Stiff equations — implicit methods, backward Euler: Question bank

4.8.26 · D5 · Maths › Numerical Methods › Stiff equations — implicit methods, backward Euler

Yahan har symbol parent note se aata hai:

$y^{'} = λ y$ test problem hai — ek single ODE jiska true solution $y (t) = y_{0} e^{λ t}$ hai.
$λ$ (lambda) ek number hai jo decay ki speed set karta hai; $λ < 0$ ka matlab hai solution zero ki taraf shrink karta hai.
$λ$ ek complex number $λ = a + bi$ bhi ho sakta hai. Iska real part $Re λ = a$ likha jaata hai — woh part jisme $i$ nahi lagaa. Geometrically, agar $λ$ ko ek plane par point ki tarah plot karo (horizontal axis = real part, vertical axis = imaginary part), toh $Re λ$ uska horizontal coordinate hai. Negative real part ( $Re λ < 0$ ) ka matlab hai point us plane ke left half mein baitha hai, aur true solution decay karta hai.
$h$ step size hai — hum time mein har step par kitna aage jump karte hain.
$y_{n}$ $n$ steps ke baad hamara numerical guess hai; $y_{n + 1}$ uske baad wala hai.
Amplification factor woh number hai jisse hum $y_{n}$ ko multiply karte hain taaki $y_{n + 1}$ mile. Forward Euler ka factor $1 + hλ$ hai; backward Euler ka $\frac{1}{1 - hλ}$ hai. Agar iska size (absolute value) $1$ se neeche hai, toh numerical solution shrink karta hai — yahi stability hai.

True ya false — justify karo

Backward Euler, forward Euler se zyada accurate hai.

False. Dono first-order hain,

O (h)

— same accuracy order. Backward Euler ka faayda stability hai, jo accuracy se bilkul alag axis hai.

Stiff problem simply woh problem hai jisme bahut chhota step size chahiye.

False. Chhota step sirf explicit methods mein, aur sirf stability ke liye chahiye, tab bhi jab true solution smooth aur slow ho. Stiffness clashing time scales ke baare mein hai, raw difficulty ke baare mein nahi.

Agar true solution zero tak decay karta hai, toh forward Euler ka numerical solution bhi zaroor zero tak decay karega.

False. Sirf tab jab

∣1 + hλ ∣ < 1

ho.

λ = - 100, h = 0.05

ke saath

1 + hλ = - 4

milta hai, toh

y_{n} = (- 4)^{n}

blow up karta hai jabki true

e^{- 100 t} \to 0

Backward Euler kisi bhi ODE ke liye, chahe kuch bhi ho, stable hai.

False. Yeh A-stable hai: unconditionally stable jab

Re λ < 0

ho (decaying problems). Ek growing problem (

Re λ > 0

) ke liye true solution grow karta hai, aur use zero force karna galat hoga.

y^{'} = λ y

ke liye

λ < 0

ke saath, backward Euler ka amplification factor

\frac{1}{1 - hλ}

hamesha

1

se chhota ek positive number hota hai.

True.

λ < 0

aur

h > 0

ke saath,

1 - hλ > 1

, toh iska reciprocal strictly

1

se neeche ek positive number hai — isliye kisi bhi step size ke liye decay hota hai.

Ek explicit method ko implicit banana automatically use A-stable bana deta hai.

False. Implicit hona bade stability regions ki permission deta hai lekin A-stability guarantee nahi karta — specific method ke amplification factor ka analyse karna padega jaanne ke liye.

Trapezoidal (Crank–Nicolson) method, implicit hone ki wajah se, ek bahut stiff mode ko backward Euler ki tarah zero tak damp karega.

False. Trapezoidal A-stable hai lekin L-stable nahi: bahut negative

λ

ke liye iska factor

\to - 1

ho jaata hai, toh stiff modes amplitude

1

par ring karte hain vanish hone ki jagah. Backward Euler ka factor

\to 0

ho jaata hai, unhe khatam kar deta hai.

Stiff problem par backward Euler ke liye

h

chunna survival ke baare mein hai, accuracy ke baare mein nahi.

False. Kyunki backward Euler kisi bhi

h > 0

par survive karta hai, step purely accuracy ke liye choose kiya jaata hai — woh exact freedom jo stiffness explicit methods ko deny karti hai.

Error dhundho

"Backward Euler amplification

1 - hλ

hai, toh

λ < 0

ke saath yeh grow karta hai — yeh sahi nahi ho sakta."

Error formula mein hai. Backward Euler se

(1 - hλ) y_{n + 1} = y_{n}

milta hai, toh factor

\frac{1}{1 - hλ}

hai — reciprocal. Woh

1

se neeche hai, toh yeh decay karta hai.

λ = - 1000

ke liye, forward Euler ko

h < 2/1000

sirf fast transient ke dauran chahiye; uske baad main

h

badha sakta hoon."

Galat. Stability cap

h < 2/∣ λ ∣

λ

par depend karta hai, jo kabhi nahi badlata. Fast mode ke die hone ke baad bhi, forward Euler hamesha tiny steps lene par majboor hai.

"Fixed-point iteration

y^{(k + 1)} = y_{n} + h f (t_{n + 1}, y^{(k)})

implicit step theek se solve karta hai, toh Newton ki kya zaroorat?"

Fixed-point iteration tabhi converge karta hai jab

∣ h \partial f / \partial y ∣ < 1

ho, jahan partial derivative

\partial f / \partial y

step ke point

(t_{n + 1}, y_{n + 1})

par evaluate hota hai — wahi exact stiff restriction jise hum escape karna chahte the. Newton ka koi aisa limit nahi hai, bade steps preserve karta hai.

"Kyunki implicit methods har step mein zyada kaam karte hain, stiff problems ke liye overall slower honge."

Overall count galat hai. Explicit methods ko hazaron tiny steps chahiye (jaise

\geq 5000

to reach

t = 10

with

λ = - 1000

); implicit ko sirf ~

100

bade steps chahiye, toh costlier steps ke bawajood implicit jeetta hai.

"Forward Euler

λ = - 100, h = 0.05

par

y_{1} = - 4

deta hai, jo negative hai, lekin phir bhi stable hai kyunki yeh

y_{0}

se neeche shrink ho raha hai."

Sign flip warning hai.

∣1 + hλ ∣ = ∣ - 4∣ = 4 > 1

; size har step par grow karta hai (

1, 4, 16, \dots

), toh yeh unstable hai aur oscillate kar raha hai — magnitude, sign nahi, stability decide karta hai.

"Backward Euler hamesha true solution se undershoot karta hai, toh koi oscillation introduce nahi karta."

"Hamesha undershoot" nahi, lekin useful sach yeh hai: iska factor

\frac{1}{1 - hλ}

λ < 0

ke liye

(0, 1)

mein ek positive number rehta hai, toh yeh monotone decay produce karta hai bina sign oscillation ke — forward Euler ke unlike.

Why questions

Hum real system ki jagah single scalar test problem

y^{'} = λ y

kyun padhte hain?

Kyunki ek linear system independent scalar equations mein diagonalise ho jaata hai, ek per eigenvalue

λ

. Poore system ki stability, har

λ

par apply kiye scalar rule tak reduce ho jaati hai.

"Zyada negative

λ

" ka matlab "stiffer" kyun hota hai?

True mode

e^{λ t}

hai; zyada negative

λ

faster decay karta hai, toh yeh shorter time scale set karta hai, jo explicit stability cap

h < 2/∣ λ ∣

tight karta hai jabki slow modes ko lamba integration abhi bhi chahiye.

Backward Euler

f

ko old point

t_{n}

ki jagah new point

t_{n + 1}

par kyun evaluate karta hai?

Taaki unknown

y_{n + 1}

f

ke andar appear kare, update ko implicit banate hue. Future point ko aim karna exactly wahi hai jo

y_{n + 1}

ko calm region mein force karta hai, unconditional stability kharidta hai.

A-stability ko "poora left half-plane cover karna" kyun kaha jaata hai?

z = hλ

ko plane par point ki tarah plot karo (horizontal = real part, vertical = imaginary part). Backward Euler exactly wahan stable hai jahan

∣1/ (1 - z) ∣ < 1

ho, yaani

∣1 - z ∣ > 1

ho. Point

1 - z

z

ko reflect aur shift karta hai taaki woh origin se

∣1 - z ∣

doori par ho; woh doori har

z

ke liye

1

se zyada hai jo vertical axis ke left mein ho. Toh poora shaded region left half-plane

Re (z) < 0

hai — precisely wahan jahan true solution decay karta hai.

Hum stiff explicit run fix karne ke liye sirf

h

shrink kyun nahi kar sakte?

Principle mein kar sakte hain, lekin required

h

absurdly small hai (millions of steps), use hopelessly slow banate hue. Stiffness precisely woh signal hai method badalne ka, step size nahi.

Newton's method yahan natural kyun hai jabki ODE step "sirf ek equation" hai?

Kyunki nonlinear

f

ke liye hum residual function

g (y) = y - y_{n} - h f (t_{n + 1}, y)

define karte hain, jiska root

g (y) = 0

hi implicit step hai. Newton phir uski slope

g^{'} (y) = 1 - h \partial f / \partial y

use karke update karta hai, stiffness se independent fast convergence ke saath.

Hum stability ki parwah accuracy se alag kyun karte hain?

Accuracy control karta hai

y_{n}

true value ke kitne close hai; stability control karta hai errors kaafi steps mein grow karenge ya shrink. Ek unstable method useless hai chahe uske formula ka order kitna bhi accurate ho.

Edge cases

Forward Euler exactly

h = 2/∣ λ ∣

par kya hota hai (jab

1 + hλ = - 1

Factor ka size exactly

1

hai, toh

y_{n} = (- 1)^{n} y_{0}

hamesha constant amplitude par oscillate karta hai — marginally stable, na decay na explosion. Strict inequality

∣1 + hλ ∣ < 1

satisfied nahi hai.

y^{'} = λ y, λ < 0

par

h \to \infty

limit mein backward Euler kya karta hai?

Factor

\frac{1}{1 - hλ} \to 0

, toh

y_{1} \to 0

ek step mein. Yeh stiff modes ko zero par slam kar deta hai — yeh strong damping L-stability hai.

Agar

λ = 0

ho (constant, non-decaying solution)?

Backward Euler ka factor

\frac{1}{1 - 0} = 1

hai aur forward Euler ka

1 + 0 = 1

; dono constant exactly preserve karte hain. Na badhta na ghatta, jo correct hai kyunki

y^{'} = 0

Agar

λ

complex ho jisme

Re λ < 0

ho (oscillatory decay)?

z = hλ

likho; phir

Re (z) < 0

, toh

1 - z = 1 - Re (z) - i Im (z)

ka real part

1 - Re (z) > 1

hai. Ek complex number jiska real part already

1

se zyada hai, woh origin se kam se kam

1

doori par hoga (uska magnitude

(real)^{2} + (imag)^{2} \geq ∣ real ∣ > 1

hai), toh

∣1 - z ∣ > 1

aur

∣1/ (1 - z) ∣ < 1

— stable. Isliye A-stability sirf real axis par nahi, poore left half-plane par state ki jaati hai.

Agar

Re λ > 0

ho — kya ek acche method ko phir bhi

y_{n} \to 0

drive karna chahiye?

Nahi. True solution grow karta hai, toh ek faithful method ko

y_{n}

bhi grow karne dena chahiye. Growing problem ke liye "stability" ka matlab us growth ko match karna hai, decay force karna nahi.

λ = - 1

aur

λ = - 1 0^{6}

eigenvalues wale system mein, kaun sa explicit step cap karta hai?

Sabse negative wala,

λ = - 1 0^{6}

h < 2/1 0^{6}

force karta hai — fastest mode hamesha explicit step dictate karta hai, yahi stiffness ka essence hai.

Recall Pure trap-set ka one-line summary

Stability aur accuracy alag axes hain; stiffness ek time-scale clash hai; backward Euler future ko aim karke jeetta hai, kisi bhi step size ke liye stiff modes zero tak damp karta hai.

Connections — har linked note yahan kya add karta hai

Forward Euler method — explicit baseline jiska factor $1 + hλ$ aur cap $h < 2/∣ λ ∣$ se hum baar baar contrast karte hain.
Region of absolute stability — plotted shaded set of $hλ$ jahan ek method bounded rehta hai; yeh poora page iske shape ke baare mein hai.
A-stability and L-stability — "good" ke do levels naam deta hai: A-stable = left half-plane cover karta hai; L-stable = stiff modes ko bhi zero kar deta hai.
Trapezoidal / Crank–Nicolson method — A-stable-but-not-L-stable cousin jiska factor vanish hone ki jagah $- 1$ par ring karta hai.
Newton's method for root finding — woh engine jo implicit step ke residual $g (y) = 0$ ko stiff step cap ke bina solve karta hai.
Runge–Kutta methods — higher-order schemes jo implicit/A-stable banaye ja sakte hain taaki accuracy aur stiff stability combine ho.
Eigenvalues and time scales of linear systems — $λ$ values supply karta hai; sabse negative eigenvalue stiffness set karta hai.

Question bank — Stiff equations — implicit methods, backward Euler

True ya false — justify karo

Error dhundho

Why questions

Edge cases

Connections — har linked note yahan kya add karta hai

Connection map