Visual walkthrough — Euler-Lagrange equation — derivation

4.10.13 · D2 · Maths › Advanced Topics (Elite Level) › Euler-Lagrange equation — derivation

Related trails: Calculus of Variations · Principle of Least Action · Lagrangian Mechanics · Integration by Parts · Beltrami Identity.

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Step 1 — Functional actually hota kya hai?

KYA HAI. Ordinary calculus mein aap ek number $x$ ko function $f$ mein daalo aur ek number $f(x)$ nikalte ho. Functional ek level upar hai: aap poori curve $y(x)$ daalo, aur ek single number $J[y]$ nikalta hai.

$J[y]={\int }_a^b\underset{\begin{array}{c}\text{a rule that}\\ \text{scores each}\\ \text{tiny slice}\end{array}}{\underbrace{L}}(\underset{\text{where}}{\underbrace{x}},\underset{\text{height}}{\underbrace{y(x)}},\underset{\text{slope}}{\underbrace{y^{\prime }(x)}})dx$

• $y^{\prime }(x)=\frac{dy}{dx}$ curve ka slope hai — $x$ par kitni steeply rise karti hai.
• $L$, yaani Lagrangian, ek scoring machine hai: har point par dekho ki aap kahan ho, kitna upar ho, aur kitna steep ho — aur ek chhota sa cost nikalte ho.
• ${\int }_a^b\dots dx$ un saare chhhote costs ko curve ke saath add up karke ek total banata hai.

YEH STEP KYUN. Pehle yeh crystal clear hona chahiye ki hum jo cheez minimize kar rahe hain wo ek number hai jo poori curve se attached hai. Curve badlo → number badlo.

PICTURE. Left panel hai "number in, number out" wali duniya $f(x)$ ki. Right panel hai nayi duniya: ek red curve andar jaati hai, aur ek single number $J[y]$ bahar aata hai.

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Step 2 — Endpoints fix karo, phir nearby rivals banao

KYA HAI. Maano ki winning curve $y(x)$ hai (red mein drawn). Hum usse competitors ke ek poore family se compare karte hain, sab same do endpoints $(a,y_a)$ aur $(b,y_b)$ par pinned hain:

$Y(x,\epsilon )=\underset{\text{champion}}{\underbrace{y(x)}}+\underset{\begin{array}{c}\text{tiny}\\ \text{dial}\end{array}}{\underbrace{\epsilon }}\underset{\begin{array}{c}\text{wiggle}\\ \text{shape}\end{array}}{\underbrace{\eta (x)}}$

• $\eta (x)$ (Greek "eta") ek arbitrary smooth wiggle hai — jo bhi bump chahiye.
• $\epsilon$ (Greek "epsilon") ek dial hai jo wiggle ki size control karta hai. $\epsilon =0$ par hum exactly champion par hain.

YEH STEP KYUN. Ordinary calculus directly "saari curves pe differentiate" nahi kar sakta. Toh hum ek ordinary number, $\epsilon$, problem mein daal dete hain: yeh "poori curve wiggle karo" ko "ek knob ghao" mein badal deta hai.

Endpoint rule. Har rival ko same start aur finish pe hit karna chahiye. Kyunki champion already wahan hai, wiggle ko ends par zero hona chahiye: $\eta (a)=\eta (b)=0.$ Warna rival galat endpoint se guzrega — disqualified.

PICTURE. Red champion curve, kaafi faint black rivals uske aaspaas fan kar rahe hain, sab do endpoints par clamp hain jahan wiggle zero par pinched hai.

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Step 3 — Curve-search ko ek single-number graph mein collapse karo

KYA HAI. Poori family ko functional mein daalo. Result sirf dial $\epsilon$ par depend karta hai:

$\Phi (\epsilon )={\int }_a^{b}L(x,Y(x,\epsilon ),Y^{\prime }(x,\epsilon ))dx.$

Ab $\Phi$ ek number ka ordinary function hai. Kyunki champion ($\epsilon =0$ par) sabse best hai, dial ko kisi bhi taraf move karna cheezein worse ya flat hi banayega. Toh $\epsilon =0$ ek stationary point hai:

${\frac{d\Phi }{d\epsilon }|}_{\epsilon =0}=0.$

• $\frac{d\Phi }{d\epsilon }$ is one-variable graph ka ordinary slope hai.
• Ise zero set karna exactly waisa hi idea hai jaise "$f^{\prime }(x)=0$ at a minimum" — bas $\Phi$ par apply kiya.

YEH STEP KYUN. Yahi pura trick hai: ek infinite-dimensional problem (saari curves) ko ek 1-D problem (ek parabola-jaisa graph in $\epsilon$) mein squeeze kar diya. Aisi graph ka bottom kaise find karna hai yeh hum jaante hain.

PICTURE. Ek U-shape curve $\Phi (\epsilon )$ jiske flat bottom par $\epsilon =0$ pe red mark hai — wahan tangent horizontal hai.

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

KYA HAI. $\frac{d}{d\epsilon }$ ko integral ke andar le jaao. Pehle, ek notation reminder taaki kuch ambiguous na ho:

• $Y(x,\epsilon )=y(x)+\epsilon \eta (x)$ perturbed curve hai — ek rival jo dial $\epsilon$ par depend karta hai.
• $y(x)$ champion hai, $\epsilon =0$ par wala special rival.
• $\frac{\partial L}{\partial Y}$ ka matlab hai "scoring machine $L$ ko uske height slot ke respect se differentiate karo, phir us slot ko $Y$ par evaluate karo". Isi tarah $\frac{\partial L}{\partial Y^{\prime }}$ uske slope slot mein derivative hai, $Y^{\prime }$ par evaluate kiya.

$L$ ko $\epsilon$ sirf $Y$ aur uske slope $Y^{\prime }$ ke through feel hota hai, toh chain rule do channels deta hai:

$\frac{d\Phi }{d\epsilon }={\int }_a^b(\underset{\begin{array}{c}\text{how }L\text{ reacts}\\ \text{to height}\end{array}}{\underbrace{\frac{\partial L}{\partial Y}}}\underset{=\eta }{\underbrace{\frac{\partial Y}{\partial \epsilon }}}+\underset{\begin{array}{c}\text{how }L\text{ reacts}\\ \text{to slope}\end{array}}{\underbrace{\frac{\partial L}{\partial Y^{\prime }}}}\underset{={\eta }^{\prime }}{\underbrace{\frac{\partial Y^{\prime }}{\partial \epsilon }}})dx.$

Ab $\epsilon =0$ par evaluate karo. Wahan $Y=y$ aur $Y^{\prime }=y^{\prime }$ hain, toh height slot $y$ se fill hota hai aur slope slot $y^{\prime }$ se — exactly yahi har uppercase $Y$ ko lowercase $y$ mein badalta hai:

${\frac{\partial L}{\partial Y}|}_{\epsilon =0}=\frac{\partial L}{\partial y},{\frac{\partial L}{\partial Y^{\prime }}|}_{\epsilon =0}=\frac{\partial L}{\partial y^{\prime }}.$

Kyunki $Y=y+\epsilon \eta$ hai, $\epsilon$ bump karne se height $\eta$ se aur slope ${\eta }^{\prime }$ se badhti hai. Sab milake:

$0={\int }_a^b\left(\frac{\partial L}{\partial y}\eta +\frac{\partial L}{\partial y^{\prime }}{\eta }^{\prime }\right)dx.$

YEH STEP KYUN. Humne "graph flat hai" ko ek concrete integral condition mein badal diya jo wiggle $\eta$, uske slope ${\eta }^{\prime }$, aur scoring machine ki sensitivities ko — sab champion curve $y$ ke saath read kiye — tie karta hai.

PICTURE. Do "sensitivity channels": ek arrow height wiggle $\eta$ se $\partial L/\partial y$ mein, ek slope wiggle ${\eta }^{\prime }$ se $\partial L/\partial y^{\prime }$ mein.

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Step 5 — Integration by parts: $\eta$ ko uske derivative se azaad karo

KYA HAI. ${\eta }^{\prime }$ wala term annoying hai — usme wiggle ka derivative hai, wiggle khud nahi. Wo tool jo ek factor se doosre factor par derivative swap karta hai wo Integration by Parts hai. Uski identity hai:

${\int }_a^{b}udv=[uv{]}_a^b-{\int }_a^{b}vdu.$

Yahan hum woh choice karte hain jo $\eta$ se derivative hatati hai:

• $u=\frac{\partial L}{\partial y^{\prime }}$ lo, toh $du=\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)dx$;
• $dv={\eta }^{\prime }dx$ lo, toh $v=\eta$ (${\eta }^{\prime }$ integrate karne se $\eta$ wapas milta hai).

Unhe identity mein daalo:

${\int }_a^b\frac{\partial L}{\partial y^{\prime }}{\eta }^{\prime }dx=\underset{\text{the }[uv{]}_a^b\text{ boundary term}}{\underbrace{{\left[\frac{\partial L}{\partial y^{\prime }}\eta \right]}_a^b}}-{\int }_a^b\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)\eta dx.$

• Boundary term sirf endpoints $a$ aur $b$ par evaluate hota hai.
• Step 2 se, $\eta (a)=\eta (b)=0$, toh yeh poora boundary term zero hai. Exactly isliye humne endpoints pin kiye the.

YEH STEP KYUN. Hum chahte hain ki sab kuch multiply karte hue ek single, bare $\eta$ ho, taaki baad mein argue kar sakein ki "$\eta$ arbitrary hai". ${\eta }^{\prime }$ ne yeh block kiya tha; integration by parts ($dv={\eta }^{\prime }dx$ choose karke taaki derivative $L$ par jaaye) use hata deta hai — ek boundary term ki chhoti si keemat par — jo hamare endpoint rule se khatam ho jaati hai.

PICTURE. Boundary term ko $x=a$ aur $x=b$ par do dots ki tarah draw kiya; unpar ek red "= 0" stamp kyunki $\eta$ wahan pinched hai.

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Step 6 — Wiggle ko factor out karo

KYA HAI. Pieces ko wapas jodo. Dono surviving terms ab ek bare $\eta$ carry karte hain, toh factor karo:

$0={\int }_a^b\underset{\text{call this }g(x)}{\underbrace{\left(\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}\right)}}\eta (x)dx.$

• $g(x)$ champion curve se determined ek fixed function hai — yeh wiggle par depend nahi karta.
• $\eta (x)$ koi bhi smooth wiggle hai jo ends par vanish kare.

YEH STEP KYUN. Humne situation isolate kar li: "koi fixed function $g$, har possible wiggle $\eta$ ke saath dot kiya, hamesha zero deta hai." Yeh bahut strong statement hai — Step 7 isko cash karta hai.

PICTURE. Ek single black landscape $g(x)$ aur uske upar ek red wiggle $\eta (x)$; shaded product $g\cdot \eta$ har red wiggle ke liye exactly zero tak integrate hona chahiye.

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Step 7 — Fundamental Lemma force karta hai $g(x)=0$

KYA HAI. Agar har admissible $\eta$ ke liye ${\int }_a^{b}g(x)\eta (x)dx=0$ hai, toh $g(x)=0$ har jagah $[a,b]$ par.

YEH KYUN SACH HONA CHAHIYE (steel-man). Maano $g$ kisi chhote interval par, say, positive hota. Toh ek wiggle $\eta$ choose karo jo exactly us interval par ek chhota positive bump ho aur baaki jagah zero. Product $g\cdot \eta$ wahan positive hai aur baaki jagah zero, toh integral strictly positive hai — "yeh hamesha zero hai" se contradict karta hai. Iska ek hi escape hai: $g\equiv 0$.

PICTURE. Ek jagah jahan $g>0$ hai (black), aur ek red bump $\eta$ exactly wahan planted; shaded positive area prove karta hai ki integral zero nahi ho sakta jab tak $g$ vanish na kare.

$g(x)=0$ set karne se result milta hai:

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Step 8 — Edge case: agar endpoints free hoon toh?

KYA HAI. Step 5 mein humne boundary term ${\left[\frac{\partial L}{\partial y^{\prime }}\eta \right]}_a^b$ ko $\eta (a)=\eta (b)=0$ use karke kill kiya. Par maano ek endpoint pinned nahi hai — curve ek vertical line par kahin bhi end ho sakti hai, toh $\eta$ ko wahan vanish karna zaroor nahi. Tab boundary term ka woh hissa survive karta hai, aur (kyunki interior pehle se E–L equation force kar chuka hai) use apne aap vanish karna padega. Teen sub-cases:

(a) Sirf right end free hai ($\eta (a)=0$ lekin $\eta (b)$ arbitrary): ${\frac{\partial L}{\partial y^{\prime }}|}_{x=b}=0.$

(b) Sirf left end free hai ($\eta (b)=0$ lekin $\eta (a)$ arbitrary): ${\frac{\partial L}{\partial y^{\prime }}|}_{x=a}=0.$

(c) Dono ends free hain ($\eta (a)$ aur $\eta (b)$ dono arbitrary aur independent hain): aapko ek saath do natural boundary conditions milti hain, ${\frac{\partial L}{\partial y^{\prime }}|}_{x=a}=0\text{aur}{\frac{\partial L}{\partial y^{\prime }}|}_{x=b}=0.$

(d) Endpoints given curves/manifolds par lie karne ke liye constrained hain (fixed points nahi, fully free bhi nahi — jaise end ko ek target curve par sit karna hai): surviving boundary term ek transversality condition produce karta hai, ek relation jo $L$ aur $\frac{\partial L}{\partial y^{\prime }}$ ko mix karta hai aur kehta hai ki extremal ko target curve se ek prescribed angle par milna chahiye. Yeh woh general umbrella hai jiske special cases (a)–(c) hain.

YEH STEP KYUN MAAYANE RAKHTA HAI. Aapko hamesha check karna chahiye ki endpoints kya kar sakte hain. Fixed points → boundary term khatam, plain E–L. Free ends → E–L plus har free end ke liye ek natural condition. Ends ek manifold par → E–L plus ek transversality condition. Surviving boundary term bhool jaana ek classic trap hai.

PICTURE. Left: dono ends pinned hain (wiggle vanish hoti hai → boundary term khatam). Right: right end ek vertical rail (red) par slide karne ke liye free hai, toh wahan ek nayi condition $\partial L/\partial y^{\prime }=0$ appear hoti hai.

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Ek-picture summary

Upar sab kuch, ek single flow mein compress: curve → number → dial → flat graph → integral condition → E–L ODE.

Recall Feynman retelling — plain words mein bol do

Hum chahte the best curve, best number nahi. Trick: har curve ko ek number $J$ se score karo. Us curve ko jo humhe lagta hai best hai, ek tiny adjustable wiggle $\epsilon \eta (x)$ se nudge karo, uske do ends nailed down rakhte hue. Ab score sirf ek dial $\epsilon$ par depend karta hai, toh yeh ek ordinary graph hai — aur agar hamaari curve truly best hai, toh woh graph $\epsilon =0$ par flat hai. "Flat" ko derivative ke roop mein likhne se ek integral milta hai jo wiggle $\eta$ aur uske slope ${\eta }^{\prime }$ ko mix karta hai. Hum wiggle ka slope idhar-udhar nahi chahte, toh hum integrate by parts karte hain usse trade karne ke liye; leftover boundary term mar jaata hai kyunki humne ends pin kiye the. Ab poora integral hai "(koi fixed function $g$) times (koi bhi wiggle) $=0$". Kyunki wiggle koi bhi bump ho sakta hai, fixed function har jagah zero hona chahiye — warna wahan ek bump plant karo jahan zero nahi hai aur equation tod do. Woh "$g=0$" hi Euler–Lagrange equation hai. Agar ek end pin karne ki bajaye free chhod diya jaaye, toh boundary term survive karta hai aur wahan ek bonus condition deta hai; agar dono ends free hain toh do aisi conditions milti hain, aur agar ek end ko ek target curve par ride karna ho toh uski jagah ek transversality condition milti hai.

Recall Quick self-test

Fixed endpoints ke liye $\eta (a)=\eta (b)=0$ kyun hona chahiye? ::: Har rival curve ko champion ke same do endpoints se pass karna chahiye, toh wiggle un points ko move nahi kar sakta. Kaunsa step zaroor endpoint condition chahta hai? ::: Step 5 — yeh integration by parts se aane wale boundary term ko vanish karta hai. Integration by parts mein, $u$ aur $dv$ ki kaunsi choice $\eta$ se derivative hatati hai? ::: $u=\partial L/\partial y^{\prime }$ aur $dv={\eta }^{\prime }dx$, toh $v=\eta$ derivative-free nikalte ho. Fundamental Lemma $\int g\eta dx=0$ se hume kya conclude karne deta hai? ::: Ki $g(x)=0$ har jagah, kyunki $\eta$ arbitrary hai. Agar dono endpoints free hain, toh kaun se extra conditions appear hote hain? ::: Do natural boundary conditions, $\partial L/\partial y^{\prime }=0$ har end par.