Existence and uniqueness theorem — Picard-Lindelöf (statement): Question bank

4.6.8 · D5 · Maths › Ordinary Differential Equations › Existence and uniqueness theorem — Picard-Lindelöf (statemen

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

Notation reminders jo throughout use honge: IVP hai $y^{'} = f (x, y), y (x_{0}) = y_{0}$ ; $M$ , rectangle $R$ par $∣ f ∣$ ko bound karta hai; $L$ Lipschitz constant hai $y$ mein; $h = min (a, b / M)$ guaranteed half-width hai interval ka.

Operator $T$ (jiske around zyaadatar traps ghoomte hain) aur Bielecki weighted norm weight — dono ki apni picture bhi hai:

True or false — justify

True or false: Agar

f

rectangle

R

par continuous hai, to IVP ka exactly ek solution hai.

False. Continuity akeli sirf existence deti hai (Peano) lekin uniqueness guarantee nahi karta; ek hi solution pin down karne ke liye tumhe

y

mein Lipschitz condition chahiye. Example:

y^{'} = y^{2/3}

continuous hai phir bhi

(0, 0)

se do solutions nikalta hai.

True or false: Lipschitz condition variable

x

mein hold karni chahiye.

False. Contraction estimate mein

∣ f (t, y_{1}) - f (t, y_{2}) ∣ \leq L ∣ y_{1} - y_{2} ∣

use hota hai, jo same

t

par do alag

y

-values compare karta hai. Lipschitz $y$ mein chahiye;

x

mein sirf continuity kaafi hai.

True or false: Agar

\partial f / \partial y

exist karta hai aur ek convex region par bounded hai, to

f

wahan

y

mein Lipschitz hai.

True. Mean Value Theorem se

f (x, y_{1}) - f (x, y_{2}) = \partial_{y} f (x, ξ) (y_{1} - y_{2})

, isliye

L = sup ∣ \partial_{y} f ∣

kaam karta hai. Yahi everyday tarika hai

L

find karne ka.

True or false:

y

mein Lipschitz hona,

y

mein continuous hone se strictly stronger hai.

True. Lipschitz (uniform) continuity imply karta hai, lekin

y^{2/3}

continuous hai aur

0

ke paas Lipschitz nahi (uska

y

-slope blow up ho jaata hai). To Lipschitz functions ki class ek proper subset hai.

True or false: Theorem poore rectangle

∣ x - x_{0} ∣ \leq a

par solution guarantee karta hai.

False. Solution sirf

∣ x - x_{0} ∣ \leq h = min (a, b / M)

par guaranteed hai, kyunki graph box ki top/bottom (height

b

) se pehle hi escape kar sakta hai jab

x

\pm a

tak pahunche. Result local hai.

True or false: Picard–Lindelöf ko

L h < 1

ek hypothesis ki zaroorat hai.

False.

L h < 1

sirf ek proof route ki convenience hai (sup-norm mein direct Banach). Factorial estimate ya Bielecki weighted norm operator ko kisi bhi

h

ke liye poore interval par contraction bana deta hai, isliye aisi koi hypothesis impose nahi hoti.

True or false: Theorem mein uniqueness continuity se aati hai, existence Lipschitz se.

False (dono halves). Yeh naive slogan ka roughly ulta hai aur woh reversal bhi imprecise hai: Picard proof mein Lipschitz estimate dono — convergence (existence) aur uniqueness — drive karta hai; continuity bound

M

supply karta hai jo iteration shuru hone deta hai.

True or false: Agar

f

x

mein continuous hai aur poore

R^{2}

par globally Lipschitz in

y

hai, to solution sab

x

ke liye exist karta hai.

True. Global Lipschitz bound linear growth deta hai

∣ f (x, y) ∣ \leq ∣ f (x, 0) ∣ + L ∣ y ∣

, jo Grönwall ke through finite-time blow-up rule out karta hai, isliye Maximal interval of existence poora

R

hai. (Subtlety sirf yeh hai ki "kisi rectangle par Lipschitz" sirf local

h

deta hai; genuinely global Lipschitz poori line deta hai.)

True or false: Bada Lipschitz constant

L

guaranteed interval

h

ko sirf shrink kar sakta hai.

False.

h = min (a, b / M)

mein

L

hai hi nahi —

L

convergence speed aur uniqueness govern karta hai, length

h

nahi. Sirf

M

(size bound)

h

ko shrink karta hai.

Spot the error

"Kyunki

∣ f (x, y) ∣ \leq M

hai, solution ka slope at most

M

hai, isliye distance

a

ke baad yeh at most

M a

utha — jo hamesha height

b

mein fit ho jaata hai."

Error:

M a

necessarily

\leq b

nahi hota. Agar

M a > b

hai to graph box se escape karta hai, aur exactly isliye hum

b / M

tak shrink karte hain aur

h = min (a, b / M)

lete hain.

T [y] = y_{0} + \int_{x_{0}}^{x} f (t, y) d t

satisfy karta hai

sup ∣ T [u] - T [v] ∣ \leq L h sup ∣ u - v ∣

, isliye

T

ek contraction hai — ho gaya."

Error: yeh contraction tabhi hai jab

L h < 1

ho. Example

y^{'} = x^{2} + y^{2}

ke liye (jahan

L = 2, h = \frac{1}{2}

) hume

L h = 1

milta hai, to yeh step fail karta hai; iske bajaaye factorial bound ya Bielecki weighted norm chahiye.

f = y^{2/3}

y

mein differentiable hai, aur differentiable functions Lipschitz hote hain, isliye uniqueness hold karta hai."

Error:

\partial_{y} f = \frac{2}{3} y^{- 1/3}

y \to 0

par unbounded hai, isliye

f

0

ke paas Lipschitz nahi hai. Unbounded derivative wali differentiability Lipschitz nahi deti, aur uniqueness wakai fail karta hai.

"Picard iterates ka convergence prove karne ke liye hum ODE ko differentiate karte hain robustness ke liye."

Error: hum integrate karte hain, differentiate nahi.

y^{'} = f

y = y_{0} + \int f

mein convert karna ek fragile derivative condition ko ek robust integral operator mein badal deta hai jiska fixed point hum dhundte hain.

"Grönwall's inequality prove karta hai ki iterates converge karte hain."

Error: Grönwall uniqueness ke liye use hota hai (

ϕ = ∣ y - \tilde{y} ∣ \equiv 0

force karne ke liye). Iterates ka convergence factorial bound par Weierstrass M-test se aata hai.

"Bound

∣ y_{n + 1} - y_{n} ∣ \leq M \frac{L ^{n} ∣ x - x _{0} ∣ ^{n + 1}}{( n + 1 )!}

sirf tabhi converge karta hai jab

L h < 1

ho."

Error: factorials kisi bhi exponential ko beat karte hain — series

\frac{M}{L} (e^{L h} - 1) < \infty

tak sum hoti hai har finite

h

ke liye. Exactly isliye proof

L h \geq 1

mein bhi survive karta hai.

Why questions

Proof ODE ko integrate kyun karta hai,

y^{'}

ke saath directly kyun nahi kaam karta?

Kyunki integration ek smoothing, contractive operation hai: operator

T [y] = y_{0} + \int_{x_{0}}^{x} f

continuous functions ko continuous functions mein map karta hai aur uske fixed points exactly solutions hain, jo hume fixed-point machinery invoke karne deta hai.

h

minimum kyun hona chahiye, sirf

b / M

kyun nahi?

Do independent constraints dono hold karni chahiye — box mein vertically raho (

∣ x - x_{0} ∣ \leq b / M

) aur horizontally (

∣ x - x_{0} ∣ \leq a

). Jo tighter hai woh jeetta hai, isliye

h = min (a, b / M)

T

ka ek fixed point aur IVP ka solution ek hi cheez kyun hai?

Agar

T [y] = y

hai to

y (x) = y_{0} + \int_{x_{0}}^{x} f (t, y) d t

; differentiate karne par

y^{'} = f (x, y)

milta hai aur

x = x_{0}

set karne par

y (x_{0}) = y_{0}

milta hai. Equivalence dono taraf kaam karta hai.

Uniqueness ke liye Lipschitz chahiye, sirf continuity kyun nahi?

Continuity kai candidate graphs ko ek point se squeeze through hone deti hai jahan

f

y

-slope unbounded ho (jaise

0

par

y^{2/3}

). Lipschitz us slope ko cap karta hai, aur Grönwall ke through do solutions alag nahi ho sakti.

Banach step ke liye

h < 1/ L

tak shrink karne ke baad bhi hum kabhi kabhi poora interval

h = min (a, b / M)

kyun recover kar lete hain?

Continuation se: ek chhote contraction interval par solve karo, phir uske endpoint se ek nayi initial condition ki tarah restart karo aur pieces patch karo, solution ko

h

tak extend karo (aur Maximal interval of existence ki taraf).

Bielecki weighted norm

∥ u ∥ = sup e^{- 2 L ∣ x - x_{0} ∣} ∣ u (x) ∣

kyun help karta hai?

Exponential weight

x_{0}

se door values ko discount karta hai, us factor ko absorb karta hai jo otherwise

L h < 1

force karta; is norm ke under

T

kisi bhi

h

ke liye poore interval par ek genuine contraction ban jaata hai.

h

ki length

L

nahi balki

M

kyun control karta hai?

M

slope bound karta hai, isliye yeh dictate karta hai ki graph box ki top ki taraf kitni tezi se climb kar sakta hai;

L

govern karta hai ki nearby solutions kitni tezi se saath aati hain (convergence/uniqueness), jo

R

ke andar rehne se alag concern hai.

Edge cases

Agar rectangle par

M = 0

ho to kya hoga?

Tab

f \equiv 0

hai, isliye

y^{'} \equiv 0

aur

y \equiv y_{0}

— ek constant. Formula

b / M

+ \infty

hai, isliye

h = min (a, \infty) = a

; trivial solution poori width par exist karta hai, theorem se consistent.

Agar starting point

(x_{0}, y_{0})

exactly

R

ki boundary par ho jahan

f

jump karta hai?

Hypotheses require karte hain ki

f

closed

R

par continuous (hence bounded) ho jisme

(x_{0}, y_{0})

contain ho; agar wahan

f

discontinuous hai, to koi

M

exist nahi karta aur theorem simply apply nahi hota — tum existence bhi guarantee nahi kar sakte.

Boundary case

L h = 1

exactly par (jaise

y^{'} = x^{2} + y^{2}

mein), kya uniqueness lost ho jaati hai?

Nahi.

L h = 1

sirf naive sup-norm contraction ko defeat karta hai; factorial/Bielecki route phir bhi ek unique solution deta hai. Uniqueness kabhi bhi

L h < 1

par depend nahi karti thi.

Agar

f

Lipschitz hai

L = 0

ke saath, iska matlab kya hai?

∣ f (x, y_{1}) - f (x, y_{2}) ∣ \leq 0

force karta hai ki

f

y

se independent ho, yaani

y^{'} = g (x)

. Tab

y = y_{0} + \int_{x_{0}}^{x} g

ek aur sirf ek solution hai — pure integration, unconditionally unique.

Degenerate rectangle

b = 0

: kya theorem meaningful hai?

Koi usable interval nahi —

b / M = 0

h = 0

milta hai, isliye "interval" sirf single point

x_{0}

hai. Rectangle ko koi interior hona chahiye jisme solution reh sake, isliye

b > 0

(aur

a > 0

) zaroori hai.

Agar

f

R

par Lipschitz hai lekin solution

x

a

tak pahunchne se pehle edge

∣ y - y_{0} ∣ = b

tak pahunch jaata hai?

Exactly yahi scenario

h = b / M < a

protect karta hai: theorem

x_{0} + h

par guarantee karna band kar deta hai. Usse aage tumhe theorem ko nayi base point se re-apply karna hoga (continuation) agar

f

wahan bhi nice hai.

Kya ek solution exist ho sakta hai aur unique bhi ho jab Picard–Lindelöf ki hypotheses fail hon?

Haan. Theorem sufficient, not necessary, conditions deta hai — kuch non-Lipschitz problems ke phir bhi unique solutions hote hain. Hypotheses fail karna matlab hai "koi guarantee nahi," "guaranteed to break" nahi.

Recall Quick self-test

Woh single fact jo zyaadatar traps resolve karta hai ::: $h = min (a, b / M)$ $M$ par depend karta hai, $L$ par nahi; Lipschitz uniqueness/convergence control karta hai, continuity existence aur bound $M$ control karti hai.