Euler-Lagrange equation — derivation

4.10.13 · Maths › Advanced Topics (Elite Level)

HUM kya optimize kar rahe hain?

Classic example: do points ke beech sabse chhota raasta mein $L=\sqrt{1+y^{\prime 2}}$ hota hai, kyunki arc length $J[y]={\int }_a^b\sqrt{1+y^{\prime 2}}dx$ hai. Hum chahte hain woh $y(x)$ jo ise sabse chhota banaye.

KAISE: variation trick (derivation from scratch)

Step 1 — Nearby curves ki ek family banao. Maan lo $y(x)$ sach mein extremal hai. Ise perturb karo: $Y(x,\epsilon )=y(x)+\epsilon \eta (x)$

Yeh step kyun? Hume poore curve ko wiggle karna hai. $\eta (x)$ ek arbitrary smooth "wiggle shape" hai aur $\epsilon$ ek chhota dial hai jo wiggle ki size control karta hai. $\epsilon =0$ rakhne par true curve wapas milti hai.

Step 2 — Fixed endpoints ka dhyan rakho. Kyunki har competitor ko same endpoints se guzarna hai, $Y(a)=y_a$ aur $Y(b)=y_b$ sab $\epsilon$ ke liye. Isse yeh enforce hota hai: $\boxed{\eta (a)=\eta (b)=0}$

Yeh step kyun? Agar $\eta$ kisi endpoint par nonzero hoti, toh perturbed curve required endpoint se chuk jaati — yeh allowed nahi hai.

Step 3 — Functional ko $\epsilon$ ki ordinary function mein badlo. $\Phi (\epsilon )={\int }_a^{b}L(x,Y,Y^{\prime })dx$ Kyunki true curve $\epsilon =0$ par hai, aur yahi $J$ ko extremize karti hai, ordinary function $\Phi (\epsilon )$ ka $\epsilon =0$ par stationary point hona chahiye: ${\frac{d\Phi }{d\epsilon }|}_{\epsilon =0}=0$

Yeh step kyun? Humne ek infinite-dimensional problem ko 1-D calculus problem mein reduce kar diya! Ab hum ordinary $\frac{d}{d\epsilon }$ use karte hain.

Step 4 — Integral ke andar differentiate karo. $\frac{d\Phi }{d\epsilon }={\int }_a^b\left(\frac{\partial L}{\partial Y}\frac{\partial Y}{\partial \epsilon }+\frac{\partial L}{\partial Y^{\prime }}\frac{\partial Y^{\prime }}{\partial \epsilon }\right)dx$ Kyunki $\frac{\partial Y}{\partial \epsilon }=\eta$ aur $\frac{\partial Y^{\prime }}{\partial \epsilon }={\eta }^{\prime }$, aur $\epsilon =0$ par evaluate karte hue (toh $Y\to y$): $0={\int }_a^b\left(\frac{\partial L}{\partial y}\eta +\frac{\partial L}{\partial y^{\prime }}{\eta }^{\prime }\right)dx$

Yeh step kyun? Chain rule: $L$ sirf $Y$ aur $Y^{\prime }$ ke through $\epsilon$ par depend karta hai.

Step 5 — Doosre term ko integration by parts se handle karo taaki $\eta$ apne derivative se azaad ho. ${\int }_a^b\frac{\partial L}{\partial y^{\prime }}{\eta }^{\prime }dx=\underset{=0\text{(endpoints!)}}{\underbrace{{\left[\frac{\partial L}{\partial y^{\prime }}\eta \right]}_a^b}}-{\int }_a^b\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)\eta dx$

Yeh step kyun? Boundary term Step 2 ki wajah se vanish ho jaata hai ($\eta (a)=\eta (b)=0$). Isliye endpoint condition matter karti thi. Ab har term mein bare $\eta$ hai.

Step 6 — $\eta$ ko factor out karo. $0={\int }_a^b\left(\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}\right)\eta (x)dx$

Step 7 — Calculus of Variations ka Fundamental Lemma apply karo.

Yeh step kyun? $\eta$ arbitrary hai. Agar bracket $g(x)$ kahin bhi nonzero hota, hum ek aisa $\eta$ choose kar sakte the jo exactly wahan bump up kare, jisse integral nonzero ho jaata — contradiction. (Steel-man ke liye neeche worked example dekho.)

Useful shortcut: Beltrami identity

Derive karo: $\frac{dL}{dx}$ ko totally compute karo: $\frac{dL}{dx}=\frac{\partial L}{\partial x}+\frac{\partial L}{\partial y}{y}^{\prime }+\frac{\partial L}{\partial y^{\prime }}{y}^{\prime \prime }$ E–L equation $\frac{\partial L}{\partial y}=\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}$ substitute karo: $\frac{dL}{dx}=\frac{\partial L}{\partial x}+y^{\prime }\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}+y^{\prime \prime }\frac{\partial L}{\partial y^{\prime }}=\frac{\partial L}{\partial x}+\frac{d}{dx}\left(y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)$ Toh $\frac{d}{dx}\left(L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=\frac{\partial L}{\partial x}$. Agar $\partial L/\partial x=0$:

Worked Examples

Common Mistakes (Steel-manned)

Active-Recall Flashcards

Connections

• Calculus of Variations — parent framework
• Principle of Least Action — $L$ ka physics origin
• Lagrangian Mechanics — E–L Newton's laws ko generalize karta hai
• Brachistochrone Problem — historic motivating example
• Geodesics — surfaces par shortest paths
• Integration by Parts — key algebraic tool
• Beltrami Identity — first-integral shortcut
• Noether's Theorem — $L$ ki symmetries conservation laws deti hain

Concept Map