Calculus of variations — functionals, functional derivative: Worked examples

4.10.12 · D3 · Maths › Advanced Topics (Elite Level) › Calculus of variations — functionals, functional derivative

Parent topic ka yeh child page ek hi kaam karta hai: Euler–Lagrange machine (jo parent mein hai) ko har tarah ke case ke against drill karta hai. Pehle hum cases list karte hain, phir har cell ke liye ek example karte hain.

Pehli line se pehle, teen reminders seedhi bhasha mein (yahan kuch assume nahi kiya — sab parent mein hai):

Scenario matrix

Is topic ka har problem inhi shapes mein se ek mein aata hai. Last column us example ka naam deta hai jo use cover karta hai.

#	Case class	Usmein kya khaas hai	Covered by
C1	$L$ mein koi $y$ nahi ( $\partial L / \partial y = 0$ )	E–L collapse hokar " $L_{y^{'}} =$ const" ban jaata hai	Ex 1
C2	$L$ mein koi explicit $x$ nahi	Beltrami use karo ( $y^{''}$ bachao)	Ex 2
C3	$L$ mein koi $y^{'}$ nahi ( $\partial L / \partial y^{'} = 0$ )	E–L algebraic ban jaata hai, koi ODE nahi	Ex 3
C4	Full $L (x, y, y^{'})$ — sabhi slots present	Total-derivative chain rule matter karta hai	Ex 4
C5	Degenerate / limiting input ( $ω \to 0$ , $c \to \infty$ , straight-line limit)	Check karo ki machine sane limit deti hai	Ex 5
C6	Free endpoint — boundary term bachta hai	Natural boundary condition appear hoti hai	Ex 6
C7	Constraint (fixed area/length)	Lagrange multiplier $L$ mein enter karta hai	Ex 7
C8	Word problem (real-world: fastest slide)	Model → $L$ identify karo → solve karo	Ex 8
C9	Exam twist (sign trap / higher derivative)	Yahan students marks kho dete hain	Ex 9

Har example ka tag [Cn] dekho — woh batata hai kaunsa cell fill ho raha hai.

Ex 1 — $L$ mein koi $y$ nahi `[C1]`

Worked example Shortest path, dhyaan se kiya

$J [y] = \int_{0}^{1} 1 + y^{'2} d x$ minimize karo, $y (0) = 0, y (1) = 2$ ke saath. Yahan $L = 1 + y^{'2}$ .

Forecast: answer padhne se pehle guess karo. Do points ke beech sabse chhoti wire → kaunsi shape? (Zor se bolo.)

Check karo kaunsa cell. $L$ mein $y^{'}$ hai lekin koi bare $y$ nahi. Yeh step kyun? Agar $\partial L / \partial y = 0$ , toh E–L ka pehla term chala jaata hai aur easy ho jaata hai.
$\frac{\partial L}{\partial y ^{'}} = \frac{y ^{'}}{1 + y ^{'2}}$ . Kyun? $1 + y^{'2}$ ko $y^{'}$ ko variable maan ke differentiate karo (chain rule: $\frac{1}{2} (1 + y^{'2})^{- 1/2} \cdot 2 y^{'}$ ).
E–L: $0 - \frac{d}{d x} \frac{y ^{'}}{1 + y ^{'2}} = 0 \Rightarrow \frac{y ^{'}}{1 + y ^{'2}} = c$ . Kyun? Derivative zero equal hone ka matlab andar wala constant hai.
$y^{'}$ ke liye solve karo: square karo → $y^{'2} = c^{2} (1 + y^{'2}) \Rightarrow y^{'2} (1 - c^{2}) = c^{2} \Rightarrow y^{'} = \frac{c}{1 - c ^{2}} \equiv m$ (constant). Kyun? Constant slope.
Integrate karo: $y = m x + b$ . Endpoints fit karo: $b = 0$ , $m = 2$ . Toh $y = 2 x$ .

Verify: $(0, 0)$ se $(1, 2)$ tak ka seedha segment ki length $1^{2} + 2^{2} = 5 \approx 2.236$ hai. Koi bhi bend length badhata hai. ✓ Forecast se match — ek seedhi line.

Figure — Calculus of variations — functionals, functional derivative

Ex 2 — Koi explicit $x$ nahi → Beltrami `[C2]`

Worked example Minimal surface of revolution (catenary)

$J [y] = \int_{a}^{b} y 1 + y^{'2} d x$ minimize karo, $y > 0$ ke saath. Toh $L = y 1 + y^{'2}$ — ek soap-film area.

Forecast: hanging-chain shape hai ya parabola? Guess karo.

Cell spot karo: $L$ mein koi explicit $x$ nahi. Yeh step kyun? Tab Beltrami $L - y^{'} L_{y^{'}} = const$ ugly $y^{''}$ se bachata hai.
$L_{y^{'}} = \frac{y y ^{'}}{1 + y ^{'2}}$ . Kyun? Sirf $1 + y^{'2}$ hi $y^{'}$ par depend karta hai; constant-in- $y^{'}$ factor $y$ ke saath product.
Beltrami: $y 1 + y^{'2} - y^{'} \cdot \frac{y y ^{'}}{1 + y ^{'2}} = \frac{y}{1 + y ^{'2}} = c$ . Kyun? Common denominator par combine karo; numerator collapse hokar $y$ ban jaata hai.
Rearrange karo: $y^{2} = c^{2} (1 + y^{'2}) \Rightarrow y^{'} = \frac{y ^{2}}{c ^{2}} - 1$ . Kyun? $y^{'}$ isolate karo taaki separable ODE mile (dekho Ordinary differential equations).
Separate karo & integrate karo: $\int \frac{d y}{y ^{2} / c ^{2} - 1} = \int d x$ se milta hai $y = c cosh (\frac{x - x _{0}}{c})$ . Kyun? $\frac{d}{d u} cosh u = sinh u$ aur $cosh^{2} - sinh^{2} = 1$ isko fit karte hain.

Verify: $y = c cosh \frac{x}{c}$ rakho (lo $x_{0} = 0$ ). Tab $y^{'} = sinh \frac{x}{c}$ aur $\frac{y}{1 + y ^{'2}} = \frac{c cosh}{1 + sinh ^{2}} = \frac{c cosh}{cosh} = c$ . Constant ✓ — Beltrami satisfied. Yeh soap films ke peeche wali geometry hai.

Ex 3 — $L$ mein koi $y^{'}$ nahi: E–L algebraic hai `[C3]`

Ex 4 — Full $L (x, y, y^{'})$ : total-derivative trap `[C4]`

Worked example Teeno slots present

$J [y] = \int_{0}^{1} (y^{'2} + 12 x y) d x$ , $y (0) = 0, y (1) = 1$ . $L = y^{'2} + 12 x y$ mein $x$ , $y$ aur $y^{'}$ teeno hain.

Forecast: resulting polynomial $y (x)$ ki degree kya hogi? Guess karo.

Cell: har slot use ho raha hai — total $x$ -derivative mein $x$ ki movement include karni hogi. Yeh step kyun? Yahi woh jagah hai jahan beginners galti se partial derivative use karte hain.
$\frac{\partial L}{\partial y} = 12 x$ , $\frac{\partial L}{\partial y ^{'}} = 2 y^{'}$ . Kyun? Partial derivatives, ek slot at a time.
$\frac{d}{d x} (2 y^{'}) = 2 y^{''}$ . Kyun? Yahan $L_{y^{'}} = 2 y^{'}$ mein koi explicit $x$ ya $y$ nahi, isliye total derivative sirf $2 y^{''}$ hai (general chain rule $L_{y^{'} x} + L_{y^{'} y} y^{'} + L_{y^{'} y^{'}} y^{''}$ sirf last term par reduce hota hai).
E–L: $12 x - 2 y^{''} = 0 \Rightarrow y^{''} = 6 x$ . Kyun? E–L expression ko zero set karo.
Do baar integrate karo: $y^{'} = 3 x^{2} + A$ , $y = x^{3} + A x + B$ . Endpoints: $B = 0$ ; $1 = 1 + A \Rightarrow A = 0$ . Toh $y = x^{3}$ .

Verify: $y = x^{3} \Rightarrow y^{''} = 6 x$ ✓ aur $y (0) = 0, y (1) = 1$ ✓. Cubic, jaisa forecast hona chahiye tha.

Ex 5 — Degenerate / limiting input `[C5]`

Worked example Oscillator energy jab

$ω \to 0$ Parent se: $J [y] = \int_{0}^{1} (\frac{1}{2} y^{'2} + \frac{1}{2} ω^{2} y^{2}) d x$ E–L deta hai $y^{''} = ω^{2} y$ . Ab limit $ω \to 0$ push karo.

Forecast: ek taut string jisme koi stiffness nahi, woh kaunsi shape prefer karegi? Guess karo.

Cell: limiting/degenerate parameter. Yeh step kyun? Ek acchi machine ko extreme inputs par bhi sane answers dene chahiye.
$y^{''} = ω^{2} y$ ka general solution (jab $ω > 0$ ho): $y = P cosh (ω x) + Q sinh (ω x)$ . Kyun? Positive coefficient → hyperbolic (growing/decaying) modes.
$ω \to 0$ set karo: $cosh (ω x) \to 1$ , $\frac{sinh ( ω x )}{ω} \to x$ . Toh $y \to$ (constant) $+$ (constant) $\cdot x$ = straight line. Kyun? Equation degenerate hokar $y^{''} = 0$ ban jaata hai, jiske solutions lines hain.
Ex 1 ke saath sanity check: koi restoring term nahi ( $ω = 0$ ) toh sirf $\frac{1}{2} y^{'2}$ bachta hai — ek "shortest-slope" energy → straight line. ✓

Verify (numeric): endpoints lo $y (0) = 1, y (1) = 3$ . Limit answer $y = 1 + 2 x \Rightarrow y (0.5) = 2$ . Ab finite case $ω = 0.001$ same endpoints ke saath $y (0.5) \approx 2.0000$ deta hai 4 dp tak — limit se match. ✓

Ex 6 — Free endpoint → natural boundary condition `[C6]`

Worked example Ek end loose hai

$J [y] = \int_{0}^{1} (\frac{1}{2} y^{'2} - y) d x$ minimize karo, $y (0) = 0$ ke saath lekin $y (1)$ free (prescribed nahi).

Forecast: right end free hone par, nature wahan kaunsi slope choose karegi? Guess karo.

Cell: free endpoint. Yeh step kyun? IBP boundary term $[L_{y^{'}} η]_{0}^{1}$ ab $x = 1$ par zero nahi hogi, kyunki $η (1)$ ko zero hone ki zaroorat nahi.
Interior E–L: $L_{y} = - 1$ , $L_{y^{'}} = y^{'}$ , toh $- 1 - y^{''} = 0 \Rightarrow y^{''} = - 1$ . Kyun? Interval ke andar same E–L.
Free end par boundary term sabhi $η (1)$ ke liye zero hona chahiye: $L_{y^{'}}_{x = 1} = 0 \Rightarrow y^{'} (1) = 0$ . Kyun? Yeh natural boundary condition hai — machine tumhe missing endpoint data deti hai.
$y^{''} = - 1$ solve karo: $y = - \frac{1}{2} x^{2} + A x + B$ . Use karo $y (0) = 0 \Rightarrow B = 0$ ; $y^{'} (1) = - 1 + A = 0 \Rightarrow A = 1$ . Toh $y = - \frac{1}{2} x^{2} + x$ .

Verify: $y^{'} = - x + 1$ , toh $y^{'} (1) = 0$ ✓ (loose end par flat slope, jaisa forecast tha) aur $y (0) = 0$ ✓. Functional analysis dekho — kyun free ends conditions generate karte hain instead of require karne ke.

Ex 7 — Constraint via Lagrange multiplier `[C7]`

Worked example Fixed length ke saath sabse badi area (isoperimetric)

$[0, 1]$ par curves $y (x)$ mein jo $y (0) = y (1) = 0$ aur fixed length $\int_{0}^{1} 1 + y^{'2} d x = ℓ$ satisfy karte hain, enclosed area $\int_{0}^{1} y d x$ maximize karo.

Forecast: circle-arc, parabola, ya catenary? Guess karo.

Cell: constrained problem. Yeh step kyun? Constraint ko multiplier $λ$ ke saath attach karo (dekho Constrained optimization & Lagrange multipliers): form karo $L = y + λ 1 + y^{'2}$ .
$L_{y} = 1$ , $L_{y^{'}} = \frac{λ y ^{'}}{1 + y ^{'2}}$ . Kyun? Augmented integrand ke ordinary partials.
E–L: $1 - \frac{d}{d x} \frac{λ y ^{'}}{1 + y ^{'2}} = 0 \Rightarrow \frac{λ y ^{'}}{1 + y ^{'2}} = x - x_{0}$ . Kyun? E–L ko ek baar integrate karo ( $1$ integrate hokar $x$ ban jaata hai).
$y^{'}$ ke liye solve karo aur integrate karo → $(x - x_{0})^{2} + (y - y_{0})^{2} = λ^{2}$ : ek circular arc. Kyun? Relation kehta hai tangent direction $x$ -displacement ke saath uniformly rotate karta hai — circle ki pehchaan.

Verify: radius $R = λ$ ke circle par, $\frac{y ^{'}}{1 + y ^{'2}} = sin θ$ jahan $θ$ tangent angle hai; aur $R sin θ$ = horizontal offset $x - x_{0}$ ✓. Classical isoperimetric answer hai hi circle. Forecast: circle-arc.

Ex 8 — Word problem: fastest slide `[C8]`

Worked example Brachistochrone, words se frame kiya

Ek bead frictionlessly $(0, 0)$ se $(x_{1}, y_{1})$ tak gravity $g$ ke under slide karta hai, rest se start karke. Woh wire shape dhundho jo sabse kam travel time deti hai. (Full context: Brachistochrone problem.)

Forecast: seedha ramp, ya ek curve jo pehle steep dive leta hai? Guess karo.

Model. Energy se speed: $\frac{1}{2} v^{2} = g y \Rightarrow v = 2 g y$ ( $y$ downward measure karo). Time $= \int \frac{d s}{v}$ jahan $d s = 1 + y^{'2} d x$ . Yeh step kyun? "Time" ko ek functional mein badlo taaki E–L apply ho sake.
Toh $J [y] = \int_{0}^{x_{1}} \frac{1 + y ^{'2}}{2 g y} d x$ , deta hai $L = \frac{1 + y ^{'2}}{2 g y}$ . Koi explicit $x$ nahi → Beltrami. Kyun? Ex 2 wala hi trick.
Beltrami: $L - y^{'} L_{y^{'}} = \frac{1}{2 g y 1 + y ^{'2}} = const$ . Kyun? Algebra bilkul Ex 2 ke collapse jaisi form mein.
Rearrange karo: $y (1 + y^{'2}) = const = k$ . Kyun? Square karo aur denominators clear karo.
Yeh ODE ek cycloid se solve hoti hai: $x = \frac{k}{2} (ϕ - sin ϕ), y = \frac{k}{2} (1 - cos ϕ)$ . Kyun? $y = \frac{k}{2} (1 - cos ϕ)$ substitute karne par $y (1 + y^{'2}) = k$ satisfy hota hai.

Verify: $ϕ = π$ par, $y = \frac{k}{2} (1 - (- 1)) = k$ ; tab $y^{'} = 0$ (arc ka bottom) toh $y (1 + y^{'2}) = k \cdot 1 = k$ ✓. Bead pehle dive karta hai — forecast confirmed. Yeh Lagrangian mechanics se link karta hai jahan same $L$ -machinery mechanics chalati hai.

Ex 9 — Exam twist: sign trap & $L$ mein $y^{''}$ `[C9]`

Worked example Second-derivative functional (Euler–Poisson)

$J [y] = \int_{0}^{1} (\frac{1}{2} y^{''2}) d x$ minimize karo — ek beam ki "bending energy" — $y (0) = 0, y^{'} (0) = 0, y (1) = 1, y^{'} (1) = 0$ ke saath. Note karo $L$ $y^{''}$ par depend karta hai!

Forecast: yahan ordinary E–L kaafi nahi hai. Governing equation ki order kya hogi? Guess karo.

Cell / twist: $L$ mein $y^{''}$ hai, isliye ordinary E–L incomplete hai. Yeh step kyun? IBP do baar repeat karne par ek $+ \frac{d ^{2}}{d x ^{2}} \frac{\partial L}{\partial y ^{''}}$ term add hota hai (Euler–Poisson equation): $\frac{\partial L}{\partial y} - \frac{d}{d x} \frac{\partial L}{\partial y ^{'}} + \frac{d ^{2}}{d x ^{2}} \frac{\partial L}{\partial y ^{''}} = 0$ .
Yahan $L_{y} = 0, L_{y^{'}} = 0, L_{y^{''}} = y^{''}$ . Toh equation hai $\frac{d ^{2}}{d x ^{2}} (y^{''}) = 0 \Rightarrow y^{(4)} = 0$ . Kyun? Sirf $y^{''}$ slot nonzero hai; use do baar differentiate karo.
Sign trap: sign plus hai $y^{''}$ term par (do IBPs = do minus signs = ek plus). Yahan sign drop karna classic exam error hai. Kyun? Har IBP sign flip karta hai; even number of flips → positive.
$y^{(4)} = 0$ solve karo: cubic $y = a x^{3} + b x^{2} + c x + d$ . Chaar conditions apply karo: $y (0) = 0 \Rightarrow d = 0$ ; $y^{'} (0) = 0 \Rightarrow c = 0$ ; $y (1) = a + b = 1$ ; $y^{'} (1) = 3 a + 2 b = 0$ . Solve karo: $b = 3, a = - 2$ . Toh $y = - 2 x^{3} + 3 x^{2}$ .

Verify: $y (0) = 0, y^{'} (0) = 0$ ✓. $y (1) = - 2 + 3 = 1$ ✓. $y^{'} (x) = - 6 x^{2} + 6 x \Rightarrow y^{'} (1) = 0$ ✓. Aur $y^{(4)} = 0$ ✓ (yeh ek cubic hai). Fourth-order equation, jaisa forecast tha. Yeh woh smooth-blend curve hai jise engineers Hermite cubic kehte hain.

Matrix ka Recap

Recall Kya humne har cell hit kiya?

Kaunsa example prove karta hai "no $y$ ⇒ $L_{y^{'}} =$ const"? ::: Ex 1 (straight line) Kaunsa Beltrami use karta hai kyunki $L$ mein explicit $x$ nahi? ::: Ex 2 (catenary) aur Ex 8 (cycloid) Kaunsa E–L ko pure algebra mein badal deta hai? ::: Ex 3 ( $y = x$ ) Kaunse ko full total-derivative chain chahiye? ::: Ex 4 ( $y = x^{3}$ ) Kaunsa degenerate limit $ω \to 0$ test karta hai? ::: Ex 5 (line) Kaunsa natural boundary condition produce karta hai? ::: Ex 6 ( $y^{'} (1) = 0$ ) Kaunsa Lagrange multiplier use karta hai? ::: Ex 7 (circle arc) Kaunsa real-world word problem hai? ::: Ex 8 (brachistochrone) Kaunsa sign trap / higher derivative ka surprise deta hai? ::: Ex 9 (Euler–Poisson, $y^{(4)} = 0$ )