ODE solvers — Euler's method (derivation, global error): Exercises

4.8.22 · D4 · Maths › Numerical Methods › ODE solvers — Euler's method (derivation, global error)

Neeche sab kuch sirf parent note Euler's Method mein banaye gaye tools use karta hai: update rule $y_{n + 1} = y_{n} + h f (x_{n}, y_{n}), x_{n + 1} = x_{n} + h,$ local truncation error $LTE = O (h^{2})$ , aur global error $O (h)$ .

Level 1 — Recognition

Goal: ek IVP ke pieces pehchano aur formula mein ek baar plug karo.

Exercise 1.1

IVP $y^{'} = 3 x - y$ diya hai jisme $y (1) = 2$ hai, inhe identify karo: function $f (x, y)$ , starting point $(x_{0}, y_{0})$ , aur start par solution curve ki slope.

Recall Solution

$f (x, y) = 3 x - y$ — yeh the right-hand side of the ODE hai, woh machine jo kisi bhi point ke liye slope return karta hai.
$(x_{0}, y_{0}) = (1, 2)$ — known starting value.
Slope at start $= f (1, 2) = 3 (1) - 2 = 1$ .

Humne kya kiya: humne kuch solve nahi kiya — sirf parts read off kiye. $f$ , seed point ko pehchaanna, aur slope ko ek baar evaluate karna — yahi poora skill hai L1 par.

Exercise 1.2

$y^{'} = x + y, y (0) = 1$ ke liye, $h = 0.1$ se ek Euler step lo. $y_{1}$ kya hai aur $x_{1}$ kahan hai?

Recall Solution

Slope at start: $f (0, 1) = 0 + 1 = 1$ . $y_{1} = y_{0} + h f (x_{0}, y_{0}) = 1 + 0.1 (1) = 1.1, x_{1} = 0 + 0.1 = 0.1.$ Kyun: "Naya = Purana + step × slope-at-old." Boxed formula ki ek application, kuch nahi iske siwa.

Level 2 — Application

Goal: recursion ko kaafi steps tak haath se chalao.

Exercise 2.1

$y^{'} = x + y, y (0) = 1$ ko $h = 0.1$ se teen steps ke liye solve karo. $y_{3} \approx y (0.3)$ nikalo.

Recall Solution

$n$	$x_{n}$	$y_{n}$	slope $f = x_{n} + y_{n}$	$y_{n + 1} = y_{n} + 0.1 f$
0	0.0	1.000	1.000	1.100
1	0.1	1.100	1.200	1.220
2	0.2	1.220	1.420	1.362

Toh $y_{3} = 1.362$ at $x_{3} = 0.3$ . Humne kya kiya: har row pichle $y$ ko reuse karti hai — yahi woh "self-bootstrapping" hai jo parent note ne describe ki thi. Yeh kaisa dikhta hai: teen chhote seedhe segments, har ek wahan se shuru hota hai jahan pichhla khatam hua (neeche figure dekho).

Figure — ODE solvers — Euler's method (derivation, global error)

Exercise 2.2

$y^{'} = - y, y (0) = 1$ ko $h = 0.25$ se chaar steps ke liye solve karo. $y_{4} \approx y (1)$ nikalo, phir true value $e^{- 1}$ se compare karo.

Recall Solution

Yahan $f (x, y) = - y$ hai, toh har step hai $y_{n + 1} = y_{n} + 0.25 (- y_{n}) = 0.75 y_{n}$ . $y_{1} = 0.75, y_{2} = 0.5625, y_{3} = 0.421875, y_{4} = 0.31640625.$ True value: $e^{- 1} = 0.367879 \dots$ Error $= 0.367879 - 0.316406 = 0.051473$ . Multiplication trick kyun? Jab $f = - y$ ho, toh update ek constant factor $0.75$ per step ban jaata hai — Euler continuous decay ko geometric decay $0.7 5^{n}$ mein badal deta hai. Yeh shrink karta hai, lekin bahut fast, kyunki har seedha step curving-up-toward-zero real solution ko undershoot karta hai.

Level 3 — Analysis

Goal: sirf numbers crunch karne ki jagah error size aur order ke baare mein reason karo.

Exercise 3.1

$y^{'} = x + y, y (0) = 1$ ke liye true solution $y = 2 e^{x} - x - 1$ hai. Parent note se $h = 0.1$ ka result $y_{2} = 1.22$ at $x = 0.2$ use karke, $x = 0.2$ par global error compute karo. Phir predict karo ki agar $h = 0.05$ use karo toh error kya hoga, aur kyun batao.

Recall Solution

True value: $y (0.2) = 2 e^{0.2} - 0.2 - 1 = 2 (1.221402 \dots) - 1.2 = 1.242806$ . Global error at $x = 0.2$ with $h = 0.1$ : $∣1.242806 - 1.22∣ = 0.022806.$ $h = 0.05$ ke liye prediction: Euler first order hai, GE $= O (h)$ . $h$ ko half karne se error bhi roughly half honi chahiye, toh hum expect karte hain $\approx 0.0114$ . Kyun: endpoint $x = 0.2$ fixed hai. $h$ ko $0.1$ se $0.05$ tak shrink karne par steps ki sankhya double ho jaati hai ( $2 \to 4$ ), lekin har local error $\frac{h ^{2}}{2} y^{''}$ factor of $4$ se shrink ho jaati hai. Net effect: $2 \times \frac{1}{4} = \frac{1}{2}$ . Ek power of $h$ survive karta hai → linear, halving behaviour.

Exercise 3.2

Global error bound hai $∣ y (x_{n}) - y_{n} ∣ \leq \frac{h M}{2 L} (e^{L (x_{n} - x_{0})} - 1)$ . $[0, 0.2]$ par $y^{'} = x + y$ ke liye $L = 1$ hai (kyunki $\partial f / \partial y = 1$ ) aur $∣ y^{''} ∣ \leq M$ . $y^{''} = 2 e^{x}$ diya hai, $[0, 0.2]$ par $M$ nikalo aur $h = 0.1$ ke saath $x = 0.2$ par bound evaluate karo. Kya actual error (Ex 3.1) usme fit hoti hai?

Recall Solution

$y^{''} = 2 e^{x}$ increasing hai, isliye $[0, 0.2]$ par iska max right end par hai: $M = 2 e^{0.2} = 2.442806$ . Bound at $x_{n} = 0.2$ : $\frac{( 0.1 ) ( 2.442806 )}{2 ( 1 )} (e^{1 (0.2)} - 1) = 0.1221403 \times (1.221403 - 1) = 0.1221403 \times 0.221403 = 0.027043.$ Actual error thi $0.022806$ . Kyunki $0.022806 \leq 0.027043$ , actual error bound ke andar hai. ✓ Bound bada kyun hai: yeh ek worst-case guarantee hai — yeh sabse bada $∣ y^{''} ∣$ use karta hai aur assume karta hai ki errors poori Lipschitz rate $e^{L (x - x_{0})}$ par amplify hoti hain. Real error usually thodi chhoti hoti hai.

Level 4 — Synthesis

Goal: stepping, error reasoning, aur ek doosre concept ko combine karo.

Exercise 4.1

$y^{'} = - 2 y, y (0) = 1$ ko $h = 0.5$ se do steps leke $x = 1$ tak solve karo. $y_{2}$ report karo, true value $e^{- 2}$ se compare karo, aur stability quantity $∣1 + hλ ∣$ (jisme $λ = - 2$ ) use karke explain karo ki answer itna galat kyun hai.

Recall Solution

Update: $y_{n + 1} = y_{n} + 0.5 (- 2 y_{n}) = y_{n} (1 + 0.5 (- 2)) = y_{n} (1 - 1) = 0$ . $y_{1} = 1 (0) = 0, y_{2} = 0 (0) = 0.$ True: $e^{- 2} = 0.135335$ . Error $= 0.135335$ — answer ke relative mein enormous. Stability reasoning: decay problem $y^{'} = λ y$ ke liye, ek Euler step factor $1 + hλ$ se multiply karta hai. Yahan $1 + hλ = 1 + 0.5 (- 2) = 0$ . Exactly $0$ ka factor solution ko ek hi step mein zero par collapse kar deta hai — start par tangent itni steep hai ki seedha axis ke paar overshoot kar jaati hai. Yeh stability failure hai, sirf accuracy nahi: Numerical Stability dekho. Real decay ke liye humein $∣1 + hλ ∣ < 1$ chahiye, yaani yahan $h < 1$ ; $h = 0.5$ bilkul us edge case par hai jahan factor vanish ho jaata hai.

Exercise 4.2

Usi $y^{'} = - 2 y, y (0) = 1$ ke liye, $h = 0.25$ choose karo aur $x = 1$ tak chaar steps lo. $y_{4}$ compute karo, $e^{- 2}$ se compare karo, aur confirm karo ki stability factor ab safely $1$ se neeche hai.

Recall Solution

Factor per step: $1 + 0.25 (- 2) = 0.5$ . Kyunki $∣0.5∣ < 1$ , numerical solution sensibly decay karta hai. $y_{4} = 1 \times 0. 5^{4} = 0.0625.$ True: $e^{- 2} = 0.135335$ . Error $= 0.072835$ — ab bhi bada, lekin ab solution exactly zero par collapse hone ki jagah oopar se zero ki taraf decay karta hai. Chhota $h$ qualitatively correct behaviour restore kar diya. Kyun: factor $∣1 + hλ ∣ = 0.5 < 1$ guarantee karta hai ki har step $y$ ko overshoot ke bina shrink karta hai, isliye numerical solution ab true exponential decay jaisa dikhta hai chahe magnitude abhi accurate na ho.

Level 5 — Mastery

Goal: ek general result derive karo aur use ek design decision ke liye use karo.

Exercise 5.1

Tumhe Euler se $[0, 1]$ par ek IVP solve karni hai, aur $h = 0.1$ ke saath $x = 1$ par tumhara global error $E_{1} = 0.040$ nikla. Tumhari tolerance $0.005$ hai. Sirf method ke order ko use karke, sabse bada $h$ nikalo jo tolerance meet kare, aur steps ki sankhya $N$ batao.

Recall Solution

First order ⇒ global error linearly scale karta hai: kisi constant $C$ ke liye $E (h) \approx C h$ . $E_{1} = C (0.1) = 0.040$ se hume milta hai $C = 0.4$ . Chahiye $C h \leq 0.005 \Rightarrow 0.4 h \leq 0.005 \Rightarrow h \leq 0.0125$ . Sabse bada convenient $h$ jo interval ko divide kare: $h = 0.0125$ lo, jisse milta hai $N = \frac{1 - 0}{0.0125} = 80 steps .$ Humne kya kiya: humne sirf yeh fact use kiya ki GE $\propto h$ hai, ek measured error se us step size tak extrapolate karne ke liye jo ek target hit kare. Yeh kyun kaam karta hai: constant $C$ (jisme $M$ , $L$ , $e^{L (x - x_{0})}$ factor chhupa hai) ratio mein cancel ho jaata hai — order akela scaling law deta hai. Isliye practically " $O (h)$ " jaanna valuable hai.

Exercise 5.2

Cost compare karo: ek hypothetical second-order method ka GE $\approx C_{2} h^{2}$ hai aur usne $h = 0.1$ par wahi $E_{1} = 0.040$ produce kiya. Tolerance $0.005$ ke liye usse jo $h$ aur step count chahiye woh nikalo, aur Euler ke 80 steps ke versus lesson batao.

Recall Solution

Second order: $E (h) = C_{2} h^{2}$ . $C_{2} (0.1)^{2} = 0.040$ se $\Rightarrow C_{2} = 4.0$ . Chahiye $4.0 h^{2} \leq 0.005 \Rightarrow h^{2} \leq 0.00125 \Rightarrow h \leq 0.035355$ . Steps: $N = \frac{1}{0.035355} = 28.28 \dots \Rightarrow$ round up karke $N = 29$ steps ( $h \approx 0.03448$ ). Lesson: accuracy mein $8$ ka factor gain karne ke liye ( $0.040 \to 0.005$ ), Euler ko $8 \times$ zyada steps chahiye ( $80$ ), lekin second-order method ko sirf $8 \approx 2.83 \times$ zyada chahiye ( $29$ ). Jaise tolerances tight hoti hain, higher order dramatically faaydemand hota hai — yahi Runge-Kutta Methods ki motivation hai.

Recall Paanch levels ka ek-line summary

L1 $f, x_{0}, y_{0}$ read off karo → L2 table crank karo → L3 global error measure karo aur dekho ki yeh half hoti hai → L4 stability accuracy ko gate karta hai → L5 order $p$ ek scaling law deta hai jo step-count decisions drive karta hai.

Connections

Taylor Series Expansion — L3 mein use ki gayi $O (h^{2})$ local error ka origin.
Numerical Stability — woh $∣1 + hλ ∣$ threshold jo L4 decide karta hai.
Runge-Kutta Methods — higher-order methods jo L5 se motivated hain.
Backward Euler & Implicit Methods — woh "new-slope" method jo L1 trap describe karta hai.
Lipschitz Continuity — Ex 3.2 ke error bound mein constant $L$ .
Finite Difference Approximations — update ke peeche forward difference.