Direction fields and Euler's method — visual - numerical intuition first: Visual walkthrough

4.6.2 · D2 · Maths › Ordinary Differential Equations › Direction fields and Euler's method — visual - numerical int

Hum koi bhi symbol use nahi karte jab tak usse draw na kar lein. Chalte hain uss ek cheez se jo ek ODE actually hamare haath mein deta hai.

Step 1 — ODE tumhe ek slope deta hai, curve nahi

KYA. Ek first-order ODE kuch aisa dikhta hai: $\frac{d y}{d x} = f (x, y) .$ Har piece padho:

$x$ ::: horizontal position (tum kitna right ho).
$y$ ::: us position par solution curve ki height.
$\frac{d y}{d x}$ ::: slope — jitni tezi se $y$ badhta hai jab $x$ right move karta hai. "Rise per unit run."
$f (x, y)$ ::: ek machine: isko point $(x, y)$ do, ye ek number ugal deti hai — aur woh number hi woh slope hai jo tumhare paas wahan honi chahiye.

KYUN. Zyaadatar equations batate hain koi cheez kya hai. Ye equation refuse karti hai. Ye kabhi nahi kehti " $y$ aisa-waisa hai." Ye sirf kehti hai: tumhara curve jo bhi ho, point $(x, y)$ par uski steepness exactly $f (x, y)$ honi chahiye. Isliye hum answer lookup nahi kar sakte — humein usse construct karna hoga, har jagah slope ko maan ke.

PICTURE. Ek single point par, ODE ek choti si tilted dash lagata hai: woh slope jo woh maangta hai.

Figure — Direction fields and Euler's method — visual - numerical intuition first

Step 2 — Har point par slope lagao: direction field

KYA. Step 1 ko ek baar nahi, balki puri grid of points par karo. Har $(x, y)$ par ek chota segment draw karo jiska tilt $f (x, y)$ ke barabar ho. Dashes ka woh carpet direction field kehlata hai (isse slope field bhi kehte hain).

Hamare running example ke liye hum choose karte hain $f (x, y) = x + y,$ toh kisi point par slope bas "tumhari right-position plus tumhari height" hai.

KYUN. Ek dash ek local rule hai. Field woh rule hai jo global ban gaya — ODE ke har command ka ek map. Tab ek solution woh koi bhi curve hai jo har instant mein us dash ke tangent par rehti hai jis par woh khadi hai. Koi bhi dash kabhi sideways cross nahi hota.

Yeh carpet haath se draw karne ka ek shortcut hai. Har grid point par ek-ek karke slope compute karne ki jagah, un curves ko dhoondo jahan slope ek fixed number hoti hai — inhe isoclines kehte hain ("iso" = same, "cline" = slope).

Isoclines ko drawing speed up karne ke liye kaise use karein: kuch values of $m$ chuno (maano $m = 0, 1, 2$ ), lines $x + y = m$ draw karo, aur har line par simply ek hi tilt (slope $m$ ) ki short dashes ki ek row copy karo. Kyunki ek poori line ek slope share karti hai, tum field ko strokes mein fill karte ho na ki point-by-point — picture almost free mein appear hoti hai.

PICTURE. Gaur karo ki dashes exactly wahan flatten hokar horizontal ho jaati hain jahan $x + y = 0$ hai (anti-diagonal): woh flat line $m = 0$ ke liye isocline hai. Upar-right move karo aur dashes bade $m$ ke isoclines ke saath steep hoti jaati hain.

Step 3 — Ek starting point se EK curve pin karo

KYA. Field akele mein infinitely many curves hain jo usse thread karti hain (har point se ek). Ek single select karne ke liye hum ek initial condition add karte hain: $y (x_{0}) = y_{0}, yahan y (0) = 1.$

$x_{0} = 0$ ::: jahan hum shuru karna choose karte hain.
$y_{0} = 1$ ::: woh height jahan se hum shuru karte hain.

KYUN. Field ko arrows ki nadi samjho. Ek jagah ek patta daalo aur woh exactly ek raah trace karta hai. Kisi aur jagah daalo aur alag raah milti hai. Starting point kaun sa patta hai. (Exactly ek path kab exist karta hai yeh Existence and Uniqueness (Picard–Lindelöf) ka kaam hai.)

PICTURE. Red dot humara patta hai $(0, 1)$ par; ghostly curve woh asli trail hai jise hum trace karne ki koshish kar rahe hain.

Step 4 — Seedha step kyun? Tangent-line idea

KYA. Hum poora curve nahi dekh sakte. Lekin apne current point par hum uski slope padh sakte hain — ODE deta hai: $f (x_{0}, y_{0}) = 0 + 1 = 1$ . Toh hum curve ko shuruat ke paas uski tangent line se approximate karte hain aur usi par chalte hain: $y (x_{0} + h) \approx y_{0} + h \cdot f (x_{0}, y_{0}) .$

$h$ ::: step size — hum kitna right chalte hain slope re-read karne se pehle (the "run").
$h \cdot f$ ::: "rise" = run × slope. Yehi poora trick hai: rise = slope times run.

Ye tool — tangent — kyun? Kyunki derivative defined hi hai tangent line ki slope ke roop mein, aur tangent ek point ke paas curve ki best straight-line copy hai. Ek tiny run par curve barely bend karti hai, toh straight step true curve ke bahut paas land karta hai. Ye exactly first-order Taylor expansion hai: $y (x_{0} + h) = y_{0} + h y^{'} (x_{0}) + throw kar diya \frac{h ^{2}}{2} y^{''} (ξ),$ aur hum ODE se $y^{'} (x_{0}) = f (x_{0}, y_{0})$ substitute karte hain. Hum straight part rakhte hain, bendy $h^{2}$ part discard karte hain.

PICTURE. Green = true curve, blue = straight tangent step. Shuruat mein agree karte hain aur dheere dheere drift hote hain.

Step 5 — PEHLA step lo (n = 0), haath se

KYA. $h = 0.5$ , $x_{0} = 0$ , $y_{0} = 1$ ke saath: $f (0, 1) = 0 + 1 = 1 \Rightarrow y_{1} = 1 + 0.5 (1) = 1.5, x_{1} = 0.5.$

KYUN. Hum run $h = 0.5$ se slope $1$ par right move karte hain, toh rise $0.5 \times 1 = 0.5$ hai, height $1.5$ par land karte hain. Humne Step 4 ki blue line ka ek straight segment walk kiya.

PICTURE. Dekho blue landing point green true curve se thoda neeche baitha hai — curve upar ki taraf bend karti hai aur humara straight step peechhe reh gaya.

Step 6 — Slope RE-READ karo aur DOOSRA step lo (n = 1)

KYA. Ab naye point $(x_{1}, y_{1}) = (0.5, 1.5)$ par khade ho aur fresh slope padho: $f (0.5, 1.5) = 0.5 + 1.5 = 2 \Rightarrow y_{2} = 1.5 + 0.5 (2) = 2.5, x_{2} = 1.0.$

KYUN. Explicit Euler hamesha wahan ki slope use karta hai jahan woh abhi khada hai — naye step ka left point, kyunki jis point par woh ja raha hai woh abhi pata nahi. Har step ka apna tilt hota hai; isliye blue path segments ki ek chain hai jo field ko chase karne ke liye bend karti hai.

PICTURE. Ab do blue segments hain, har ek alag tilt ke saath, neeche field arrows ko track karte hue.

Abhi tak ka answer: $y (1) \approx 2.5$ . Exact solution hai $y = 2 e^{x} - x - 1$ , toh $y (1) = 2 e - 2 \approx 3.44$ . Euler under-shoot karta hai — agle step mein reason dekhte hain.

Step 7 — Euler yahan under-shoot kyun karta hai (convexity picture)

KYA. Yahan true curve convex hai — ek fancy word jiska matlab hai "upar ki taraf bend karti hai, jaise muskaan." Concretely, convex ka matlab hai second derivative positive hai, $y^{''} > 0$ : slope khud right move karte hue badhti rehti hai. Aise upar-bending curve ki tangent line curve ke neeche rehti hai. Toh har straight step bahut neeche land karta hai, aur errors pile up hote hain.

KYUN. Discard kiya hua Taylor term $\frac{h ^{2}}{2} y^{''} (ξ)$ tha. Jab $y^{''} > 0$ tab woh term positive hai — exactly woh height hai jo humne drop ki. Positive dropped height ⇒ hum neeche land karte hain. Agar curve concave hoti (upar ki taraf bend nahi, $y^{''} < 0$ ) toh woh term negative hota aur Euler over-shoot karta. Agar curve ek straight line hoti ( $y^{''} = 0$ ) toh Euler exact hota. Teen cases, ek rule: $y^{''}$ ka sign batata hai curve straight step se kis taraf bend hoti hai.

PICTURE. Vertical red gaps woh missing $\frac{h ^{2}}{2} y^{''}$ pieces hain, ek per step, sab ek hi taraf.

Step 8 — Chhote steps, paas ka walk: error $O (h)$ hai

KYA. Step ko $h = 0.25$ tak halve karo (ab $x = 1$ tak pahunchne ke liye 4 steps):

y_{1} y_{2} y_{3} y_{4} = 1 + 0.25 (0 + 1) = 1.25 = 1.25 + 0.25 (0.25 + 1.25) = 1.625 = 1.625 + 0.25 (0.5 + 1.625) = 2.15625 = 2.15625 + 0.25 (0.75 + 2.15625) = 2.883.

Toh $y (1) \approx 2.88$ — coarse $2.50$ se $3.44$ ke zyada paas.

$O (h)$ kyun, $O (h^{2})$ kyun nahi. Har step $\sim \frac{h ^{2}}{2} y^{''}$ error karta hai. Length $L = 1$ ka interval cross karne ke liye $N = L / h$ steps lagte hain. Total $\sim N \cdot h^{2} = \frac{L}{h} \cdot h^{2} = L h$ . Step-count $N$ $h$ ki ek power kha jaata hai, global error $\propto h$ rehta hai: $h$ halve karo, roughly error halve hota hai. Iska matlab "first-order method" yahi hai. (Isse beat karne ke liye har step mein kaafi slopes sample karna Runge-Kutta Methods (RK4) ka idea hai.)

PICTURE. Same field par do Euler walks: coarse (kam, lambe segments, green se door) vs fine (zyada, chhote segments, green se chipke hue).

Ek-picture summary

Upar ki saari cheez ek single frame mein: dashes ki field (ODE ke commands), smooth true curve (exact solution), aur blue chain of tangent steps jo Euler khade hokar, slope padh ke, aur step lete hue banata hai — hamesha thoda neeche kyunki curve upar bend karti hai.

Recall Feynman retelling — poora walk plain words mein

Tum fog mein ek hillside par khoye hue ho. Ek magic compass ( $f$ ) tumhe steepness batata hai exactly jahan tum khade ho, lekin baaki hill ke baare mein kuch nahi. Toh tum ek tiny straight step us direction mein commit karte ho (rise = slope × step). Phir apne naye spot par compass se dobara poochhte ho, fresh steepness milti hai, aur ek aur tiny straight step lete ho. Jo zig-zag trail tum chhodte ho woh asli path ka tumhara best guess hai. Kyunki asli trail upar curve karti hai aur tumhare steps seedhe hain, tum hamesha thoda neeche land karte ho — woh gaps $\frac{h ^{2}}{2} y^{''}$ ke woh crumbs hain jo tumne drop kiye. Chhote steps lo aur tum chhote crumbs drop karte ho aur unhe zyada lete ho, toh total shortfall step size ke proportion mein shrink hoti hai. Yehi Euler's method hai: Stand, Slope, Step — first-order accurate, honest, aur har fancier ODE solver ka ancestor.

Recall Quick self-test

Step 6 kaun se point ki slope use karta hai? ::: Current/left point $(0.5, 1.5)$ , jo $f = 2$ deta hai. Humara walk true curve se neeche kyun land karta hai? ::: Solution convex hai ( $y^{''} > 0$ ), toh tangent steps curve ke neeche rehte hain. Coarse $y (1)$ vs fine $y (1)$ vs exact? ::: $2.50$ vs $2.88$ vs $\approx 3.44$ ; chhota $h$ zyada paas hai. Global error $O (h)$ kyun hai? ::: Per-step error $\sim h^{2}$ times $N = L / h$ steps deta hai $\sim L h$ .

Connections

Taylor's Theorem — tangent step aur discarded $\frac{h ^{2}}{2} y^{''}$ crumb.
Separable First-Order ODEs — humne exact $y = 2 e^{x} - x - 1$ compare karne ke liye kaise nikala.
Runge-Kutta Methods (RK4) — crumb ko tezi se shrink karne ke liye har step mein kaafi slopes sample karo.
Existence and Uniqueness (Picard–Lindelöf) — ek starting point exactly ek curve kyun pin karta hai, aur Lipschitz hypothesis.
Stability and Stiff Equations — jab tiny steps bhi explicit Euler ko save nahi kar sakte, aur implicit Euler kyun help karta hai.