Exercises — Euler-Lagrange equation — derivation

KHAATA KYA HAI: teeno slots $x$, $y$, $y^{\prime }$ ko independent labels maano aur slot by slot differentiate karo. KYU: partial $\partial /\partial y$ ka matlab hai "$x$ aur $y^{\prime }$ ko freeze karo, is step ke liye woh sirf constants hain".

• $\frac{\partial L}{\partial y}$: sirf $y^2$ term mein $y$ hai, isliye $\frac{\partial L}{\partial y}=2y$.
• $\frac{\partial L}{\partial y^{\prime }}$: sirf $x{y}^{\prime 2}$ term mein $y^{\prime }$ hai; $x$ frozen hai, isliye $\frac{\partial L}{\partial y^{\prime }}=2x{y}^{\prime }$. Answer: $\boxed{\frac{\partial L}{\partial y}=2y,\frac{\partial L}{\partial y^{\prime }}=2x{y}^{\prime }}$

KHAATA KYA HAI: ek akela $x$ dhuundo jo $y$ ya $y^{\prime }$ ke andar chhupta nahi ho.

• (a) $\sqrt{1+y^{\prime 2}}$ — koi akela $x$ nahi. ✅ Beltrami apply hota hai.
• (b) $x{y}^{\prime 2}$ — factor $x$ explicitly wahan baitha hai. ❌ Beltrami nahi apply hota.
• (c) $\sqrt{(1+y^{\prime 2})/y}$ — sirf $y$ aur $y^{\prime }$ dikh rahe hain. ✅ Beltrami apply hota hai. KYU matter karta hai: Beltrami yeh kehta hai ki "agar rules $x$ ke saath nahi badte, toh ek khaas combination conserved hai". Visible $x$ ka matlab hai rules position ke saath badal rahe hain, toh kuch conserved nahi hoga.

KHAATA KYA HAI / KYU: $\frac{\partial L}{\partial y}=0$ (koi $y$ nahi dikh raha), toh E–L sirf $\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}=0$ reh jaata hai. Agar kisi cheez ki derivative zero hai, toh woh cheez constant hai.

• $\frac{\partial L}{\partial y^{\prime }}=\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}=c$.
• Square karna aur solve karna — branches ka dhyan rakho. Squaring deta hai $y^{\prime 2}=c^2(1+y^{\prime 2})$, isliye $y^{\prime 2}=\frac{c^2}{1-c^2}$. Root lene se $\pm$ aata hai: $y^{\prime }=\pm \frac{c}{\sqrt{1-c^2}}$. KYU sign problem nahi hai yahan: right side ek fixed constant hai (poori cheez ko $m$ bolo, sign samet), kyunki $c$ constant hai. Hum jis bhi branch par hain, $y^{\prime }$ apni value kabhi nahi badalta curve ke saath — squaring ek aisi curve nahi bana sakta jiska slope flip ho, kyunki $y^{\prime }$ ek constant par fixed hai. Toh $y^{\prime }=m=\text{const}$, aur $m$ ka sign baad mein sirf endpoints se fix hota hai (yahan positive aata hai).
• DIKHTA KAISA HAI (Fig. s01): constant slope ka matlab hai extremal ek seedhi ramp hai. Figure mein orange line $y=2x$ woh extremal hai; ek teal dashed competitor $y+\epsilon \eta$ bahar bulge karta hai aur visibly longer hai, aur plum dots woh do fixed endpoints hain jinhe wiggle $\eta$ disturb nahi kar sakta ($\eta =0$ dono ends par).
• Toh $y=mx+b$. Endpoints: $y(0)=0\Rightarrow b=0$; $y(1)=2\Rightarrow m=2$ (positive branch data se select hua). Answer: $\boxed{y=2x}$.

KHAATA KYA HAI: seedha E–L apply karo. $\frac{\partial L}{\partial y}=2y$, $\frac{\partial L}{\partial y^{\prime }}=2y^{\prime }$.

• E–L: $2y-\frac{d}{dx}(2y^{\prime })=0\Rightarrow y^{\prime \prime }-y=0$.
• KYU solution $e^x$ aur $e^{-x}$ se banta hai (grounded, asserted nahi): equation $y^{\prime \prime }=y$ kehti hai "doosri derivative original function wapas de deti hai". Socho: kaun si functions apni doosri derivative khud hain? Exponential $e^{kx}$ mein $\frac{d^2}{d{x}^2}{e}^{kx}=k^2{e}^{kx}$ hai, toh $y^{\prime \prime }=y$ force karta hai $k^2=1$, matlab $k=+1$ ya $k=-1$. Isliye exactly do building blocks aate hain, $e^x$ (ek curve jo uthti jaati hai aur steep hoti jaati hai) aur $e^{-x}$ (ek curve jo decay karti hai), aur har solution $y=P{e}^x+Q{e}^{-x}$ ka blend hai.
• KYU $\sinh ,\cosh$ mein likhein: do natural symmetric blends hain $\cosh x=\frac{1}{2}(e^x+e^{-x})$ (even, $x=0$ par value $1$) aur $\sinh x=\frac{1}{2}(e^x-e^{-x})$ (odd, $x=0$ par value $0$). Kyunki humara left endpoint demand karta hai $y(0)=0$, toh $\cosh$ piece (jo $0$ par $1$ hai) ko switch off karna hoga — isliye $\sinh$ yahan tailor-made basis hai.
• DIKHTA KAISA HAI (Fig. s02): figure mein $e^x$ (teal) aur $e^{-x}$ (plum) plot hain, aur unka difference $\sinh x$ (orange) — dhyaan do ki orange curve origin se thiktak guzarti hai ($y(0)=0$) exactly jaise boundary condition chahta hai, jabki uska gentle upward bow extremal shape hai.
• General solution $y=A\sinh x+B\cosh x$. $y(0)=0\Rightarrow B=0$. $y(1)=1\Rightarrow A=\frac{1}{\sinh 1}$. Answer: $\boxed{y=\frac{\sinh x}{\sinh 1}}$.

KHAATA KYA HAI: $\partial L/\partial x=0$, toh Beltrami deta hai $L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}=C$.

• $\frac{\partial L}{\partial y^{\prime }}=\sqrt{y}\cdot \frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}$.
• Beltrami: $\sqrt{y}\sqrt{1+y^{\prime 2}}-\sqrt{y}\frac{y^{\prime 2}}{\sqrt{1+y^{\prime 2}}}=\sqrt{y}\cdot \frac{(1+y^{\prime 2})-y^{\prime 2}}{\sqrt{1+y^{\prime 2}}}=\frac{\sqrt{y}}{\sqrt{1+y^{\prime 2}}}=C$.
• KYU beech wala step: common denominator, aur $(1+y^{\prime 2})-y^{\prime 2}=1$ — $y^{\prime 2}$ terms cancel ho jaate hain, yahi Beltrami ka poora point hai. Answer (first-order ODE): $\boxed{\frac{y}{1+y^{\prime 2}}=C^2}$.

KHAATA KYA HAI: Beltrami ek valid first integral produce karta hai E–L ka jab bhi $\partial L/\partial x=0$ ho. Toh consistency derivation se hi guaranteed hai — hum ise re-trace karte hain sure hone ke liye.

• Beltrami derive kiya gaya tha $\frac{d}{dx}\left(L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=\frac{\partial L}{\partial x}$ se.
• Yahan $\partial L/\partial x=0$, toh $L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}=C$ (constant). Yeh E–L ka ek integrated form hai; ise ek baar differentiate karne par $y^{\prime }\left(\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}\right)=0$ milta hai.
• KYU consistent hai — aur $y^{\prime }=0$ edge case explicitly bata rahe hain. Product $y^{\prime }\cdot (\text{E–L expression})=0$ do tarike se satisfy ho sakta hai: ya toh E–L bracket zero ho, ya $y^{\prime }=0$ ho (ek horizontal point/segment). Ek isolated turning point par jahan $y^{\prime }=0$ momentarily ho, equation $y^{\prime }=0$ factor se trivially satisfy hoti hai, toh us ek instant par Beltrami E–L bracket ke baare mein kuch naya nahi bolta — lekin E–L ek continuous condition hai, aur jahan bhi curve genuinely slide kar rahi hai ($y^{\prime }\ne 0$) bracket zero hone par force hoti hai; continuity se bracket isolated $y^{\prime }=0$ point ke through bhi zero rahti hai. Genuinely degenerate case sirf woh hai jab curve poore interval par horizontal ho ($y^{\prime }\equiv 0$ wahan): tab Beltrami $L=C$ tak degenerate ho jaata hai aur us stretch par koi E–L information nahi deta, aur tumhe directly full E–L equation par wapas jaana padta hai. Brachistochrone ke liye curve strictly descend karti hai sirf start ko chhodkar, toh yeh degenerate stretch kabhi nahi aati.
• Ab is $L$ ke liye explicitly algebra karo taaki kuch chupta nahi rahe. $L=\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{y}}$ likho.
  • Step 1 — momentum term: $\frac{\partial L}{\partial y^{\prime }}=\frac{1}{\sqrt{y}}\cdot \frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}=\frac{y^{\prime }}{\sqrt{y}\sqrt{1+y^{\prime 2}}}$. KYU: $\sqrt{1+y^{\prime 2}}$ par chain rule, jahan $1/\sqrt{y}$ is partial ke liye frozen constant hai.
  • Step 2 — $L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}$ banao: $\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{y}}-\frac{y^{\prime 2}}{\sqrt{y}\sqrt{1+y^{\prime 2}}}=\frac{(1+y^{\prime 2})-y^{\prime 2}}{\sqrt{y}\sqrt{1+y^{\prime 2}}}=\frac{1}{\sqrt{y}\sqrt{1+y^{\prime 2}}}.$ KYU: common denominator $\sqrt{y}\sqrt{1+y^{\prime 2}}$, phir $(1+y^{\prime 2})-y^{\prime 2}=1$ — wohi $y^{\prime 2}$-cancellation jo Beltrami ko useful banati hai.
  • Step 3 — Beltrami constant $C$ ke barabar rakho aur square karo: $\frac{1}{\sqrt{y}\sqrt{1+y^{\prime 2}}}=C\Rightarrow y(1+y^{\prime 2})=\frac{1}{C^2}=\text{const}$. Answer: first integral $y(1+y^{\prime 2})=\text{const}$ E–L ka ek legitimate consequence hai; koi contradiction nahi.

KHAATA KYA HAI: free right endpoint ka matlab hai boundary term ${\left[\frac{\partial L}{\partial y^{\prime }}\eta \right]}_0^1$ jo Integration by Parts se aata hai, woh $x=1$ par automatically zero nahi hota; woh ek natural boundary condition force karta hai.

• Interior E–L: $\frac{\partial L}{\partial y}=1$, $\frac{\partial L}{\partial y^{\prime }}=y^{\prime }$. Toh $1-\frac{d}{dx}(y^{\prime })=0\Rightarrow y^{\prime \prime }=1$.
• Integrate karo: $y=\frac{1}{2}{x}^2+Ax+B$. Fixed left end $y(0)=0\Rightarrow B=0$.
• $x=1$ par Natural BC: ${\frac{\partial L}{\partial y^{\prime }}|}_{x=1}=y^{\prime }(1)=0$. KYU: $\eta (1)$ free hone ke saath, boundary term sirf tab zero hogi jab $\partial L/\partial y^{\prime }=0$ wahan ho.
• $y^{\prime }=x+A$, toh $y^{\prime }(1)=1+A=0\Rightarrow A=-1$. Answer: $\boxed{y=\frac{1}{2}{x}^2-x}$, aur $y(1)=-\frac{1}{2}$.

KHAATA KYA HAI: kaafi saare unknown functions $q_i(t)$ hone par, har function ke liye ek E–L equation hoti hai — tum har ek ko independently wiggle karte ho.

• $q_1$ ke liye: $\frac{\partial L}{\partial q_1}=-q_1$, $\frac{\partial L}{\partial {\dot{q}}_1}={\dot{q}}_1$. E–L: $-q_1-\frac{d}{dt}{\dot{q}}_1=0\Rightarrow {\ddot{q}}_1=-q_1$.
• Same algebra se: ${\ddot{q}}_2=-q_2$.
• KYU / DIKHTA KAISA HAI: $\ddot{q}=-q$ simple harmonic motion hai; har coordinate $\cos t,\sin t$ jaisa oscillate karta hai. Do independent oscillators. Answer: $\boxed{{\ddot{q}}_1=-q_1,{\ddot{q}}_2=-q_2}$ — do uncoupled simple-harmonic oscillators.

KHAATA KYA HAI: candidate conserved quantity ki total derivative compute karo aur dikhao ki woh zero hai.

• $L$ ki total derivative: $\frac{dL}{dx}=\frac{\partial L}{\partial x}+\frac{\partial L}{\partial y}{y}^{\prime }+\frac{\partial L}{\partial y^{\prime }}{y}^{\prime \prime }$.
• Ab compute karo $\frac{d}{dx}\left(y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=y^{\prime \prime }\frac{\partial L}{\partial y^{\prime }}+y^{\prime }\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}$.
• Subtract karo: $\frac{d}{dx}\left(L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=\frac{\partial L}{\partial x}+y^{\prime }\underset{=0\text{by E–L}}{\underbrace{\left(\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}\right)}}$.
• KYU bracket zero hota hai: woh bracket hi Euler–Lagrange expression hai, har extremal par zero.
• Toh $\frac{d}{dx}\left(L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=\frac{\partial L}{\partial x}$. Agar $\partial L/\partial x=0$, toh left side constant hai. Answer: $\boxed{L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}=\text{const}}$ (dekho Beltrami Identity).

KHAATA KYA HAI: $L=y\sqrt{1+y^{\prime 2}}$, koi explicit $x$ nahi → Beltrami.

• $\frac{\partial L}{\partial y^{\prime }}=\frac{y{y}^{\prime }}{\sqrt{1+y^{\prime 2}}}$.
• Beltrami: $y\sqrt{1+y^{\prime 2}}-\frac{y{y}^{\prime 2}}{\sqrt{1+y^{\prime 2}}}=\frac{y}{\sqrt{1+y^{\prime 2}}}=a$. KYU: $y^{\prime 2}$ terms exactly cancel ho jaate hain, jaise L2.3 mein. $a$ kahan se aata hai? Yeh simply Beltrami constant hai — woh fixed number jiske barabar left side hoti hai. L2.3 mein hum ise $C$ bolte the; yahan usi constant ko $a$ naam dete hain kyunki, jaise do lines baad dikhega, $a$ ek length nikalta hai (neck radius), aur ise $a$ bolne se woh geometric meaning obvious ho jaata hai. Toh $a$ define hota hai $a:=\frac{y}{\sqrt{1+y^{\prime 2}}}$ se, curve ke saath conserved Beltrami value.
• Rearrange karo: $\sqrt{1+y^{\prime 2}}=\frac{y}{a}\Rightarrow y^{\prime 2}=\frac{y^2}{a^2}-1\Rightarrow y^{\prime }=\pm \sqrt{\frac{y^2}{a^2}-1}$.
• Integrate karo — key step, poora dikha rahe hain. Variables separate karo: $\frac{dy}{\sqrt{y^2/a^2-1}}=\pm dx$. Upar neeche $a$ se multiply karo: $\frac{ady}{\sqrt{y^2-a^2}}=\pm dx$. KYU yeh integral: $\frac{1}{\sqrt{y^2-a^2}}$ ki shape inverse hyperbolic cosine ka tell-tale signature hai, kyunki $\frac{d}{dy}\mathrm{arcosh}(y/a)=\frac{1}{\sqrt{y^2-a^2}}$. Dono sides integrate karo: $a\mathrm{arcosh}(y/a)=\pm (x-b)$, jahan $b$ integration constant hai. $y$ ke liye solve karo: $\mathrm{arcosh}(y/a)=\pm \frac{x-b}{a}$, aur kyunki $\cosh$ even hai toh $\pm$ absorb ho jaata hai, milta hai $\boxed{y=a\cosh \left(\frac{x-b}{a}\right)}$.
• Kaun sa branch? Dono legitimate hain — aur dono chahiye. $+$ branch film ko throat se dur uthte hue describe karta hai (increasing $y$), $-$ branch use throat ki taraf girte hue describe karta hai. KYU answer nahi badalta: $y=a\cosh \left(\frac{x-b}{a}\right)$ mein $y^{\prime }=\sinh \left(\frac{x-b}{a}\right)$ hai, jo throat $x=b$ ke baayein negative hai ($-$ branch) aur daayein positive ($+$ branch). Ek single symmetric catenary automatically dono branches ko stitch karta hai, throat par milta hai jahan $y^{\prime }=0$ hai. Toh "positive branch choose karna" bas right half describe karne ka choice hai.
• DIKHTA KAISA HAI (Fig. s03): orange curve woh catenary hai; bottom par plum dot throat hai jahan $y^{\prime }=0$ aur $y=a$ (yahan $a=1$); do teal vertical bars woh rings hain jahan film ko pahunchna hai $x=\pm 1$ par. Is profile ko horizontal axis ke around revolve karne par catenoid soap film banta hai.
• $a=1,b=0$ ke saath: $y=\cosh x$. Endpoints check karo: $y(\pm 1)=\cosh 1\approx 1.5431$. ✓ Answer: $\boxed{y=a\cosh \left(\frac{x-b}{a}\right)}$; symmetric case $y=\cosh x$.

KHAATA KYA HAI: do symmetries → do conserved quantities.

• Time-translation ($\partial L/\partial t=0$): Beltrami-in-$t$ deta hai $E={\dot{q}}_1\frac{\partial L}{\partial {\dot{q}}_1}+{\dot{q}}_2\frac{\partial L}{\partial {\dot{q}}_2}-L=\frac{1}{2}({\dot{q}}_1^2+{\dot{q}}_2^2)+\frac{1}{2}(q_1^2+q_2^2)$ = total energy, conserved.
• $(q_1,q_2)$ plane mein rotational symmetry: Noether's Theorem conserved angular momentum $\ell =q_1{\dot{q}}_2-q_2{\dot{q}}_1$ deta hai.
• KYU: action ki har continuous symmetry ek conserved quantity yield karti hai — yahi precisely Noether's theorem hai, jiska Beltrami (time symmetry ke liye) special case hai. Answer: energy $E=\frac{1}{2}({\dot{q}}_1^2+{\dot{q}}_2^2+q_1^2+q_2^2)$ aur angular momentum $\ell =q_1{\dot{q}}_2-q_2{\dot{q}}_1$.

Beltrami tab apply hota hai jab :: $L$ mein koi explicit $x$ na ho ($\partial L/\partial x=0$). Ek free endpoint fixed height ki jagah replace karta hai :: natural boundary condition $\partial L/\partial y^{\prime }=0$ se wahan. $n$ unknown functions se kitne E–L equations bante hain :: exactly $n$, ek har function ke liye. Catenoid neck radius barabar hoti hai :: Beltrami constant $a$ ke, jahan $y^{\prime }=0$.