Exercises — Stiff equations — implicit methods, backward Euler

KYA: bas substitute karo. $z=h\lambda =(0.05)(-20)=-1$. YEH numbers kyun: $z$ step size aur decay ko ek dimensionless quantity mein bundle karta hai — stability test sirf isi ko dekhta hai. $R(z)=\frac{1}{1-z}=\frac{1}{1-(-1)}=\frac{1}{2}=\mathrm{0.5.}$ Kyunki $|0.5|<1$, yeh step $y$ ko shrink karta hai: stable. ✅

$z=-1$, toh $R(z)=1+(-1)=0$. $|0|<1$ → stable (actually yeh true value ke fast decay ko bilkul theek se pakadta hai). Yeh pehchanna ki forward Euler ka factor $1+z$ hai (na ki $1/(1-z)$) — yahi is rung ka poora point hai.

Backward Euler. Uski stable region (violet) right par ek chhoti disk ke bahar ka poora area hai, jo poori negative real axis ko contain karti hai — har negative $z$ stable hai. Forward Euler ki stable region (magenta) $-1$ par centered radius $1$ ki ek disk hai; yeh sirf $-2<z<0$ cover karti hai. $z=-2$ ke baad forward Euler unstable ho jaata hai. Yahi backward Euler ki A-stability hai jo picture ke roop mein dikhayi gayi hai.

KYA: linear test problem par backward Euler $y_{n+1}=\frac{1}{1-h\lambda }{y}_n$ deta hai. Rearrange kyun: $y_{n+1}$ $f$ ke andar aata hai, isliye hum use left side par le jaate hain — yahi implicit move ki defining step hai. $z=h\lambda =(0.5)(-4)=-2,R=\frac{1}{1-(-2)}=\frac{1}{3}.$ $y_1=\frac{1}{3}\cdot 1=0.3333,y_2=\frac{1}{3}\cdot y_1=\mathrm{0.1111.}$ True $y(1)=e^{-4}=0.0183$ se compare karo — yeh bhi decay kar raha hai, direction sahi hai (first order hai, toh exact nahi).

$z=h\lambda =(0.6)(-4)=-2.4$. Forward: $R=1+z=1-2.4=-1.4$. $y_1=-1.4$, $y_2=1.96$. $|-1.4|>1$ → badhta hai aur sign flip karta hai: unstable. Backward: $R=1/(1-(-2.4))=1/3.4=0.2941$. $y_1=0.2941$, $y_2=0.0865$ → smoothly decay karta hai: stable. Fark poora factor mein hai: forward radius $1$ ki disk se bahar chala gaya; backward kabhi nahi ja sakta.

KYA: $y_1=y_0+hf(t_1,y_1)=0+0.1(-30y_1+3)$. KYun: $f$ naye $y_1$ par evaluate hota hai, toh $y_1$ dono sides par hai — use collect karo. $y_1=-3y_1+0.3\Rightarrow y_1+3y_1=0.3\Rightarrow 4y_1=0.3\Rightarrow y_1=\mathrm{0.075.}$ (Sanity check: steady state hai $y^{\prime }=0\Rightarrow y=3/30=0.1$; hum $0$ se $0.1$ ki taraf move kiye. ✅)

Stability: $|1+h\lambda |<1\Rightarrow h<\frac{2}{|\lambda |}=\frac{2}{250}=0.008$. $t=5$ tak pahunchne ke liye: $\text{steps}=\frac{5}{0.008}=625$ steps, sirf stable rehne ke liye, accuracy ke liye nahi. Yeh bound kyun: real axis par stable interval $-2<z<0$ hai, yaani $-2<h\lambda <0$, jo $h<2/|\lambda |$ deta hai.

Maano $z=h\lambda$. Kyunki $h>0$ aur $\lambda <0$, hame $z<0$ milta hai, toh $1-z=1-(\text{negative})=1+|z|>1$. Isliye $|R(z)|=\left|\frac{1}{1-z}\right|=\frac{1}{1-z}<1\text{kyunki denominator }1\text{ se zyada hai.}$ $h$ par koi upper bound nahi aata — yahi exactly unconditional stability ka matlab hai. Step sirf accuracy ke liye choose kiya jaata hai.

Forward Euler ko fastest mode ke liye stability satisfy karni hai: $h<2/|{\lambda }_2|=2/1000=0.002$. Steps $=10/0.002=5000$. Backward Euler ka koi stability cap nahi; $h=0.1$ use karke steps $=10/0.1=100$ milte hain. Ratio $=5000/100=50\times$ kam steps. Fast mode kyun dominate karta hai: explicit method ki stability smallest time scale $1/|{\lambda }_2|$ se set hoti hai, chahe woh component numerically zero ho chuka ho — yahi stiffness ki definition hai.

KYA: implicit equation hai $g(y)=y-y_0-h(-y^3)=y-1+0.5y^3=0$. Newton kyun: simple iteration ko $|h\partial f/\partial y|<1$ chahiye hogi, wahi stiff restriction jisse hum bhaag rahe hain; Newton ka aisa koi cap nahi. $g^{\prime }(y)=1+1.5y^2.$ $y^{(0)}=1$ par: $g(1)=1-1+0.5=0.5$, $g^{\prime }(1)=1+1.5=2.5$. $y^{(1)}=y^{(0)}-\frac{g(y^{(0)})}{g^{\prime }(y^{(0)})}=1-\frac{0.5}{2.5}=1-0.2=\mathrm{0.8.}$ Direction ka ek aur check: $y$ ko $1$ se decay karna chahiye, aur indeed $0.8<1$. ✅ (Aur Newton steps $g$ ke exact root ki taraf refine karte hain.)

$z=h\lambda =(0.5)(-2)=-1$. Backward Euler: $y_1=\frac{1}{1-z}=\frac{1}{2}=0.5$. Error $=0.5-0.3679=0.1321$. Trapezoidal: factor $R=\frac{1+z/2}{1-z/2}=\frac{1-0.5}{1+0.5}=\frac{0.5}{1.5}=0.3333$. Error $=0.3333-0.3679=-0.0346$ — lagbhag $4\times$ chhota. Kyun: trapezoidal second-order $O(h^2)$ hai; backward Euler first-order $O(h)$ hai. Dono yahan stable hain, par trapezoidal zyada accurate hai — yeh yaad dilata hai ki stability aur accuracy alag-alag axes hain.

(a) Stiff? Stiffness ratio $=\frac{|{\lambda }_{\max }|}{|{\lambda }_{\min }|}=\frac{40000}{1}=40000$ — modes time scale ke 4 orders of magnitude span karte hain. Haan, strongly stiff. (b) Forward Euler steps: fastest mode se governed: $h<2/40000=5\times 10^{-5}$. Steps $=20/(5\times 10^{-5})=400000$. Prohibitive hai. (c) Choice: ek L-stable implicit method jaise backward Euler. Reason: fastest mode ka $z=h\lambda \to -\infty$ kisi bhi reasonable $h$ ke liye; backward Euler ka factor $1/(1-z)\to 0$ use ek step mein damp kar deta hai, jabki explicit method nahi kar sakta. $h$ sirf slow $\lambda =-1$ mode ki accuracy ke liye choose karo (jaise $h\approx 0.05$–$0.1$), Newton se implicit equations solve karke.

Define karo $E(h)=\frac{1/(1+h)-e^{-h}}{e^{-h}}=\frac{e^h}{1+h}-1$ ($e^h$ se multiply karke). $h=0.2$ par: $e^{0.2}=1.2214$, $1+h=1.2$, ratio $=1.01784$, toh $E=0.01784\le 0.05$ ✅. $h=0.3$ par: $e^{0.3}=1.34986$, $1+h=1.3$, ratio $=1.03835$, toh $E=0.03835\le 0.05$ ✅. Dono pass karte hain; tolerance $h=0.3$ (aur thoda aage bhi) permit karta hai. Yeh kyun matter karta hai: stability free hone par, $h$ ko size karne wali sirf yahi accuracy budget hai — exactly woh situation jo poore topic ne promise ki thi.

A-stability L3.2 mein dikhayi gayi thi (poora left half-plane stable). Limit ke liye: ${\lim }_{|z|\to \infty }\left|\frac{1}{1-z}\right|={\lim }_{|z|\to \infty }\frac{1}{|1-z|}=0.$ Denominator bina bound ke badhta hai, toh factor vanish ho jaata hai. Dono conditions hold hoti hain → backward Euler L-stable hai. Trapezoidal se contrast karo, jiska factor $|-1|=1$ ki taraf jaata hai (A-stable par L-stable nahi).

Sab kuch ek number par reduce hota hai ::: $z=h\lambda$: $R(z)$ padho, $|R(z)|<1$ check karo, aur yaad rakho backward Euler ka $|R|<1$ left half-plane ke sabhi $z$ ke liye hota hai jabki forward Euler ka sirf $-1$ ke around radius $1$ ki disk ke andar hota hai.