Runge-Kutta 4th order (RK4) — derivation: Question bank

4.8.24 · D5 · Maths › Numerical Methods › Runge-Kutta 4th order (RK4) — derivation

Reminder un cheezon ka jo test ho rahi hain (hamare convention mein $k_{i} = h f$ , isliye har $k_{i}$ mein step size $h$ ka ek factor already hai):

True ya false — justify karo

RK4 ke liye aapko

f_{x}

f_{y}

(yani

f

ke partial derivatives) aur

f

ke higher derivatives haath se compute karne padte hain.

False. Poora point yahi hai ki RK4 un partial derivatives ki jagah interior sample points par $f$ ke extra evaluations leta hai, phir bhi Taylor series ko

h^{4}

tak match karta hai.

RK4 bilkul exact hoti hai jab bhi

f

sirf

x

par depend kare (yaani

y^{'} = f (x)

False in general. Tab ye Simpson's rule mein reduce ho jaati hai, jo sirf degree 3 tak ke polynomials ke liye exact hai; zyada degree ya non-polynomial

f

ke liye ab bhi

O (h^{5})

per-step error hoti hai.

Weights

\frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{6}

ka sum

1

hona chahiye.

True. Ye slopes ka ek weighted average hain; agar inका sum

1

na ho to method pehla Taylor term

y_{n} + h f

bhi reproduce nahi kar sakta, isliye consistent nahi hoga.

h

ko half karne se global error lagbhag

16

guna chhoti ho jaati hai.

True. Global error

O (h^{4})

hai, aur

(\frac{1}{2})^{4} = \frac{1}{16}

, isliye step ko har baar half karne par total error roughly 16 ke factor se cut hoti hai.

Kyunki har step ki error

O (h^{5})

hai, poore interval cross karne ke baad bhi error

O (h^{5})

hi rehti hai.

False. Ek fixed interval cross karne ke liye

\sim 1/ h

steps chahiye, isliye

O (h^{5})

per-step errors accumulate hokar

O (h^{4})

globally ban jaati hain — ek power lost ho jaati hai.

RK4 unconditionally accurate hai, isliye koi bhi step size achha answer deti hai.

False. "4th order" sirf yeh describe karta hai ki error kitni tezi se ghatti hai jab $h \to 0$ ; bada

h

phir bhi kharab answer de sakta hai, aur stiff problems mein solution blow up bhi kar sakta hai.

k_{2}

aur

k_{3}

dono same

x

-location par evaluate hote hain.

True. Dono midpoint

x_{n} + \frac{h}{2}

par baithe hain; fark sirf $y$ -value mein hai (

k_{2}

mein

k_{1}

use hota hai,

k_{3}

mein refined

k_{2}

Agar aap "slope-only" convention use karo jahan

k_{i} = f (\dots)

, to update formula unchanged rehta hai.

False. Tab aapko

h

wapas laana padta hai:

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

, kyunki

k_{i}

mein ab

h

ka factor nahi hai.

RK4 same step size

h

ke liye hamesha Euler se better hoti hai.

Sirf qualifiers ke saath. Smooth problem ke liye, per step RK4 kaafi zyada accurate hai (aur 4 evaluations vs Euler ke 1 ke bawajood per unit work bhi jeet jaati hai). Lekin non-smooth ya stiff problems mein order advantage khatam ho sakta hai ya RK4 diverge bhi kar sakti hai — isliye bina "smooth, non-stiff" ke "hamesha" galat hai.

Error dhundho

Ek student likhta hai

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

. Kya galat hai?

x

-argument galat hai:

k_{2}

midpoint

x_{n} + \frac{h}{2}

par hota hai, endpoint

x_{n} + h

par nahi. Sirf

k_{4}

mein

x_{n} + h

use hota hai.

Ek student likhta hai

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

. Kya galat hai?

k_{3}

mein refined midpoint value

y_{n} + \frac{k _{2}}{2}

use honi chahiye, na ki

\frac{k _{1}}{2}

k_{1}

dobara use karne se bas

k_{2}

recompute hoga aur refinement waste ho jaayegi.

Ek student likhta hai

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

aur

k_{i} = h f

use karta hai. Kya galat hai?

h

ka double factor aa gaya. Kyunki

k_{i}

mein pehle se

h

hai, blend

\frac{1}{6} (\dots)

hai jisme koi extra $h$ nahi;

\frac{h}{6}

wala form sirf slope-only convention par apply hota hai.

Ek student likhta hai

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

. Kya galat hai?

k_{4}

mein pura

k_{3}

use hota hai, yaani

y_{n} + k_{3}

, na ki uska aadha — kyunki ye midpoint se right edge tak pura

h

step leta hai.

Ek student weights

1, 1, 1, 1

divided by

4

use karta hai

1, 2, 2, 1

divided by

6

ki jagah. Accuracy kyun collapse hoti hai?

Equal weighting order conditions ko satisfy nahi karti; ye ab Simpson's integration ko match nahi karti, isliye accuracy 4th order se roughly 1st–2nd order par aa jaati hai.

Ek student compute karta hai

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

lekin har

k_{i}

ke andar

h

ka factor bhool gaya (sirf

k_{i} = f

use kiya). Kya symptom dikhega?

Har

k_{i}

1/ h

times too large hai (ek

f

instead of

h f

), isliye update overshoot karta hai —

h = 0.1

ke liye step

\sim 10 \times

too big hoga aur

y

true curve se kaafi door chali jaayegi.

Why questions

k_{2}, k_{3}

dono midpoint samples ko double weight kyun milta hai?

Midpoint slope interval ke average behaviour ko sabse achhe se represent karti hai (Simpson intuition), isliye zyada weight dene se blended slope true curved path ko kaafi accurately match karti hai.

RK4 slope ko step ke andar sample kyun karta hai, Euler ki tarah sirf left edge pe karne ki bajaye?

Slope

f

step ke across badlati hai; sirf left-edge slope se yeh variation miss ho jaati hai, jabki interior samples se slope ke modnay ka andaza ho jaata hai aur kaafi better average slope milti hai.

RK4 weights

1, 2, 2, 1

Simpson's rule jaisi kyun hain?

Jab

f = f (x)

only ho, update literally

\int f d x

ke liye Simpson's rule ban jaata hai; RK4, Simpson's rule ka generalization hai jo

y

-dependence allow karta hai, isliye wahi weights inherit karta hai.

RK4 ko "4th order" kyun kehte hain na ki "5th order", jabki per-step error

O (h^{5})

hai?

Naam global order

O (h^{4})

refer karta hai;

\sim 1/ h

steps mein error accumulation se

h

ki ek power khatam ho jaati hai.

RK4 bina

f

ko differentiate kiye true Taylor expansion ko

h^{4}

tak match kaise karta hai?

Interior function evaluations, jab Taylor-expand kiye jaate hain, exactly wahi partial-derivative combinations (

f_{x} + f_{y} f

, etc.) reproduce karte hain jo

y^{''}, y^{'''}, \dots

mein aati hain — isliye extra evaluations derivatives ki jagah le lete hain.

Classical RK4 ek hi 4th-order 4-stage method kyun nahi hai?

Order conditions ek aise underdetermined system banate hain jinke infinitely many solutions hain; classical

c_{2} = c_{3} = \frac{1}{2}

choice sirf ek elegant symmetric option hai, lekin doosre valid constant sets bhi exist karte hain.

Zyada, chhote steps lene par eventually RK4 ke liye bhi diminishing returns kyun aata hai?

Jaise

h

chhhota hota hai, truncation error

h^{4}

ke roop mein ghatti rehti hai, lekin finite-precision arithmetic se rounding error badhti hai; kuch

h

ke neeche roundoff dominant ho jaata hai aur accuracy kharab hone lagti hai.

Edge cases

Agar

f \equiv 0

ho (solution constant hai) to charon

k_{i}

ka kya hoga?

Har

k_{i} = 0

, isliye

y_{n + 1} = y_{n}

exactly — RK4 sahi tarike se constant solution ko unchanged chhod deta hai.

Linear autonomous problem

y^{'} = y

y (0) = 1

ke liye ek step

h = 1

y_{1}

term-by-term kya hoga?

y_{1} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} = \frac{65}{24} \approx 2.70833

, yaani

e^{x}

ki series ke pehle paanch terms — yeh seedha demonstration hai ki RK4 Taylor expansion ko

h^{4}

tak reproduce karta hai.

Agar

f

kisi step ke andar discontinuous ho, kya RK4 ki

O (h^{5})

local error phir bhi valid hai?

Nahi. Order analysis assume karti hai ki

f

step ke across smooth hai (kaafi continuous derivatives ke saath); koi kink ya jump Taylor matching tod deta hai aur accuracy bahut kharab ho jaati hai.

Trivial case

y^{'} = 1

(constant slope 1) par RK4 kya karta hai?

Saare

k_{i} = h

, isliye

y_{n + 1} = y_{n} + \frac{1}{6} (h + 2 h + 2 h + h) = y_{n} + h

— bilkul straight-line answer, jaisa hona chahiye.

Stiff equation jaise

y^{'} = - 1000 y

ke liye, moderate

h

se RK4 diverge kyun kar sakti hai?

RK4 ek explicit method hai jiska bounded stability region hai; agar

∣ - 1000 h ∣

us region se bahar chala jaaye to numerical solution decay karne ki bajaye amplify hone lagta hai, chahe equation ka true solution shrink kar raha ho. Isse Adaptive Step Size (RKF45) ki zaroorat samjh aati hai.

Agar exact solution

x

mein ek cubic polynomial hai (aur

f = f (x)

only), to RK4 ki error kitni badi hogi?

Zero (rounding tak): Simpson's rule cubics ko exactly integrate karta hai, aur RK4 yahan Simpson mein reduce ho jaata hai, isliye true value bilkul sahi milti hai.

Recall Traps ka ek-line summary

Teen sabse khatarnak hain: (1) $k_{2}, k_{3}$ ko midpoint ki jagah endpoint par rakhna, (2) conventions mix karke $h$ double-count karna, aur (3) local $O (h^{5})$ aur global $O (h^{4})$ mein confusion.

Connections

Parent derivation (Hinglish) — jahan ye saare facts build hote hain.
Euler's Method — single-slope baseline jisse traps compare karte hain.
Modified Euler / Heun's Method — 2nd-order sibling; midpoint logic se contrast karo.
Simpson's Rule — $1, 2, 2, 1$ weights ka source.
Taylor Series Methods — wo target jo RK4 bina derivatives ke match karta hai.
Local vs Global Truncation Error — order-lowering argument.
Adaptive Step Size (RKF45) — stiffness/step-size edge cases ko fix karta hai.