Question bank — Calculus of variations — functionals, functional derivative

Jhooth. Yeh sirf $\delta J=0$ enforce karti hai (ek stationary point); solution maximum ya saddle bhi ho sakta hai, bilkul jaise $f^{\prime }(x)=0$ maxima aur inflections bhi dhundta hai. Minimum confirm karne ke liye tumhe second-variation test chahiye.

Jhooth. Agar $L$ mein bare $y$ nahi hai, toh $\partial L/\partial y=0$ hai, isliye E–L reduce ho jaata hai $\frac{d}{dx}(\partial L/\partial y^{\prime })=0$ mein, matlab $\partial L/\partial y^{\prime }$ constant hai — $y$ nahi. $L=\sqrt{1+y^{\prime 2}}$ ke liye yeh nonzero slope wali straight line deta hai.

Jhooth. Yeh puri function of $x$ hai — har point par ek "gradient component," $\delta J=\int \frac{\delta J}{\delta y(x)}\delta y(x)dx$ se defined. Sirf kisi specific $x$ par evaluate karne ke baad tumhe ek number milta hai.

Partly jhooth. E–L ka interior equation dono cases mein hold karta hai, lekin free endpoints ke saath boundary term $[L_{y^{\prime }}\eta {]}_a^b$ survive karta hai aur natural boundary conditions $L_{y^{\prime }}=0$ free end par add karta hai. Isliye full stationarity condition genuinely change hoti hai.

Jhooth. Yeh E–L ka ek first integral hai, valid sirf tab jab $\partial L/\partial x=0$ ho. Isme E–L contained hai (isse differentiate karo aur tum $y^{\prime }(L_y-\frac{d}{dx}{L}_{y^{\prime }})=0$ recover kar lete ho), isliye yeh same information carry karta hai lekin ek kam derivative ke saath.

Jhooth. $\eta$ interior mein arbitrary hai lekin $\eta (a)=\eta (b)=0$ satisfy karne ke liye forced hai, kyunki trial curve $y+\epsilon \eta$ ko endpoints pinned rakhne chahiye. Wohi constraint boundary term ko khatam karta hai.

Jhooth. E–L mein outer $\frac{d}{dx}$ ek total derivative hai jo $y(x)$ aur $y^{\prime }(x)$ ko bhi chain rule ke through dekhta hai, $L_{y^{\prime }x}+L_{y^{\prime }y}{y}^{\prime }+L_{y^{\prime }{y}^{\prime }}{y}^{\prime \prime }$ produce karta hai. Operations ke yeh do orders yahan commute nahi karte.

Error: $\partial L/\partial y=0$ sirf pehli term ko kill karta hai. E–L phir bhi $\frac{d}{dx}(y^{\prime }/\sqrt{1+y^{\prime 2}})=0$ demand karta hai, jo constant slope force karta hai. Har curve qualify nahi karti — sirf straight lines.

Error: $y$ aur $y^{\prime }$ coupled hain — $y(x)$ ko perturb karne se automatically $y^{\prime }(x)$ bhi perturb hota hai. $-\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}$ term drop karna aadhi condition throw away kar deta hai (aur mechanics mein, saari dynamics).

Error: Boundary term "small" nahi hai, yeh yahan exactly zero hai — aur sirf isliye kyunki $\eta (a)=\eta (b)=0$. Ise "small" kehna yeh fact chhupa deta hai ki free endpoints ke saath yeh ek genuine, nonzero condition hai jise tumhe rakhna hai.

Error: Fundamental Lemma ko integral har smooth $\eta$ ke liye zero chahiye jo ends par vanish kare, na ki sirf ek ke liye. Ek $\eta$ tumhe almost kuch nahi bataata; "all $\eta$" wala condition hi $f$ ko pointwise pin karta hai.

Error: Doosre $L$ mein explicit $x$ hai, isliye $\partial L/\partial x\ne 0$ aur Beltrami apply nahi hota. Beltrami sirf tab licensed hai jab $x$, $L$ mein absent ho.

Error: Minimum ${\Phi }^{\prime }(0)=0$ force karta hai lekin ${\Phi }^{\prime \prime }(0)\ge 0$, na ki $=0$. ${\Phi }^{\prime \prime }(0)=0$ set karna degenerate/inflection case ki condition hogi, generic minimum ki nahi.

Kyunki unknown ek function $y(x)$ hai, isliye stationarity har point $x$ par wiggles ke against hold karni chahiye; conditions ka woh infinite family (Fundamental Lemma ke via) collapse hokar E–L ban jaata hai jo har $x$ par hold karta hai — ek differential equation.

$\Phi (\epsilon )$ single number $\epsilon$ ki ordinary function hai, isliye hum ordinary calculus use kar sakte hain; $x$-integral ke andar $\partial /\partial \epsilon$ move karna (smooth integrands ke liye legal) ek mushkil variational problem ko familiar chain-rule computation mein badal deta hai.

Hum ek single common factor $\eta$ extract karna chahte hain taaki Fundamental Lemma act kar sake. Jab tak $\eta$ aur ${\eta }^{\prime }$ dono appear karte hain, hum factor nahi kar sakte aur conclude nahi kar sakte ki bracket zero hai; IBP derivative ko ${\eta }^{\prime }$ se hat kar $L_{y^{\prime }}$ par le jaata hai.

Fundamental Lemma ko ek "bump" chahiye jo concentrate ho jahan $f\ne 0$ ho, aur derivation mein ${\eta }^{\prime }$ ko integral ke andar use kiya gaya tha — dono ke liye enough differentiability chahiye. Smoothness $y$ ki assumed smoothness se bhi match karti hai.

Yeh ek first-order relation hai (ek first integral), isliye usne already ek integration use kar li hai; tum $y^{\prime \prime }$ compute karne aur messy total derivative se bachte ho, ek 2nd-order ODE ko 1st-order mein badal deta hai.

Finite dimensions mein gradient ka har coordinate $x_i$ ke liye ek component hota hai; yahan "coordinates" har point $x$ par $y(x)$ ki values hain, isliye gradient ka ek component per point hota hai — ek vector ki jagah ek function.

Fixed endpoint par $\eta$ zero hone ke liye forced hai, jo boundary term ko automatically kill kar deta hai. Free endpoint par $\eta$ wahan nonzero ho sakta hai, isliye $\delta J=0$ still hold karne ke liye coefficient $L_{y^{\prime }}$ ka vanish hona zaroori hai.

$y^{\prime \prime }$-term coefficient $L_{y^{\prime }{y}^{\prime }}$ vanish ho jaata hai, isliye E–L degenerate hokar algebraic (non-differential) constraint $P_y-Q_x=0$ mein aa jaata hai; generally koi free curve solve karne ke liye nahi hoti — ek sign ki problem ill-posed hai ya usme hidden divergence structure hai.

Tab $\tilde{y}=y$ aur ${\Phi }^{\prime }(0)$ trivially $0=0$ deta hai — koi information nahi. Yahi reason hai ki hum condition sabhi nonzero admissible $\eta$ ke liye demand karte hain; zero perturbation useless degenerate case hai.

Catenary solution $y=c\cosh (\frac{x-x_0}{c})$ ki minimum value $c>0$ hai; endpoints ko axis ke bahut paas (small $y$) force karne se koi smooth solution exist nahi kar sakta, aur true minimiser "Goldschmidt" discontinuous solution mein degenerate ho jaata hai — smooth theory ki ek genuine failure.

Haan — E–L solution $y=mx+b$ hai jisme slope endpoints se fixed hai; equal endpoints sirf $m=0$ dete hain, ek horizontal segment, jo abhi bhi shortest path hai. Koi degeneracy nahi hoti.

Tab ek integration by parts kaafi nahi hai; tum do baar integration by parts karte ho, higher Euler–Lagrange equation $L_y-\frac{d}{dx}{L}_{y^{\prime }}+\frac{d^2}{d{x}^2}{L}_{y^{\prime \prime }}=0$ produce hoti hai aur har end par do boundary terms, jिनके लिए extra endpoint data chahiye.

$L=y$ deta hai $\partial L/\partial y=1$ aur $\partial L/\partial y^{\prime }=0$, isliye E–L $1=0$ padhta hai — koi stationary curve exist nahi karti. Functional ka simply koi interior extremum nahi hai; yeh jo bhi constraints tum add karo unki boundary par push ho jaata hai.

Saare partials vanish ho jaate hain, isliye E–L $0=0$ hai: har admissible curve stationary hai kyunki $J$ unke liye same constant $\times (b-a)$ hai. Variational problem degenerate hai — optimise karne ke liye kuch nahi.