $\eta (a)=\eta (b)=0$ ki requirement hata do. Tab Step 5 ka boundary term vanish nahi karta. ${\Phi }^{\prime }(0)=0$ set karne ke liye ab interior integral aur boundary term dono ko vanish karna hoga.

KYUN. Free end ke saath, wiggles endpoint ko bhi move karne dete hain, isliye wahan $\eta =0$ nahi milta. $[\frac{\partial L}{\partial y^{\prime }}\eta {]}_a^b$ arbitrary end-wiggles ke liye tabhi vanish ho sakta hai jab $\frac{\partial L}{\partial y^{\prime }}$ khud free end pe zero ho. Yeh natural boundary conditions hain.

PICTURE. Left panel: pinned end — nail pe bump flat, boundary term $0$. Right panel: free end — edge pe bump non-zero (orange), jo $\frac{\partial L}{\partial y^{\prime }}{|}_{\text{end}}=0$ force karta hai.

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Ek-picture summary

Upar saab kuch, ek single flow mein: wire pin karo → use wiggle karo → 1-D dial tak collapse karo → bottom pe flat → chain rule → integration by parts boundary ko khatam karta hai → $\eta$ factor out karo → Fundamental Lemma → Euler–Lagrange.

Recall Feynman: poora walk ek 12-saal ke bache ko batao

Socho ek bendy wire jo ek board pe do nails se pin ki gayi hai. Tum chahte ho ki wire ki woh shape mile jo kuch total cost — uski length, ya ek bead ke neeche slide karne ka time — ko minimum kare. Tum kaise jaante ho ki tumne best shape pa li? Tum use kahin bhi beech mein ek tiny wiggle dete ho (nails ko wahi rakhte ho) aur check karte ho ki cost drop nahi hoti. Agar koi bhi wiggle ise sasta bana sakti, tum abhi best shape pe nahi the. Ab, "kisi bhi shape ki tiny wiggle, kahin bhi" test karna impossible lagta hai — lekin ek slick move hai. Pehle tum wiggle ko ek single dial mein badal do (dial ghuma o, wire zyada bulge karti hai), toh cost ab ek ordinary curve hai ek bottom ke saath, aur bottom wahan hai jahan dial zero padh raha hai. Phir tum do standard tricks use karte ho: chain rule (wire ko raise karne aur tilt karne pe cost ka har chhota slice kaise react karta hai?), aur integration by parts (ek tarika hai derivative ko wiggle se cost par dhakelne ka, jo nails pe ek leftover bhi produce karta hai — lekin woh leftover zero hai kyunki wiggle nails pe zero hai). Us cleanup ke baad, cost-change ek saaf sum hai: (ek khaas quantity) times (tumhari wiggle), wire ke saath joda gaya. Agar yeh har wiggle ke liye zero hai jo tum kabhi bhi draw kar sako, toh ek hi possibility hai ki woh "khaas quantity" har single point pe zero hai. Woh quantity Euler–Lagrange expression hai, aur use zero set karna woh rule hai jo perfect curve choose karta hai.

See also: Lagrangian mechanics, Geodesics and differential geometry, Brachistochrone problem, Functional analysis, Constrained optimization & Lagrange multipliers.