Modified Euler (Heun's method): Question bank — Hinglish deep dive

4.8.23 · D5 · Maths › Numerical Methods › Modified Euler (Heun's method)

Ek picture apne dimag mein rakho: ek step ke dono siron par do slopes khinchi hain, aur unki average direction mein ek akela walk liya gaya hai.

Sach ya jhooth — justify karo

Heun's method same

h

ke liye plain Euler se hamesha zyada accurate hoti hai.

Aam taur par, hamesha nahi. Smooth problems ke liye yeh bahut behtar hai (

O (h^{2})

O (h)

), lekin bahut zyada stiff ya badly-scaled ODE par bahut bade

h

ke saath, predictor itna aage nikal sakta hai ki corrected step Euler se bhi buri ho jaaye — accuracy order sirf us limit ko describe karta hai jab

h \to 0

Heun's method har step mein function

f

ko exactly do baar evaluate karti hai.

Sach hai basic Heun ke liye: ek baar

k_{1} = f (x_{0}, y_{0})

ke liye aur ek baar

k_{2} = f (x_{1}, y_{1}^{(p)})

ke liye. Iterated corrector variant har extra sweep ke liye

f

ko dobara evaluate karta hai, isliye uska cost zyada hota hai.

Predictor value

y_{1}^{(p)}

step ka final answer hai.

Jhooth hai.

y_{1}^{(p)}

sirf ek scratch guess hai jiska ek hi maqsad hai — hume end-slope

k_{2}

compute karne dena; ise pheink diya jaata hai aur corrected

y_{1}

se replace kar diya jaata hai.

Heun's method ek Runge–Kutta method hai.

Sach hai. Yeh RK2 hai — apne do slopes par equal weights

\frac{1}{2}, \frac{1}{2}

wala ek 2-stage Runge–Kutta scheme.

Agar

f (x, y)

sirf

x

par depend karta hai (

y

par nahi), toh Heun's method ek plain integral ke liye trapezoidal rule mein reduce ho jaati hai.

Sach hai. Jab

f = f (x)

ho, predicted

y

kabhi

f

mein enter nahi karta, isliye

y_{1} = y_{0} + \frac{h}{2} [f (x_{0}) + f (x_{1})]

— exactly Trapezoidal Rule

\int f d x

par apply hota hai.

Step size

h

ko double karne se Heun ka global error roughly double ho jaata hai.

Jhooth hai. Global error

O (h^{2})

hai, isliye

h

ko double karne se error roughly

4

guna hota hai,

2

nahi. (Double karne se plain Euler ka

O (h)

error roughly double hota hai.)

Corrector mein

\frac{h}{2}

ek rounding convenience hai jiska koi gehri meaning nahi hai.

Jhooth hai.

\frac{1}{2}

do slopes ka average hai: har slope step ka aadha contribute karta hai. Ise

h

se replace karo aur tum double-length step lete ho — yeh ek genuine error hai, koi style choice nahi.

Heun ka local (per-step) error aur global error same order ke hain.

Jhooth hai. Local error

O (h^{3})

per step hai;

\sim 1/ h

steps par accumulate hone se ek power drop hoti hai, global

O (h^{2})

milta hai. Yeh ek-power drop one-step methods ke liye generic hai.

Error dhundho

Ek student corrector likhta hai

y_{1} = y_{0} + \frac{h}{2} [f (x_{0}, y_{0}) + f (x_{1}, y_{0})]

. Kya galat hai?

End-slope predicted

y_{1}^{(p)} = y_{0} + h k_{1}

ki jagah purane

y_{0}

par evaluate ki gayi hai.

y

step ke across kaise move hua isko ignore karna second-order accuracy ko khatam kar deta hai — tum wahi "look-ahead" kho dete ho jo Heun ko define karta hai.

Ek student

k_{2} = f (x_{0}, y_{0} + h k_{1})

compute karta hai — same

x_{0}

, moved

y

. Theek karo.

End-slope end point

x_{1} = x_{0} + h

par sample honi chahiye, isliye

k_{2} = f (x_{0} + h, y_{0} + h k_{1})

y

ko advance karna lekin

x

ko nahi,

f

ko ek aisi jagah evaluate karta hai jo step par kabhi exist nahi karti.

Ek student step khatam hone se pehle corrected

y_{1}

k_{1}

ke andar wapas use karta hai. Yeh circular kyun hai?

k_{1} = f (x_{0}, y_{0})

step ke start se fixed hota hai aur baaki sab se pehle jaana jaata hai. Isme baad ki value feed karna matlab ab tumhare paas start-slope nahi hai — method apna predictor kho deti hai aur Heun rehti nahi.

Koi likhta hai

y_{1} = y_{0} + h f (x_{0}, y_{0}) + \frac{h}{2} f (x_{1}, y_{1}^{(p)})

. Diagnose karo.

Unhone ek full Euler step

h k_{1}

aur aadha end-slope add kar diya — Euler aur Heun ka mix. Do slopes par sahi total weight

h

tak sum hona chahiye: yeh

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

hona chahiye.

Example 3 mein ek student corrector ko iterate karta hai lekin har sweep ke liye wahi pehla predictor

y^{(p)} = 1.1

k_{2}

ke liye use karta rehta hai. Kya galat hai?

Iterate karne ka matlab hai latest corrected

y

se end-slope re-predict karna (jaise

1.11

), taaki

k_{2}

har sweep mein update ho.

k_{2}

ko original guess par freeze karna sirf pehla corrector repeat karta hai aur kuch naya nahi milta.

Why questions

Hum

y_{1}

ko correct karne se pehle predict kyun karte hain?

Trapezoidal corrector ko

f (x_{1}, y_{1})

chahiye, lekin

y_{1}

wahi unknown hai jise hum solve kar rahe hain. Ek sasta Euler predictor is chicken-and-egg deadlock ko pehli guess se todta hai.

Do slopes ko average karne ki jagah sirf end-slope

k_{2}

use kyun nahi karte?

Sirf

k_{2}

use karna over-correct karta — true slope poore interval mein vary karti hai, aur entry aur exit slopes ka mean interval ke behaviour se akele kisi bhi endpoint se ek order zyada match karta hai.

Heun ko "predictor–corrector" method kyun kaha jaata hai, sirf "Euler twice" nahi?

Do stages alag roles play karti hain: predictor ek rough forward guess karta hai, corrector us guess ka use ek better-averaged step banane ke liye karta hai. Predictor-Corrector Methods dekho — yeh asymmetry defining feature hai.

Heun

h^{2}

term tak true Taylor series se kyun match karta hai lekin

h^{3}

par fail hota hai?

Start aur end slopes ko average karna

y^{'}

aur

y^{''}

ko exactly reproduce karta hai, lekin

h^{3}

term ko

y^{'''}

(curvature-of-curvature) chahiye jise do slope samples capture nahi kar sakte.

h^{3}

ko sahi karna hi tumhe RK4 ki taraf push karta hai. Taylor Series Methods dekho.

h

ko aadha karna Heun ki zyada madad kyun karta hai Euler se?

Heun ka error

h^{2}

scale karta hai, Euler ka

h

;

h

ko aadha karne se yeh

4

aur

2

se divide hote hain respectively — zyada power refinement ko zyada reward deti hai. Yahi Order of Accuracy and Step Size ka poora point hai.

Corrector ko iterate karna answer improve kyun kar sakta hai, phir bhi exact solution tak kabhi nahi pahunch sakta?

Iteration implicit trapezoidal equation ko zyada precisely solve karta hai, predictor error hatata hai — lekin trapezoidal rule mein

O (h^{3})

discretisation error hai jo itni bhi iteration se erase nahi ho sakti.

Heun exact kyun hai jab true solution ek straight line ho (

y^{''} = 0

Seedhi solution mein constant slope hoti hai, isliye

k_{1} = k_{2}

aur unka average har jagah true slope ke equal hota hai. Trapezoid tab exact hota hai jab integrand constant ho, isliye koi error introduce nahi hoti.

Edge cases

Jab

f (x_{0}, y_{0}) = 0

ho (flat start slope) toh Heun kya karta hai?

Predictor ruka rehta hai (

y_{1}^{(p)} = y_{0}

), lekin

k_{2} = f (x_{1}, y_{0})

nonzero ho sakta hai, isliye corrector phir bhi move karta hai. Yahi wajah hai ki Heun woh "curvature notice" karta hai jo plain Euler miss karta hai jab yeh flat start karta hai (Example 2 dekho).

Agar step size

h = 0

hai, toh formula kya deta hai?

y_{1} = y_{0}

aur

x_{1} = x_{0}

— koi step nahi liya jaata. Predictor aur corrector dono identity mein collapse ho jaate hain; method consistent hai (zero step kuch nahi badalta).

Maan lo

f (x, y)

y

mein linear hai, jaise

f = a y + b (x)

. Kya predictor phir bhi zaroori hai?

Practice mein haan, lekin note karo corrector ek explicit linear equation ban jaata hai jise tum

y_{1}

ke liye seedha solve kar sakte ho. Heun ise predict karke sidestep karta hai — exactness ki jagah ek fixed, sasta recipe trade karta hai.

Agar predictor massively overshoot kare (fast-decaying ODE par huge

h

) toh corrector ka kya hota hai?

k_{2}

bilkul galat

y

par sample hota hai, isliye averaged slope corrupt ho jaata hai aur

y_{1}

Euler se bhi bura ho sakta hai — ek stability failure. Order-2 accuracy ek small-

h

promise hai, large-

h

guarantee nahi.

y^{'} = - y

jaisi purely oscillatory problem ke liye ek step mein, Heun ka answer true decay se upar hai ya neeche?

Heun ka averaged slope large

h

ke liye thoda under-decay karta hai kyunki trapezoid rule curved integrands ko over/undershoot karta hai; small

h

ke liye discrepancy tiny hai (

O (h^{3})

per step) aur Heun true exponential ko closely track karta hai.

Agar do alag smooth ODEs ek step par same

y_{0}

k_{1}

aur

k_{2}

share karein, toh kya woh same Heun step lete hain?

Haan — Heun ka step sirf

y_{0}

aur do slope samples par depend karta hai, underlying formula par nahi. Recipe mein identical inputs se identical output milta hai, construction se.

Connections

Euler's Method — predictor hai ek plain Euler step.
Trapezoidal Rule — corrector disguise mein yahi rule hai (aur equals karta hai jab $f = f (x)$ ).
Runge-Kutta Methods — Heun hai RK2; RK4 $h^{3}$ gap fix karta hai.
Order of Accuracy and Step Size — kyun $O (h^{2})$ $h$ halving ko reward karta hai.
Predictor-Corrector Methods — Heun sabse simple member hai.
Taylor Series Methods — woh tool jo prove karta hai Heun second order hai.