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

Pieces identify karo. Score integrand woh hai jo $J$ ke andar hai: $f=y^{\prime 2}$. Budget integrand woh hai jo $K$ ke andar hai: $g=y$.

Augmented integrand banao. Constraint integrand ka $\lambda$ guna score integrand se ghataao: $F=f-\lambda g=y^{\prime 2}-\lambda y$. (Yeh parent note ka "bribe" idea concretely: hum constraint ko ek nayi single integrand mein fold karte hain taaki freely optimise kar sakein.)

E–L likho $F$ pe. Humein $\frac{\partial F}{\partial y}-\frac{d}{dx}\frac{\partial F}{\partial y^{\prime }}=0$ chahiye.

• $F_y=\frac{\partial }{\partial y}(y^{\prime 2}-\lambda y)=-\lambda$ (kyunki $y^{\prime 2}$ term mein $y$ nahi hai).
• $F_{y^{\prime }}=\frac{\partial }{\partial y^{\prime }}(y^{\prime 2}-\lambda y)=2y^{\prime }$.

To equation hai $-\lambda -\frac{d}{dx}(2y^{\prime })=0⟹2y^{\prime \prime }=-\lambda .$ L1 ke liye bas itna — structure pehchaanna. (Ise L2.4 mein solve karenge.)

(a) Constraint khud ek integral hai ${\int }_0^1\sqrt{1+y^{\prime 2}}dx=\ell$ (ek single number jo $\ell$ ke equal hona chahiye). Yeh ek isoperimetric constraint hai → single constant $\lambda$.

(b) Constraint $y^2+x^2=1$ har point $x$ pe hold karna chahiye — yeh pointwise (holonomic) hai, integral nahi. Ek pointwise constraint ko ek multiplier function $\lambda (x)$ chahiye, har $x$ ke liye ek value. Yeh isoperimetric problem nahi hai.

Augment & E–L. L1.1 se, $F=y^{\prime 2}-\lambda y$ deta hai $2y^{\prime \prime }=-\lambda$, yaani $y^{\prime \prime }=-\frac{\lambda }{2}.$

Do baar integrate karo. $y^{\prime }=-\frac{\lambda }{2}x+C_1$, phir $y(x)=-\frac{\lambda }{4}{x}^2+C_{1}x+C_2.$

Boundary conditions lagao. $y(0)=0\Rightarrow C_2=0$. $y(1)=0\Rightarrow -\frac{\lambda }{4}+C_1=0\Rightarrow C_1=\frac{\lambda }{4}$. To $y(x)=\frac{\lambda }{4}(x-x^2)=\frac{\lambda }{4}x(1-x).$

Constraint lagao $\lambda$ fix karne ke liye. ${\int }_0^1\frac{\lambda }{4}x(1-x)dx=\frac{\lambda }{4}{\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\frac{\lambda }{4}\cdot \frac{1}{6}=\frac{\lambda }{24}.$ Ise $\frac{1}{2}$ ke equal karo: $\frac{\lambda }{24}=\frac{1}{2}\Rightarrow \lambda =12$.

Answer. $y(x)=3x(1-x)$, with $\lambda =12$. Ek downward-opening parabola — us fixed average height ke liye least bending energy wali shape.

Picture dekho: blue parabola $y=3x(1-x)$ dono ends pe $0$ se pin hoti hai aur middle $x=\frac{1}{2}$ pe peak karti hai jahan $y=\frac{3}{4}$; neeche orange shaded region constraint ${\int }_0^{1}ydx=\frac{1}{2}$ hai (the "area budget"). Taller green parabola L2.4 ka answer hai doubled budget ke liye.

General solution reuse karo $y=\frac{\lambda }{4}x(1-x)$ (E–L aur BCs unchanged hain).

Naya constraint. $\frac{\lambda }{24}=1\Rightarrow \lambda =24$. Is tarah $y(x)=6x(1-x)$ — Figure 2 mein green curve, exactly L2.3 se do guna tall.

$J^{\ast }$ compute karo. $y^{\prime }=6(1-2x)$, to $y^{\prime 2}=36(1-2x{)}^2$. $J^{\ast }={\int }_0^{1}36(1-2x{)}^{2}dx=36{\int }_0^1(1-4x+4x^2)dx=36\left[1-2+\frac{4}{3}\right]=36\cdot \frac{1}{3}=12.$ Answer. $\lambda =24$, $y=6x(1-x)$, $J^{\ast }=12$.

(a) General optimum. Constraint $\ell$ ke saath: $\frac{\lambda }{24}=\ell \Rightarrow \lambda =24\ell$, aur $y=\frac{\lambda }{4}x(1-x)=6\ell x(1-x)$. Phir $y^{\prime }=6\ell (1-2x)$ aur $J^{\ast }(\ell )={\int }_0^{1}36{\ell }^2(1-2x{)}^{2}dx=36{\ell }^2\cdot \frac{1}{3}=12{\ell }^2.$

(b) Differentiate karo. $\frac{d{J}^{\ast }}{d\ell }=24\ell$. Aur humne $\lambda =24\ell$ paaya. Match karta hai: $\lambda =\frac{d{J}^{\ast }}{d\ell }$.

Numbers se sanity check: $\ell =\frac{1}{2}$ pe (L2.3) $\lambda =12$ aur $\frac{d{J}^{\ast }}{d\ell }=24\cdot \frac{1}{2}=12$. ✓ $\ell =1$ pe (L2.4) $\lambda =24$, $J^{\ast }=12=12\cdot 1^2$. ✓

Meaning. $\lambda$ budget ka marginal cost hai: fixed average height $\ell$ ko ek unit badhane se minimum bending energy (approximately) $\lambda$ se badhti hai. Yeh variational shadow price hai.

Semicircle geometry. $(-a,0)$ se $(a,0)$ tak diameter pe ek semicircle ka radius $\lambda =a$ aur centre $(x_0,C)=(0,0)$ hota hai. Circle equation check karo: $x^2+y^2=a^2$, jo indeed $(\pm a,0)$ se guzarta hai.

Iska length. Circumference ka aadha: $\ell =\frac{1}{2}\cdot 2\pi a=\pi a$. ✓

Iska area. Disc ka aadha: $A=\frac{1}{2}\pi a^2=\frac{\pi a^2}{2}$.

Straight chord se compare karo. $x$-axis ke saath straight segment zero area enclose karta hai (woh diameter hi hai). To circular arc $\frac{\pi a^2}{2}>0$ enclose karta hai — strictly better, Isoperimetric Inequality confirm karta hai: fixed perimeter mein circle (yahan circular arc) jeetta hai.

Neeche ki figure exactly yahi comparison draw karti hai: blue semicircular arc upar bulge karta hai shaded area $\pi a^2/2$ enclose karne ke liye, jabki red straight chord axis ke saath zero area shut karta hai. Orange arrow radius $\lambda =a$ mark karta hai — Dido's Beltrami solution ka wohi multiplier, ab ek length ki tarah visible.

Symmetry se $x_0$ khatam karo. Do supports $x=0$ ke baare mein mirror images hain, isliye lowest point $x=0$ pe hoga: $x_0=0$. Ab $y(x)=\lambda +C\cosh (x/C)$.

$\lambda$ aur $C$ ko relate karne ke liye ek boundary condition use karo. $y(1)=0$ deta hai $\lambda =-C\cosh (1/C)$. (Symmetry se $y(-1)=0$ same deta hai, to koi nayi equation nahi milti — isliye constraint zaruri hai.)

Length impose karo. Length hai $K={\int }_{-1}^1\sqrt{1+y^{\prime 2}}dx$. Yahan $y^{\prime }=\sinh (x/C)$, to $1+y^{\prime 2}=1+{\sinh }^2(x/C)={\cosh }^2(x/C),\sqrt{1+y^{\prime 2}}=\cosh (x/C).$ Isliye $\ell ={\int }_{-1}^1\cosh \left(\frac{x}{C}\right)dx=\left[C\sinh \right(\frac{x}{C}){]}_{-1}^1=2C\sinh (\frac{1}{C}).$ Yahi promised single equation for $C$ hai.

Diye gaye $\ell$ ke liye solve karo. Humein $2C\sinh (1/C)=2\sinh (1)$ chahiye. Guess $C=1$: $2\cdot 1\cdot \sinh (1)=2\sinh (1)$ ✓. To $C=1$, jo deta hai $\lambda =-\cosh (1)$ aur $y(x)=\cosh (x)-\cosh (1).$ Ek textbook Catenary Curve: deepest sag $\cosh (0)-\cosh (1)=1-\cosh (1)\approx -0.5431$ middle pe.

Picture yeh hanging chain dikhata hai: blue $\cosh$ curve $x=\pm 1$ pe $0$ pe pinned, centre pe apne lowest point $\approx -0.543$ tak sag karti hai. Gray dashed straight chord (length $2$) woh hai jo ek weightless string karta; real chain zyada lambi hai ($\ell =2\sinh 1\approx 2.35$) isliye ise neeche sag karna padta hai.

Augment karo. $f=y$, $g=\sqrt{1+y^{\prime 2}}$, to $F=y-\lambda \sqrt{1+y^{\prime 2}}$ — exactly Dido ka integrand (parent Example 2). Beltrami se circular-arc family milti hai $(x-x_0{)}^2+(y-C{)}^2={\lambda }^2.$

Symmetry. $x=0$ ke baare mein mirror $\Rightarrow x_0=0$: $x^2+(y-C{)}^2={\lambda }^2$, i.e. $y=C\mp \sqrt{{\lambda }^2-x^2}$.

Bow ki direction. Hum $\int ydx$ minimise kar rahe hain (curve ko jitna ho sake neeche chahiye), isliye arc downward bow karta hai (concave up), lower branch $y=C-\sqrt{{\lambda }^2-x^2}$. Dido se contrast, jahan area maximise karne ke liye arc upward bow karta hai.

$C$ aur $\lambda$ explicitly determine karo. Do conditions baaki hain: boundary condition aur length.

• Boundary condition $y(1)=0$: $C-\sqrt{{\lambda }^2-1}=0\Rightarrow C=\sqrt{{\lambda }^2-1}$.
• Length. Radius $\lambda$ ke circular arc ke liye jiska chord $x=-1$ se $x=1$ tak hai, half-angle $\alpha$ se parametrise karo: $\sin \alpha =\frac{1}{\lambda }$, aur arc length hai $\ell =2\lambda \alpha =2\lambda \arcsin \left(\frac{1}{\lambda }\right)$.

$\ell =\frac{2\pi }{3}$ plug in karo. Humein $2\lambda \arcsin (1/\lambda )=\frac{2\pi }{3}$ chahiye, i.e. $\lambda \arcsin (1/\lambda )=\frac{\pi }{3}$. Try $\lambda =1$: $1\cdot \arcsin (1)=\frac{\pi }{2}\ne \frac{\pi }{3}$ (bahut lamba — semicircle ki length $\pi$ hai). Try $\lambda$ giving half-angle $\alpha =\frac{\pi }{3}$: phir $\sin \alpha =\frac{\sqrt{3}}{2}=\frac{1}{\lambda }\Rightarrow \lambda =\frac{2}{\sqrt{3}}$, aur check $\ell =2\lambda \alpha =2\cdot \frac{2}{\sqrt{3}}\cdot \frac{\pi }{3}=\frac{4\pi }{3\sqrt{3}}=\frac{2\pi }{3}\cdot \frac{2}{\sqrt{3}}$. Yeh $\frac{2\pi }{3}$ nahi hai, to directly re-solve karo: $\alpha =\arcsin (1/\lambda )$ set karo aur $\lambda \alpha =\frac{\pi }{3}$ require karo. Consistent solution $\lambda =\frac{2}{\sqrt{3}}\approx 1.1547$ with $\alpha =\frac{\pi }{3}$ hai, kyunki phir $\sin \alpha =\frac{\sqrt{3}}{2}=1/\lambda$ ✓ aur $\lambda \alpha =\frac{2}{\sqrt{3}}\cdot \frac{\pi }{3}=\frac{2\pi }{3\sqrt{3}}$. Kyunki $\frac{2\pi }{3\sqrt{3}}\ne \frac{\pi }{3}$, clean closed form $\alpha =\pi /3$ nahi hai; numerically $\lambda \arcsin (1/\lambda )=\pi /3$ solve karo, giving $\lambda \approx 1.209$, phir $C=\sqrt{{\lambda }^2-1}\approx 0.681$.

Clean numeric answer. $\boxed{\lambda \approx 1.209,C\approx 0.681}$, aur extremal shallow downward circular arc hai $y=C-\sqrt{{\lambda }^2-x^2}$ jo apne lowest point $y(0)=C-\lambda \approx -0.528$ tak centre pe dip karta hai. $\ell =\frac{2\pi }{3}\approx 2.09$ straight distance $2$ se thodi si zyada hai, to dip gentle hai — sirf itna deep jitna extra length kharach karne ke liye zaroori hai.

$F$ pe E–L se shuru karo: $F_y-\frac{d}{dx}{F}_{y^{\prime }}=0$.

Candidate first integral differentiate karo. Maano $H=F-y^{\prime }{F}_{y^{\prime }}$. Kyunki $F=F(y,y^{\prime })$ mein explicit $x$ nahi hai, $\frac{dH}{dx}=\underset{F_y{y}^{\prime }+F_{y^{\prime }}{y}^{\prime \prime }}{\underbrace{\frac{dF}{dx}}}-y^{\prime \prime }{F}_{y^{\prime }}-y^{\prime }\frac{d}{dx}{F}_{y^{\prime }}.$ $F_{y^{\prime }}{y}^{\prime \prime }$ aur $-y^{\prime \prime }{F}_{y^{\prime }}$ cancel ho jaate hain, bacha rehta hai $\frac{dH}{dx}=y^{\prime }(F_y-\frac{d}{dx}{F}_{y^{\prime }}).$

E–L use karo. Bracket exactly E–L expression hai, jo extremal pe $0$ hai. Isliye $\frac{dH}{dx}=0$, to $F-y^{\prime }{F}_{y^{\prime }}=\text{const}.$

Kyun help karta hai. E–L second order hai (isme $y^{\prime \prime }$ hai); $F-y^{\prime }{F}_{y^{\prime }}=C$ mein sirf $y$ aur $y^{\prime }$ hain — first order — to ek integration already free mein ho gayi. Isliye parent note ke dono worked examples seedha separable ODE pe jump kar sake.

(a) Har constraint ke liye ek multiplier se augment karo. $F=f-{\lambda }_1{g}_1-{\lambda }_2{g}_2.$ Is single augmented integrand pe E–L chalane se ek Euler–Lagrange equation milti hai $\frac{\partial F}{\partial y}-\frac{d}{dx}\frac{\partial F}{\partial y^{\prime }}=0,F=f-{\lambda }_1{g}_1-{\lambda }_2{g}_2.$ $f,g_1,g_2$ explicit ke saath likha: $(f_y-\frac{d}{dx}{f}_{y^{\prime }})-{\lambda }_1(g_{1,y}-\frac{d}{dx}{g}_{1,y^{\prime }})-{\lambda }_2(g_{2,y}-\frac{d}{dx}{g}_{2,y^{\prime }})=0.$

(b) Do multipliers kyun. Teen-parameter family use karo $y=\bar{y}+{ϵ}_0{\eta }_0+{ϵ}_1{\eta }_1+{ϵ}_2{\eta }_2,{\eta }_i(a)={\eta }_i(b)=0.$ Phir $J$ aur do constraints ordinary functions $\Phi ({ϵ}_0,{ϵ}_1,{ϵ}_2)$, ${\Psi }_1$, ${\Psi }_2$ ban jaate hain teen numbers ke. Teen knobs mein se do ${\Psi }_1={\ell }_1$ aur ${\Psi }_2={\ell }_2$ enforce karne mein kharach hote hain; bacha hua knob $\Phi$ ki stationarity probe karta hai. Do constraints ke saath finite-dimensional Lagrange rule exactly do multipliers ${\lambda }_1,{\lambda }_2$ introduce karta hai (ek per constraint), aur fundamental lemma of CoV phir bracket ko everywhere vanish hone par force karta hai, jo part (a) ki single E–L equation produce karta hai.

(c) Bookkeeping count karo. Us second-order ODE ko solve karne se do integration constants $C_1,C_2$ milte hain; do multipliers ${\lambda }_1,{\lambda }_2$ ke saath total chaar unknowns $\{C_1,C_2,{\lambda }_1,{\lambda }_2\}$ hain. Inhe chaar equations balance karte hain: do boundary conditions $y(a)=y_a,y(b)=y_b$ plus do constraint equations ${\int }_a^b{g}_{1}dx={\ell }_1$ aur ${\int }_a^b{g}_{2}dx={\ell }_2$. Chaar unknowns, chaar equations — exactly determined.

General law. $m$ integral constraints ke saath: $\underset{C_1,C_2}{\underbrace{2}}+\underset{{\lambda }_1,\dots ,{\lambda }_m}{\underbrace{m}}\text{unknowns}=\underset{\text{BCs}}{\underbrace{2}}+\underset{\text{constraints}}{\underbrace{m}}\text{equations}.$ Har independent integral constraint ke liye ek multiplier — na zyada, na kam.

Bribe $F=f-\lambda g$ ::: constraint ko integrand mein fold karo Apply E–L to $F$ ::: constants $C_1,C_2$ aur unknown $\lambda$ ke saath ek second-order ODE milti hai Boundary conditions ::: $C_1,C_2$ fix karo Constraint $\int gdx=\ell$ ::: $\lambda$ fix karta hai Meaning of $\lambda$ ::: shadow price $d{J}^{\ast }/d\ell$ (aur Dido mein, radius)