Exercises — Calculus of variations — functionals, functional derivative

$L$ KYA hai? Integrand: $L(x,y,y^{\prime })=3x{y}^{\prime 2}-y^2+5y$. Dono partials $y$ aur $y^{\prime }$ ko independent slots ki tarah treat karte hain (ek ko freeze karo jab doosre ko differentiate karo): $\frac{\partial L}{\partial y}=-2y+5,\frac{\partial L}{\partial y^{\prime }}=6x{y}^{\prime }.$ Note karo ki $\partial L/\partial y^{\prime }$ mein abhi bhi $x$ hai; jab hum baad mein iska $\frac{d}{dx}$ lete hain tab woh matter karta hai.

Beltrami ko koi explicit $x$ nahi chahiye — matlab $x$ sirf $y(x)$ aur $y^{\prime }(x)$ ke through aaye, kabhi akela nahi.

• (i) $\sqrt{1+y^{\prime 2}}$: koi akela $x$ nahi → apply hoti hai.
• (ii) $x{y}^{\prime 2}$: akela $x$ saamne hai → apply nahi hoti.
• (iii) $y^2+y^{\prime 2}$: koi akela $x$ nahi → apply hoti hai.
• (iv) $y\sqrt{1+y^{\prime 2}}$: koi akela $x$ nahi → apply hoti hai.

E–L KYUN yahan: $L=\sqrt{1+y^{\prime 2}}$ mein koi akela $y$ nahi hai, toh $\partial L/\partial y=0$; E–L reduce ho jaati hai $\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}=0$ mein, yaani $\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}=\text{const}$. $\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}$ ka constant hona force karta hai $y^{\prime }=\text{const}=m$ ko (ratio slope ka ek monotonic function hai, toh isko fix karna slope ko fix karta hai). Isliye $y=mx+b$, ek straight line — exactly Figure 2 mein red line. Endpoints fit karo: $y(0)=1\Rightarrow b=1$. $y(3)=7\Rightarrow 3m+1=7\Rightarrow m=2$. Toh $y=2x+1$. Length: slope $2$ over run $3$ se rise $6$ milti hai; length $=\sqrt{3^2+6^2}=\sqrt{45}=3\sqrt{5}\approx 6.708$.

$L=y^{\prime 2}+4y$. Partials: $\frac{\partial L}{\partial y}=4$, $\frac{\partial L}{\partial y^{\prime }}=2y^{\prime }$. E–L: $4-\frac{d}{dx}(2y^{\prime })=0\Rightarrow 4-2y^{\prime \prime }=0\Rightarrow y^{\prime \prime }=2$. Do baar integrate karo: $y^{\prime }=2x+C_1$, phir $y=x^2+C_{1}x+C_2$. Endpoints: $y(0)=0\Rightarrow C_2=0$. $y(1)=1\Rightarrow 1+C_1=1\Rightarrow C_1=0$. $\boxed{y=x^2}$.

KYA change hota hai: integration-by-parts step mein boundary term $[\frac{\partial L}{\partial y^{\prime }}\eta {]}_0^1$ ab $x=1$ par nahi marti, kyunki $\eta (1)$ forced nahi hai $0$ par (endpoint move karne ke liye free hai).

• $x=0$ par: $\eta (0)=0$ (woh end fixed hai) → term wahan vanish ho jaata hai.
• $x=1$ par: $\eta (1)$ arbitrary hai → sab aise $\eta$ ke liye ${\Phi }^{\prime }(0)$ ko kill karne ke liye humein chahiye $\frac{\partial L}{\partial y^{\prime }}{|}_{x=1}=0$. Yeh natural boundary condition hai. Interior equation (E–L): $L=\frac{1}{2}{y}^{\prime 2}+y$, toh $\partial L/\partial y=1$, $\partial L/\partial y^{\prime }=y^{\prime }$. E–L: $1-y^{\prime \prime }=0\Rightarrow y^{\prime \prime }=1$. Integrate karo: $y^{\prime }=x+C_1$, $y=\frac{1}{2}{x}^2+C_{1}x+C_2$. Conditions apply karo: fixed left end $y(0)=0\Rightarrow C_2=0$. Natural condition $y^{\prime }(1)=0\Rightarrow 1+C_1=0\Rightarrow C_1=-1$. $\boxed{y=\frac{1}{2}{x}^2-x}$, aur waqai $y^{\prime }(1)=0$: curve free end par horizontally aati hai.

$L(y,y^{\prime })$ ka total $x$-derivative (chain rule, koi akela $x$ nahi): $\frac{dL}{dx}=L_y{y}^{\prime }+L_{y^{\prime }}{y}^{\prime \prime }$. Ab product $y^{\prime }{L}_{y^{\prime }}$ differentiate karo: $\frac{d}{dx}(y^{\prime }{L}_{y^{\prime }})=y^{\prime \prime }{L}_{y^{\prime }}+y^{\prime }\frac{d}{dx}{L}_{y^{\prime }}$. Subtract karo: $\frac{d}{dx}(L-y^{\prime }{L}_{y^{\prime }})=\underset{dL/dx}{\underbrace{L_y{y}^{\prime }+L_{y^{\prime }}{y}^{\prime \prime }}}-\underset{\text{cancels}}{\underbrace{y^{\prime \prime }{L}_{y^{\prime }}}}-y^{\prime }\frac{d}{dx}{L}_{y^{\prime }}.$ Dono $y^{\prime \prime }{L}_{y^{\prime }}$ terms cancel hote hain — ek $dL/dx$ se, ek product rule se — bacha rehta hai $y^{\prime }(L_y-\frac{d}{dx}{L}_{y^{\prime }}).$ E–L se bracket $0$ hai, toh $L-y^{\prime }{L}_{y^{\prime }}=\text{const}$. $y^{\prime \prime }$ term ki cancellation exactly wahi reason hai kyun Beltrami ko kabhi second derivative ki zaroorat nahi hoti.

Koi explicit $x$ nahi, toh Beltrami apply hoti hai. $L=\frac{1}{2}{y}^{\prime 2}-\frac{1}{2}{\omega }^2{y}^2$, $L_{y^{\prime }}=y^{\prime }$. $L-y^{\prime }{L}_{y^{\prime }}=(\frac{1}{2}{y}^{\prime 2}-\frac{1}{2}{\omega }^2{y}^2)-y^{\prime }\cdot y^{\prime }=-\frac{1}{2}{y}^{\prime 2}-\frac{1}{2}{\omega }^2{y}^2=\text{const}.$ $-2$ se multiply karo: $y^{\prime 2}+{\omega }^2{y}^2=\text{const}=E$. Yeh ek conserved "energy" hai. Consistency check: $y^{\prime 2}+{\omega }^2{y}^2=E$ ko $x$ ke w.r.t. differentiate karo: $2y^{\prime }{y}^{\prime \prime }+2{\omega }^{2}y{y}^{\prime }=0\Rightarrow y^{\prime }(y^{\prime \prime }+{\omega }^{2}y)=0$. Non-constant $y$ ke liye, $y^{\prime \prime }+{\omega }^{2}y=0$, yaani $y^{\prime \prime }=-{\omega }^{2}y$ — exactly E–L equation. ✓

Multiplier KYUN: hum ek functional minimise kar rahe hain subject to ek aur constant hone ke condition mein — yeh Lagrange multipliers ka functional analogue hai. $\tilde{L}=y^{\prime 2}-\lambda y^2$ form karo aur $\tilde{L}$ par E–L apply karo. $\frac{\partial \tilde{L}}{\partial y}=-2\lambda y$, $\frac{\partial \tilde{L}}{\partial y^{\prime }}=2y^{\prime }$. E–L: $-2\lambda y-\frac{d}{dx}(2y^{\prime })=0\Rightarrow y^{\prime \prime }=-\lambda y$. $y(0)=y(\pi )=0$ ke saath yeh classic eigenvalue problem hai: $y=\sin (\sqrt{\lambda }x)$ ko chahiye $\sqrt{\lambda }\pi =n\pi$, toh $\lambda =n^2$, $y_n=A\sin (nx)$. Sabse chhota $\int y^{\prime 2}$ sabse chhota eigenvalue hai: $n=1$, deta hai $y=A\sin x$. Constraint se $A$ fix karo: ${\int }_0^{\pi }{A}^2{\sin }^{2}xdx=A^2\cdot \frac{\pi }{2}=1\Rightarrow A=\sqrt{2/\pi }$. Minimal value: ${\int }_0^{\pi }{y}^{\prime 2}dx={\int }_0^{\pi }{A}^2{\cos }^{2}xdx=A^2\cdot \frac{\pi }{2}=1$. Toh $\boxed{y=\sqrt{2/\pi }\sin x,\int y^{\prime 2}=1}$.

Ek nayi equation KYUN: ab $L$ second derivative par depend karta hai, toh variation ek ${\eta }^{\prime \prime }$ term produce karta hai jisko integration by parts do baar karna padta hai. $y\to y+\epsilon \eta$ vary karo $\eta (a)=\eta (b)={\eta }^{\prime }(a)={\eta }^{\prime }(b)=0$ ke saath: ${\Phi }^{\prime }(0)={\int }_a^b(L_y\eta +L_{y^{\prime }}{\eta }^{\prime }+L_{y^{\prime \prime }}{\eta }^{\prime \prime })dx.$ ${\eta }^{\prime }$ term par ek baar IBP aur ${\eta }^{\prime \prime }$ term par do baar IBP karo; saare boundary terms vanish ho jaate hain kyunki $\eta$ aur ${\eta }^{\prime }$ dono ends par pinned hain. Common $\eta$ extract karo: $\boxed{L_y-\frac{d}{dx}{L}_{y^{\prime }}+\frac{d^2}{d{x}^2}{L}_{y^{\prime \prime }}=0}(\text{Euler–Poisson}).$ Beam: $L=\frac{1}{2}{y}^{\prime \prime 2}$, toh $L_y=0$, $L_{y^{\prime }}=0$, $L_{y^{\prime \prime }}=y^{\prime \prime }$. Equation: $\frac{d^2}{d{x}^2}(y^{\prime \prime })=0\Rightarrow y^{(4)}=0$. General solution: $y=c_0+c_{1}x+c_2{x}^2+c_3{x}^3$ — ek cubic, ideal bent elastic beam ki shape. ✓

Theorem miniature mein: "$L$ ki $x$-independence" (ek symmetry) force karti hai ki $H\equiv y^{\prime }{L}_{y^{\prime }}-L$ kisi bhi extremal ke saath constant rahe — yeh conserved Hamiltonian/energy hai, sign flip ke saath Beltrami constant. Beltrami: $L-y^{\prime }{L}_{y^{\prime }}=\text{const}\iff H=y^{\prime }{L}_{y^{\prime }}-L=\text{const}$. Free particle $L=\frac{1}{2}{y}^{\prime 2}$: $L_{y^{\prime }}=y^{\prime }$, toh $H=y^{\prime }\cdot y^{\prime }-\frac{1}{2}{y}^{\prime 2}=\frac{1}{2}{y}^{\prime 2}=\text{const}$. Constant $\frac{1}{2}{y}^{\prime 2}$ ⇒ constant $|y^{\prime }|$ ⇒ straight line. Symmetry (koi preferred $x$ nahi) ne ek conservation law produce kiya (constant "speed"), woh pattern jo Lagrangian mechanics mein energy conservation ban jaata hai. ✓

Full E–L dono $L_y$ AUR $-\frac{d}{dx}{L}_{y^{\prime }}$ use karta hai ::: haan — kabhi sirf $L_y=0$ nahi. Beltrami condition ::: $L$ mein koi explicit (akela) $x$ nahi. Free endpoint deta hai ::: natural condition $\partial L/\partial y^{\prime }=0$ wahan. Constraint $\int g=$const handle hoti hai ::: functional Lagrange multiplier se, $L-\lambda g$ extremise karo. $L$ $y^{\prime \prime }$ par depend karta hai ⇒ equation ::: $L_y-\frac{d}{dx}{L}_{y^{\prime }}+\frac{d^2}{d{x}^2}{L}_{y^{\prime \prime }}=0$. $L$ ki $x$-independence ⇒ ::: conserved $H=y^{\prime }{L}_{y^{\prime }}-L$ (Beltrami / baby Noether).