ODE solvers — Euler's method (derivation, global error): Question bank

4.8.22 · D5 · Maths › Numerical Methods › ODE solvers — Euler's method (derivation, global error)

Poora page ek cheez test karta hai: kya tum jaante ho kaun sa error kaun sa hai, method kab theek kaam karta hai, aur geometry mein ek curving path pe straight step kyun hota hai. Yahan koi arithmetic grinding nahi hai — woh worked-example pages mein hai.

True or false — justify karo

Euler's method second order hai kyunki uska local truncation error

O (h^{2})

hai.

False. Ek step ka local error

O (h^{2})

hota hai, lekin ek fixed endpoint tak global error

N \propto 1/ h

steps mein accumulate hota hai, ek power ghata ke

O (h)

de deta hai — isliye Euler first order hai.

Step

h

ko half karne se global error bhi roughly half ho jaata hai.

True. Global error

\propto h^{1}

hai, isliye

h

\frac{1}{2}

se scale karne par error bhi roughly

\frac{1}{2}

ho jaata hai. (Agar quarter hota, toh method second order hota.)

Agar hum har step ki shuruaat exactly true curve par karein, toh sirf local truncation error bachta hai.

True — yahi LTE ki exact definition hai: ek single step ka error, assuming perfect start. Global error tab hota hai jab tum har baar curve par restart nahi karte aur galtiyan compound hoti hain.

h

ko apne computer ke allowed minimum tak shrink karna hamesha sabse accurate answer deta hai.

False.

h

ko chhota karne se truncation error toh ghatta hai, lekin steps ki sankhya badh jaati hai, isliye round-off error accumulate hota hai aur cost bhi explode karti hai. Ek sweet spot hota hai, zero par limit nahi.

Linear ODE

y^{'} = λ y

ke liye jahan

λ < 0

ho, Euler ka approximation hamesha true solution ki tarah decay karta hai.

False. Euler deta hai

y_{n + 1} = (1 + hλ) y_{n}

; decay tabhi hoti hai jab

∣1 + hλ ∣ < 1

ho. Bahut bada

h

is factor ko magnitude mein

\geq 1

bana deta hai aur numerical solution badhne ya oscillate karne lagta hai — yeh ek stability failure hai, accuracy ka nahi. Dekho Numerical Stability.

Global error bound

\frac{h M}{2 L} (e^{L (x_{n} - x_{0})} - 1)

kehta hai ki jaise-jaise hum aur integrate karte hain, error bina limit ke badhta jaata hai.

True worst case mein: ek fixed

h

ke liye, endpoint

x_{n}

ko aur aage le jaane se

e^{L (x_{n} - x_{0})}

factor bada hota jaata hai, isliye accumulated error interval ke saath exponentially grow kar sakta hai.

Euler's method ko chalane ke liye ODE ka closed-form solution chahiye.

False. Isse sirf slope function

f (x, y)

aur ek starting point chahiye. Poora point hi yahi hai ki yeh tab kaam karta hai jab koi closed form exist hi na kare.

Agar

y^{''} = 0

har jagah ho (true solution ek straight line hai), toh Euler's method exact hai.

True. Discarded term

\frac{h ^{2}}{2} y^{''} (ξ) = 0

hai, isliye kuch bhi throw away nahi karna padta — tangent step exactly line par land karta hai.

Galti dhundho

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

— yahi Euler's update hai."

Galat — yeh naye point par slope use karta hai, jo Backward (implicit) Euler hai aur

y_{n + 1}

ke liye solve karna padta hai. Forward Euler

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

use karta hai, yaani current point par slope.

"Global error

= N \times

LTE

= \frac{b - x _{0}}{h} \cdot \frac{h ^{2}}{2} y^{''} = \frac{( b - x _{0} ) h}{2} y^{''}

, isliye error

O (h^{2})

hai."

Algebra last claim tak sahi hai:

\frac{( b - x _{0} ) h}{2} y^{''}

h

mein linear hai, yaani

O (h)

hai,

O (h^{2})

nahi. Student ne

h^{1}

h^{2}

padh liya.

"Step formula hai

y_{n + 1} = y_{n} + f (x_{n}, y_{n})

Step size

h

missing hai. Iske bina update

h

ki actual distance chahe jo bhi ho, ek full unit-length rise le leta hai — yeh hona chahiye

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

N = (b - x_{0}) / h

steps hain, aur kyunki har naya step independent hai, errors sirf add hote hain aur total

O (h^{2})

rehta hai."

Do galtiyan hain: errors

O (h^{2})

nahi rehte kyunki

N

badhta hai jab

h

ghatta hai, aur woh independent nahi hain — pehle ki errors baad ke steps se amplify hoti hain, aur yahi woh jagah hai jahaan se

e^{L (x - x_{0})}

factor aata hai.

(x_{n}, y_{n})

par tangent paane ke liye hume

y (x)

ko wahan ke aas-paas jaanna padega."

Nahi — ODE khud hi slope directly

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

ke roop mein deta hai. Yahi poora trick hai: curve jaane bina tangent ka slope mil jaata hai.

"Kyunki LTE

= \frac{h ^{2}}{2} y^{''} (ξ)

hai jahan

ξ

unknown hai, hum error ke baare mein kuch useful nahi keh sakte."

Keh sakte hain: interval par

∣ y^{''} ∣ \leq M

bound karne se unknown

ξ

ek usable inequality mein badal jaata hai

∣ LTE ∣ \leq \frac{h ^{2}}{2} M

, jo poore global bound ko drive karta hai.

Why questions

Global error, local error ke mukable mein exactly ek power of

h

kyun khota hai?

Kyunki ek fixed endpoint tak pahunchne mein

N = (b - x_{0}) / h

steps lagte hain, aur

N

1/ h

ki tarah badhta hai. Per-step

O (h^{2})

O (1/ h)

steps se multiply karne par ek power mar jaati hai:

O (h^{2}) \cdot O (h^{- 1}) = O (h)

Error bound mein ek sada sum ke bajaye exponential

e^{L (x - x_{0})}

kyun hota hai?

Kyunki pehle ki gayi galti wahan baith nahi jaati — woh har baad ke step mein

f

ke through feed hoti rehti hai, aur nearby solutions rate

L

par diverge kar sakti hain (Lipschitz constant). Aise growth ka worst-case compounding exponential hota hai.

Hum exact derivative ki jagah forward difference

\frac{y _{n + 1} - y _{n}}{h}

kyun use karte hain?

Ek computer woh limit

h \to 0

nahi le sakta jo derivative ko define karta hai.

h

ko finite rakhne se ek usable, computable approximation milta hai — ki mat hai woh truncation error jo hum phir analyse karte hain.

Error ke liye Taylor's theorem sahi tool kyun hai, sirf guess karne ki jagah?

Taylor's theorem exact hai: remainder term

\frac{h ^{2}}{2} y^{''} (ξ)

koi approximation nahi balki precisely woh cheez hai jo hum discard karte hain. Isse hum error ko name aur bound kar sakte hain, hand-waving ki jagah. Dekho Taylor Series Expansion.

Euler ke liye

f

par Lipschitz condition kyun matter karti hai?

Yeh guarantee karta hai ki nearby solutions rate

L

se zyada fast diverge nahi kar sakti, aur yahi cheez accumulated errors ko bounded rakhti hai. Iske bina,

e^{L (x - x_{0})}

factor — aur poori error guarantee — exist hi nahi kar sakti.

Ek chhota step hume curving true solution ke paas kyun rakhta hai jab hum straight lines mein chalte hain?

Ek tiny horizontal distance

h

par, curve barely bend karta hai, isliye uska tangent aur curve almost coincide karte hain. Per step gap

O (h^{2})

hai —

h

ghataane se har straight step curve ko tighter hug karta hai.

Edge cases

Euler kya karta hai jab ek step mein

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

ho (slope flat ho)?

Update

y_{n + 1} = y_{n}

ban jaata hai: point us step ke liye wahan ka wahan rehta hai. Yeh ek bilkul valid step hai — ek horizontal tangent — koi error nahi.

y^{'} = - 2 y

y (0) = 1

ke liye

h = 0.5

ke saath, Euler

y_{1} = 0

deta hai phir

0

par rehta hai. Kya galat hua?

Factor

1 + hλ = 1 + 0.5 (- 2) = 0

value ko exactly zero kar deta hai, aur true

e^{- 1} \approx 0.37

bahut zyada miss ho jaata hai. Fast decay ke liye step bahut bada hai — ek instability,

h

ghata ke theek ho sakti hai.

Euler ka kya hoga agar true solution ka interval ke andar vertical asymptote (blow up) ho?

Euler finite tangent steps leta rehta hai aur finite numbers produce karta hai even jahan true solution infinite hai, isliye blow-up ke baad silently nonsense deta hai. Bound ne

∣ y^{''} ∣ \leq M

assume kiya tha, jo wahan fail karta hai.

Agar

h

aisa choose kiya gaya ki

x_{0} + N h

target

b

se aage nikal jaaye, toh kya danger hai?

Final step

b

se aage land karta hai, isliye tum solution ko galat point par evaluate karte ho.

h

aisa choose karo ki

N = (b - x_{0}) / h

ek whole number ho, ya ek chhota final step lo.

Steps ki sabse chhoti meaningful sankhya,

N = 1

kya hai?

x_{0}

b = x_{0} + h

tak ek single Euler step: yeh sirf ek tangent jump hai. Global aur local error yahan coincide karte hain kyunki accumulate karne ke liye kuch hai hi nahi.

Kya Euler differently behave karta hai agar slope function

f

sirf

x

par depend kare (yaani

y^{'} = g (x)

Tab Euler integral

\int g d x

ke liye left-endpoint rectangle rule mein reduce ho jaata hai — ek jaana-pehchana first-order quadrature. Yeh

O (h)

order ko ek alag angle se confirm karta hai.

Connections

Euler's Method — Derivation & Global Error — woh parent jise yeh bank drill karta hai
Taylor Series Expansion — exact remainder term ka source
Numerical Stability — decay/blow-up edge cases
Backward Euler & Implicit Methods — implicit look-alike trap
Lipschitz Continuity — exponential error factor kyun exist karta hai
Runge-Kutta Methods — first-order limit ke liye higher-order fixes
Finite Difference Approximations — forward-difference derivation route