Euler-Lagrange equation — derivation: Worked examples

4.10.13 · D3 · Maths › Advanced Topics (Elite Level) › Euler-Lagrange equation — derivation

Yeh page ek cases ka catalogue hai. Parent derivation ne tumhe machine di thi: $\frac{\partial L}{\partial y} - \frac{d}{d x} (\frac{\partial L}{\partial y ^{'}}) = 0.$ Yahan hum isme har tarah ka Lagrangian feed karenge jo tum kabhi bhi dekh sakte ho, taaki koi bhi input aisa na ho jo humne show na kiya ho. Pehle ek map, phir worked cases.

The scenario matrix

Calculus of Variations ka har problem iss baat se decide hota hai ki $L$ actually kin variables par depend karta hai. Yeh ek akela fact batata hai ki kaun sa shortcut apply hoga. Neeche diya matrix har case class list karta hai.

Cell	$L$ depend karta hai…	Kya simplify hota hai	Tool of choice	Example
A	sirf $y^{'}$ par ( $L = L (y^{'})$ )	$\partial L / \partial y = 0$	$\partial L / \partial y^{'} = const$	Ex 1
B	$y, y^{'}$ par (koi $x$ nahi)	$\partial L / \partial x = 0$	Beltrami Identity	Ex 2
C	$x, y^{'}$ par (koi $y$ nahi)	$\partial L / \partial y = 0$	$\partial L / \partial y^{'} = const$	Ex 3
D	poore $x, y, y^{'}$ par	kuch skip nahi hota	full E–L ODE	Ex 4
E	degenerate $L$ jo $y^{'}$ mein linear hai, YA bilkul $y^{'}$ nahi	E–L collapse ho jaata hai	consistency check karo	Ex 5
F	free endpoint (ek end $y$ ya $x$ mein loose hai)	boundary term bachta hai	natural BC / transversality	Ex 6
G	word problem (real world)	model banao phir B apply karo	Brachistochrone Problem-style	Ex 7
H	exam twist (higher $y^{''}$ / symmetry)	generalized E–L / Noether's Theorem	modified rule	Ex 8

Ex 1 — Cell A: $L$ sirf $y^{'}$ par depend karta hai

Worked example Shortest path, cleanly re-derive kiya gaya

$J [y] = \int_{0}^{1} 1 + y^{'2} d x$ ko extremize karo with $y (0) = 0$ , $y (1) = 2$ .

Forecast: answer ki shape pehle guess karo. (Straight line? Ya curve?)

Steps.

Note karo ki $\partial L / \partial y = 0$ kyunki $y$ kabhi appear nahi karta. Yeh step kyun? E–L equation ka aadha hissa instantly kill ho jaata hai — Cell A ka gift.
E–L ban jaata hai $\frac{d}{d x} (\frac{\partial L}{\partial y ^{'}}) = 0$ , toh $\frac{\partial L}{\partial y ^{'}} = c_{1}$ (constant). Kyun? Jis quantity ka $x$ -derivative $0$ hai woh $x$ ke saath change nahi kar sakti.
Compute karo $\frac{\partial L}{\partial y ^{'}} = \frac{y ^{'}}{1 + y ^{'2}} = c_{1}$ . Kyun? $1 + y^{'2}$ par chain rule.
Solve karo: dono sides square karo, $y^{'2} = c_{1}^{2} (1 + y^{'2})$ , toh $y^{'2} (1 - c_{1}^{2}) = c_{1}^{2}$ , jisse milta hai $y^{'} = \pm \frac{c _{1}}{1 - c _{1}^{2}} =: m$ ek constant. $\pm$ kyun? Squaring ne sign lose kar diya; dono branches candidate slopes hain. Endpoints decide karte hain: humein $y$ ko $0$ se $2$ tak badhna hai, toh hum positive branch lete hain. (Agar target $y (1) = - 2$ hota, toh negative branch lete.)
Integrate karo: $y = m x + b$ . Endpoints $y (0) = 0 \Rightarrow b = 0$ ; $y (1) = 2 \Rightarrow m = 2 > 0$ (positive branch ke saath consistent). Toh $y = 2 x$ .

Verify: slope har jagah $2$ hai, $\frac{\partial L}{\partial y ^{'}} = \frac{2}{5} \approx 0.894$ — path ke saath genuinely constant. Straight line ✓ (curved-space version ke liye Geodesics dekho).

Figure — Euler-Lagrange equation — derivation

Ex 2 — Cell B: koi explicit $x$ nahi, Beltrami use karo

Worked example Minimal surface of revolution (soap film)

Curve $y (x)$ ko $x$ -axis ke around spin karo; iska surface area $\propto \int y 1 + y^{'2} d x$ . Toh $L = y 1 + y^{'2}$ — ismein $y$ aur $y^{'}$ hain, koi bare $x$ nahi ⇒ Cell B.

Forecast: answer ek famous curve hai. Guess karo kaun si "hanging" shape.

Steps.

Kyunki $\partial L / \partial x = 0$ , Beltrami Identity use karo: $L - y^{'} \frac{\partial L}{\partial y ^{'}} = C_{B}$ . Kyun? Ek poora $\frac{d}{d x}$ skip ho jaata hai — parent note ne yeh shortcut prove kiya tha.
$\frac{\partial L}{\partial y ^{'}} = \frac{y y ^{'}}{1 + y ^{'2}}$ . Kyun? $y 1 + y^{'2}$ ko differentiate karo $y$ ko constant treat karte hue w.r.t. $y^{'}$ .
Substitute karo: $y 1 + y^{'2} - \frac{y y ^{'2}}{1 + y ^{'2}} = \frac{y}{1 + y ^{'2}} = C_{B}$ . Kyun? Common denominator: $(1 + y^{'2}) - y^{'2} = 1$ .
Slope isolate karne ke liye rearrange karo: $y = C_{B} 1 + y^{'2} \Rightarrow y^{'2} = \frac{y ^{2}}{C _{B}^{2}} - 1 \Rightarrow y^{'} = \pm \frac{1}{C _{B}} y^{2} - C_{B}^{2}$ . $\pm$ kyun? $y^{'2}$ ka square-root; sign baad mein fix hota hai is baat se ki curve kis taraf climb karti hai. Rising half par $+$ branch lo.
Variables separate karo (yahi ODE solve ka WHY hai): equation ab hai $\frac{d y}{y ^{2} - C _{B}^{2}} = \frac{d x}{C _{B}}$ . Kyun separate? $y^{'} = d y / d x$ sirf $y$ par depend karta hai, toh hum saare $y$ ek side aur $x$ doosri side collect kar ke dono sides independently integrate kar sakte hain.
Dono sides integrate karo: $\int \frac{d y}{y ^{2} - C _{B}^{2}} = arccosh (\frac{y}{C _{B}})$ aur $\int \frac{d x}{C _{B}} = \frac{x - x _{0}}{C _{B}}$ (jahan $x_{0}$ integration constant hai). Arccosh kyun? Kyunki $\frac{d}{d u} arccosh u = \frac{1}{u ^{2} - 1}$ — exactly humara integrand $u = y / C_{B}$ ke baad.
Equate karo aur invert karo: $arccosh (\frac{y}{C _{B}}) = \frac{x - x _{0}}{C _{B}} \Rightarrow y = C_{B} cosh (\frac{x - x _{0}}{C _{B}})$ — ek catenary. $cosh$ kyun? Yeh $arccosh$ ka inverse hai; dono sides par $cosh$ apply karne se arc-function undo ho jaata hai.

Verify (numeric): $C_{B} = 1, x_{0} = 0$ lo. Phir $y = cosh x$ , $y^{'} = sinh x$ , aur $y^{2} = cosh^{2} x = 1 + sinh^{2} x = 1 + y^{'2}$ . Isliye $\frac{y ^{2}}{1 + y ^{'2}} = 1$ , yaani $\frac{y}{1 + y ^{'2}} = 1 = C_{B}$ ✓ har $x$ ke liye.

Ex 3 — Cell C: koi explicit $y$ nahi, lekin $x$ present hai

Worked example Weighted straight-ish path

$J = \int_{1}^{2} x 1 + y^{'2} d x$ ko extremize karo. Yahan $L = x 1 + y^{'2}$ : ismein $x$ aur $y^{'}$ hain, koi $y$ nahi ⇒ Cell C.

Forecast: kya answer phir bhi straight line hoga, ya $x$ -weight use bend kar deta hai?

Steps.

$\partial L / \partial y = 0$ , toh E–L deta hai $\frac{\partial L}{\partial y ^{'}} = c_{3}$ (constant). Kyun? Cell C ka gift — same conserved-quantity logic jaise Cell A mein, chahe $x$ appear kare.
$\frac{\partial L}{\partial y ^{'}} = \frac{x y ^{'}}{1 + y ^{'2}} = c_{3}$ . Kyun? $x$ $y^{'}$ -derivative ke dauran constant ki tarah ride karta hai.
$y^{'}$ ke liye solve karo: $x^{2} y^{'2} = c_{3}^{2} (1 + y^{'2}) \Rightarrow y^{'2} (x^{2} - c_{3}^{2}) = c_{3}^{2} \Rightarrow y^{'} = \pm \frac{c _{3}}{x ^{2} - c _{3}^{2}}$ . $\pm$ kyun, aur domain note karo: square-root demand karta hai $x^{2} > c_{3}^{2}$ , yaani $x > ∣ c_{3} ∣$ (for $x > 0$ ) — solution sirf $x \geq c_{3}$ par valid hai. Increasing curve ke liye $+$ branch lo.
Integrate karo: $y = c_{3} arccosh (\frac{x}{c _{3}}) + b$ valid range $x \geq c_{3}$ par. Kyun? $\int \frac{d x}{x ^{2} - c _{3}^{2}} = arccosh (x / c_{3})$ , jiski domain $x / c_{3} \geq 1$ exactly step 3 mein mili restriction hai.

Verify (numeric): $c_{3} = 1, b = 0$ pick karo, toh $y = arccosh (x)$ (valid for $x \geq 1$ ; humara interval $[1, 2]$ uske andar hai), $y^{'} = \frac{1}{x ^{2} - 1}$ . $x = 2$ par: $\frac{x y ^{'}}{1 + y ^{'2}} = \frac{2/ 3}{1 + 1/3} = \frac{2/ 3}{2/ 3} = 1 = c_{3}$ ✓.

Ex 4 — Cell D: full dependence, poora ODE run karo

Worked example Kuch skip nahi hota

$J = \int_{0}^{1} (y^{'2} + y^{2} + 2 x y) d x$ ko extremize karo. $L = y^{'2} + y^{2} + 2 x y$ mein $x, y, y^{'}$ hain ⇒ Cell D, full E–L.

Forecast: extremal ek 2nd-order linear ODE solve karta hai — guess karo ki woh oscillate karta hai ya badhta hai.

Steps.

$\frac{\partial L}{\partial y} = 2 y + 2 x$ . Kyun? $y^{2} + 2 x y$ ko $y$ mein differentiate karo, $x$ frozen rakhke.
$\frac{\partial L}{\partial y ^{'}} = 2 y^{'}$ , toh $\frac{d}{d x} \frac{\partial L}{\partial y ^{'}} = 2 y^{''}$ . Kyun? Phir ek total $x$ -derivative lo — yahan $2 y^{'} \to 2 y^{''}$ .
E–L: $2 y + 2 x - 2 y^{''} = 0 \Rightarrow y^{''} - y = x$ . Kyun? Dono pieces ko $\partial_{y} L - \frac{d}{d x} \partial_{y^{'}} L = 0$ mein plug karo.
Solve karo: homogeneous $y_{h} = A e^{x} + B e^{- x}$ ; particular $y_{p} = - x$ (kyunki $y_{p}^{''} - y_{p} = 0 - (- x) = x$ ). Toh $y = A e^{x} + B e^{- x} - x$ . Kyun? Linear ODE ⇒ complementary + particular. Yeh badhta hai (exponentials), oscillate nahi karta.

Verify: $y = - x$ (particular part) substitute karo: $y^{''} - y = 0 - (- x) = x$ ✓. Koi bhi $A e^{x} + B e^{- x}$ add karo: har ek $y^{''} - y = 0$ satisfy karta hai, toh sum phir bhi $x$ deta hai ✓.

Ex 5 — Cell E: degenerate — do flavours

Worked example 5a:

$L$ linear in $y^{'}$ (E–L ek identity par collapse ho jaata hai) $J = \int_{a}^{b} y^{'} g (y) d x$ ko extremize karo jahan $g$ smooth hai — $y^{'}$ sirf linearly appear karta hai.

Forecast: normal ODE, ya kuch strange?

Steps.

$\frac{\partial L}{\partial y} = y^{'} g^{'} (y)$ . Kyun? $y^{'}$ ko $y$ ke w.r.t. constant treat karo.
$\frac{\partial L}{\partial y ^{'}} = g (y)$ , toh $\frac{d}{d x} \frac{\partial L}{\partial y ^{'}} = g^{'} (y) y^{'}$ . Kyun? Chain rule — $g (y)$ $y (x)$ ke through change hota hai.
E–L: $y^{'} g^{'} (y) - g^{'} (y) y^{'} = 0$ , yaani $0 = 0$ identically. Kyun? Terms cancel ho jaate hain — $y$ ke baare mein koi information nahi.
Interpretation: $L = y^{'} g (y) = \frac{d}{d x} G (y)$ with $G^{'} = g$ , toh $J = G (y (b)) - G (y (a))$ sirf endpoints par depend karta hai. Har curve extremal hai (Lagrangian Mechanics mein gauge freedom ka seed).

Verify (numeric): $g (y) = y$ , $L = y y^{'} = \frac{d}{d x} (y^{2} /2)$ . $y (0) = 0$ se $y (1) = 1$ tak do curves: line $y = x$ aur parabola $y = x^{2}$ . $\int_{0}^{1} y y^{'} d x = [y^{2} /2]_{0}^{1} = 1/2$ dono ke liye ✓.

Worked example 5b:

$L$ mein bilkul koi $y^{'}$ nahi — $L = L (x, y)$ $J = \int_{0}^{1} (y^{2} - 2 x y) d x$ ko extremize karo. Yahan $L = y^{2} - 2 x y$ mein koi $y^{'}$ nahi.

Forecast: koi $y^{'}$ nahi toh integrate karne ke liye koi ODE nahi — toh $y$ ko kya pin karta hai?

Steps.

$\frac{\partial L}{\partial y ^{'}} = 0$ (koi $y^{'}$ hai hi nahi), toh $\frac{d}{d x} \frac{\partial L}{\partial y ^{'}} = 0$ . Kyun? Jo hai hi nahi use differentiate nahi kar sakte.
E–L purely algebraic condition par collapse ho jaata hai $\frac{\partial L}{\partial y} = 0$ : $2 y - 2 x = 0 \Rightarrow y = x$ . Kyun? Koi derivative term survive nahi karta — extremal point-by-point milta hai, integrate karke nahi.
Degeneracy warning: $y = x$ pointwise forced hai aur generally imposed endpoint values meet nahi kar sakta (yahan $y (0) = 0, y (1) = 1$ fit ho jaate hain, lekin koi bhi aur endpoints problem ko inconsistent bana denge — koi smooth extremal exist nahi karta). Kyun? Ek algebraic (order-0) condition mein boundary data satisfy karne ke liye free constants nahi hote.

Verify (numeric): $y = x$ par, $\partial L / \partial y = 2 x - 2 x = 0$ ✓ saare $x$ ke liye; aur yeh $y (0) = 0, y (1) = 1$ ✓ satisfy karta hai (ek lucky-fit case).

Ex 6 — Cell F: free endpoints ( $y$ aur $x$ mein)

Worked example 6a: ek end

$y$ mein loose — natural boundary condition $J = \int_{0}^{1} (\frac{1}{2} y^{'2} - y) d x$ ko extremize karo with $y (0) = 0$ lekin $y (1)$ free (right end vertically slide kar sakta hai).

Forecast: fixed end ke saath tumhe ek second endpoint value chahiye — uski jagah kya aata hai?

Steps.

Interior E–L: $\frac{\partial L}{\partial y} = - 1$ , $\frac{\partial L}{\partial y ^{'}} = y^{'}$ , toh $- 1 - y^{''} = 0 \Rightarrow y^{''} = - 1$ . Kyun? Free end sirf boundary change karta hai, interior ODE nahi.
General solution $y = - \frac{1}{2} x^{2} + A x + B$ . Apply karo $y (0) = 0 \Rightarrow B = 0$ .
Variation $η$ recall karo (upar define kiya gaya): parent mein Integration by Parts step ek boundary term $[\frac{\partial L}{\partial y ^{'}} η]_{0}^{1}$ chod gaya tha, jahan $η$ $y$ ka wiggle hai. $x = 0$ par end fixed hai, toh $η (0) = 0$ woh piece kill kar deta hai. $x = 1$ par end free hai, toh $η (1)$ arbitrary aur non-zero hai — term drop nahi ki ja sakti. Yeh kyun matter karta hai: ek maatra tarika hai ki $\frac{\partial L}{\partial y ^{'}}_{1} η (1) = 0$ saare aise $η (1)$ ke liye, agar coefficient vanish ho: natural boundary condition $\frac{\partial L}{\partial y ^{'}}_{x = 1} = 0$ , yaani $y^{'} (1) = 0$ .
Apply karo $y^{'} (1) = 0$ : $y^{'} = - x + A$ , toh $- 1 + A = 0 \Rightarrow A = 1$ . Final: $y = - \frac{1}{2} x^{2} + x$ .

Verify: $y^{''} = - 1$ ✓, $y (0) = 0$ ✓, $y^{'} (1) = - 1 + 1 = 0$ ✓ — curve loose end par flat pahunchti hai, natural BC ka geometric meaning.

Ex 7 — Cell G: real-world word problem

Worked example Fastest ski run (numbers ke saath Brachistochrone)

Ek frictionless skier $(0, 0)$ par rest se start karta hai aur $(π, 2)$ tak pahunchta hai, jahan $y$ downward measure kiya jaata hai taaki gravity help kare.

Physics setup karna (har constant define karna). Energy conservation se, rest se vertical drop $y$ ke through slide karne par speed milti hai $v = 2 g y$ , jahan $g$ acceleration due to gravity hai. Arc-length element hai $d s = 1 + y^{'2} d x$ , toh use traverse karne ka time hai $d t = d s / v = \frac{1 + y ^{'2}}{2 g y} d x$ . Sum karne par, $T = \int_{0}^{π} \frac{1 + y ^{'2}}{2 g y} d x = \frac{1}{2 g} \int_{0}^{π} \frac{1 + y ^{'2}}{y} d x .$ Prefactor $1/ 2 g$ ek constant hai aur is baat par affect nahi karta ki kaun sa curve $T$ minimise karta hai, toh hum $L = \frac{1 + y ^{'2}}{y}$ extremize karte hain — koi bare $x$ nahi ⇒ Cell B.

Forecast: straight ramp, ya koi curve jo pehle steeply dive karta hai?

Steps.

Beltrami (kyunki $\partial L / \partial x = 0$ ): $L - y^{'} \frac{\partial L}{\partial y ^{'}} = C_{B}$ . Kyun? Kisi bhi $x$ -free problem ka energy-like first integral.
Compute karo $\frac{\partial L}{\partial y ^{'}} = \frac{y ^{'}}{y 1 + y ^{'2}}$ . Substitute karo: $\frac{1 + y ^{'2}}{y} - \frac{y ^{'2}}{y 1 + y ^{'2}} = \frac{( 1 + y ^{'2} ) - y ^{'2}}{y 1 + y ^{'2}} = \frac{1}{y 1 + y ^{'2}} = C_{B} .$ Kyun? Common denominator; numerator $1$ par collapse ho jaata hai.
Square karo aur first integral tak rearrange karo: $\frac{1}{y ( 1 + y ^{'2} )} = C_{B}^{2} \Rightarrow y (1 + y^{'2}) = \frac{1}{C _{B}^{2}} =: k$ . Yeh "first integral" kyun hai: yeh E–L se ek order lower hai aur poore path par hold karta hai — physical WHY yeh hai ki koi bare $x$ nahi hone ka matlab conserved quantity hai, yahan $k$ .
$y (1 + y^{'2}) = k$ ko standard substitution $y^{'} = cot (θ /2)$ se solve karo; yeh parametric cycloid deta hai $x = r (θ - sin θ)$ , $y = r (1 - cos θ)$ with $k = 2 r$ . Cycloid kyun? Yeh unique curve hai jo $y (1 + y^{'2}) = const$ satisfy karti hai; start ke paas yeh almost vertically plunge karta hai, speed early bank karke straight ramp ko beat karta hai.

Verify (numeric): $r = 1$ toh $k = 2$ ke liye, use karo $y^{'} = \frac{d y / d θ}{d x / d θ} = \frac{sin θ}{1 - cos θ}$ . $θ = π /2$ par: $y = 1$ , $y^{'} = 1$ , toh $y (1 + y^{'2}) = 1 \cdot 2 = 2 = k$ ✓. $θ = π$ par: $y = 2$ , $y^{'} = 0$ , $y (1 + y^{'2}) = 2 \cdot 1 = 2 = k$ ✓ — constant, cycloid confirmed.

Ex 8 — Cell H: exam twist, higher derivative

Worked example Jab

$L$ mein $y^{''}$ ho Beam ki bending energy: $J = \int_{0}^{1} (\frac{1}{2} (y^{''})^{2}) d x$ ko extremize karo. Lagrangian mein second derivative $y^{''}$ hai — standard E–L kaafi nahi hai.

Forecast: answer woh shape hai jo ek thin ruler leta hai — iska polynomial degree guess karo.

Steps.

Generalized Euler–Poisson equation use karo: $\frac{\partial L}{\partial y} - \frac{d}{d x} \frac{\partial L}{\partial y ^{'}} + \frac{d ^{2}}{d x ^{2}} \frac{\partial L}{\partial y ^{''}} = 0$ . Yeh step kyun? $L$ mein har extra derivative ek extra $\pm \frac{d ^{k}}{d x ^{k}}$ term contribute karta hai (repeat karo Integration by Parts $k$ times, exactly parent ka Step 5 twice done).
Yahan $\partial L / \partial y = 0$ , $\partial L / \partial y^{'} = 0$ , $\partial L / \partial y^{''} = y^{''}$ . Kyun? $L$ sirf $y^{''}$ par depend karta hai.
Equation: $\frac{d ^{2}}{d x ^{2}} (y^{''}) = y^{''''} = 0$ . Kyun? Sirf teesra term survive karta hai.
Solve karo: $y^{''''} = 0 \Rightarrow y = c_{3} x^{3} + c_{2} x^{2} + c_{1} x + c_{0}$ — ek cubic, classic beam-deflection shape. Kyun? Fourth derivative zero ⇒ degree-3 polynomial.

Verify: $y = x^{3}$ deta hai $y^{''''} = 0$ ✓, aur $y^{''} = 6 x$ toh bending energy $\int_{0}^{1} \frac{1}{2} (6 x)^{2} d x = \int_{0}^{1} 18 x^{2} d x = 6$ (finite, well-defined) ✓.

Recall Kaun se cell mein hoon? (self-quiz)

$L = 1 + y^{'2}$ — kaun sa cell aur kaun sa shortcut? ::: Cell A (sirf $y^{'}$ ) → $\partial L / \partial y^{'} =$ const. $L = y 1 + y^{'2}$ — kaun sa cell? ::: Cell B (koi bare $x$ nahi) → Beltrami identity. $L = x 1 + y^{'2}$ — kaun sa cell? ::: Cell C (koi bare $y$ nahi) → $\partial L / \partial y^{'} =$ const, lekin path bend karta hai (sirf $x \geq c_{3}$ par valid). $L = y^{'} g (y)$ E–L $0 = 0$ deta hai. Iska kya matlab hai? ::: $L$ ek total derivative hai — har path extremal hai (Cell E-5a, degenerate). $L = L (x, y)$ mein koi $y^{'}$ nahi — E–L kya ban jaata hai? ::: Ek algebraic condition $\partial L / \partial y = 0$ ; koi ODE nahi, endpoints se clash ho sakta hai (Cell E-5b). Right end $y$ mein free hai. Missing value ki jagah kya aata hai? ::: Natural BC $\partial L / \partial y^{'} = 0$ wahan (Cell F-6a). Endpoint ka $x$ bhi free hai — kaun si condition? ::: Transversality $L - y^{'} \partial L / \partial y^{'} = 0$ us end par (Cell F-6b). $L$ mein $y^{''}$ hai. E–L ki jagah kaun si equation aati hai? ::: Euler–Poisson: add karo $+ \frac{d ^{2}}{d x ^{2}} \frac{\partial L}{\partial y ^{''}}$ (Cell H).