Existence and uniqueness theorem — Picard-Lindelöf (statement): Worked examples

4.6.8 · D3 · Maths › Ordinary Differential Equations › Existence and uniqueness theorem — Picard-Lindelöf (statemen

Shuru karne se pehle, un tools ka ek-line refresher jo hum use karte hain, taaki koi symbol bina explanation ke na aaye:

Recall Wo teen numbers jo hum hamesha extract karte hain

Question ::: Ek rectangle $R = {∣ x - x_{0} ∣ \leq a, ∣ y - y_{0} ∣ \leq b}$ ke liye, $M$ , $L$ , $h$ ka kya matlab hai? $M$ = $R$ par $∣ f ∣$ ki sabse badi value (steepest slope jo solution le sakta hai). ::: $L$ = Lipschitz constant, cap jo batata hai ki $f$ kitni jaldi $y$ ke change ke saath change hoti hai. $h = min (a, b / M)$ = wo guaranteed half-width jahan unique solution exist karti hai.

Scenario matrix

Is theorem se related har ODE inhi cells mein se ek mein fit hoti hai. Neeche ke examples un cell(s) ke saath labelled hain jo wo hit karte hain — mil ke puri table cover ho jaati hai.

Cell	Situation	Kya galat ho sakta hai	Example
A	$f$ har jagah Lipschitz, $h$ ko $a$ limit karta hai	kuch nahi — clean	Ex 1
B	$f$ $R$ par Lipschitz, $h$ ko $b / M$ limit karta hai (box-height jeet jaati hai)	interval $a$ se chhoti ho jaati hai	Ex 2
C	Boundary case $L h = 1$	naive contraction fail, factorial bound bachata hai	Ex 3
D	Continuous lekin start par NOT Lipschitz	uniqueness kho jaati hai, kaafi solutions	Ex 4
E	Lipschitz locally lekin solution finite time mein $\infty$ tak escape	interval genuinely finite hai	Ex 5
F	Degenerate: $f$ $y$ se independent ( $L = 0$ )	kya theorem ki zaroorat bhi hai?	Ex 6
G	Real-world word problem (cooling / mixing)	words → IVP translate karo, phir apply karo	Ex 7
H	Exam twist: same $f$ par Peano vs Picard compare karo	kaun sa theorem kya deta hai?	Ex 8

Example 1 — Cell A: clean, interval width $a$ se limited

Worked example

$y^{'} = sin (x y)$ , $y (0) = 0$ , $R = {∣ x ∣ \leq 1, ∣ y ∣ \leq 3}$ par Forecast (pehle guess karo): kya interval $h$ box ki width $a = 1$ se limited hogi ya box ki height $b = 3$ se? Ek guess likho.

Step 1 — $M$ bound karo. Kyunki $∣ sin (kuch bhi) ∣ \leq 1$ , hume milta hai $∣ f ∣ = ∣ sin (x y) ∣ \leq 1 = M$ . Yeh step kyun? $M$ woh steepest slope hai jitna solution climb kar sakta hai; ek bounded sine ise instantly deta hai.

Step 2 — Lipschitz constant $L$ dhundho. $y$ mein differentiate karo: $\partial_{y} f = x cos (x y)$ , toh $∣ \partial_{y} f ∣ = ∣ x ∣∣ cos (x y) ∣ \leq 1 \cdot 1 = 1$ $R$ par (kyunki $∣ x ∣ \leq 1$ ). Ek convex box par bounded $y$ -derivative, Mean Value Theorem se Lipschitz deta hai $L = sup ∣ \partial_{y} f ∣ = 1$ ke saath. Yeh step kyun? Lipschitz woh hypothesis hai jo solution ko unique banata hai; finite $L$ ke bina hume sirf Peano ka existence milta.

Step 3 — $h$ compute karo. $h = min (a, \frac{b}{M}) = min (1, \frac{3}{1}) = min (1, 3) = 1.$ Yeh step kyun? Jo tighter constraint hogi wahi jeetegi. Yahan box height ( $b / M = 3$ ) generous hai, toh width $a = 1$ bottleneck hai — yeh forecast ka jawab hai.

Verify: $L h = 1 \cdot 1 = 1$ . Naive contraction chahta hai $L h < 1$ , toh yeh exactly edge par hai — lekin factorial/Bielecki bound phir bhi convergence deta hai. $∣ x ∣ \leq 1$ par unique solution exist karti hai. ✔

Example 2 — Cell B: interval box height $b / M$ se limited

Worked example

$y^{'} = x^{2} + y^{2}$ , $y (0) = 0$ , $R = {∣ x ∣ \leq 2, ∣ y ∣ \leq 1}$ par Forecast: humne jaanbujhkar $a = 2$ bada aur $b = 1$ chhota rakha. Kaun sa $h$ ko throttle karega?

Step 1 — $M$ bound karo. $R$ par: $∣ x^{2} + y^{2} ∣ \leq 2^{2} + 1^{2} = 4 + 1 = 5 = M$ . Yeh step kyun? Har square box ke corner par sabse bada hota hai; dono maxima add karo.

Step 2 — Lipschitz $L$ . $\partial_{y} f = 2 y$ , aur $∣2 y ∣ \leq 2$ $R$ par, toh $L = 2$ .

Step 3 — $h$ compute karo. $h = min (2, \frac{1}{5}) = \frac{1}{5} = 0.2.$ Yeh step kyun? Ab $b / M = 1/5$ bahut chhota hai, $a = 2$ se kaafi neeche: box height bottleneck hai. Solution height- $1$ box se jaldi bahar nikal sakti hai, toh hum ise sirf $∣ x ∣ \leq 0.2$ tak trust karte hain.

Verify: slope-line check — rate $M = 5$ par, $∣ x ∣ \leq 0.2$ tak solution zyada se zyada $5 \times 0.2 = 1 = b$ move karti hai, exactly box ki lid touch karte hue. Consistent. ✔ (Parent note ke Ex 1 se compare karo jahan $b = 1, M = 2$ ne $h = 1/2$ diya — same $f$ , alag box, alag $h$ .)

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

Example 3 — Cell C: boundary case $L h = 1$

Worked example

$y^{'} = 2 y$ , $y (0) = 1$ , $R = {∣ x ∣ \leq \frac{1}{2}, ∣ y ∣ \leq 1}$ $y_{0} = 1$ ke around Yahan $R$ ka matlab hai $∣ x ∣ \leq \frac{1}{2}$ aur $∣ y - 1∣ \leq 1$ (toh $0 \leq y \leq 2$ ). Forecast: kya naive " $T$ ek contraction hai" argument kaam karega, ya hume heavier machinery chahiye?

Step 1 — $M$ bound karo. $∣ f ∣ = ∣2 y ∣ \leq 2 \cdot 2 = 4 = M$ (box par sabse bada $∣ y ∣$ hai $2$ ).

Step 2 — Lipschitz $L$ . $\partial_{y} f = 2$ constant, toh $L = 2$ exactly, har jagah.

Step 3 — $h$ compute karo aur $L h$ test karo. $h = min (\frac{1}{2}, \frac{1}{4}) = \frac{1}{4}, L h = 2 \cdot \frac{1}{4} = \frac{1}{2} < 1.$ Ruko — is box ke saath, $L h < 1$ , toh Banach directly apply hota hai. True boundary $L h = 1$ hit karne ke liye, widen karo: $a = \frac{1}{2}$ lo lekin crudely bound karo taaki $h = \frac{1}{2}$ ho; tab $L h = 2 \cdot \frac{1}{2} = 1$ . Yeh step kyun? Yeh dikhata hai ki $L h$ theorem ka hissa nahi — yeh is par depend karta hai ki aap box kaise kaatte ho, ODE par nahi.

Verify: true solution hai $y = e^{2 x}$ . Picard iterates $y_{0} (x) = 1$ se shuru hote hue: $y_{1} = 1 + \int_{0}^{x} 2 d t = 1 + 2 x$ , $y_{2} = 1 + \int_{0}^{x} 2 (1 + 2 t) d t = 1 + 2 x + 2 x^{2}$ . Ye $e^{2 x} = 1 + 2 x + 2 x^{2} + \dots$ ke partial sums hain. $L h = 1$ ke bawajood sabhi $x$ ke liye convergent — factorial bound se proof, Banach se nahi. ✔

Example 4 — Cell D: continuous lekin Lipschitz nahi → uniqueness kho gayi

Worked example

$y^{'} = 3 y^{2/3}$ , $y (0) = 0$ Forecast: $f = 3 y^{2/3}$ continuous hai. Kya Picard–Lindelöf apply hota hai? Guess karo kitne solutions hain.

Step 1 — Continuity check karo. $y^{2/3} = (3 y)^{2} \geq 0$ har jagah continuous hai. Peano isliye kam se kam ek solution guarantee karta hai. Yeh step kyun? Continuity existence ke liye entry fee hai — lekin uniqueness ke liye nahi.

Step 2 — $y = 0$ ke paas Lipschitz condition test karo. $\partial_{y} f = 3 \cdot \frac{2}{3} y^{- 1/3} = 2 y^{- 1/3} \to + \infty$ jab $y \to 0^{+}$ . Koi finite $L$ ise cap nahi kar sakta, toh $f$ initial point par Lipschitz nahi hai. Yeh step kyun? Jis moment Lipschitz $(x_{0}, y_{0})$ par fail hota hai, theorem ki uniqueness guarantee void ho jaati hai.

Step 3 — Do solutions dikhao.

Flat wala: $y \equiv 0$ satisfy karta hai $y^{'} = 0 = 3 \cdot 0^{2/3}$ . ✔
Rising wala: $y = x^{3}$ try karo. Tab $y^{'} = 3 x^{2}$ aur $3 y^{2/3} = 3 (x^{3})^{2/3} = 3 x^{2}$ . ✔ Dono $(0, 0)$ se guzarte hain. Yeh step kyun? Concretely us failure ko demonstrate karta hai jiske baare mein theorem warn karta hai.

Verify: $x = 2$ par: solution 1 deta hai $y = 0$ ; solution 2 deta hai $y = 2^{3} = 8$ . Ek IVP se do alag values — uniqueness genuinely broken. ✔

Example 5 — Cell E: Lipschitz locally lekin finite time mein blow-up

Worked example

$y^{'} = 1 + y^{2}$ , $y (0) = 0$ Forecast: $f = 1 + y^{2}$ $y$ mein ek smooth polynomial hai. Kya solution forever live karti hai, ya guaranteed interval ka ek hard finite edge hai?

Step 1 — Local Lipschitz. $\partial_{y} f = 2 y$ ; kisi bhi bounded box $∣ y ∣ \leq b$ par hume milta hai $∣2 y ∣ \leq 2 b < \infty$ . Toh $f$ har finite box par Lipschitz hai — theorem har point par locally apply hota hai. Yeh step kyun? Local Lipschitz start ke paas unique solution deta hai; yeh kuch nahi kehta ki yeh kitni door tak extend hoti hai.

Step 2 — Edge dekhne ke liye exactly solve karo. Yeh separable hai: $\frac{d y}{1 + y ^{2}} = d x \Rightarrow arctan y = x + C$ . $y (0) = 0$ ke saath, $C = 0$ , toh $y = tan x$ . Yeh step kyun? $arctan$ ( $tan$ ka inverse) woh tool hai jo integral $\int \frac{d y}{1 + y ^{2}} = arctan y$ ke under $1 + y^{2}$ ko "undo" karta hai — woh ek elementary antiderivative jo fit baitta hai.

Step 3 — Maximal interval dhundho. $tan x \to + \infty$ jab $x \to \frac{π}{2}^{-}$ . Solution $(- \frac{π}{2}, \frac{π}{2})$ par exist karta hai aur unique hai aur $x = \frac{π}{2} \approx 1.5708$ se aage continue nahi ho sakta. Yeh step kyun? "Local" ka honest matlab yahi hai: nice $f$ , phir bhi finite lifespan kyunki $y$ infinity tak bhaag jaati hai.

Verify: check karo $y = tan x$ isko solve karta hai: $\frac{d}{d x} tan x = sec^{2} x = 1 + tan^{2} x = 1 + y^{2}$ . ✔ Aur $y (0) = tan 0 = 0$ . ✔

Example 6 — Cell F: degenerate case, $f$ $y$ se independent ( $L = 0$ )

Worked example

$y^{'} = cos x$ , $y (0) = 2$ Forecast: $f = cos x$ mein bilkul bhi $y$ nahi hai. Lipschitz constant kya hai, aur kya theorem kuch kaam bhi karta hai yahan?

Step 1 — Lipschitz constant. $∣ f (x, y_{1}) - f (x, y_{2}) ∣ = ∣ cos x - cos x ∣ = 0 \leq 0 \cdot ∣ y_{1} - y_{2} ∣$ , toh $L = 0$ . Yeh step kyun? $L = 0$ sabse strong possible Lipschitz condition hai — $f$ literally nahi hilta jab $y$ hilta hai. Uniqueness automatic hai.

Step 2 — $h$ kya banta hai. Bound $M = max ∣ cos x ∣ = 1$ phir bhi $h = min (a, b / M)$ limit karta hai, lekin contraction constant $L h = 0 < 1$ trivially — Banach bahut room ke saath apply hota hai. Yeh step kyun? Dikhata hai ki machinery obvious mein collapse ho jaati hai: ek baar integrate karo.

Step 3 — Directly solve karo. $y = y_{0} + \int_{0}^{x} cos t d t = 2 + sin x - sin 0 = 2 + sin x$ . Yeh step kyun? Jab $f$ mein koi $y$ nahi hota, Picard operator ek step mein answer deta hai — koi iteration nahi chahiye. Yeh sirf antidifferentiation hai, puri theory ka base case.

Verify: $y (0) = 2 + sin 0 = 2$ ✔ aur $y^{'} = cos x$ ✔. Unique kyunki $L = 0$ . ✔

Example 7 — Cell G: real-world word problem (Newton cooling)

Worked example Ek cup coffee

$9 0^{\circ}$ C par ek $2 0^{\circ}$ C wale room mein rakha hai; yeh temperature ke gap ke proportional rate se thanda hota hai constant $k = 0.1$ per minute ke saath. Kya temperature history unique hai? Forecast: physics ko ek answer hona chahiye. Kya Picard–Lindelöf ise confirm kar sakta hai?

Step 1 — Words ko ek IVP mein translate karo. "Cooling rate $\propto$ gap" deta hai $y^{'} = - k (y - 20)$ $y (0) = 90$ ke saath, $k = 0.1$ . Yahan $y (t)$ temperature hai, $t$ minutes mein. Yeh step kyun? Theorem tab tak apply nahi ho sakta jab tak sentence $y^{'} = f (t, y)$ na ban jaaye.

Step 2 — $f$ extract karo, hypotheses check karo. $f (t, y) = - 0.1 (y - 20)$ continuous hai, aur $\partial_{y} f = - 0.1$ constant hai, toh $L = 0.1$ har jagah. Linear ODE, globally Lipschitz. Yeh step kyun? Constant bounded $y$ -slope ⇒ Picard–Lindelöf poore time axis par existence AUR uniqueness deta hai.

Step 3 — Solve karo aur numbers read karo. Standard linear first-order solution: $y (t) = 20 + (90 - 20) e^{- 0.1 t} = 20 + 70 e^{- 0.1 t}$ . $t = 10$ par: $y = 20 + 70 e^{- 1} = 20 + 70 (0.367879 \dots) = 45.7515 \dots^{\circ}$ C. Yeh step kyun? $e^{- 0.1 t}$ exponential decay tool hai — woh unique function jiska derivative apne aap ke proportional hai, exactly wahi jo "rate $\propto$ gap" demand karta hai.

Verify: units — $[k] = min^{- 1}$ , toh $k (y - 20)$ ke units hain $^{\circ}$ C/min = temperature rate. ✔ $t = 0$ par: $20 + 70 = 9 0^{\circ}$ C start se match karta hai. ✔ Jab $t \to \infty$ : $y \to 2 0^{\circ}$ C (room temperature), physically correct. Uniqueness confirmed kyunki $L = 0.1 < \infty$ . ✔

Example 8 — Cell H: exam twist — same $f$ par Peano vs Picard

Worked example

$y^{'} = ∣ y ∣$ , $y (0) = 0$ ke liye: precisely batao ki har theorem kya deta hai. Forecast: ek theorem "haan" kahega, doosra "main promise nahi kar sakta". Kaun kaun sa hai?

Step 1 — Continuity. $f = ∣ y ∣$ har jagah continuous hai (root ke andar absolute value sign problem remove kar deta hai). Toh Peano guarantee karta hai ki ek solution exist karti hai. Yeh step kyun? Peano ko sirf continuity chahiye — yeh weaker lekin broader net hai.

Step 2 — Origin par Lipschitz test. $y > 0$ ke liye: $\partial_{y} f = \frac{1}{2 y} \to \infty$ jab $y \to 0^{+}$ . $(0, 0)$ par Lipschitz nahi. Toh Picard–Lindelöf uniqueness conclude nahi kar sakta yahan. Yeh step kyun? Picard–Lindelöf stronger claim hai (uniqueness) lekin stronger hypothesis demand karta hai (Lipschitz), jo fail hoti hai.

Step 3 — Explicit solutions se ambiguity confirm karo. $y \equiv 0$ kaam karta hai. Saath hi $y = \frac{x ^{2}}{4}$ $x \geq 0$ ke liye: tab $y^{'} = \frac{x}{2}$ aur $∣ y ∣ = x^{2} /4 = \frac{x}{2}$ ( $x \geq 0$ ke liye). ✔ Dono $(0, 0)$ se guzarte hain. Yeh step kyun? Exam point cement karta hai: Peano kehta hai "≥1 solution", Picard kehta hai "exactly 1 sirf tab jab Lipschitz ho" — aur yahan nahi hota.

Verify: $x = 2$ par: solution 1 deta hai $0$ ; solution 2 deta hai $2^{2} /4 = 1$ . Do answers ⇒ genuinely non-unique, Peano-yes / Picard-no verdict se match karta hai. ✔

Recall Self-test: kaun sa cell?

Question ::: Aapke paas $y^{'} = e^{y}$ , $y (0) = 0$ hai. Locally Lipschitz? Finite lifespan? $\partial_{y} e^{y} = e^{y}$ kisi bhi finite box par bounded ⇒ locally Lipschitz (locally unique). Lekin solve karne par, $y = - ln (1 - x)$ $x = 1$ par blow up hota hai ⇒ Cell E, finite maximal interval $(- \infty, 1)$ . :::