Euler-Lagrange equation — derivation: Question bank

4.10.13 · D5 · Maths › Advanced Topics (Elite Level) › Euler-Lagrange equation — derivation

Un symbols ki reminder jo tumhe milenge (sab parent mein build kiye gaye hain): $J [y] = \int_{a}^{b} L (x, y, y^{'}) d x$ functional hai (poori curve ko assign kiya gaya ek number), $η (x)$ arbitrary wiggle shape hai, $ε$ chhota dial hai, aur Euler–Lagrange (E–L) equation hai $\frac{\partial L}{\partial y} - \frac{d}{d x} \frac{\partial L}{\partial y ^{'}} = 0$ .

True or false — justify

E–L minimum ke liye ek sufficient condition deta hai

False — yeh sirf necessary hai (stationarity, "curves ke liye gradient zero"); ek solution minimum, maximum, ya saddle ho sakta hai. Yeh jaanne ke liye second variation test karni padti hai.

E–L equation ka har solution us

L

ke liye shortest/least curve hota hai

False — E–L stationary curves dhundta hai. Shortest-path

L

ke liye woh minima hote hain, lekin doosre

L

ke liye ek extremal, functional ka maximum ya inflection ho sakta hai.

Agar

L

mein

y

explicitly nahi hai, toh extremal ke saath

\frac{\partial L}{\partial y ^{'}}

constant hota hai

True — E–L reduce ho jaata hai

\frac{d}{d x} \frac{\partial L}{\partial y ^{'}} = 0

mein, toh woh quantity conserved hai. Yeh "

y

is a cyclic coordinate" first integral hai.

Agar

L

mein

x

explicitly nahi hai, toh

L

khud extremal ke saath constant hota hai

False —

L - y^{'} \frac{\partial L}{\partial y ^{'}}

(Beltrami quantity) constant hota hai,

L

nahi. Dekho Beltrami Identity.

Endpoint condition

η (a) = η (b) = 0

sirf ek convenience hai jise hum drop kar sakte hain

False — yahi hai jo Integration by Parts mein boundary term ko khatam karta hai. Ise drop karo aur tumhe extra natural boundary conditions milte hain

\frac{\partial L}{\partial y ^{'}}_{a, b} = 0

Φ (ε) = \int_{a}^{b} L (x, Y, Y^{'}) d x

ek real variable ka ordinary function hai

True — jab wiggle shape

η

fix ho jaati hai, sirf dial

ε

vary karta hai, toh

Φ

ek plain function hai jiska derivative hum zero set kar sakte hain.

Fundamental Lemma ko integral ek cleverly chosen

η

ke liye zero hone ki zaroorat hai

False — yeh har admissible

η

ke liye zero hona chahiye. Ek single

η

positive aur negative parts ke cancellation se nonzero integrand chhupa sakta hai.

L

mein constant add karne se extremal curve badal jaati hai

False — ek constant

C (b - a)

mein integrate hota hai,

y

se independent, toh

\frac{\partial L}{\partial y}

aur

\frac{\partial L}{\partial y ^{'}}

dono unchanged hain; E–L same curve deta hai.

L

ko ek positive constant se multiply karne par extremal unchanged rehta hai

True — E–L,

L

mein linear/homogeneous hai, toh

k > 0

se scale karne par poori equation

k

se scale hoti hai aur zero set (extremal) identical hota hai.

Spot the error

\frac{d}{d x}

aur

\frac{\partial}{\partial y ^{'}}

commute karte hain, toh E–L hai

\frac{\partial L}{\partial y} - \frac{\partial L}{\partial y ^{'}} = 0

Wrong —

\frac{d}{d x} \frac{\partial L}{\partial y ^{'}}

ek total

x

-derivative hai jo chain rule se

x

y

, aur

y^{'}

ko hit karta hai:

\frac{\partial ^{2} L}{\partial x \partial y ^{'}} + y^{'} \frac{\partial ^{2} L}{\partial y \partial y ^{'}} + y^{''} \frac{\partial ^{2} L}{\partial y ^{'2}}

. Yeh

\frac{\partial L}{\partial y ^{'}}

nahi hai.

"Step 4 mein chain rule

\frac{\partial L}{\partial x} \cdot \frac{\partial x}{\partial ε}

bhi deta hai."

Wrong —

x

independent variable hai aur

ε

par depend nahi karta, toh

\frac{\partial x}{\partial ε} = 0

. Sirf

Y

aur

Y^{'}

mein

ε

-dependence hai.

"Kyunki shortest-path case mein

\frac{\partial L}{\partial y ^{'}} = \frac{y ^{'}}{1 + y ^{'2}} = c

, hum conclude karte hain

y^{'} = c

Wrong —

\frac{y ^{'2}}{1 + y ^{'2}} = c^{2}

solve karne par

y^{'}

kisi constant

m

ke barabar milta hai,

c

ke nahi. Point yeh hai ki

y^{'}

constant hai, isliye ek straight line;

m

ek alag constant hai.

"Beltrami step ke liye hum

\frac{\partial L}{\partial y} = \frac{d}{d x} \frac{\partial L}{\partial y ^{'}}

substitute karte hain jo hamesha true hai."

Wrong — woh substitution E–L hai, sirf extremal par valid hai. Beltrami E–L ke solutions ke baare mein ek statement hai, arbitrary curves ke liye identity nahi.

\int_{a}^{b} g η d x = 0

jab

g

continuous hai aur kahin

g (x_{0}) > 0

hai, phir bhi possible hai."

Wrong —

η

x_{0}

ke paas concentrated smooth positive bump lo (endpoints par zero); integral strictly positive ho jaata hai, ek contradiction. Isliye

g \equiv 0

"Newton's law example:

\frac{\partial L}{\partial q ˙} = m \overset{q}{˙}

, aur

\frac{d}{d t} (m \overset{q}{˙}) = 0

Wrong —

\frac{d}{d t} (m \overset{q}{˙}) = m \overset{q}{¨}

hai, jo generally nonzero hai. E–L deta hai

- V^{'} (q) - m \overset{q}{¨} = 0 \Rightarrow m \overset{q}{¨} = - V^{'} (q) = F

Why questions

Kyun

Φ (ε)

par reduce karna "infinitely many unknowns" ki problem solve karta hai

Kyunki wiggle shape

η

fix karne ke baad, poora perturbation sirf single real dial

ε

se govern hota hai, jo infinite-dimensional search ko ordinary 1-D calculus mein convert karta hai.

Kyun

Φ^{'} (0) = 0

ki demand har

η

ke liye hold karni chahiye, sirf ek ke liye nahi

True curve

J

ko sab competitors ke against extremize karta hai; har admissible

η

competition ki ek alag direction hai, toh stationarity har aisi direction mein hold karni chahiye.

Kyun integration by parts crucial move hai, sirf algebra nahi

Yeh derivative ko

η^{'}

se hata kar

\frac{\partial L}{\partial y ^{'}}

par transfer karta hai, har jagah ek bare

η

chhod kar, taaki hum ise factor out kar sakein aur Fundamental Lemma invoke kar sakein. Iske bina,

η

aur

η^{'}

ko alag nahi kar sakte.

Kyun boundary term specifically vanish hota hai

Kyunki

[\frac{\partial L}{\partial y ^{'}} η]_{a}^{b}

mein

η (a)

aur

η (b)

hain, dono Step 2 ki fixed-endpoint requirement se zero forced hain.

Kyun Principle of Least Action mein Newton's laws "contained" hain

\int (\frac{1}{2} m \overset{q}{˙}^{2} - V) d t

ko E–L se extremize karne par milta hai

m \overset{q}{¨} = - V^{'} (q) = F

; variational principle

F = ma

ko apni E–L equation ke roop mein reproduce karta hai. Dekho Lagrangian Mechanics.

Kyun

L

ki symmetries conserved quantities ka hint deti hain

Agar

L

kisi coordinate ko ignore karta hai (koi explicit

x

nahi ya cyclic

y

), toh E–L ek first integral deta hai (Beltrami ya

\frac{\partial L}{\partial y ^{'}} =

const). Yeh Noether's Theorem ka beej hai.

Kyun shortest path ke liye

L = 1 + y^{'2}

sahi integrand hai

Ek curve ki arc length

\int d x^{2} + d y^{2} = \int 1 + y^{'2} d x

hai; total length minimize karna exactly is functional ko extremize karne ka matlab hai, jo Geodesics ka flat-space case hai.

Kyun Brachistochrone Problem raw E–L ki jagah Beltrami use karta hai

Iska

L = (1 + y^{'2}) / y

mein koi explicit

x

nahi hai, toh Beltrami directly ek first-order ODE deta hai, jo messier second-order E–L equation se bachata hai.

Edge cases

Agar

η

ek endpoint par nonzero hai, toh kya perturbed curve still admissible hai

Nahi — woh required fixed endpoint

y (a)

y (b)

miss kar degi, toh woh valid competitor nahi hai; exactly isliye hum

η (a) = η (b) = 0

force karte hain.

Agar endpoints free hain fixed nahi

Boundary term automatically vanish nahi hota; ise vanish hone ki demand karne par natural boundary conditions milte hain

\frac{\partial L}{\partial y ^{'}}_{a} = \frac{\partial L}{\partial y ^{'}}_{b} = 0

Kya hota hai jab

L

y^{'}

mein linear ho, jaise

L = P (x, y) + Q (x, y) y^{'}

Toh

\frac{\partial L}{\partial y ^{'}} = Q

mein koi

y^{'}

nahi, toh

\frac{\partial ^{2} L}{\partial y ^{'2}} = 0

; E–L collapse ho jaata hai ek algebraic condition

P_{y} - Q_{x} = 0

mein,

y

ke liye koi genuine ODE nahi deta (ek degenerate variational problem).

Kya hota hai agar E–L identically satisfy ho (E–L expression sabhi curves ke liye

0

ho)

Toh

L

ek "null Lagrangian" hai (ek total derivative

\frac{d}{d x} F

), toh

J

sirf endpoints par depend karta hai aur har admissible curve stationary hai.

Kya hoga agar brachistochrone ka

y \to 0

(start point)

L = (1 + y^{'2}) / y

y \to 0

par blow up hota hai; cycloid ise handle karta hai kyunki wahan slope

y^{'} \to \infty

exactly sahi tarike se hota hai taaki

y (1 + y^{'2})

finite (constant) rahe.

Kya E–L tab bhi apply hota hai jab

y

ek vector

(y_{1}, \dots, y_{n})

Haan — tumhe har component ke liye ek E–L equation milta hai,

\frac{\partial L}{\partial y _{i}} - \frac{d}{d x} \frac{\partial L}{\partial y _{i}^{'}} = 0

, kyunki har

η_{i}

independently arbitrary hai.

Agar

L

mein koi

y

aur koi explicit

x

dono nahi hain, toh kaun sa shortcut zyada strong hai

Dono apply hote hain:

\frac{\partial L}{\partial y ^{'}} =

const (cyclic

y

) aur Beltrami

L - y^{'} \frac{\partial L}{\partial y ^{'}} =

const. Milke woh aksar ODE integrate kiye bina algebraically curve solve kar dete hain.

Recall Poore trap set ka one-line summary

E–L necessary hai, sufficient nahi; $\frac{d}{d x}$ ek total derivative hai; boundary term sirf isliye marta hai kyunki endpoints fixed hain; aur Fundamental Lemma ko sabhi $η$ chahiye, sirf ek nahi.