Exercises — Existence and uniqueness theorem — Picard-Lindelöf (statement)

Yahan $f(x,y)=3y+\sin x$ hai. Starting point hai $(x_0,y_0)=(0,2)$. $\frac{\partial f}{\partial y}=3.$ $y$-derivative constant $3$ hai, toh $|f(x,y_1)-f(x,y_2)|=|3y_1-3y_2|=3|y_1-y_2|$. Yeh exactly Lipschitz inequality hai $L=3$ ke saath, aur yeh sab $(x,y)$ ke liye hold karta hai. Toh $f$ globally Lipschitz hai $y$ mein. ($\sin x$ term sirf $x$ par depend karta hai, toh difference mein cancel ho jaata hai — Lipschitz $y$ ke baare mein hai, $x$ ke nahin.)

$f(x,y)=\sqrt{|y|}=|y{|}^{1/2}$ har jagah continuous hai, toh existence theek hai (Peano, Peano existence theorem). Lekin $\partial f/\partial y=\frac{1}{2}|y{|}^{-1/2}\to \infty$ jab $y\to 0$, toh $f$ $y$ mein Lipschitz nahi hai $y=0$ ke paas. Isliye Picard–Lindelöf uniqueness guarantee nahi karta. Existence abhi bhi hold karti hai; uniqueness woh cheez hai jo khatam hoti hai.

Rectangle hai $R=\{(x,y):|x|\le 1,|y|\le 2\}$ (kyunki $x_0=y_0=0$).

• $M={\max }_R|x^2+y^2|=1^2+2^2=5$ (dono terms corner par max hote hain).
• ${\partial }_{y}f=2y$, toh $L={\max }_R|2y|=2\cdot 2=4$.
• $h=\min \left(a,\frac{b}{M}\right)=\min \left(1,\frac{2}{5}\right)=\frac{2}{5}=\mathrm{0.4.}$ Toh $|x|\le 0.4$ par ek unique solution guaranteed hai. (Parent example 1 se $b=1$ ke saath compare karo: $b$ badhane se $h$ shrink ho sakta hai kyunki $M$ tez badh jaata hai — L2 trap dekho.)

Rectangle yahan hai $R=\{(x,y):|x|\le 10,|y-1|\le b\}$. $|y-1|\le b$ par sabse bada $|y|$ hai $1+b$, toh $M=(1+b{)}^2$. $b/M$ constraint hai: $g(b)=\frac{b}{(1+b{)}^2}.$ Maximise karo: $g^{\prime }(b)=\frac{(1+b{)}^2-b\cdot 2(1+b)}{(1+b{)}^4}=\frac{(1+b)-2b}{(1+b{)}^3}=\frac{1-b}{(1+b{)}^3}.$ $g^{\prime }(b)=0$ set karne par $b=1$ milta hai. Kyun yeh maximum hai (minimum nahin): denominator $(1+b{)}^3>0$ $b\ge 0$ ke liye hai, toh $g^{\prime }$ ka sign numerator $1-b$ ka sign hai. $0<b<1$ ke liye $1-b>0$ toh $g^{\prime }>0$ (rising); $b>1$ ke liye $1-b<0$ toh $g^{\prime }<0$ (falling). $b=1$ par sign change $+\to -$ exactly ek maximum hai. Phir $M=(1+1{)}^2=4$ aur $b/M=\frac{1}{4}$. Kyunki $a=10\gg \frac{1}{4}$, min $b/M$ hai, toh: $h_{\max }=\frac{1}{4}=0.25\text{at }b=1.$ Beauty check: true solution $y=\frac{1}{1-x}$ $x=1$ par blow up karta hai, toh koi bhi finite rectangle $h$ ko usse aage push nahi kar sakta. Hamaara $h=0.25$ ek safe local guarantee hai, maximal interval nahin.

$f(t,y)=2ty$, $y_0(x)=1$. $y_1=1+{\int }_0^{x}2t\cdot 1dt=1+x^2.$ $y_2=1+{\int }_0^{x}2t(1+t^2)dt=1+{\int }_0^x(2t+2t^3)dt=1+x^2+\frac{x^4}{2}.$ $y_3=1+{\int }_0^{x}2t\left(1+t^2+\frac{t^4}{2}\right)dt=1+{\int }_0^x\left(2t+2t^3+t^5\right)dt=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}.$ Pattern $\sum _{k=0}^{\infty }\frac{(x^2{)}^k}{k!}=e^{x^2}$ ke partial sums hai. Toh iterates unique solution $y=e^{x^2}$ par converge karte hain — exactly wahi jo separation of variables deta hai ($\ln y=x^2$). Yeh Picard iteration (method of successive approximations) literally Taylor series build kar raha hai.

$y\equiv 0$: toh $y^{\prime }=0$ aur $y^{2/3}=0$; dono sides $0$ ke barabar hain, aur $y(0)=0$. ✔ (sab $x$ ke liye valid.) $y=(x/3{)}^3$: $y^{\prime }=3\cdot \frac{1}{3}{\left(\frac{x}{3}\right)}^2={\left(\frac{x}{3}\right)}^2$. Ab right side check karo $x$ ke har sign ke liye:

• $x\ge 0$: toh $(x/3{)}^3\ge 0$ aur $y^{2/3}={\left[(x/3{)}^3\right]}^{2/3}=(x/3{)}^2$. $y^{\prime }$ se match karta hai. ✔
• $x<0$: toh $(x/3{)}^3<0$. $y^{2/3}$ ko $((x/3{)}^3{)}^{2/3}=|x/3{|}^2=(x/3{)}^2$ padhne par (squaring sign ko khatam karta hai), yeh abhi bhi $y^{\prime }=(x/3{)}^2$ ke barabar hai. Toh wahi analytic formula IVP ko dono sides par solve karta hai — cubic curve origin se below-left se above-right tak jaati hai, aur $y\equiv 0$ wahan usse overlap karta hai.

Ek hi point se do alag solutions ⟹ uniqueness fail hoti hai. Cause: ${\partial }_y(y^{2/3})=\frac{2}{3}{y}^{-1/3}\to \infty$ jab $y\to 0$, toh $y=0$ ke paas koi finite $L$ kaam nahi karta.

Neeche wala figure dono solution curves plot karta hai. Blue flat line $y\equiv 0$ hai; orange curve $y=(x/3{)}^3$ hai $x<0$ aur $x>0$ dono ke liye draw ki gayi. Woh red dot $(0,0)$ par cross karte hain — woh single shared point jismein do solutions hain exactly woh non-uniqueness hai jise theorem forbid karta hai jab Lipschitz hold karta hai. Gray arrow mark karta hai jahan ${\partial }_{y}f$ blow up karta hai ($y=0$ par), woh missing ingredient.

General uniqueness. Dono integral form solve karte hain, toh subtract karne par, $y(x)-\tilde{y}(x)={\int }_{x_0}^x[f(t,y)-f(t,\tilde{y})]dt.$ $\varphi (x)=|y(x)-\tilde{y}(x)|\ge 0$ lo. Absolute values lo aur Lipschitz use karo $|f(t,y)-f(t,\tilde{y})|\le L\varphi (t)$: $\varphi (x)\le L{\int }_{x_0}^x\varphi (t)dt.$ Grönwall (zero constant term ke saath) $\varphi (x)\le 0\cdot e^{L(x-x_0)}=0$ deta hai. Kyunki $\varphi \ge 0$, $\varphi \equiv 0$, toh $y\equiv \tilde{y}$. Uniqueness. ✔ Perturbation bound. $f(x,y)=2y+g(x)$ ke liye, $L=2$. Do solutions $y,\tilde{y}$ jinka $|y(x_0)-\tilde{y}(x_0)|=\epsilon$ hai, general Grönwall estimate $\varphi (x)\le \epsilon e^{L(x-x_0)}$ follow karte hain. $x=x_0+1$ par: $\varphi (x_0+1)\le \epsilon e^{2\cdot 1}=\epsilon e^2\approx 7.389\epsilon .$ (Yeh sharp hai: $y-\tilde{y}$ satisfy karta hai $(y-\tilde{y}{)}^{\prime }=2(y-\tilde{y})$, exactly $\epsilon e^{2(x-x_0)}$.)

Linear ODEs ke liye Lipschitz constant $L=|p|\le P$ hai, ek finite constant jo $y$ se independent hai — box height $b$ kabhi $L$ ko upar nahin force karta. Kyun finite-time blow-up hi ek obstruction hai (continuation argument). Picard–Lindelöf pehle ek small interval par local solution deta hai. Phir tum theorem ko right endpoint par restart kar sakte ho, wahan ki value ko new initial condition ki tarah use karke, solution ko thoda aur extend karte ho; repeat. Yeh process ya toh hamesha continue karta hai (poori interval cover karke) ya ruk jaata hai — aur jo iska ruk sakne ka ek hi tarika hai kisi finite $x^{\ast }$ par woh yeh hai ki solution $\pm \infty$ ki taraf run kare jab $x\to x^{\ast }$ (yeh har box se escape karta hai, toh koi naya box build nahi ho sakta). Symbols mein: ek maximal solution jo finite $x^{\ast }$ par ruk jaata hai uske liye $|y(x)|\to \infty$ wahan hona zaroori hai (Maximal interval of existence). Toh agar hum $|y|$ ko har finite interval par bound kar sakein, toh blow-up impossible hai aur continuation kabhi nahi rukta. Grönwall exactly woh bound supply karta hai: $|y(x)|\le (|y_0|+\int |q|)e^{P|x-x_0|}$, har finite $x$ ke liye finite. Koi blow-up available nahin hone se, continuation poori domain tak jaati hai jahan $p,q$ defined hain. Contrast: $y^{\prime }=y^2$ globally Lipschitz nahi hai (${\partial }_y=2y$ unbounded); Grönwall aisa koi bound nahi deta, aur solution $y=\frac{1}{1-x}$ $y(0)=1$ se genuinely $|y|\to \infty$ satisfy karti hai $x=1$ par — continuation wahan ruk jaati hai. Unbounded ${\partial }_{y}f$ hi woh cheez hai jo finite-time blow-up ko permit karta hai.

(a) $R$ par, $|x|\le 1$ aur $|y|\le 2$, toh $M=\max |x+y|=1+2=3$. ${\partial }_{y}f=1\Rightarrow L=1$. Phir: $h=\min \left(1,\frac{1}{3}\right)=\frac{1}{3}.$ (b) Yaad karo (upar ke reminders se) Picard operator $T[u](x)=y_0+{\int }_{x_0}^{x}f(t,u(t))dt$, jo continuous functions par sup-norm $\Vert u\Vert ={\max }_{|x|\le h}|u(x)|$ ke saath act karta hai. Uska contraction constant $Lh$ hai: do candidates $u,v$ ke liye, $\Vert T[u]-T[v]\Vert \le Lh\Vert u-v\Vert .$ Yahan $Lh=1\cdot \frac{1}{3}=\frac{1}{3}<1$, toh $T$ distances shrink karta hai — yeh $|x|\le \frac{1}{3}$ par sup-norm mein ek genuine contraction hai, aur Banach fixed-point theorem directly ek unique fixed point (solution) deta hai. (c) $y_0=1$. Pehla iterate: $y_1=1+{\int }_0^x(t+y_0)dt=1+{\int }_0^x(t+1)dt=1+x+\frac{x^2}{2}.$ Doosra iterate: yahan $f(t,y_1(t))=t+y_1(t)=t+\left(1+t+\frac{t^2}{2}\right)$. Like terms collect karne par, constant $1$ hai, $t$-terms $t+t=2t$, aur quadratic $\frac{t^2}{2}$ hai, toh integrand hai $1+2t+\frac{t^2}{2}$. Isliye: $y_2=1+{\int }_0^x\left(1+2t+\frac{t^2}{2}\right)dt=1+x+x^2+\frac{x^3}{6}.$ (d) Linear ODE $y^{\prime }-y=x$ solve karo. Integrating factor $e^{-x}$: $(y{e}^{-x}{)}^{\prime }=x{e}^{-x}$, toh right side integrate karke (by parts) $\int x{e}^{-x}dx=-(x+1)e^{-x}$, jo deta hai $y{e}^{-x}=-(x+1)e^{-x}+C$, yaani $y=C{e}^x-x-1$. $y(0)=1$ apply karo: $x=0$ par term $-x-1$ equals $-(0)-1=-1$ hai, toh $y(0)=C\cdot 1-1=C-1$. $C-1=1$ set karne par $C=2$ milta hai. Isliye: $y(x)=2e^x-x-1.$ $x=h=\frac{1}{3}$ par: $y(\frac{1}{3})=2e^{1/3}-\frac{1}{3}-1=2e^{1/3}-\frac{4}{3}\approx \mathrm{1.4600.}$

Rectangle hai $R=\{(x,y):|x|\le a,|y|\le b\}$ jahan $a$ large hai. $|y|\le b$ par, $M=\max |1+y^2|=1+b^2$. $g(b)=\frac{b}{1+b^2}$ optimise karo: $g^{\prime }(b)=\frac{(1+b^2)-b\cdot 2b}{(1+b^2{)}^2}=\frac{1-b^2}{(1+b^2{)}^2}.$ $g^{\prime }(b)=0$ set karne par $b^2=1$ milta hai, toh $b=1$ ($b>0$ lete hue). Kyun maximum hai: denominator $(1+b^2{)}^2>0$, toh $\mathrm{sign}{g}^{\prime }=\mathrm{sign}(1-b^2)$. $0<b<1$ ke liye, $1-b^2>0\Rightarrow g^{\prime }>0$ (rising); $b>1$ ke liye, $1-b^2<0\Rightarrow g^{\prime }<0$ (falling). $b=1$ par sign change $+\to -$ ⟹ maximum. Phir $M=1+1=2$ aur $h=\frac{b}{M}=\frac{1}{2}=0.5$. Kyunki $\pi /2\approx 1.5708$, hamaara best local guarantee $h=0.5$ true blow-up time ke andar hai — safe lekin conservative, jaise ek local theorem se expected hai.

Neeche wala figure yeh geometrically dikhata hai. Dashed gray box $R$ hai ($b=1$ ke saath). Green wedge start se $\pm M$ slope ki do lines ki pair hai — solution uske andar trapped hai, aur yeh box ki height se exactly $x=h=0.5$ (blue vertical line) par exit karta hai. Orange curve true solution $y=\tan x$ hai, jo climb karta rehta hai aur sirf red dotted line $x=\pi /2\approx 1.57$ par blow up karta hai. Blue aur red lines ke beech ka gap woh difference hai jo theorem guarantee karta hai aur jo actually hota hai.

::: answer) Kaun si quantity slope bound karti hai? ::: $M={\max }_R|f|$ Kaun si quantity convergence aur uniqueness control karti hai? ::: Lipschitz constant $L$ Guaranteed half-interval ka formula? ::: $h=\min (a,b/M)$ Sirf continuity kaun sa conclusion deti hai? ::: sirf existence (Peano) Woh tool jo "do solutions" ko "ek solution" mein convert karta hai? ::: Grönwall's inequality