Question bank — Implementing ODE solvers from scratch — Euler, RK4

True — ye IVP solvers hain; starting value ke bina aage step karne ke liye kuch nahi hoga, kyunki har step current point par slope padhti hai. Dekho scipy.integrate.solve_ivp jo y0 bhi demand karta hai.

True — Euler ki global error $O(h)$ hai, isliye error $h$ ke saath linearly scale hoti hai; $h$ ko do mein kaato aur error lagbhag $2$ se gir jaati hai.

False — RK4 $O(h^4)$ hai, isliye $h$ ko half karne se error lagbhag $2^4=16\times$ kam hoti hai, na ki $2\times$.

False — yeh fourth-order accurate hai, exact nahi; yeh $O(h^5)$ local terms ko discard karta hai aur global $O(h^4)$ error accumulate karta rehta hai, jab tak ki true solution degree $\le 4$ ka polynomial na ho jise RK4 exactly integrate kar sake.

True function evaluations mein — RK4, $f$ ko har step mein chaar baar call karta hai vs Euler ka ek, lekin yeh galat metric hai: equal accuracy par RK4 ko bahut kam steps chahiye aur overall jeetta hai.

False — $f$ yahan $t$ par depend karta hai (non-autonomous), isliye $k_2,k_3$ ko $t+\frac{h}{2}$ use karna chahiye aur $k_4$ ko $t+h$; unhe skip karna answer ko corrupt kar deta hai.

False — Simpson teen points par $(1,4,1)$ use karta hai; RK4 chaar stages par $(1,2,2,1)$ use karta hai, do midpoint evaluations ke saath jinke weights $2+2=4$ sum karte hain — analogous hain, identical nahi. Dekho Simpson's Rule.

True — forward difference $\frac{y_{n+1}-y_n}{h}\approx f(t_n,y_n)$ ko rearrange karne par exactly Euler update milta hai. Dekho Finite Difference Approximation of Derivatives.

False — stability limit se aage, bada $h$ explicit Euler aur RK4 ko blow up (infinity tak oscillate) kara deta hai, khaaskar stiff problems par. Dekho Numerical Stability and Stiff ODEs.

Galat — $k_3$ ko pichle stage $k_2$ use karna chahiye: $k_3=f(t+\frac{h}{2},y+\frac{h}{2}{k}_2)$. Stages ek chain hain; $k_1$ reuse karna RK4 ko lower order par collapse kar deta hai aur silently $O(h^4)$ accuracy destroy kar deta hai.

Galat — weights $1,2,2,1$ hain (sum $6$), isliye $6$ se divide karte hain: $y_{n+1}=y_n+\frac{h}{6}(k_1+2k_2+2k_3+k_4)$. 4 se plain average galat weighted mean deta hai aur accuracy kho deta hai.

Galat — Euler sirf left endpoint $(t_n,y_n)$ par sample karta hai aur assume karta hai ki yeh poore taraf tak hold karta hai; midpoint par sampling ek alag (better) method hoga.

Galat — aap $\frac{h}{2}{k}_1$ add karte hain, na ki $\frac{h}{2}$; $y$ ka nudge slope $k_1$ se scale hona chahiye, kyunki $y$ (slope × step) se change hota hai, sirf step se nahi.

Galat — Euler deta hai $y_1=1+1\cdot 1=2$, lagbhag $\approx 0.718$ ki error ke saath; sirf true solution $e$ ke barabar hai, aur Euler ki straight-line hop curving exponential ke neeche reh jaati hai.

Galat — RK4 ki local truncation error $O(h^5)$ hai; global error ($N=T/h$ steps mein accumulate hoti hui) $O(h^4)$ hai. Dono ko confuse mat karo.

Kyunki yeh poori step ke liye ek single left-endpoint slope par trust karta hai; agar curve upar bend karti hai toh straight line uske neeche hoti hai (aur vice versa), isliye errors ek direction mein accumulate hoti hain cancel hone ki bajaye.

Midpoint interval par average slope ko best represent karta hai; pehla midpoint estimate $k_2$ ($k_1$ se bana) ek doosre $k_3$ ($k_2$ se bana) se refine hota hai, aur dono ko weight $2$ dena Taylor series ko $h^4$ tak match karta hai.

Kyunki RK4 update ko Taylor series ke roop mein expand karna aur demand karna ki yeh true $y(t_n+h)$ se $h^4$ term tak agree kare, ye coefficients uniquely produce karta hai — ye free choice nahi hain. Dekho Taylor Series Expansion.

Target error $\epsilon$ par, Euler ki cost $1/\epsilon$ jaisi scale hoti hai jabki RK4, $1/{\epsilon }^{1/4}$ jaisi; small $\epsilon$ ke liye RK4 ko astronomically kam total evaluations chahiye.

Zyaadatar real $f(t,y)$ ka koi closed-form integral nahi hota, lekin slope $f$ khud hamesha computable hoti hai, isliye hum small computable hops mein aage badhte hain.

Kyunki $\frac{h}{6}(k_1+2k_2+2k_3+k_4)$ ek genuine weighted mean slope hai (weights sum to $6$) times $h$; koi bhi aur divisor step ko galat scale kar deta hai aur order break kar deta hai.

Koi progress nahi — $t_{n+1}=t_n$ aur $y_{n+1}=y_n$; solver hamesha ke liye ruk jaata hai, isliye positive step size zaroori hai.

Nahi — kyunki $f$, $t$ ko ignore karta hai, $t$, $t+\frac{h}{2}$, ya $t+h$ par evaluate karne se koi farak nahi padta; $t$-shifts sirf non-autonomous $f(t,y)$ ke liye matter karte hain.

Yeh exact ban jaata hai — true solution ek straight line $y_0+ct$ hai, aur Euler ki straight-line hop kisi bhi $h$ ke liye zero truncation error ke saath perfectly match karti hai.

Explicit methods ki ek bounded stability region hoti hai; agar $h$ times (bada negative) rate uske bahar hai, toh numerical solution oscillate karta hai aur grow karta hai bina bound ke, chahein true solution decay kare. Fix karo implicit methods ya adaptive stepping se. Dekho Numerical Stability and Stiff ODEs.

Yeh exact hota hai — RK4, polynomials ko degree $4$ tak bina error ke integrate karta hai, kyunki discarded terms $h^5$ se start hote hain aur ek cubic mein aise koi terms nahi hote.

Fixed-step Euler/RK4 singularity ke upar khushi se step kar lega aur garbage return karega; aapko blow-up detect karne aur use cross karne se mana karne ke liye adaptive step control chahiye. Dekho Adaptive Step Size (RK45 / Dormand–Prince).

Indefinitely nahi — ek point ke baad, bahut saare tiny increments sum karne se floating-point round-off dominate karta hai, isliye error bottom out ho jaati hai aur phir $h$ aur chota hone par slowly badh jaati hai.