Foundations — Isoperimetric problems — constraints (Lagrange multipliers in variational sense)

4.10.16 · D1 · Maths › Advanced Topics (Elite Level) › Isoperimetric problems — constraints (Lagrange multipliers i

Isse pehle ki tum parent note Isoperimetric problems padh sako, us page ka har symbol tumhare liye kuch maane rakhna chahiye. Yeh page har ek symbol ko scratch se build karta hai, usi order mein jis order mein woh ek doosre pe depend karte hain.

1. Ek function $y(x)$ — ek curve jo tum mod sako

Sabse basic object se shuru karte hain. Jab hum $y(x)$ likhte hain toh matlab hai: ==mujhe ek horizontal position $x$ do, aur main tumhe height $y$ wapas doonga==. $x$ ko left se right sweep karo aur heights mil ke ek curve trace karti hain.

Humein iska kyun zaroorat hai: is poore chapter mein, woh unknown jiske liye hum solve kar rahe hain woh ek number nahi hai — woh ek poori curve hai. Yeh ordinary calculus se sabse bada jump hai. "Woh $x$ dhundho jo $f(x)$ minimise kare" mein answer ek point hai; yahan answer ek rope ya fence ki shape hai.

2. Interval $[a,b]$ aur boundary conditions

Symbols $a$ aur $b$ woh do ground positions hain jahan curve start aur end hoti hai — woh do posts jinke beech ek chain hang karti hai, ya woh do pins jo ek fence ko thaame rakhte hain.

Picture: Figure s01 mein do red dots frozen hain. Jo bhi curve hum consider karein woh dono se pass karni chahiye.

Humein iska kyun zaroorat hai: bina pinned ends ke "best" curve infinity tak chali jaayegi. Ends fix karna hi woh cheez hai jo problem ka answer finite banata hai.

3. Derivative $y^{\prime }$ — har point pe steepness

$y^{\prime }$ (padho "y-prime") curve ka slope hai ek point pe: ek unit right pe kitne units upar, wahan curve ko touch karne wali tiny straight piece pe measure kiya hua.

Humein iska kyun zaroorat hai: parent note mein har integrand ($f$ aur $g$ dono) $y^{\prime }$ pe depend karta hai, kyunki length aur energy dono steepness ki parwah karte hain.

4. Arc length aur symbol $\sqrt{1+y^{\prime 2}}$

Yahan ek aisa symbol hai jo scary lagta hai lekin pure Pythagoras hai. Zameen ke saath $dx$ width ka ek tiny step lo. Curve us step mein $dy=y^{\prime }dx$ se upar jaati hai. Curve ka actual slanted piece ek right triangle ka hypotenuse hai jiske legs $dx$ (across) aur $dy$ (upar) hain.

Triangle se padho: hypotenuse${}^2=$ base${}^2+$ height${}^2=d{x}^2+(y^{\prime }dx{)}^2$. $d{x}^2$ factor out karo aur square root lo — $dx$ bahar aa jaata hai, $\sqrt{1+y^{\prime 2}}$ reh jaata hai.

Humein iska kyun zaroorat hai: har isoperimetric example mein constraint — "string ki fixed length hai" — $\int \sqrt{1+y^{\prime 2}}dx=\ell$ likha jaata hai. Yahi woh symbol hai sum up hua.

5. Integral ${\int }_a^b\dots dx$ — saare tiny pieces add karna

Tall S sign $\int$ ka matlab hai tiny contributions ka continuous pile sum karo. Hum interval $[a,b]$ ko countless tiny widths $dx$ mein kaatein, har ek pe andar wali cheez evaluate karein, aur sab add kar dein.

Picture: curve ke neeche infinitely many thin slabs side by side rakhdo aur glue kardo — integral woh glued total hai.

Humein iska kyun zaroorat hai: length, area, energy sab whole-curve quantities hain, point quantities nahi. Sirf ek integral hi "shape" ko "ek number jo optimise ya constrain karna ho" mein badal sakta hai.

6. Ek functional $J[y]$ — ek machine jo poori curve khaata hai

Ab key naya object. Ek ordinary function ek number khaata hai. Ek functional poori curve khaata hai aur ek single number return karta hai.

Andar wala chhota function $f(x,y,y^{\prime })$ integrand hai: ek recipe jo tumhe batata hai, har $x$ pe, yeh curve kitna contribute karti hai — tum kahan ho ($x$), kitna upar ($y$), aur kitna steep ($y^{\prime }$) use karke.

7. Variation $\delta y$, bump $\eta (x)$, aur knob $ϵ$

Best curve dhundne ke liye humein ek candidate curve ko nudge karna hoga aur dekhna hoga ki number $J$ upar jaata hai ya neeche. Nudge $\delta y$ likha jaata hai ("delta y"), curve ka ek chhota deformation jo endpoints ko pinned rakhta hai.

Knob $ϵ$ ghumao aur Figure s04 mein saari red dashed curves extremes ke beech swing karti hain, lekin hamesha dono fixed dots se hoke.

Humein iska kyun zaroorat hai: "best curve" ka matlab hai koi bhi chhota legal nudge $J$ ko behtar nahi banata. Yahi condition hai $\delta J=0$, aur use likhne ke liye humein $\eta$ aur $ϵ$ chahiye.

8. Euler–Lagrange expression $f_y-\frac{d}{dx}{f}_{y^{\prime }}$

Jab tum $J$ ko nudge karo aur demand karo ki woh har bump $\eta$ ke liye na badle, toh algebra (integration by parts + Fundamental Lemma) ek master equation squeeze out karti hai.

Isse ek balance law ki tarah padho: har point pe height ka direct pull ($f_y$) steepness-sensitivity ke running change ($\frac{d}{dx}{f}_{y^{\prime }}$) ko exactly cancel karna chahiye. Full derivation ke liye Euler–Lagrange Equation dekho jiske liye yeh page tumhe prepare karta hai.

Humein iska kyun zaroorat hai: yeh woh tool hai jo "$J$ stationary hai" ko ek concrete differential equation mein badalta hai jise tum curve ke liye solve kar sako.

9. Multiplier $\lambda$ — budget ki keemat

Aakhir mein, parent note ka star. $\lambda$ ("lambda") ek single fixed number hai — fine per unit of budget. Jab bhi problem mein ek constraint $K[y]=\ell$ ho, hum ise directly nahi maante. Iski jagah hum augmented recipe $F=f-\lambda g$ banate hain aur use freely optimise karte hain.

Exactly ek number kyun (function $\lambda (x)$ nahi)? Kyunki ek integral constraint ek single scalar equation hai ($\int gdx=\ell$), toh woh sirf ek fine "charge" kar sakta hai. (Pointwise constraints ko puri function $\lambda (x)$ chahiye hogi — parent note mein last wali galti.) Yeh ordinary finite-dimensional Lagrange multipliers ko exactly mirror karta hai.

Yeh sab topic ko kaise feed karta hai

Trace karo: curves aur slopes length banate hain; length sum hoke integral hai; ek poori curve ka integral ek functional hai; functional ko nudge karna Euler–Lagrange deta hai; priced budget add karna isoperimetric rule deta hai. Baaki sab downstream — Beltrami Identity, Catenary Curve, Isoperimetric Inequality, Brachistochrone Problem, aur Calculus of Variations — Fundamental Lemma — exactly inhi nodes pe plug in hote hain.

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