Runge-Kutta 4th order (RK4) — derivation — Hinglish notes

4.8.24 · Maths › Numerical Methods

$\frac{d y}{d x} = f (x, y)$ , $y (x_{0}) = y_{0}$ ko solve karna, bina closed-form solution jaane.

The Big Picture

WHAT: hum kya karne ki koshish kar rahe hain?

HOW: general RK4 skeleton derive karna

Ek 4-stage explicit RK method ki form hoti hai $y_{n + 1} = y_{n} + (a_{1} k_{1} + a_{2} k_{2} + a_{3} k_{3} + a_{4} k_{4})$ jahan har stage pichle wale use karti hai:

k_{1} k_{2} k_{3} k_{4} = h f (x_{n}, y_{n}) = h f (x_{n} + c_{2} h, y_{n} + b_{21} k_{1}) = h f (x_{n} + c_{3} h, y_{n} + b_{31} k_{1} + b_{32} k_{2}) = h f (x_{n} + c_{4} h, y_{n} + b_{41} k_{1} + b_{42} k_{2} + b_{43} k_{3})

Ye shape kyun? Har $k_{i}$ ek sample point pe $h \cdot (slope)$ ka estimate hai. Hum constants $a_{i}, c_{i}, b_{ij}$ is tarah tune karte hain ki poori cheez true solution ki Taylor series se $h^{4}$ tak agree kare.

Step 1 — True solution ka Taylor expansion (target)

Kyunki $y^{'} = f$ hai, baar baar differentiate karo (chain rule, yaad rakho $y$ depend karta hai $x$ pe): $y^{'} = f, y^{''} = f_{x} + f_{y} f, y^{'''} = f_{xx} + 2 f_{x y} f + f_{y y} f^{2} + f_{y} (f_{x} + f_{y} f) .$

Ye step kyun? Ye exact quantities hain jo true solution carry karti hai. Hamara method inhe reproduce karna chahiye.

To exact step hai

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

Step 2 — Har $k_{i}$ ko Taylor expand karo aur match karo

$k_{2}, k_{3}, k_{4}$ ko 2-variable Taylor series ke roop mein expand karke aur weighted sum ko upar diye exact expansion ke barabar force karke order conditions ka ek system milta hai. Classic solution (infinitely many hain; ye elegant symmetric wala hai):

$c_{2} = c_{3} = \frac{1}{2}, c_{4} = 1, a_{1} = a_{4} = \frac{1}{6}, a_{2} = a_{3} = \frac{1}{3},$ aur $b_{21} = b_{32} = \frac{1}{2}$ , $b_{43} = 1$ , baaki sab zero.

Ye numbers kyun? Weights $(\frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{6})$ exactly Simpson's rule weights hain — kyunki jab $f$ sirf $x$ pe depend karta hai, RK4 reduce ho jaata hai Simpson's $\frac{h}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4})$ -style integration mein, jo $O (h^{5})$ accurate hai. RK4 Simpson's rule ka $y$ -dependence ke liye generalization hai!

Step 3 — Classical RK4 formulas

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

Worked Example 1

$y^{'} = x + y$ , $y (0) = 1$ solve karo, $h = 0.1$ se $y (0.1)$ nikalo.

$k_{1} = h f (0, 1) = 0.1 (0 + 1) = 0.1$ — Kyun? Start pe slope.
$k_{2} = 0.1 f (0.05, 1.05) = 0.1 (0.05 + 1.05) = 0.11$ — Kyun? $k_{1}$ ka aadha use karke midpoint.
$k_{3} = 0.1 f (0.05, 1.055) = 0.1 (0.05 + 1.055) = 0.1105$ — Kyun? $k_{2}$ ka aadha use karke refined midpoint.
$k_{4} = 0.1 f (0.1, 1.1105) = 0.1 (0.1 + 1.1105) = 0.12105$ — Kyun? Pura $k_{3}$ use karke endpoint.

$y_{1} = 1 + \frac{1}{6} (0.1 + 2 (0.11) + 2 (0.1105) + 0.12105) = 1.11034.$

Exact solution $y = 2 e^{x} - x - 1 \Rightarrow y (0.1) = 1.110342$ . Error $\approx 1 0^{- 6}$ ! Itna achha kyun? Fourth-order accuracy ki wajah se.

Worked Example 2 (Forecast-then-Verify)

$y^{'} = y$ , $y (0) = 1$ , ek step $h = 1$ solve karo. Forecast: true value $e = 2.71828$ hai; Euler $2$ deta hai. Predict karo ki RK4 $e$ ke kaafi kareeb aayega.

$k_{1} = 1 \cdot 1 = 1$
$k_{2} = 1 \cdot (1 + 0.5) = 1.5$
$k_{3} = 1 \cdot (1 + 0.75) = 1.75$
$k_{4} = 1 \cdot (1 + 1.75) = 2.75$ $y_{1} = 1 + \frac{1}{6} (1 + 3 + 3.5 + 2.75) = 1 + \frac{10.25}{6} = 2.70833.$

Verify: $2.70833$ vs $2.71828$ , error $\approx 0.01$ even with huge step $h = 1$ . Gaur karo $y_{1} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24}$ = $e^{x}$ ki series ke pehle paanch terms — direct proof ki RK4 Taylor se $h^{4}$ tak match karta hai.

Common Mistakes

Recall Feynman: ek 12-saal ke bacche ko explain karo

Imagine karo tum fog mein ek pahaadi par chal ke sahi height pe utarna chahte ho. Euler sirf wahan slope check karta hai jahan se wo shuru hota hai aur seedha chalta hai — wo overshoot kar deta hai. RK4 zyada samajhdar hai: wo start pe slope dekhta hai, phir beech mein guess karta hai, check karta hai, beech ko dobara check karta hai, phir door wale end ko check karta hai — aur charon peeks ka cleverly weighted average leta hai (do beech wali peeks double count hoti hain). Wo average real curved path ko almost perfectly match karta hai. To bade steps mein bhi, wo almost exactly target pe land karta hai.

Active Recall

What IVP does RK4 solve?

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

with

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

, ek baar mein ek step

h

aage badhte hue.

Write the four RK4 slope formulas (with

k_{i} = h f

k_{1} = h f (x_{n}, y_{n})

;

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

;

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

;

k_{4} = h f (x_{n} + h, y_{n} + k_{3})

What is the RK4 update formula?

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

What are the local and global error orders?

Local

O (h^{5})

, global

O (h^{4})

Why do

k_{2}, k_{3}

get weight 2?

Ye midpoint slope estimate karte hain (interval ko best represent karte hain); Simpson's

1, 2, 2, 1

weighting se match karta hai.

Where is each

k_{i}

sampled?

k_{1}

left edge pe,

k_{2}, k_{3}

midpoint pe,

k_{4}

right edge pe.

What does RK4 reduce to when

f = f (x)

only?

Integral

\int f d x

ke liye Simpson's rule.

How does RK4 avoid computing derivatives of

f

Ye derivatives ko interior points pe extra function evaluations se replace karta hai, Taylor series ko match karte hue.

Why is global error one order lower than local?

1/ h

steps per-step

O (h^{5})

error ko accumulate karke

O (h^{4})

bana dete hain.

Connections

Euler's Method — 1st-order ancestor (single left-edge slope).
Modified Euler / Heun's Method — 2nd-order sibling (edges ka average).
Simpson's Rule — $1, 2, 2, 1$ weights ka source.
Taylor Series Methods — jo RK4 bina derivatives ke mimic karta hai.
Local vs Global Truncation Error — globally 4th order kyun hai.
Adaptive Step Size (RKF45) — error control ke saath RK4.