Runge-Kutta 4th order (RK4) — derivation: Visual walkthrough

4.8.24 · D2 · Maths › Numerical Methods › Runge-Kutta 4th order (RK4) — derivation

Yeh note RK4 method ko picture-by-picture, ground up se rebuild karta hai. Har symbol pehle earn kiya jaata hai, tab use hota hai. Agar tumne sirf ek straight-line "guess and step" method jaise Euler's Method dekha hai, toh yahan se shuru karo — hum kuch aur assume nahi karte.

Question hai kya actually?

Hume ek rule diya gaya hai jo kisi bhi location par steepness (yaani slope) batata hai:

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

Hamara kaam: $(x_{n}, y_{n})$ se ek step of width $h$ daayein lena aur agley position par nayi height estimate karna. Hum positions ko ek chain mein label karenge aur fixed step se aage badhenge:

$x_{n + 1} = x_{n} + h$

Step 1 — Ek slope kyun kaafi nahi

KYA HAI. Euler bilkul seedhi line mein chalta hai, steepness sirf step ke left edge par padhta hai.

KYUN FAIL HOTA HAI. Asli pahaad curve karta hai. Left slope se khinchi seedhi line poore width $h$ tak wahi slope rakhti hai — lekin asli slope to middle tak pahunchte-pahunchte badal chuka hota hai. Tum curve se door ho jaate ho.

PICTURE. Blue curve sach hai. Orange dashed line Euler ka seedha guess hai jo left slope se shuru hota hai. $x_{n} + h$ par vertical red gap error hai — aur yeh $h$ ke saath badhta hai.

Figure — Runge-Kutta 4th order (RK4) — derivation

$y_{n + 1}^{Euler} = y_{n} + step width h \cdot slope at left edge only f (x_{n}, y_{n})$

Sabak: ek reading, bahut jaldi li gayi, poore interval ko galat represent karti hai. Hume zyada readings chahiye, step ke andar li gayi.

Step 2 — Honest target: exact step ek area ke roop mein

KYA HAI. Guess karne se pehle, exact answer likhte hain jo hum dhundh rahe hain.

KYUN. Agar hum jaante hain "perfect" kaisa dikhta hai, toh hum apna guess tune kar sakte hain. Exact nayi height hai purani height plus step bhar accumulated slope:

$y (x_{n} + h) = y_{n} + \int_{x_{n}}^{x_{n} + h} f (x, y (x)) d x$

PICTURE. Slope-curve ke neeche shaded green area exact height gain ke barabar hai. Koi bhi accha method sirf ek clever tarika hai kuch slope samples se is area ko estimate karne ka.

Yeh sab kuch reframe kar deta hai: RK4 ek rule hai area ko kuch sample heights se approximate karne ka — exactly wahi jo Simpson's Rule karta hai. Yeh thought yaad rakho.

Step 3 — Ek step ke andar chaar sample slopes

KYA HAI. Hum chaar slope readings chosen spots par lete hain aur har ek ko " $h$ times slope" ke roop mein package karte hain.

$h \cdot f$ ke roop mein package kyun? Kyunki slope $\times$ width = ek rise. $k_{i} = h f (\dots)$ likhne ka matlab hai har $k_{i}$ already ek height-jump hai, toh hum inhe directly jod sakte hain bina koi stray $h$ beech mein rakh ke.

k_{1} k_{2} k_{3} k_{4} = h f (x_{n}, y_{n}) = h f (x_{n} + \frac{h}{2}, y_{n} + \frac{k _{1}}{2}) = h f (x_{n} + \frac{h}{2}, y_{n} + \frac{k _{2}}{2}) = h f (x_{n} + h, y_{n} + k_{3}) left edge midpoint, guessed via k_{1} midpoint, refined via k_{2} right edge, via k_{3}

PICTURE. Chaar chote slope-arrows: left edge par (blue), midpoint par do baar (orange, phir green), aur right edge par (red). Har baad wala arrow pichle jump ka use karke locate karta hai kahan khada hona hai.

Step 4 — Inhe weights $1, 2, 2, 1$ se combine karo

KYA HAI. Charon jumps ko ek final jump mein blend karo:

$y_{n + 1} = y_{n} + \frac{1}{6} (left 1 k_{1} + mid 2 k_{2} + mid 2 k_{3} + right 1 k_{4})$

YEH WEIGHTS KYUN? Step 2 par wapas dekho — answer ek area hai. Simpson's rule ek area ko endpoints ek baar aur midpoint chaar baar use karke approximate karta hai, sab $6$ se divide hokar: $\frac{h}{6} (left + 4 \cdot mid + right)$ . Yahan midpoint slope do estimates mein split hai $k_{2}$ aur $k_{3}$ , toh " $4 \times$ middle" ban jaata hai " $2 \times k_{2} + 2 \times k_{3}$ ." Divisor $6 = 1 + 2 + 2 + 1$ weights ko true average rakhta hai.

PICTURE. Ek weighing-scale image: do orange/green midpoint arrows $\times 2$ wale pans par; blue aur red edge arrows $\times 1$ wale pans par. Balance point final averaged slope hai (purple arrow), aur us slope ke saath chalna hume blue curve par land karata hai.

Step 5 — Ise "4th order" kyun kehte hain (Taylor se matching, explicitly dikhaya gaya)

KYA HAI. Exact height, Taylor series ke roop mein expand ki gayi, $h$ -powers ka sum hai:

$y (x_{n} + h) = y_{n} + h y^{'} + \frac{h ^{2}}{2} y^{''} + \frac{h ^{3}}{6} y^{'''} + \frac{h ^{4}}{24} y^{''''} + O (h^{5})$

KYUN. Constants $\frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{6}$ aur midpoints decorative nahi hain — yeh matching se forced hain. Aao hum sabse simple revealing case par mechanics dikhate hain, $f (x, y) = y$ (toh $y^{'} = y$ , jiska exact solution har step mein $e^{h}$ se multiply hota hai). Har $k_{i}$ ko $h$ mein polynomial ke roop mein expand karo ( $y_{n} = y$ use karke):

k_{1} k_{2} k_{3} k_{4} = h y = h (y + \frac{k _{1}}{2}) = h (y + \frac{h y}{2}) = h y + \frac{h ^{2}}{2} y = h (y + \frac{k _{2}}{2}) = h (y + \frac{h y}{2} + \frac{h ^{2}}{4} y) = h y + \frac{h ^{2}}{2} y + \frac{h ^{3}}{4} y = h (y + k_{3}) = h y + h^{2} y + \frac{h ^{3}}{2} y + \frac{h ^{4}}{4} y

Ab weighted sum term-by-term banao ( $y$ factor out karke):

\frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = y (h^{1} h + h^{2} \frac{h ^{2}}{2} + h^{3} \frac{h ^{3}}{6} + h^{4} \frac{h ^{4}}{24}) .

$y (x_{n} + h) = y e^{h} = y (1 + h + \frac{h ^{2}}{2} + \frac{h ^{3}}{6} + \frac{h ^{4}}{24} + \frac{h ^{5}}{120} + \dots)$ ke exact expansion se compare karo:

$y_{n + 1} = y + y (h + \frac{h ^{2}}{2} + \frac{h ^{3}}{6} + \frac{h ^{4}}{24}) matches the truth through h^{4}, differing first at \frac{h ^{5}}{120} y .$

$h, h^{2}, h^{3}, h^{4}$ coefficients exactly agree karte hain kyunki weights $\frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{6}$ solve kiye gaye the inhe agree karaane ke liye — yahi "order conditions" ka poora content hai. Pehla mismatch $h^{5}$ term par hai, isliye local error $O (h^{5})$ hai.

PICTURE. $h$ -powers ka ek bar chart: RK4 ke bars $h^{4}$ tak exact match karte hain true Taylor bars se (green mein dikhaya), aur sirf $h^{5}$ par alag hote hain (chhota red bar) — "4th order" ka visual proof.

Step 6 — Degenerate case: jab $f$ sirf $x$ par depend karta hai

KYA HAI. Maano slope machine height bilkul ignore kar deti hai: $f (x, y) = g (x)$ . Tab do midpoint readings ek hi hoti hain ( $k_{2} = k_{3}$ ek hi $x_{n} + \frac{h}{2}$ use karte hain aur height-guess ab matter nahi karta).

KYUN MATTER KARTA HAI. Poora update collapse ho jaata hai:

$y_{n + 1} - y_{n} = \frac{h}{6} (g (x_{n}) + 4 g (x_{n} + \frac{h}{2}) + g (x_{n} + h))$

jo $\int_{x_{n}}^{x_{n} + h} g d x$ ke liye exactly Simpson's Rule hai. Toh RK4 koi nayi invention nahi hai — yeh Simpson's rule hai jo generalise ho gaya hai un slopes ke liye jo height par bhi depend karte hain. Yeh Step 2 mein khola gaya loop band karta hai.

PICTURE. $f = g (x)$ ke saath, charon arrows ek flat curve par aate hain; do midpoint arrows ek hi spot par land karte hain aur ek single " $\times 4$ " pan mein merge ho jaate hain — Simpson's rule, revealed.

Step 7 — Ek asli curve par sanity check

KYA HAI. $y^{'} = y$ , $y (0) = 1$ par ek bade step mein RK4 vs Euler dekho (sach: $y = e^{x}$ , toh $y (1) = e = 2.71828$ ).

KYUN. Ek mota step $h = 1$ sabse harsh test hai — yeh dikhata hai curvature kitna hurt karti hai.

Euler: ek left slope $\Rightarrow y_{1} = 1 + 1 \cdot 1 = 2$ (bahut kam).
RK4: $k_{1} = 1, k_{2} = 1.5, k_{3} = 1.75, k_{4} = 2.75 \Rightarrow y_{1} = 1 + \frac{1}{6} (1 + 3 + 3.5 + 2.75) = 2.70833$ .

Note karo $2.70833 = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24}$ — $e^{1}$ ki series ke pehle paanch terms — Step 5 ke $h^{4}$ -match ka direct darshan.

PICTURE. Blue $e^{x}$ curve; Euler ka endpoint bahut neeche $2$ par; RK4 ka endpoint curve ke paas $2.708$ par, aur sach wala $2.718$ thoda upar marked.

Step 8 — Wahi chaar-slope idea systems ke liye (vector $f$ )

KYA HAI. Asli problems aksar kai quantities ek saath track karti hain (position aur velocity, kai chemicals, etc.). Tab $y$ ek vector $y = (y^{(1)}, y^{(2)}, \dots)$ ban jaata hai aur slope machine ek vector $f (x, y)$ return karta hai.

KYUN YEH BAS KAAM KARTA HAI. Derivation mein kahin bhi "sirf ek $y$ hai" use nahi kiya gaya. Har $k_{i}$ ek vector $k_{i}$ ban jaata hai, aur wahi formulas component-by-component apply hote hain:

k_{1} k_{2} k_{3} k_{4} y_{n + 1} = h f (x_{n}, y_{n}) = h f (x_{n} + \frac{h}{2}, y_{n} + \frac{k _{1}}{2}) = h f (x_{n} + \frac{h}{2}, y_{n} + \frac{k _{2}}{2}) = h f (x_{n} + h, y_{n} + k_{3}) = y_{n} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4})

Addition aur scaling ordinary vector operations hain. Kyunki ek single high-order ODE (maano $y^{''} = g (x, y, y^{'})$ ) ko hamesha first-order ODEs ke system ke roop mein rewrite kiya ja sakta hai, yeh ek upgrade RK4 ko practically har ODE jo tum miloge solve karne deta hai — chaar-slope picture wahi rehti hai; sirf arrows ab higher dimensions mein hote hain.

Ek-picture summary

Sab ek saath: exact area jo hum chase kar rahe hain (green), chaar sampled slope-arrows (blue/orange/green/red) left–mid–mid–right par, unke weights $1, 2, 2, 1$ , averaged purple slope, aur asli curve par landing point.

Recall Feynman retelling — poora walk simple words mein

Tum ek foggy pahaad par ho aur ek step aage apni height guess karni hai. Euler zameen sirf wahan feel karta hai jahan se shuru hota hai, seedha chalta hai, aur overshoot karta hai kyunki pahaad curve karta hai. RK4 patient hai. Pehle woh apne pairon ke neeche slope feel karta hai ( $k_{1}$ ). Usi se woh middle kahan hoga guess karta hai aur wahan slope feel karta hai ( $k_{2}$ ). Woh us guess par trust nahi karta, toh use middle fir se locate karne ke liye use karta hai aur dobara feel karta hai ( $k_{3}$ ). Aakhir mein woh far edge tak stride karta hai aur wahan slope feel karta hai ( $k_{4}$ ). Ab uske paas chaar honest readings hain: do edges par, do middle mein. Woh ek weighted average leta hai — middles double count hote hain — kyunki middle poore stretch ko best represent karta hai. Woh average slope asli curve ke itna faithful hai ki ek giant step ke saath bhi woh almost exactly target par land karta hai. Aur middles double kyun count karte hain wahi reason hai jo Simpson's rule middle ko heavily weight karta hai: RK4 Simpson's rule hai differential-equation costume pehne hue.

Active Recall

Step 2 mein integral kya represent karta hai?

slope-curve ke neeche ka area = step mein exact height gain.

k_{2}

ke liye midpoint height guess kyun?

slope machine

f (x, y)

ko ek

y

chahiye; hum ise pehle jump

k_{1}

ka aadha lekar estimate karte hain.

1, 2, 2, 1

weights ka source?

Simpson's rule (midpoint heavily weighted), do midpoint slopes

k_{2}, k_{3}

mein split.

O (h^{p})

" ka matlab kya hai?

error chote

h

ke liye ek constant times

h^{p}

se bounded hai — yeh kam se kam

h^{p}

ki speed se shrink hoti hai.

Global vs local error?

local

O (h^{5})

per step;

N \sim 1/ h

steps accumulate hokar global

O (h^{4})

bante hain.

Jab

f = g (x)

sirf ho toh RK4 kya ban jaata hai?

exactly Simpson's rule

\int g d x

ke liye.

ODEs ke system ke liye kya badalta hai?

y, f, k_{i}

vectors ban jaate hain; identical formulas component-wise apply hote hain.

Connections

Euler's Method — single-left-slope method jo Steps 1 aur 7 mein haara.
Modified Euler / Heun's Method — do edge slopes average karta hai; RK4 ka 2nd-order sibling.
Simpson's Rule — Steps 4 aur 6 mein chhupi area rule.
Taylor Series Methods — Step 5 mein match kiya gaya exact target.
Local vs Global Truncation Error — Step 5 ka one-power drop.
Adaptive Step Size (RKF45) — RK4 error control ke saath upgrade hua.