Existence and uniqueness theorem — Picard-Lindelöf (statement): Visual walkthrough

4.6.8 · D2 · Maths › Ordinary Differential Equations › Existence and uniqueness theorem — Picard-Lindelöf (statemen

Intuition Yeh page kya karta hai

Parent note ne theorem state kiya tha aur proof ka sketch diya tha. Yahan hum use draw karte hain — har step ko ek picture milti hai. Hum ek blank sheet se shuru karte hain: tumhe bas itna pata hona chahiye ki graph kya hota hai, slope kya hota hai, aur "area under a curve" ka kya matlab hai. Baki sab — box, operator, contraction, convergence — hum tumhare saamne banate hain.

Hum ek precise claim ko unpack kar rahe hain. Hypotheses: $f$ closed rectangle $R = {(x, y) : ∣ x - x_{0} ∣ \leq a, ∣ y - y_{0} ∣ \leq b}$ par defined aur continuous hai, aur $f$ ==Lipschitz in $y$ on $R$ == hai (Step 6 mein define kiya gaya hai). Conclusion: IVP $y^{'} = f (x, y), y (x_{0}) = y_{0}$ ka ek unique solution $∣ x - x_{0} ∣ \leq h$ par hai jahan $h = min (a, b / M)$ . Informal slogan "not-too-wild near the start" exactly yahi do hypotheses hain.

Step 1 — Problem ek starting dot plus slope ke liye ek rule hai

KYA HAI. Hume plane mein ek point $(x_{0}, y_{0})$ aur ek rule $f$ diya gaya hai. Rule kehta hai: kisi bhi point $(x, y)$ par, usse guzarne wale graph ki slope $f (x, y)$ honi chahiye.

Dono facts symbols mein: $graph ki slope at x y^{'} (x) = rule jo slope demand karta hai yahan f (x, y), graph start dot se guzarna chahiye y (x_{0}) = y_{0} .$

KYO. Is pair ko initial value problem (IVP) kehte hain. Slope rule akele infinitely many curves de sakta hai jo use obey karti hain; start dot us tangle mein se ek thread pick karta hai. Poora theorem is baare mein hai ki "ek thread" sach mein ek hai ya nahi.

PICTURE. Slope field dekho: chote-chote line segments, har ek us slope par drawn jise rule $f$ demand karta hai. Ek solution wo curve hai jo har us segment ko touch karte waqt tangent ho, aur amber dot se guzre.

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

Step 2 — Problem ko ek box $R$ ke andar fence karo

KYA HAI. Hum poore plane ko dekhne se mana karte hain. Hum start ke around ek closed rectangle draw karte hain: $R = {(x, y) : width 2 a ∣ x - x_{0} ∣ \leq a, height 2 b ∣ y - y_{0} ∣ \leq b} .$ Yahan $a > 0$ aur $b > 0$ hum khud choose karte hain — ki hum rule par left/right aur up/down kitna trust karne ko taiyaar hain.

KYO. Do reasons hain. Pehla, $f$ door par misbehave kar sakta hai; ek chhote box ke andar hum use control kar sakte hain. Doosra, ek closed bounded box par continuous $f$ bounded hota hai: ek number $M$ milta hai jahan $∣ f (x, y) ∣ \leq M$ $R$ mein har jagah ho. Yeh number $M$ sabse tej slope hai jo koi bhi solution box ke andar kisi bhi waqt le sakta hai — yahi key ceiling hai jise hum aage exploit karte hain.

PICTURE. Center mein amber dot, uske around cyan box, aur steepest allowed lines (slope $+ M$ aur $- M$ ) jo ek wedge ki tarah drawn hain — koi bhi solution in se zyada steep nahi ho sakta.

Step 3 — Safe interval $h = min (a, b / M)$ kyun hai

KYA HAI. Height $y_{0}$ se shuru karte hue, agar slope kabhi $M$ se zyada nahi, toh horizontal distance $∣ x - x_{0} ∣$ move karne ke baad graph vertically zyada se zyada $M ∣ x - x_{0} ∣$ move kar sakta hai: $vertical travel ∣ y (x) - y_{0} ∣ \leq max slope M \cdot horizontal travel ∣ x - x_{0} ∣ .$

KYO. Box ke andar rehne ke liye humein vertical travel $\leq b$ rakhna hoga: $M ∣ x - x_{0} ∣ \leq b ⟹ ∣ x - x_{0} ∣ \leq \frac{b}{M} .$ Hum sides se bhi bahar nahi ja sakte: $∣ x - x_{0} ∣ \leq a$ . Dono hold karne chahiye, toh tighter wala jeetta hai: $h = min (a, \frac{b}{M}) .$ min kyun aur max kyun nahi? Kyunki ek constraint ek wall hai — tum pehli wall par rukते ho, jo choti number hai.

PICTURE. Slope- $\pm M$ ki do lines start dot se fan out karti hain. Jahan wo box ki top/bottom se paar hoti hain wo $b / M$ deta hai; left/right walls $a$ deti hain. Safe interval $h$ woh crossing hai jo pehle hoti hai (amber mein drawn).

Step 4 — ODE ko "aisi curve dhundho jo tum hila nahi sako" mein rewrite karo

KYA HAI. $y^{'} = f (x, y)$ ke dono sides ko $x_{0}$ se $x$ tak integrate karo. Left side Fundamental Theorem of Calculus se telescope karta hai, aur $y (x_{0}) = y_{0}$ boundary term ko khatam kar deta hai: $y (x) = start height y_{0} + accumulated slope = total climb \int_{x_{0}}^{x} f (t, y (t)) d t .$

KYO. Differentiation fragile hai (ye chote wiggles amplify kar sakta hai); integration robust hai (ye smooth karta hai). Differential equation ko is integral equation se trade karna slopes ki jagah areas se estimate karne deta hai. Operator $T$ define karo jo ek candidate curve $y$ khaata hai aur ek naya curve ugalta hai: $T [y] (x) = y_{0} + \int_{x_{0}}^{x} f (t, y (t)) d t .$ IVP ka solution ab exactly woh curve hai jise $T$ unchanged chhod deta hai — ek fixed point: $T [y] = y$ . (Yeh Banach fixed-point theorem (contraction mapping) ka darwaza hai.)

PICTURE. Ek candidate curve baaye box mein jaati hai; $T$ shaded area (integral) measure karta hai aur daayein ek naya curve banata hai jo $y_{0}$ se shuru hota hai aur us area se utha. Fixed point woh curve hai jahan "in" aur "out" ek saman hain.

Step 5 — $T$ ko guesses feed karke curve banao (Picard iteration)

KYA HAI. Sabse aasaan guess se shuru karo: flat line $y_{0} (x) = y_{0}$ . Phir $T$ baar baar apply karo: $y_{n + 1} (x) = y_{0} + \int_{x_{0}}^{x} f (t, y_{n} (t)) d t .$ Har pass pehle guess ko slope rule manne ki taraf thoda aur bend karta hai. Yeh Picard iteration (method of successive approximations) hai.

KYO. Hum answer likh nahi sakte, toh hum use approach karte hain. Flat guess galat hai lekin safe hai — Step 6 dikhata hai ki har guess box ke andar rehti hai, toh $f$ hamesha kisi aisi jagah evaluate hoti hai jahan wo actually defined hai.

PICTURE. $y^{'} = y, y (0) = 1$ dekhte hain: flat line $1$ , phir $1 + x$ , phir $1 + x + \frac{x ^{2}}{2}$ , … har naya cyan curve $e^{x}$ (amber) ke zyada paas. Iterates $e^{x}$ ke partial sums hain — machine literally answer spell out kar rahi hai.

Step 6 — Lipschitz define karo, phir dikhao ki guesses box mein rehti hain aur paas aati hain

KYA HAI (missing definition — $L$ use karne se pehle state karo).

KYA HAI (andar rehna). $∣ x - x_{0} ∣ \leq h$ par, kyunki $∣ f ∣ \leq M$ : $∣ y_{n + 1} (x) - y_{0} ∣ \leq \int_{x_{0}}^{x} \leq M ∣ f (t, y_{n}) ∣ d t \leq M h \leq M \cdot \frac{b}{M} = b .$ Har iterate height $b$ ke andar rehti hai — toh har $y_{n} \in C_{h}$ , aur yahi reason hai ki Step 3 mein humne $h \leq b / M$ choose kiya.

KYA HAI (ek step paas). Consecutive iterates subtract karo; integral ke andar $y_{1} = y_{n} (t), y_{2} = y_{n - 1} (t)$ ke saath Lipschitz bound apply karo: $∣ y_{n + 1} (x) - y_{n} (x) ∣ \leq \int_{x_{0}}^{x} ∣ f (t, y_{n}) - f (t, y_{n - 1}) ∣ d t \leq L \int_{x_{0}}^{x} ∣ y_{n} (t) - y_{n - 1} (t) ∣ d t .$

KYO (factorial bound, induction se prove kiya). $d_{n} (x) = ∣ y_{n + 1} (x) - y_{n} (x) ∣$ likho. Hum claim karte hain $d_{n} (x) \leq M \frac{L ^{n} ∣ x - x _{0} ∣ ^{n + 1}}{( n + 1 )!} .$

Base $n = 0$ : $d_{0} (x) = ∣ y_{1} - y_{0} ∣ \leq \int_{x_{0}}^{x} ∣ f (t, y_{0}) ∣ d t \leq M ∣ x - x_{0} ∣ = M \frac{L ^{0} ∣ x - x _{0} ∣ ^{1}}{1 !}$ . ✔
Induction step: maan lo yeh $n - 1$ ke liye sach hai. Usse one-step inequality mein feed karo: $d_{n} (x) \leq L \int_{x_{0}}^{x} d_{n - 1} (t) d t \leq L \int_{x_{0}}^{x} M \frac{L ^{n - 1} ∣ t - x _{0} ∣ ^{n}}{n !} d t = M \frac{L ^{n} ∣ x - x _{0} ∣ ^{n + 1}}{( n + 1 )!} .$ $∣ t - x_{0} ∣^{n}$ ka integral power ek se badha aur $n!$ ko $(n + 1)!$ mein badal diya. ✔

Denominator mein har factor kya karta hai: $T$ ka har application ek aur integration contribute karta hai, aur $∣ x - x_{0} ∣^{k}$ ko integrate karna new exponent $k + 1$ se divide karta hai; yeh $n + 1$ baar karne se saare divisors multiply hokar $(n + 1)!$ ban jaate hain. Isliye factorial — sirf power nahi — aata hai, aur isliye yeh upar ke $L^{n}$ ko out-grow karta hai.

CONSEQUENCE. Supremum norm mein gaps sum karo: $\sum_{n \geq 0} ∥ y_{n + 1} - y_{n} ∥_{\infty} \leq \sum_{n \geq 0} M \frac{L ^{n} h ^{n + 1}}{( n + 1 )!} = \frac{M}{L} (e^{L h} - 1) < \infty.$ Finite total travel ka matlab hai ki curves ek limit curve $y$ par pile up ho jaati hain (Weierstrass M-test) — ye uniformly converge karti hain. Woh limit $T [y] = y$ satisfy karti hai: existence complete hai, aur isne already Lipschitz use ki.

PICTURE. Gap sizes $d_{n}$ ka ek bar chart: pehle bars chunky hain ( $L^{n}$ unhe badhane ki koshish karta hai) lekin factorial unhe zero ki taraf slam karta hai. Stacked heights (total travel) ek finite amber line mein sum hote hain.

Step 7 — Sirf ek thread: uniqueness Grönwall se

KYA HAI. Maano ki do curves $y$ aur $\tilde{y}$ dono IVP solve karti hain. $ϕ (x) = ∣ y (x) - \tilde{y} (x) ∣ \geq 0$ unke beech ka gap ho. Dono integral equation satisfy karti hain, toh subtract karke aur Lipschitz use karke: $ϕ (x) \leq L \int_{x_{0}}^{x} ϕ (t) d t .$ Ise zor se padho: gap abhi us total gap se bounded hai jo $L$ times abhi tak accumulate hua hai.

KYO. Grönwall's inequality kehti hai ki ek non-negative function jo apne khud ke running integral ke constant times se bounded ho — bina kisi head-start term ke — identically zero hona chahiye. Kyunki $ϕ \geq 0$ aur $ϕ \leq L \int ϕ$ kuch add kiye bina, $ϕ \equiv 0$ . Toh $y = \tilde{y}$ : do threads ek hi thread the hamesha. Dhyaan do ki is ke liye sirf Lipschitz chahiye tha — koi shrinking nahi, koi $L h < 1$ nahi.

PICTURE. Do candidate solution curves amber dot par glued shuru hoti hain. Grönwall squeeze (ek shrinking cyan envelope) unhe glued rehne par force karta hai — wo kabhi alag nahi ho sakti.

Step 8 — Degenerate aur edge cases (reader ko kabhi akela mat chhodna)

Neeche wali figure ek four-panel breakdown hai — har corner case ke liye ek panel, har ek alag annotated.

Panel A — $M = 0$ (flat rule). Agar $f \equiv 0$ box mein, koi bhi vertical travel kabhi nahi hoti: $b / M = \infty$ , toh $h = a$ . Solution horizontal line $y = y_{0}$ hai. Koi wedge nahi, koi shrinking nahi — sabse wide possible interval.
Panel B — $L h \geq 1$ . Naive contraction fail ho jaata hai, lekin Steps 6–7 (factorial bound / Bielecki norm / Grönwall) ko kabhi $L h < 1$ ki zaroorat nahi thi. Existence aur uniqueness phir bhi hold karti hain — jaise parent ke Example 1 mein jahan exactly $L h = 1$ tha.
Panel C — $f$ continuous but NOT Lipschitz. Convergence-and-uniqueness machinery toot jaati hai. Existence Peano existence theorem ke zariye survive kar sakti hai, lekin uniqueness mar sakti hai: $y^{'} = y^{2/3}, y (0) = 0$ ke dono $y \equiv 0$ aur $y = (x /3)^{3}$ solutions hain. Ek dot se do threads.
Panel D — $h$ se aage. Theorem local hai; interval badhane ke liye endpoint se re-solve karo aur glue karo — continuation Maximal interval of existence tak.

Ek-picture summary

Sab kuch ek blueprint par: box $R$ , $\pm M$ wedge jo $h = min (a, b / M)$ fix karta hai, Picard iterates jo unique fixed-point curve mein spiral karte hain, aur Grönwall clamp jo kisi bhi rival curve ko split hone se rokta hai.

Recall Feynman retelling — plain words mein wapas bolo

Hume ek starting dot aur har jagah slope ke liye ek rule mila. Pehle humne ek box fence kiya taaki rule misbehave na kar sake; box ke andar slope kisi ceiling $M$ se zyada nahi ho sakta. Woh ceiling matlab hai ki solution box se $\pm M$ wedge se tez bahar nahi nikal sakti, jo ek safe width $h$ fix karta hai — jo bhi wall (side $a$ ya top $b / M$ ) pehle aaye. Phir humne "slope match karo" ko "area match karo" mein badla: ek solution wo curve hai jise ek certain area-machine $T$ (box mein rehne wali continuous functions par rehti hai, sabse bade vertical gap se measure ki jaati hai) unchanged chhod deta hai. Humne $T$ ko ek lazy flat guess di, phir uska apna output, baar baar; har pass curve ko aur bend karta hai, aur kyunki har pass ek integration add karta hai jo ek bade number se divide karta hai, ek factorial errors ko crush kar deta hai, toh guesses ek curve par pile ho jaate hain — yahi existence hai, aur isne already Lipschitz grip $L$ use ki. Aakhir mein, agar do curves dono solve karti hain, unka gap "gap $\leq L \times$ apna khud ka running total" obey karta, aur Grönwall kehta hai aisa gap zero hai — toh do curves ek thi. Edge cases: flat rule sabse wide interval deta hai; ek merely-continuous-but-not-Lipschitz rule jaise $y^{2/3}$ do threads mein split ho sakta hai; aur $h$ ke baad tum re-solve karo aur glue karo.

Recall Quick self-test

Safe interval min kyun hai aur max kyun nahi? ::: Kyunki do constraints (side wall $a$ , top/bottom $b / M$ ) dono walls hain; tum pehli par rukते ho, jo choti number hai. Lipschitz constant $L$ exactly kya hai? ::: Sabse chhota $L \geq 0$ jiske saath $∣ f (x, y_{1}) - f (x, y_{2}) ∣ \leq L ∣ y_{1} - y_{2} ∣$ $R$ par — ek ceiling ki $f$ kitna change hota hai jab $y$ move karta hai. Kaunsi hypothesis iterates ko converge karati hai, aur kaunsi ek answer pin karta hai? ::: Lipschitz dono karta hai — factorial bound ke zariye (convergence) aur Grönwall ke zariye (uniqueness); continuity bound $M$ supply karta hai jo cheezein shuru karta hai. $y^{'} = y^{2/3}, y (0) = 0$ ke liye kya toot jaata hai? ::: $f$ continuous hai lekin $y = 0$ par Lipschitz nahi, toh uniqueness fail hoti hai: $y \equiv 0$ aur $y = (x /3)^{3}$ dono ise solve karte hain.