Systems of ODEs — RK4 for systems: Question bank

4.8.27 · D5 · Maths › Numerical Methods › Systems of ODEs — RK4 for systems

Shuru karne se pehle, ek reminder notation ki jo poori taraf use hoti hai, taaki kuch unexplained na rahe:

$y = (y_{1}, \dots, y_{n})^{T}$ sabhi unknowns ka stacked column hai ek moment in time par.
$f (t, y)$ stacked column of slopes $(y_{1}^{'}, \dots, y_{n}^{'})$ return karta hai.
$k_{1}, k_{2}, k_{3}, k_{4}$ chaar slope-vectors hain (har ek length $n$ ka) jo width $h$ ke ek step mein sample kiye jaate hain.
$h$ step size hai (kitna $t$ mein jump karte hain har step mein).

Spot the error

Har item ek plausible-sounding statement ya step dikhata hai. Batao kya galat hai aur kyun.

"Maine pehle

y_{1}

ke saare

k_{1}, k_{2}, k_{3}, k_{4}

compute kiye, phir

y_{2}

ke liye repeat kiya."

Error hai stage synchronization kho dena:

y_{1}

k_{2}

chahiye

y_{2}

ki advanced state, jo exist nahi karti agar

y_{2}

saath mein step nahi ki gayi; pehle poora vector

k_{1}

compute karo, phir poora

k_{2}

, aur aise hi aage.

k_{3}

ke liye maine state

y_{n} + \frac{h}{2} k_{1}

use ki."

Galat slope feed ki —

k_{3}

ko refined midpoint state

y_{n} + \frac{h}{2} k_{2}

use karni chahiye; wahan

k_{1}

reuse karna

k_{3}

k_{2}

mein collapse kar deta hai aur fourth-order accuracy destroy kar deta hai.

y^{''} = g (t, y, y^{'})

reduce karne ke liye maine

u_{1} = y^{'}

aur

u_{2} = y

set kiya."

Labels readability ke liye ulte hain lekin asli trap consistency hai: chahe tum unhe jo bhi name do, tumhe

u_{1}^{'} = u_{2}

(ya matching pair) chahiye — agar tum

u_{1}^{'} = g

likhte ho jabki

u_{1} = y^{'}

hai, to tumhare paas

y^{''} = g

galat second equation ke saath paired hai aur system original ODE encode nahi karta.

"Mera final update tha

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

Middle do slopes ka

2

wala weight missing hai; sahi combination hai

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

, jo Simpson ke

1, 4, 1

pattern ko mimic karta hai (do midpoint samples milke

4

dete hain).

"Maine

\frac{h}{6}

factor drop kar diya kyunki

h

aise bhi chhota tha."

\frac{h}{6}

optional smallness nahi hai — ye hi step ki length scaling hai; iske bina tum time ke koi units nahi wala raw slope add kar rahe ho, jo ek wildly wrong (aur

h

-independent) jump deta hai.

k_{2}

mein maine time

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

par advance kiya lekin state

y_{n}

par hi rakhi."

Tumhe dono time aur state advance karni chahiye; purani state

y_{n}

ko naye time ke saath use karna

k_{2}

ko ek inconsistent slope banata hai aur midpoint prediction ko tod deta hai jis par method rely karti hai.

"Kyunki

k_{1}

already ek achha slope deta hai, maine time bachane ke liye

k_{2}, k_{3}, k_{4}

skip kar diye."

Sirf

k_{1}

use karna bas Euler's method hai (dekho Euler's method for systems), jo globally sirf order-1 accurate hai; teen extra samples ka poora point order 4 tak pahunchna hai.

Why questions

Underlying reason se jawab do, sirf restatement nahi.

Kyun scalars ko vectors se replace karne par RK4 weights unchanged rehte hain?

Kyunki order-4 derivation ne sirf Taylor terms ko addition,

h

se scalar multiplication, aur function evaluation use karke match kiya — ye sab har vector component par identically act karte hain, isliye kisi bhi weight ko re-tune karne ki zaroorat nahi padi.

RK4 midpoint par slope do baar kyun sample karta hai (

k_{2}

aur

k_{3}

deta hua)?

Ye Simpson's rule ko mimic karta hai, jiska accuracy midpoint ko heavily weight karne se aata hai; do successive midpoint estimates (

k_{3}

improved

k_{2}

use karke) method ko zyada Taylor error cancel karne dete hain jo ek single midpoint se nahi ho sakta tha.

Har component ko har stage par saath mein kyun advance karna chahiye?

Kyunki components coupled hain —

f_{i}

doosre unknowns padhta hai — isliye kisi stage ka slope tabhi meaningful hai jab woh ek consistent snapshot dekhe jahan sabhi components ek hi intermediate time par move ho chuke hon. (Dekho Simpson's Rule us quadrature ke liye jo ye imitate karta hai.)

Ek high-order ODE ko first-order system mein reduce karna kyun worthwhile hai?

Ye ek single algorithm (RK4-for-systems) ko kisi bhi order ke ODEs solve karne deta hai, isliye tumhe kabhi alag-alag order ke liye separate integrator nahi chahiye; derivatives ko new unknowns ke roop mein name karna universal adapter hai (dekho Reducing higher-order ODEs to first-order systems).

Local error

O (h^{5})

kyun hai lekin global error sirf

O (h^{4})

kyun?

Ek fixed interval cross karne ke liye roughly

1/ h

steps accumulate hote hain, isliye per-step

O (h^{5})

errors milke approximately

(1/ h) \cdot O (h^{5}) = O (h^{4})

ho jaate hain (detail mein Local vs Global Truncation Error mein hai).

Kuch systems par apni high order ke bawajood explicit RK4 badly kyun fail kar sakta hai?

Stiff systems par explicit RK4 ki stability region practically bahut chota

h

force karti hai, isliye accuracy theek hoti hai lekin stable rehne ke liye enormous computation chahiye (dekho Stiff systems and stability).

2 \times 2

system

y_{1}^{'} = y_{2}, y_{2}^{'} = - y_{1}

aur reduced

y^{''} + y = 0

ke liye same numerical answer kyun aata hai?

Kyunki ye same equations hain —

y^{''} + y = 0

ka reduction exactly

u_{1}^{'} = u_{2}, u_{2}^{'} = - u_{1}

produce karta hai, isliye RK4 identical right-hand sides dekhta hai aur identical steps produce karta hai.

Edge cases

Boundary aur degenerate scenarios jo tumhe surprise nahi karne chahiye.

Kya hota hai agar system

1 \times 1

ho (ek single equation) RK4-for-systems se run karte waqt?

Ye exactly scalar RK4 for a single ODE mein reduce ho jaata hai — length one ka vector bas ek number hai, isliye dono methods koi special-casing ke bina coincide karte hain.

Agar

f

t

par depend nahi karta (ek autonomous system), to kya hum phir bhi shifted times

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

pass karte hain?

Time arguments formula mein phir bhi exist karte hain lekin

f

unhe ignore karta hai, isliye effectively sirf state shifts matter karti hain; recipe unchanged aur correct hai.

Ek linear system

y^{'} = A y

ke liye, kya

k_{i}

phir bhi chahiye, ya koi shortcut hai?

RK4 verbatim apply hota hai (har

k_{i} = A \times (state

)); matrix exponential ke through ek closed-form bhi exist karta hai, lekin RK4 ko aisi koi special structure nahi chahiye aur ye identically kaam karta hai chahe

A

constant ho ya naa ho.

Agar ek component ki initial value exactly

0

hai — kya ye kisi stage ko break karta hai?

Nahi — ek zero entry ek perfectly valid state value hai; slopes aur averages usi tarah compute hote hain, isliye zeros bina kisi special handling ke arithmetic mein propagate ho jaate hain.

Agar ek component ek fixed point tak pahunch jaaye (wahan uska slope

0

hai), kya RK4 use unchanged chodta hai?

Agar us component ke saare chaar

k_{i}

vanish ho jaayein to woh jaisi hai waisi rehti hai; lekin agar coupling kisi midpoint slope ko nonzero banaata hai, to woh phir bhi drift kar sakti hai, isliye ek true equilibrium ke liye slope ko poore predicted step mein zero hona chahiye, sirf start mein nahi.

h \to 0

lene par accuracy aur cost dono par kya hota hai?

Accuracy

h^{4}

ki tarah improve hoti hai lekin steps ki sankhya (aur

f

-evaluations, chaar per step)

1/ h

ki tarah badhti hai, isliye ek practical trade-off hai —

h

ko hamesha shrink karte rehne par eventually sirf round-off error accumulate hoti hai, precision nahi.

Agar do stages ko same state accidentally milti hai (jaise

k_{1}

zero ho jaata hai), kya

k_{2}

aur

k_{3}

phir bhi valid hain?

Haan — zero

k_{1}

ka matlab sirf ye hai ki midpoint state

y_{n}

ke barabar hai, isliye

k_{2}

wahan legitimately evaluate karta hai; kuch degenerate nahi hota, method is step mein sirf kam distinct states sample karta hai.

Connections

Parent topic — RK4 for systems
RK4 for a single ODE — scalar case jin traps ko ye generalise karte hain.
Euler's method for systems — "sirf $k_{1}$ use karo" kya collapse karta hai.
Reducing higher-order ODEs to first-order systems — edge cases ke peeche ka adapter.
Local vs Global Truncation Error — $O (h^{5}) \to O (h^{4})$ reasoning.
Stiff systems and stability — kyun order 4 akela kaafi nahi hai.
Simpson's Rule — $1, 2, 2, 1$ weights ka source.

Question bank — Systems of ODEs — RK4 for systems

True or false — justify