Stiff equations — implicit methods, backward Euler: Worked examples

4.8.26 · D3 · Maths › Numerical Methods › Stiff equations — implicit methods, backward Euler

Shuru karne se pehle, dono amplification factors ka ek-line refresher — woh numbers jo $n$ -th power tak uthaye jaate hain:

Hum $z = hλ$ ko stability variable kehte hain. Neeche har example asal mein yahi pooch raha hai: yeh $z$ kahan land karta hai, aur har $R$ wahan kya karta hai?

Scenario matrix

Har woh case jo ek stiff-ODE problem de sakta hai, aur woh example jo use hit karta hai:

Cell	Case class	Kya special hai	Example
A	$λ < 0$ , chhota $∣ z ∣$	dono methods stable, mild step	Ex 1
B	$λ < 0$ , $z$ FE danger band mein ( $- 2 < z < - 1$ )	FE oscillate karta hai par bounded	Ex 2
C	$λ < 0$ , $z < - 2$	FE blow up karta hai, BE theek	Ex 3
D	Boundary $z = - 2$ exactly	FE marginal ( $∣ R ∣ = 1$ )	Ex 4
E	Degenerate $λ = 0$ ( $z = 0$ )	koi decay nahi, dono constant dete hain	Ex 5
F	$λ > 0$ (sach mein growth)	dono ko $\to 0$ force nahi karna chahiye; check karo ki dono faithfully grow karte hain	Ex 6
G	Complex $λ$ (oscillatory decay)	left half-plane, magnitude test	Ex 7
H	Nonlinear $f$ → Newton needed	implicit solve by iteration	Ex 8
I	Real-world word problem (do time scales)	method + step choose karo	Ex 9
J	Exam twist: sabse bada stable $h$ dhundho	stability test ko invert karo	Ex 10

Stability variable $z = hλ$ ek number line par rehta hai (real $λ$ ) ya ek plane mein (complex $λ$ ). Yeh picture har method ke safe zones dikhati hai — isko baar baar refer karo:

Figure — Stiff equations — implicit methods, backward Euler

Worked examples

Worked example Example 2 — Cell B: forward Euler oscillate karta hai par bounded rehta hai

$y^{'} = - 15 y$ , $h = 0.1$ , $y_{0} = 1$ . Forward Euler kya karta hai?

Forecast: stable ya unstable? Terms ka sign kya hoga?

$z = 0.1 \times (- 15) = - 1.5$ . Yeh step kyun? Figure s01 ki number line par $z$ locate karo — yeh $(- 2, - 1)$ mein baithta hai.
$R_{FE} = 1 + z = 1 - 1.5 = - 0.5$ . Negative! Yeh step kyun? Ek negative factor ka matlab hai $y_{n}$ ki sign har step par flip hoti hai — spurious oscillation.
$y_{1} = - 0.5, y_{2} = 0.25, y_{3} = - 0.125, \dots$ Yeh step kyun? $(- 0.5)^{n}$ sign alternate karta hai par uska size shrink karta hai.

Verify: $∣ R_{FE} ∣ = 0.5 < 1$ toh yeh stable hai (0 tak decay karta hai), phir bhi oscillate karta hai — true solution $e^{- 15 t}$ kabhi negative nahi jaata, toh yeh sign flips pur numerical artefact hain. Backward Euler: $R_{BE} = 1/ (1 + 1.5) = 0.4 > 0$ , smoothly decay karta hai, koi flip nahi. ✓

Worked example Example 4 — Cell D: exact boundary

$z = - 2$ $y^{'} = - 20 y$ , $h = 0.1$ . Forward Euler exactly edge par kya karta hai?

Forecast: decay, blow up, ya kuch nahi?

$z = 0.1 \times (- 20) = - 2$ exactly. Yeh step kyun? Yeh FE stability interval ki boundary hai.
$R_{FE} = 1 + (- 2) = - 1$ , toh $∣ R ∣ = 1$ . Yeh step kyun? $∣ R ∣ = 1$ razor's edge hai — na $< 1$ (stable) na $> 1$ (unstable) case.
$y_{n} = (- 1)^{n} = 1, - 1, 1, - 1, \dots$ Yeh step kyun? Constant magnitude, hamesha ke liye sign flip karta hai — yeh kabhi decay nahi karta, toh yeh stable nahi hai.

Verify: strict test hai $∣ R ∣ < 1$ ; yahan hum $= 1$ hit karte hain, toh numerical solution na marta hai na badhta hai — ek permanent oscillation jabki true $e^{- 20 t}$ vanish ho jaata hai. Boundary unstable side ki taraf belong karti hai. Backward Euler: $R_{BE} = 1/ (1 + 2) = 1/3 < 1$ — abhi bhi theek. ✓

Worked example Example 6 — Cell F: genuine growth

$λ > 0$ $y^{'} = + 3 y$ , $h = 0.1$ , $y_{0} = 1$ . True solution badhta hai; kya methods growth ko honestly track karte hain?

Forecast: numbers bade hone chahiye ya chhote?

$z = 0.1 \times 3 = + 0.3 > 0$ . Yeh step kyun? Positive $z$ ka matlab hai hum figure s01 ke right half mein hain — dono stability regions ke bahar, jo sahi hai kyunki true solution decay nahi honi chahiye.
$R_{FE} = 1.3$ , $R_{BE} = \frac{1}{1 - 0.3} = \frac{1}{0.7} = 1.4286$ . Yeh step kyun? Dono factors $> 1$ , toh dono badhte hain — accha hai.
$y_{2}$ : FE $= 1. 3^{2} = 1.69$ ; BE $= 1.428 6^{2} = 2.0408$ . Yeh step kyun? Growth ke do steps track karo.

Verify: true $e^{0.3} = 1.3499$ per step, $e^{0.6} = 1.8221$ at $t = 0.2$ . Dono methods badhte hain (jaise hona chahiye); yahan "unstable" sahi jawaab hai. Lesson: A-stability tabhi desirable hai jab true solution decay kare. ✓

Worked example Example 7 — Cell G: complex

$λ$ , oscillatory decay $y^{'} = λ y$ with $λ = - 1 + 4 i$ , $h = 0.5$ . Kya backward Euler stable hai? (magnitude test)

Forecast: kya $Re λ < 0$ BE stability guarantee karta hai?

$z = hλ = 0.5 (- 1 + 4 i) = - 0.5 + 2 i$ . Yeh step kyun? Ab $z$ plane mein ek point hai, line par nahi.
$1 - z = 1 - (- 0.5 + 2 i) = 1.5 - 2 i$ ; uska magnitude $∣1 - z ∣ = 1. 5^{2} + 2^{2} = 2.25 + 4 = 6.25 = 2.5$ . Yeh step kyun? $∣ R_{BE} ∣ = \frac{1}{∣1 - z ∣}$ , toh denominator ka size chahiye.
$∣ R_{BE} ∣ = \frac{1}{2.5} = 0.4 < 1$ → stable. Yeh step kyun? 1 se neeche magnitude ka matlab oscillation decay karta hai, $∣ e^{λ t} ∣ = e^{- t} \to 0$ se match karta hai.

Verify: $Re λ = - 1 < 0$ , aur A-stability guarantee karta hai $∣ R_{BE} ∣ < 1$ pure left half-plane ke liye — numerically confirm hua ( $0.4$ ). Forward Euler yahan: $∣1 + z ∣ = ∣0.5 + 2 i ∣ = 0.25 + 4 = 4.25 = 2.06 > 1$ → FE unstable hai chahe true solution decay kare. ✓ (Dekho Region of absolute stability.)

Worked example Example 8 — Cell H: nonlinear

$f$ , ek Newton solve Backward Euler on $y^{'} = - y^{2}$ , $y_{0} = 2$ , $h = 0.5$ . Newton's method for root finding use karke $y_{1}$ nikalo.

Forecast: implicit equation quadratic hai — kya Newton do steps mein land karega?

$g (y) = y - y_{0} - h (- y^{2}) = y - 2 + 0.5 y^{2} = 0$ setup karo. Yeh step kyun? Backward Euler hai $y_{1} = y_{0} + h f (y_{1})$ ; sab terms ek taraf karne se root problem milta hai.
Derivative $g^{'} (y) = 1 + h \cdot 2 y = 1 + y$ (kyunki $h = 0.5$ ). Yeh step kyun? Newton ko $g^{'}$ chahiye; yahan $g^{'} = 1 - h \partial f / \partial y = 1 - 0.5 (- 2 y) = 1 + y$ .
Start $y^{(0)} = y_{0} = 2$ . $g (2) = 2 - 2 + 0.5 (4) = 2$ , $g^{'} (2) = 3$ . $y^{(1)} = 2 - 2/3 = 1.3333$ . Yeh step kyun? Ek Newton step: $g / g^{'}$ subtract karo.
$g (1.3333) = 1.3333 - 2 + 0.5 (1.7778) = 0.2222$ , $g^{'} = 2.3333$ . $y^{(2)} = 1.3333 - 0.09524 = 1.2381$ . Yeh step kyun? Iterate karo; residual tezi se shrink ho raha hai (quadratic convergence).
Ek aur: $g (1.2381) = 1.2381 - 2 + 0.5 (1.5330) = 0.00460$ , $g^{'} = 2.2381$ , $y^{(3)} = 1.2381 - 0.002056 = 1.23605$ .

Verify: $0.5 y^{2} + y - 2 = 0$ ka exact root hai $y = \frac{- 1 + 1 + 4}{1} = \frac{- 1 + 5}{1} = 1.23607$ (positive root). Newton teen steps mein $1.23605$ tak pahuncha — match karta hai. Aur $g (1.23607) = 0$ . ✓

Worked example Example 9 — Cell I: real-world do-time-scale problem

Ek chemical mix karta hai: fast species $e^{- 500 t}$ ki tarah decay karta hai, slow $e^{- t}$ ki tarah, model $y^{'} = - 500 y$ (transient) ek slow drift ke saath coupled. Fast part ke khatam hone ke baad $t = 5$ tak integrate karne ke liye, kaun sa method aur roughly kitne steps?

Forecast: FE aur BE ke beech step-count ratio guess karo.

Fastest rate $∣ λ ∣ = 500$ , toh FE ko chahiye $h < \frac{2}{500} = 0.004$ . Yeh step kyun? FE step fastest scale se cap hota hai stability ke liye, chahe woh scale ja chuka ho.
FE ke liye $t = 5$ tak steps: $\frac{5}{0.004} = 1250$ steps. Yeh step kyun? Total time ÷ max step.
BE ka koi stability cap nahi; slow scale (rate 1) accuracy ke liye $h \approx 0.1$ allow karta hai → $\frac{5}{0.1} = 50$ steps. Yeh step kyun? A-stability hume $h$ sirf accuracy ke liye choose karne deti hai.

Verify: FE $1250$ vs BE $50$ steps — $25 \times$ ki saving, aur ratio $\approx ∣ λ_{fast} ∣/∣ λ_{slow} ∣ \times (accuracy factor)$ stiffness ratio $500$ ke saath scale karta hai. ✓

Worked example Example 10 — Cell J: exam twist, sabse bada stable

$h$ " $y^{'} = - 40 y$ ke liye, sabse bada step $h$ nikalo jiske liye forward Euler stable ho, aur confirm karo ki backward Euler ka aisa koi limit nahi."

Forecast: stability inequality ko invert karo — cutoff kya hai?

FE stable hone ke liye chahiye $∣1 + hλ ∣ < 1$ , yaani $- 1 < 1 - 40 h < 1$ . Yeh step kyun? $λ = - 40$ test mein plug karo aur window padho.
Left inequality $1 - 40 h > - 1 \Rightarrow 40 h < 2 \Rightarrow h < 0.05$ . Right inequality $1 - 40 h < 1$ $h > 0$ ke liye automatic hai. Yeh step kyun? Binding constraint hai $h < 2/∣ λ ∣$ .
Toh supremum hai $h = 0.05$ (strict; $h = 0.05$ par, $∣ R ∣ = 1$ , marginal — dekho Ex 4). Yeh step kyun? Equality boundary hai, stable nahi.
Backward Euler: $∣1/ (1 - hλ) ∣ = ∣1/ (1 + 40 h) ∣ < 1$ sab $h > 0$ ke liye kyunki $1 + 40 h > 1$ . Yeh step kyun? Denominator decaying $λ$ ke liye hamesha 1 se zyada hota hai — $h$ par koi upper bound nahi.

Verify: $2/∣ λ ∣ = 2/40 = 0.05$ match karta hai. $h = 0.05$ par: $R_{FE} = 1 - 2 = - 1$ (marginal). BE even $h = 10$ par: $1/ (1 + 400) = 0.0025 < 1$ abhi bhi stable. ✓

Recall Matrix par quick self-test

Kaun sa cell forward Euler ko blow up karta hai par backward Euler ko nahi? ::: Cell C ( $z < - 2$ ), jaise Ex 3. $z = - 2$ exactly par, forward Euler kya karta hai? ::: Constant magnitude $∣ R ∣ = 1$ ke saath oscillate karta hai — stable nahi (Cell D). "Unstable growth" actually sahi behaviour kab hota hai? ::: Jab $λ > 0$ (Cell F): true solution badhta hai, toh dono methods ko bhi badhna chahiye. Complex $λ$ ke liye, " $z < 0$ " ki jagah kya aata hai? ::: Magnitude test $∣ R ∣ < 1$ , jaise $∣1/ (1 - z) ∣ < 1$ pure left half-plane par (Cell G).

Connections

4.8.26 Stiff equations — implicit methods, backward Euler (Hinglish)
Forward Euler method
Region of absolute stability
A-stability and L-stability
Trapezoidal / Crank–Nicolson method
Newton's method for root finding
Runge–Kutta methods
Eigenvalues and time scales of linear systems