4.8.22 · D3ODE solvers — Euler's method (derivation, global error)
Intuition Yeh page kyun exist karti haiParent note ne Euler's method derive kiya tha aur uska error law bataya tha. Yeh page woh mushkil kaam karti hai: hum method ko har tarah ke input se haath se chalate hain — badhte solutions, ghatte wale, negative slopes, ek slope jo zero hai, ek itna bada step ki method bigad jaata hai, aur ek real-world word problem. Agar tum yeh sab follow kar sako, toh koi bhi exam version tumhe surprise nahi kar sakta.
Poori page ke liye ek rule, parent se: ==New = Old + step × slope-at-old==, yaani
y n + 1 = y n + h f ( x n , y n ) , x n + 1 = x n + h .
Koi bhi number se pehle, ek reminder ki har symbol ka kya matlab hai, taaki kuch bina explain ke use na ho:
Definition Woh chaar symbols jo hum har example mein use karenge
x n current horizontal position (socho: hum road par kitna chal chuke hain).y n humara current estimate (hum sochte hain curve kahan hai).h step size : woh fixed horizontal distance jitna hum har baar jump karte hain.f ( x n , y n ) slope theek wahan jahan hum khade hain , jo differential equation hamare haath mein deta hai.
"Slope" yahan matlab hai rise per unit run — har unit horizontal travel mein kitne units height milti hai. Ise apne run h rise milti hai, h f y n
Definition Ek aur symbol jo hum milenge: growth rateλ y ′ = λ y λ linear test equation kehte hain, aur hum ise isliye use karte hain kyunki iska true solution exactly y = y 0 e λ x Euler ko ek known answer se check kar sakte hain.
Agar λ < 0 0 decay karta hai (jaise cooling ya radioactive decay).
Agar λ > 0 grow karta hai.
f = λ y y n + 1 = y n + hλ y n = ( 1 + hλ ) y n 1 + hλ multiply karta hai. Woh ek number — ise stability factor kaho — sab kuch decide karta hai: agar ∣1 + hλ ∣ < 1 ∣1 + hλ ∣ ≥ 1
Yeh topic jo bhi problem throw kar sakta hai woh in cells mein se ek (ya blend) hai. Table mein cell, kya cheez use distinct banati hai, aur neeche konsa example use cover karta hai — yeh sab likha hai.
Cell
Kya cheez use special banati hai
Covered by
A. Growing solution, positive slope f > 0 y Example 1
B. Decaying solution, negative slope f < 0 y Example 2
C. Step-halving / order check global error ∝ h
Example 3
D. Degenerate: slope = 0 f = 0 flat step, y Example 4
E. Limiting / unstable: h $
1+h\lambda
F. Autonomous non-linear f y Example 6
G. Real-world word problem physics → IVP → Euler translate karo
Example 7
H. Exam twist: back-solve for h y 0 answer diya hua hai, input dhundho
Example 8
Ab hum har cell chalte hain. Poori tarah, derivation aur error law parent topic se aate hain aur Taylor Series Expansion par based hain.
y ′ = x + y , y ( 0 ) = 1 h = 0.1 y ( 0.3 ) Forecast: true solution 2 e x − x − 1 upar ki taraf curve karta hai (woh convex hai). Euler seedhi tangents follow karta hai, isliye use har step mein true curve ke neeche aana chahiye. Andaza: humara y ( 0.3 ) ≈ 1.4
Step 1 — shuru. x 0 = 0 , y 0 = 1 f ( 0 , 1 ) = 0 + 1 = 1 Kyun? ODE hi slope-machine hai ; current point use karo.
y 1 = 1 + 0.1 ⋅ 1 = 1.1 , x 1 = 0.1
Step 2. Slope f ( 0.1 , 1.1 ) = 0.1 + 1.1 = 1.2 Kyun? Naye point par slope phir se measure karo — road bend ho gayi, toh purana slope stale hai.
y 2 = 1.1 + 0.1 ⋅ 1.2 = 1.22 , x 2 = 0.2
Step 3. Slope f ( 0.2 , 1.22 ) = 0.2 + 1.22 = 1.42 Kyun? Wahi rule teesri baar — agle step se pehle freshly landed point ( x 2 , y 2 ) f y 3 = 1.22 + 0.1 ⋅ 1.42 = 1.362 , x 3 = 0.3
Verify: true value 2 e 0.3 − 0.3 − 1 = 1.39972 1.362 neeche hai (error ≈ 0.038
Neeche ka figure hamare teen orange Euler points ko teal true curve ke against plot karta hai. Dekho ki orange polyline har x neeche baithti hai, aur x = 0.3
y ′ = − y , y ( 0 ) = 2 h = 0.2 y ( 0.4 ) f = − y λ = − 1 Forecast: true solution 2 e − x 0 subtract karta hai. Andaza: 2 e − 0.4 ≈ 1.34 neeche baithega.
Step 1. f ( 0 , 2 ) = − 2 y 1 = 2 + 0.2 ( − 2 ) = 1.6 x 1 = 0.2 Kyun? Slope negative hai, isliye rise h f = − 0.4 fall hai.
Step 2. f ( 0.2 , 1.6 ) = − 1.6 y 2 = 1.6 + 0.2 ( − 1.6 ) = 1.28 x 2 = 0.4 Kyun? Naye, chhoti height y 1 = 1.6 − 1.6 − 2
Verify: true 2 e − 0.4 = 1.34064 1.28 ≈ 0.061 stable hai kyunki stability factor ∣1 + hλ ∣ = ∣1 + 0.2 ( − 1 ) ∣ = 0.8 < 1 0
y ′ = x + y , y ( 0 ) = 1 x = 0.2 h = 0.1 h = 0.05 Forecast: parent ne prove kiya tha ki global error = O ( h ) . Toh h ≈ 0.5
h = 0.1 y ( 0.2 ) = 1.22 2 e 0.2 − 1.2 = 1.24281 E 0.1 = 0.02281
h = 0.05
y 1 = 1 + 0.05 ( 0 + 1 ) = 1.05 0.05 f = 0.05 + 1.05 = 1.1 y 2 = 1.05 + 0.05 ( 1.1 ) = 1.105 0.10 f = 0.10 + 1.105 = 1.205 y 3 = 1.105 + 0.05 ( 1.205 ) = 1.16525 0.15 f = 0.15 + 1.16525 = 1.31525 y 4 = 1.16525 + 0.05 ( 1.31525 ) = 1.2310125 0.20
Error E 0.05 = 1.24281 − 1.23101 = 0.01180
Verify: ratio E 0.05 / E 0.1 = 0.01180/0.02281 = 0.517 ≈ 2 1 Exactly 0.5 O ( h ) leading-order statement hai; tiny higher-order terms use exact half se thoda hata deti hain. Jab h → 0 0.5
y ′ = x 2 − 1 , y ( 0 ) = 0.5 h = 1 y ( 2 ) bilkul zero hoti hai, toh y
Forecast: x = 1 1 2 − 1 = 0 x = 1 y
Step 1. f ( 0 , 0.5 ) = 0 2 − 1 = − 1 y 1 = 0.5 + 1 ⋅ ( − 1 ) = − 0.5 x 1 = 1 Girne ki wajah? Slope negative thi, isliye height giri.
Step 2. f ( 1 , − 0.5 ) = 1 2 − 1 = 0 y 2 = − 0.5 + 1 ⋅ 0 = − 0.5 x 2 = 2 Unchanged kyun? ==Zero slope ⇒ zero rise ⇒ y n + 1 = y n
Verify: exactly solve karo: y = 3 x 3 − x + 0.5 y ( 2 ) = 3 8 − 2 + 0.5 = 1.16 6 h = 1 − 0.5 h mechanics correct hai: y f = 0
Figure do orange steps trace karta hai. Pehla segment neeche slope karta hai (slope − 1 horizontal hai kyunki waahan f = 0 y 2 = y 1 hai .
y ′ = − 2 y , y ( 0 ) = 1 h = 1.5 y ( 3 ) λ = − 2 Forecast: true solution e − 2 x ≈ 0 stability factor hai 1 + hλ = 1 + 1.5 ( − 2 ) = − 2 ∣ − 2∣ = 2 > 1 1 grow karenge aur sign flip karenge . Forecast: oscillating, exploding values — true decay ke bilkul ulta.
Step 1. y 1 = 1 + 1.5 ( − 2 ) ( 1 ) = 1 − 3 = − 2 x = 1.5 Step 2. y 2 = − 2 + 1.5 ( − 2 ) ( − 2 ) = − 2 + 6 = 4 x = 3
Verify: har step stability factor 1 + hλ = − 2 1 → − 2 → 4 y n = ( − 2 ) n e − 6 = 0.00248 4 stability failure hai, sirf accuracy error nahi. ✓
Ilaaj: ∣1 + hλ ∣ < 1 ∣1 − 2 h ∣ < 1 ⇒ 0 < h < 1 h < 1 h = 1.5 Numerical Stability aur Backward Euler & Implicit Methods ka gateway hai, jo yahan kisi bhi h
Figure smooth teal decay curve e − 2 x 0 1 → − 2 → 4 2 > 1 sign-flipping dikhata hai. Jab tumhare numbers alternate sign karte hain aur badhte hain, yahi picture ho rahi hai.
Common mistake "Example 5 mein error ka matlab sirf answer ke paas chhotah Kyun sahi lagta hai: O ( h ) h Fix: stability threshold ke neeche, haan. Upar it, iteration diverge karta hai chahe accuracy law kuch bhi predict kare — accuracy aur stability alag guarantees hain.
y ′ = y 2 , y ( 0 ) = 1 h = 0.1 y ( 0.2 ) f y square par depend karta hai — ek non-linear slope machine. (True solution y = 1 − x 1 x = 1
Forecast: y 2 y pehle ki chhoti slope use karke, true steep climb ko undershoot karega.
Step 1. f ( 0 , 1 ) = 1 2 = 1 y 1 = 1 + 0.1 ( 1 ) = 1.1 0.1 Square kyun? Bas f ( y 0 ) 2 = 1
Step 2. f ( 0.1 , 1.1 ) = 1. 1 2 = 1.21 y 2 = 1.1 + 0.1 ( 1.21 ) = 1.221 0.2 Kyun? Naye height y 1 = 1.1 f 1.21 > 1
Verify: true 1 − 0.2 1 = 1.25 1.221 ≈ 0.029 recipe mein kuch nahi badlati — tum bas current point f
Worked example Ek cup coffee9 0 ∘ C 2 0 ∘ C d t d T = − 0.1 ( T − 20 ) t h = 10 t = 20 Translate: f ( t , T ) = − 0.1 ( T − 20 ) T 0 = 90 taraf cool hoti hai.
Forecast: true cooling 20 + 70 e − 0.1 t t = 20 20 + 70 e − 2 ≈ 29. 5 ∘ h = 10 29.5 neeche value expect karo.
Step 1. f ( 0 , 90 ) = − 0.1 ( 90 − 20 ) = − 7 T 1 = 90 + 10 ( − 7 ) = 2 0 ∘ t = 10 Itna drastic kyun? Shuru ki slope steep hai aur hum ne use poore 10
Step 2. f ( 10 , 20 ) = − 0.1 ( 20 − 20 ) = 0 T 2 = 20 + 10 ( 0 ) = 2 0 ∘ t = 20 Flat kyun? 2 0 ∘ 0
Verify: Euler 2 0 ∘ 29. 5 ∘ ≈ 9. 5 ∘ h = 10 ∣1 + hλ ∣ = ∣1 + 10 ( − 0.1 ) ∣ = ∣0∣ = 0 ∘ h h = 2 29. 5 ∘
y ′ = x + y , y ( 0 ) = 1 h y 1 = 1.15 h Forecast: hum recipe jaante hain, toh use ulta karo. Shuru mein slope f ( 0 , 1 ) = 1 y 1 = 1 + h ⋅ 1 y 1 = 1.15 h 0.15
Step 1. Update likho: y 1 = y 0 + h f ( x 0 , y 0 ) = 1 + h ( 1 ) Kyun? Ek step, slope jaani hui — h
Step 2. Solve karo 1 + h = 1.15 ⇒ h = 0.15
Verify: plug back karo: 1 + 0.15 ( 1 ) = 1.15 equation samajhte ho, black box nahi.
Recall Konsa cell sabse mushkil tha — aur kyun?Cell E (instability). Har doosra cell sirf accuracy error produce karta hai, jo h O ( h ) qualitative failure produce karta hai: numbers ek aisi truth se diverge karte hain jo decay kar rahi hai. Lesson: h accuracy aur stability dono constraints satisfy karni chahiye.
Mnemonic Compute karne se pehle scenario checklist"Sign, Zero, Size." f Sign check karo (chadh raha hai ya gir raha hai?), koi bhi Zero slope spot karo (flat step), aur confirm karo ki tumhara Size of h ∣1 + hλ ∣ < 1
Positive slope wala Euler estimate ek convex true curve ke against — upar ya neeche? Neeche (tangents ek convex curve ko undercut karte hain).
y ′ = − y h Koi bhi h ∣1 − h ∣ < 1 0 < h < 2
f = 0 y Kuch nahi — y n + 1 = y n
h Lagbhag 2 1 O ( h )
y 1 = y 0 + h f y 1 h Linear equation h = ( y 1 − y 0 ) / f ( x 0 , y 0 )
Parent topic — derivation & error law jo in examples mein use hote hainTaylor Series Expansion — har example ko milne wala local O ( h 2 ) Runge-Kutta Methods — upar dekhe gaye errors ko drastically kam kar deta (O ( h 4 ) Backward Euler & Implicit Methods — Example 5 mein kisi bhi h Numerical Stability — ∣1 + hλ ∣ < 1 Lipschitz Continuity — Example 3 ke peeche error bound guarantee karta haiFinite Difference Approximations — yahan har step ka forward-difference view
Euler update new = old + h times slope
Cell A growing undershoot
Cell D flat step y unchanged
accuracy error only order h
Cell E instability diverges
