Exercises — Adaptive step-size — RK45, error control

(a) Ek embedded Runge–Kutta pair (RK45 / Fehlberg / Dormand–Prince). (b) $\epsilon =\Vert y^{(5)}-y^{(4)}\Vert$. Ek system ke liye yeh tolerance-scaled RMS norm hai (accept karo jab $\epsilon \le 1$); single equation ke liye bas $|y^{(5)}-y^{(4)}|$ hai. Yeh free hai kyunki yeh har expensive $f$-evaluation reuse karta hai aur sirf ek sasta weighted difference $h{\sum }_i(b_i-b_i^{\ast })k_i$ add karta hai. (c) Controller $p=4$ use karta hai: error estimate essentially 4th-order result ke local error ko measure karta hai. Aap fir bhi $y^{(5)}$ se advance karte ho (local extrapolation), lekin update mein exponent $1/(p+1)=1/5$ hai $p=4$ use karte hue. Dekho Order of a Numerical Method.

Rule: accept karo agar aur sirf agar $\epsilon \le \text{tol}$.

• A: $2\times 10^{-7}\le 10^{-6}$ → accept.
• B: $10^{-6}\le 10^{-6}$ → accept (equality pass hoti hai).
• C: $9\times 10^{-6}>10^{-6}$ → reject, same $t_n$ se step redo karo.

(a) $\epsilon =5\times 10^{-8}<10^{-6}$ → accept. (b) Ratio $\frac{\text{tol}}{\epsilon }=\frac{10^{-6}}{5\times 10^{-8}}=20$. $h_{\text{new}}=0.9\times 0.20\times 20^{1/5}=0.18\times 1.8206=\mathrm{0.3277.}$ Hamare paas error spare tha, isliye hum grow karte hain — lekin sirf $20^{1/5}\approx 1.82$ se, 20 se nahi, kyunki local error $h^5$ ki tarah scale hota hai.

(a) $8\times 10^{-6}>10^{-6}$ → reject. (b) Ratio $=\frac{10^{-6}}{8\times 10^{-6}}=0.125$. $h_{\text{new}}=0.9\times 0.15\times 0.125^{1/5}=0.135\times 0.6598=\mathrm{0.08908.}$ Same $t_n$ se $h_{\text{new}}\approx 0.0891$ ke saath retry karo.

(a) $\text{tol}=\text{atol}+\text{rtol}|y_n|=10^{-8}+10^{-6}\times 40=10^{-8}+4\times 10^{-5}\approx 4.001\times 10^{-5}$. (b) $\epsilon =3\times 10^{-5}<4.001\times 10^{-5}$ → accept (relative part ne bachaya). (c) Ratio $=4.001\times 10^{-5}/3\times 10^{-5}=1.3337$. $h_{\text{new}}=0.9\times 0.10\times 1.3337^{1/5}=0.09\times 1.0593=\mathrm{0.09534.}$ Note karo ki sirf atol ($10^{-8}$) use karne se yeh perfectly good step reject ho jaata — dekho Local vs Global Truncation Error ki kyun absolute-only tolerances misbehave karti hain jab $|y|$ bada ho.

$\epsilon =\text{tol}$ set karo. Toh raw update factor hai $r=S{\left(\frac{\text{tol}}{\epsilon }\right)}^{1/5}=S\cdot 1=S=0.9,h_{\text{new}}=0.9h.$ Predicted error $h^5$ ki tarah scale hota hai, isliye ${\epsilon }_{\text{next}}\approx C{h}_{\text{new}}^5=C(0.9h{)}^5=0.9^5(C{h}^5)=0.9^5\epsilon =0.9^5\text{tol}.$ $0.9^5=0.59049$, isliye ${\epsilon }_{\text{next}}\approx 0.590\text{tol}$ — comfortably budget ke neeche. Yahi safety factor khareedta hai: jab hum tol par dead-on bhi hit karein, agla step ~59% of tol par land karne ka prediction hai, toh pehli try mein accept hone ki bahut zyada probability hai. Yeh wahi trade-off hai jo opening figure mein draw kiya gaya hai.

Shrink factor $=(\text{tol}/\epsilon {)}^{\text{exp}}=(1/4{)}^{\text{exp}}=0.25^{\text{exp}}$.

• Student X: $0.25^{1/5}=0.75786$ → $h$ ~76% tak girta hai.
• Student Y: $0.25^{1/2}=0.5$ → $h$ half ho jaata hai. Kyunki true local error $\propto h^5$ hai, factor-4 error overshoot ko fix karne ke liye sirf $h\to 4^{-1/5}h\approx 0.758h$ chahiye. Student Y ka $1/2$ exponent over-shrinks karke $0.5h$ par le jaata hai, steps waste karta hai: retried step $\epsilon \approx C(0.5h{)}^5=4\cdot 0.5^5\text{tol}=0.125\text{tol}$ ke saath wapas aayega — 8× under budget, matlab bahut agla step turant grow karne ki koshish karega. Galat exponent ⇒ oscillating, inefficient stepping. Dekho Taylor Series Expansion ki $h^5$ kahan se aata hai.

(a) Ratio $=10^{-6}/2\times 10^{-13}=5\times 10^6$. $r_{\text{raw}}=0.9\times (5\times 10^6{)}^{1/5}=0.9\times 21.867=\mathrm{19.68.}$ (b) $19.68>f_{\max }=5$ → clamp fire karta hai (yaad karo $f_{\max }\approx 5$ opening definitions se). (c) $h_{\text{new}}=h\cdot f_{\max }=0.04\times 5=0.20$. $h^5$ error model sirf locally trustworthy hai; ~20× jump ek aise region mein leap kar sakta hai jahan completely alag curvature ho, isliye hum growth ko 5× par cap karte hain.

Attempt 1: ${\epsilon }_1=6.25\times 10^{-6}>10^{-6}$ → reject. $t=0$ par raho. Ratio $=10^{-6}/6.25\times 10^{-6}=0.16$. Factor $=0.9\times 0.16^{1/5}=0.9\times 0.69314=0.62383$ (clamps ke andar). $h\to 0.10\times 0.62383=\mathrm{0.062383.}$ Attempt 2 ($t=0$ se, $h=0.062383$ ke saath): ${\epsilon }_2=4\times 10^{-7}<10^{-6}$ → accept. Advance: $t=0+0.062383=0.062383$. Ab agले interval ke liye step pick karo: ratio $=10^{-6}/4\times 10^{-7}=2.5$, factor $=0.9\times 2.5^{1/5}=0.9\times 1.2011=1.0810$. $h_{\text{next}}=0.062383\times 1.0810=\mathrm{0.067437.}$ Summary: ek baar reject (t 0 par raha), doosri try mein accept (t ≈0.06238 par advance hua), phir agले interval ke liye thoda grow karke ≈0.06744 hua.

(a) Attempts $=80+12=92$. (b) $f$ calls $=92\times 6=552$. (c) Wasted $=12\times 6=72$ calls; fraction $=72/552=0.13043\dots \approx 13.0\%$. ~10–15% rejection rate healthy hai — iska matlab controller $h$ ko tolerance ke edge tak push kar raha hai, timidly chhota nahi reh raha.

Attempt 1 ($h=0.20$): $\epsilon =3.2\times 10^{-5}>10^{-6}$ → reject. Ratio $=10^{-6}/3.2\times 10^{-5}=0.03125$. Factor $=0.9\times 0.03125^{1/5}=0.9\times 0.5=0.45$ (clamps ke andar, kyunki $0.45>f_{\min }=0.2$). $h\to 0.20\times 0.45=\mathrm{0.09.}$ Attempt 2 ($h=0.09$): $\epsilon =3.5\times 10^{-6}>10^{-6}$ → reject. Ratio $=10^{-6}/3.5\times 10^{-6}=0.285714$. Factor $=0.9\times 0.285714^{1/5}=0.9\times 0.778087=0.700278$. $h\to 0.09\times 0.700278=\mathrm{0.063025.}$ Attempt 3 ($h=0.063025$): $\epsilon =5\times 10^{-7}<10^{-6}$ → accept. Advance: $t=0.063025$. Final: do rejections aur ek acceptance ke baad $t\approx 0.06303$. Note karo ki clamp yahan kabhi fire nahi hua; har shrink previous $h$ ke $f_{\min }=0.2$ se kaafi upar raha.

(a) Ratio $=10^{-6}/0.3=3.3\overline{3}\times 10^{-6}$. Raw factor $=0.9\times (3.333\times 10^{-6}{)}^{1/5}$. $(3.333\times 10^{-6}{)}^{1/5}=0.080274$, toh raw $=0.9\times 0.080274=0.072247$. (b) $0.072247<f_{\min }=0.2$ → lower clamp fire karta hai. (c) $h_{\text{new}}=h\times f_{\min }=0.30\times 0.2=0.06$. (d) $f_{\min }$ se kaafi neeche raw factor baar baar aana stiff region ki fingerprint hai — yahan ek explicit adaptive method thrash kar sakta hai; implicit solver consider karo (dekho Stiff ODEs and Stability).

Ratio $=\text{tol}/\epsilon =32$.

• Galat ($p=5$, exponent $1/6$): factor $=32^{1/6}=1.7818$, $h_{\text{new}}=0.17818$.
• Sahi ($p=4$, exponent $1/5$): factor $=32^{1/5}=2.0$, $h_{\text{new}}=0.20$. Sahi controller zyada grow karta hai (factor 2 vs 1.78). Reason: error estimate $\epsilon$ 4th-order solution ke error ko track karta hai, jo $h^5$ ki tarah scale hota hai; 32× error surplus $32^{1/5}=2$× step increase allow karta hai. $p=5$ use karne se underestimate hota hai ki aap kitni tezi se grow kar sakte ho — safe hai lekin wasteful. Mnemonic wahi rehta hai: "Tol Over Error, to the Fifth-root."

(a) Maximum shrink: $h_{\text{raw}}=h\cdot f_{\min }=8\times 10^{-6}\times 0.2=1.6\times 10^{-6}$. (b) $1.6\times 10^{-6}<h_{\min }=2\times 10^{-6}$ → yeh floor violate karta hai. (c) Solver ko silently sub-floor step nahi lena chahiye. Standard behaviour (jaise scipy ka solve_ivp) ya toh $h$ ko $h_{\min }$ tak clamp karta hai aur thoda-sa-zyada error warning ke saath accept karta hai, ya abort karta hai "step size too small" failure ke saath. Blindly shrink karte rehna dangerous hai kyunki (i) $h$ machine $\epsilon \approx 2.2\times 10^{-16}$ ki taraf underflow kar sakta hai, jahan $t_n+h$ round back ho jaata hai $t_n$ par aur integration forever stall ho jaata hai; aur (ii) hajaaron micro-steps mein round-off accumulate hota hai jo us accuracy ko swamp kar deta hai jiska aap peecha kar rahe the. Repeatedly $h_{\min }$ hit karna implicit method mein switch karne ka classic signal hai jo stiff problems ke liye built ho.