Worked examples — Runge-Kutta 4th order (RK4) — derivation

4.8.24 · D3 · Maths › Numerical Methods › Runge-Kutta 4th order (RK4) — derivation

Is deep-dive mein parent RK4 derivation ko practice mein drill kiya gaya hai. Hum $y^{\prime }=f(x,y)$ ka ek step solve karte hain, chaar slopes $k_1,k_2,k_3,k_4$ aur update $y_{n+1}=y_n+\frac{1}{6}(k_1+2k_2+2k_3+k_4)$ use karke. Kuch naya assume nahi kiya gaya — neeche har symbol parent note mein build kiya gaya tha. Yahan hum ensure karte hain ki koi bhi case aapko surprise na kare.

Kisi bhi number se pehle, chaliye woh machine yaad kar lete hain jisme hum data feed kar rahe hain, taaki cold reader bhi follow kar sake.

Ise dobara kyun list kiya? Kyunki RK4 ki 90% galtiyan kisi stage mein galat $x$ ya $y$ plug karne ki hoti hain. Is box ko padhte waqt khula rakhein.

Kyunki hamara promise visual-first hai, neeche har worked example apne step-diagram ke saath aata hai: actual curve par slope samples ki ek picture, taaki arithmetic kabhi geometry se alag na ho. Figure pehle dekhein, phir numbers padhein.

---

Scenario matrix

Har RK4 problem jo aap kabhi dekhenge woh neeche ke kisi cell mein fit hoga (ya unka mix hoga). Jo examples follow karte hain unhe us cell ke saath label kiya gaya hai jisme woh fit hote hain, taaki saath mein poora grid cover ho jaye.

Cell	Kya cheez ise alag banati hai	Example
A. Pure-$x$ integrand $f=f(x)$	RK4 Simpson's rule mein collapse ho jata hai	Ex 1
B. Linear mixed $f=x+y$	Dono variables move karte hain; smooth growth	Ex 2
C. Negative / decaying slope $f=-y$	Signs flip hote hain; $y$ shrink karta hai, overshoot nahi hona chahiye	Ex 3
D. Nonlinear in $y$ $f=y^2$ etc.	Stages bahut differ karte hain; sensitivity	Ex 4
E1. Zero slope everywhere $f\equiv 0$	Kuch nahi hilta; sanity floor	Ex 5a
E2. Zero only at start $k_1=0$, rest $\ne 0$	Motion phir bhi hota hai; jaldi mat ruko	Ex 5b
F. Multi-step march ($h$ ke do steps)	Errors chain hoti hain; $y_2$ from $y_1$	Ex 6
G. Large step / limiting $h$ bada	Jahan accuracy strain karti hai; Taylor view	Ex 7
H. Word problem (real world) cooling / motion	Story ko $f$ mein translate karo, units rakho	Ex 8
I. Exam twist $f$ nonlinearly $x$ aur $y$ par depend karta hai, awkward numbers	Careful arithmetic, no calculator luxuries	Ex 9

---

Cell A — pure-$x$ integrand (RK4 = Simpson)

Forecast: Exact answer hai ${\int }_0^1{x}^{2}dx=\frac{1}{3}$. RK4 ise exactly nail karna chahiye — padhne se pehle guess karo kyun. (Hint: Simpson cubics par exact hai.)

Figure 1 (Cell A): parabola $y=\frac{x^3}{3}$ jisme teen sample $x$-values (left, midpoint, right) mark hain; kyunki $f$ ko $y$ ignore karta hai, dono midpoint samples coincide karte hain.

1. $k_1=hf(0)=1\cdot 0^2=0$. Yeh step kyun? Left-edge slope. Kyunki $f$, $y$ ko ignore karta hai, bas $x=0$ plug karo.
2. $k_2=1\cdot f(0.5)=0.25$. Kyun? Midpoint $x=0+\frac{1}{2}$. $y$-shift $y_n+\frac{k_1}{2}$ irrelevant hai kyunki $f$ mein $y$ nahi hai.
3. $k_3=1\cdot f(0.5)=0.25$. Kyun? Same midpoint, same value — pure-$x$ problems mein $y$-dependence na hone se $k_2=k_3$ hamesha hoga.
4. $k_4=1\cdot f(1)=1$. Kyun? Right edge $x=1$.
5. $y_1=0+\frac{1}{6}(0+2(0.25)+2(0.25)+1)$. Bracket hai $0+0.5+0.5+1=2$, toh $y_1=\frac{2}{6}=\frac{1}{3}.$

Verify: Exact ${\int }_0^1{x}^{2}dx=\frac{1}{3}$. RK4 error $=0$. Yahi Cell A ka poora point hai: jab $f=f(x)$, RK4 hai hi Simpson's rule $\frac{h}{6}(f_0+4f_{mid}+f_1)$ aur degree 3 tak ke polynomials par exact hai. Dekho Simpson's Rule.

---

Cell B — linear mixed slope

Forecast: Exact solution hai $y=2e^x-x-1$, toh $y(0.2)=2e^{0.2}-1.2\approx 1.2428$. Predict karo ki RK4 ~5 decimals tak match karta hai.

Figure 2 (Cell B): true curve $y=2e^x-x-1$ from $x=0$ to $0.2$, chaar slope samples $k_1..k_4$ as short line segments apne sample points par — dekho kaise har successive segment thoda aur steep hota ja raha hai.

1. $k_1=0.2f(0,1)=0.2(0+1)=0.2$. Kyun? Left slope.
2. $k_2=0.2f(0.1,1+0.1)=0.2(0.1+1.1)=0.24$. Kyun? Midpoint $x=0.1$, $y$ shifted by $\frac{k_1}{2}=0.1$.
3. $k_3=0.2f(0.1,1+0.12)=0.2(0.1+1.12)=0.244$. Kyun? Refined midpoint, $y$ shifted by $\frac{k_2}{2}=0.12$.
4. $k_4=0.2f(0.2,1+0.244)=0.2(0.2+1.244)=0.2888$. Kyun? Right edge, $y$ shifted by full $k_3$.
5. $y_1=1+\frac{1}{6}(0.2+2(0.24)+2(0.244)+0.2888)=1+\frac{1.4568}{6}=\mathrm{1.2428.}$

Verify: $2e^{0.2}-1.2=1.242806\dots$; hamara $1.24280$ match karta hai. Error $\sim 10^{-6}$ — textbook 4th-order behaviour.

---

Cell C — negative / decaying slope

Forecast: Exact hai $y=e^{-x}$, toh $y(0.5)=e^{-0.5}\approx 0.6065$. Naive worry: kya RK4 zero se neeche overshoot kar sakta hai? Guess: nahi, positive rahega.

Figure 3 (Cell C): decaying curve $y=e^{-x}$; har slope sample neeche point karta hai (negative), lekin arrows chote hote jaate hain jaise $y$ drop karta hai — ek self-limiting decay ki picture jo zero se neeche kabhi nahi jaati.

1. $k_1=0.5\cdot (-1)=-0.5$. Kyun? Slope $-y=-1$ start par; negative sign kehta hai "neeche jao".
2. $k_2=0.5\cdot -\left(1+\frac{-0.5}{2}\right)=0.5\cdot -(1-0.25)=0.5(-0.75)=-0.375$. Kyun? Midpoint $y$ start se neeche hai, toh downward slope gentler hai.
3. $k_3=0.5\cdot -(1+\frac{-0.375}{2})=0.5\cdot -(1-0.1875)=0.5(-0.8125)=-0.40625$. Kyun? Refined midpoint.
4. $k_4=0.5\cdot -(1+(-0.40625))=0.5\cdot -(0.59375)=-0.296875$. Kyun? Right edge $y$ aur zyada drop ho gaya hai.
5. $y_1=1+\frac{1}{6}(-0.5+2(-0.375)+2(-0.40625)+(-0.296875))=1+\frac{-2.359375}{6}=\mathrm{0.606771.}$

Verify: $e^{-0.5}=0.606531$; error $\approx 2\times 10^{-4}$ (bada step, phir bhi tiny). Crucially $y_1>0$ — RK4 ka averaged slope decaying curve ko track karta hai rather than Euler ki tarah seedha neeche fire karne ke (jo $1-0.5=0.5$ deta, bahut kam). Compare karo Euler's Method se.

---

Cell D — nonlinear in $y$

Forecast: Exact hai $y=\frac{1}{1-x}$ (yeh $x=1$ par blow up karta hai!). Toh $y(0.1)=\frac{1}{0.9}=1.11111$. Predict karo RK4 uske close hoga.

Figure 4 (Cell D): curve $y=\frac{1}{1-x}$ $x=1$ par apne blow-up ki taraf steep hoti hui; chaar slope arrows visibly $k_1$ se $k_4$ tak lengthen hote hain kyunki squaring har $y$-shift ko amplify karta hai.

1. $k_1=0.1\cdot 1^2=0.1$. Kyun? Slope $y^2=1$.
2. $k_2=0.1\cdot (1+0.05{)}^2=0.1(1.1025)=0.11025$. Kyun? Midpoint $y=1.05$; squaring shift ko amplify karta hai.
3. $k_3=0.1\cdot (1+0.055125{)}^2=0.1(1.055125{)}^2=0.1(1.113289)=0.111329$. Kyun? Refined midpoint $y=1.0551$.
4. $k_4=0.1\cdot (1+0.111329{)}^2=0.1(1.111329{)}^2=0.1(1.235052)=0.123505$. Kyun? Right edge $y=1.1113$.
5. $y_1=1+\frac{1}{6}(0.1+2(0.11025)+2(0.111329)+0.123505)=1+\frac{0.666663}{6}=\mathrm{1.111111.}$

Verify: $\frac{1}{0.9}=1.111111$; error $\sim 10^{-7}$. Singularity ke paas bhi (yahan dur hai), RK4 shine karta hai kyunki midpoint refinements ke through woh steepening slope ko sense karta hai.

---

Cell E — zero / degenerate inputs (do flavours)

Forecast: Zero slope ke saath, $y$ change nahi kar sakta. Answer exactly $7$ hona chahiye.

1. $k_1=2\cdot 0=0$. Kyun? Slope har jagah $0$ hai.
2. $k_2=2\cdot 0=0$. Kyun? Kisi bhi $y$-shift ke baad bhi zero, kyunki $f\equiv 0$.
3. $k_3=0$, $k_4=0$. Kyun? Same reasoning — sample karne ko kuch nahi.
4. $y_1=7+\frac{1}{6}(0+0+0+0)=7.$

Verify: Exact $y\equiv 7$. Error $=0$. Sanity floor: agar true solution constant hai, RK4 use constant rakhta hai.

Forecast: Exact hai $y=5+\frac{x^2}{2}$, toh $y(1)=5.5$. Trap yeh hai: $k_1=0$ dekh ke galti se conclude karna ki "$y$ nahi badhta". Badhta hai — later stages nonzero slopes sample karte hain.

Figure 5 (Cell E2): parabola $y=5+\frac{x^2}{2}$; left-edge arrow $k_1$ perfectly flat hai ($=0$), phir bhi midpoint aur right-edge arrows upar tilt karte hain — curve $5$ se $5.5$ tak flat start ke bawajood climb karta hai.

1. $k_1=1\cdot f(0)=1\cdot 0=0$. Kyun? Slope $=x=0$ left edge par — yahan genuinely flat.
2. $k_2=1\cdot f(0.5)=1\cdot 0.5=0.5$. Kyun? Midpoint $x=0.5$; $y$-shift irrelevant kyunki $f$ mein $y$ nahi, lekin $x$ move ho gaya toh slope ab $0.5$ hai.
3. $k_3=1\cdot f(0.5)=0.5$. Kyun? Same midpoint $x$; $k_2$ ke barabar (pure-$x$ integrand).
4. $k_4=1\cdot f(1)=1$. Kyun? Right edge $x=1$, slope $=1$.
5. $y_1=5+\frac{1}{6}(0+2(0.5)+2(0.5)+1)=5+\frac{3}{6}=\mathrm{5.5.}$

Verify: Exact $5+\frac{1}{2}=5.5$. Error $=0$ (pure-$x$, toh Simpson-exact). Lesson: sirf isliye march band mat karo ki $k_1=0$ hai — hamesha charon stages compute karo.

---

Cell F — multi-step march

Forecast: Same ODE as Ex 2 lekin do chhote steps mein reach kiya. $h=0.1$ ke do steps $h=0.2$ ke ek step se behtar hone chahiye (Ex 2 result $1.242806$). Predict karo $y_2\approx 1.24281$ aur thoda zyada accurate.

Figure 6 (Cell F): $y=2e^x-x-1$ ke saath do consecutive RK4 steps; pehla step $(0.1,1.11034)$ par land karta hai, aur doosra step usi point se start hoke $(0.2,1.24281)$ reach karta hai — errors ka step se step chain hone ki picture.

Step 1 (parent note ke Worked Example 1 se): $y_1=1.11034$ at $x=0.1$. Kyun reuse karein? Arithmetic parent se identical hai; hum doosra step yahan se start karte hain.

Step 2 (from $x=0.1$, $y=1.11034$, $h=0.1$):

1. $k_1=0.1f(0.1,1.11034)=0.1(0.1+1.11034)=0.121034$. Kyun? Naya left edge.
2. $k_2=0.1f(0.15,1.11034+0.060517)=0.1(0.15+1.170857)=0.132086$. Kyun? Midpoint $x=0.15$.
3. $k_3=0.1f(0.15,1.11034+0.066043)=0.1(0.15+1.176383)=0.132638$. Kyun? Refined midpoint.
4. $k_4=0.1f(0.2,1.11034+0.132638)=0.1(0.2+1.242978)=0.144298$. Kyun? Right edge $x=0.2$.
5. $y_2=1.11034+\frac{1}{6}(0.121034+2(0.132086)+2(0.132638)+0.144298)=1.11034+0.132463=\mathrm{1.242806.}$

Verify: Exact $2e^{0.2}-1.2=1.242806$. Do chhote steps essentially bang-on land karte hain. Lesson: $h$ ko half karne se global error $\approx 2^4=16\times$ shrink hota hai — "4th order" ka promise (dekho Local vs Global Truncation Error).

---

Cell G — large step / limiting behaviour

Forecast: Exact hai $e=2.71828$. Euler $2$ dega. Huge step ke saath RK4 — guess karo ki phir bhi ~1% ke andar aata hai.

1. $k_1=1\cdot 1=1$. Kyun? Slope $=y=1$.
2. $k_2=1\cdot (1+0.5)=1.5$. Kyun? Midpoint $y=1.5$.
3. $k_3=1\cdot (1+0.75)=1.75$. Kyun? Refined midpoint.
4. $k_4=1\cdot (1+1.75)=2.75$. Kyun? Right edge $y=2.75$.
5. $y_1=1+\frac{1}{6}(1+3+3.5+2.75)=1+\frac{10.25}{6}=\mathrm{2.708333.}$

Verify: $e=2.718282$; error $\approx 0.0099$. Aur yeh beautiful identity: $y_1=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}=2.708\overline{3},$ $e^x=\sum \frac{x^n}{n!}$ ke pehle paanch terms $x=1$ par. RK4 Taylor series ko exactly $h^4$ term tak reproduce karta hai — isliye woh "4th order" hai (compare karo Taylor Series Methods se). Missing piece $\frac{1}{120}+\cdots \approx 0.0099$ exactly woh $O(h^5)$ local error hai, visible kyunki $h=1$ bada hai.

Figure 7 (Cell G): true $e^x$ (dashed navy) vs Euler ki single left-slope line (orange, $2$ par landing) aur RK4 ke chaar sampled slopes (magenta dots) jo path ko $2.708$ tak bend karte hain — ek labelled snapshot ki kyun zyada slopes sample karna better hai.

---

Cell H — real-world word problem

Forecast: Coffee $20^{\circ }$ ki taraf cool hogi. Exact: $T=20+70e^{-0.1t}$, toh $T(2)=20+70e^{-0.2}=20+57.31=77.31^{\circ }\text{C}$. Predict karo RK4 $77.3$ ke close hoga.

Figure 8 (Cell H): cooling curve $T=20+70e^{-0.1t}$ $90^{\circ }$ se dashed room-temperature line $20^{\circ }$ ki taraf drop karti hui; chaar slope samples dikhate hain ki cooling slow hoti jaati hai jaise $T$ room ke paas aata hai — curve kabhi $20^{\circ }$ se neeche cross nahi karti.

Yahan $f(t,T)=-0.1(T-20)$, units ${}^{\circ }\text{C}/\text{min}$; har $k_i$ ka unit ${}^{\circ }\text{C}$ hai (slope $\times$ minutes).

1. $k_1=2\cdot (-0.1)(90-20)=2(-7)=-14$. Kyun? Left slope: garam coffee jaldi cool hoti hai.
2. $k_2=2(-0.1)((90-7)-20)=2(-0.1)(63)=-12.6$. Kyun? Midpoint $T=90+\frac{k_1}{2}=83$.
3. $k_3=2(-0.1)((90-6.3)-20)=2(-0.1)(63.7)=-12.74$. Kyun? Refined midpoint $T=83.7$.
4. $k_4=2(-0.1)((90-12.74)-20)=2(-0.1)(57.26)=-11.452$. Kyun? Right edge $T=77.26$.
5. $T_1=90+\frac{1}{6}(-14+2(-12.6)+2(-12.74)+(-11.452))=90-12.6887=\mathrm{77.3113.}$

Verify: Exact $20+70e^{-0.2}=77.3128^{\circ }\text{C}$; error $\approx 0.0015^{\circ }$. Units check: har $k_i$ ${}^{\circ }\text{C}$ mein hai, update ${}^{\circ }\text{C}$ mein ${}^{\circ }\text{C}$ add karta hai ✓. Physically sound: temperature drop hua lekin room ke $20^{\circ }$ se kaafi upar raha.

---

Cell I — exam-style twist

Forecast: Slope positive hai (dono $x,\sqrt{y}>0$), toh $y$ grow karega. Rough guess: $y^{\prime }\approx 1\cdot 2=2$, toh $0.2$ ke over $y$ ~$0.4$-ish rise karega simple estimate se aage — expect karo $y\approx 4.4$.

Figure 9 (Cell I): exact solution $y={\left(\frac{x^2}{4}+\frac{7}{4}\right)}^2$ $(1,4)$ se $(1.2,4.4527)$ tak climb karta hua; chaar slope arrows (saare upar) grow karte hain kyunki step mein dono $x$ aur $\sqrt{y}$ increase karte hain.

1. $k_1=0.2\cdot 1\cdot \sqrt{4}=0.2(1)(2)=0.4$. Kyun? Left edge $x=1$, $y=4$.
2. $k_2=0.2\cdot 1.1\cdot \sqrt{4+0.2}=0.2(1.1)\sqrt{4.2}=0.22(2.04939)=0.450866$. Kyun? Midpoint $x=1.1$, $y=4.2$.
3. $k_3=0.2\cdot 1.1\cdot \sqrt{4+0.225433}=0.22\sqrt{4.225433}=0.22(2.055586)=0.452229$. Kyun? Refined midpoint $y=4.22543$.
4. $k_4=0.2\cdot 1.2\cdot \sqrt{4+0.452229}=0.24\sqrt{4.452229}=0.24(2.110030)=0.506407$. Kyun? Right edge $x=1.2$, $y=4.45223$.
5. $y_1=4+\frac{1}{6}(0.4+2(0.450866)+2(0.452229)+0.506407)=4+\frac{2.712597}{6}=\mathrm{4.452100.}$

Verify: Separable exact solution: $\int y^{-1/2}dy=\int xdx\Rightarrow 2\sqrt{y}=\frac{x^2}{2}+C$. $x=1,y=4$ par: $2(2)=\frac{1}{2}+C\Rightarrow C=\frac{7}{2}$. $x=1.2$ par: $2\sqrt{y}=\frac{1.44}{2}+3.5=4.22\Rightarrow \sqrt{y}=2.11\Rightarrow y=4.4521$. Hamara RK4 $4.45210$ deta hai — 4 decimals tak match karta hai. $\approx 4.4$ ka forecast sahi ballpark mein tha. ✓

---

Matrix, sab green

Har cell A–I ka ab ek fully worked, verified example hai (E ko E1 aur E2 mein split kiya gaya). Sign flips (C), zeros dono total aur partial (E1, E2), nonlinearity (D, I), chaining (F), stress steps (G), aur units-bearing story (H) — sab cover ho gaye hain — aapko koi bhi RK4 scenario kabhi aise nahi milna chahiye jo is page ne rehearse na kiya ho.

Recall Close karne se pehle self-test

Kaun sa cell test karta hai ki RK4 = Simpson? ::: Cell A (pure-$x$ integrand, Ex 1). Jab $k_1=0$ ho toh computing kyun jaari rakhni chahiye? ::: Later stages ($k_2,k_3,k_4$) nonzero slopes sample kar sakte hain — dekho Ex 5b jahan $y$ $5\to 5.5$ climb karta hai despite $k_1=0$. Coffee $20^{\circ }$ se upar kyun rahi? ::: Slope $-0.1(T-20)\to 0$ jaise $T\to 20$; room temperature ke paas cooling slow ho jaati hai. Ex 7 mein $1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}$ kya represent karte hain? ::: $e^1$ ke pehle paanch Taylor terms — proof ki RK4 Taylor ko $h^4$ tak match karta hai. $h$ ko half karne se global error kitne factor se improve hota hai? ::: Approximately $2^4=16\times$ (4th-order convergence).

Connections

• Euler's Method — undershooting single-slope baseline (Ex 3, Ex 7).
• Modified Euler / Heun's Method — comparison ke liye 2nd-order sibling.
• Simpson's Rule — exactly woh jo Cell A / Ex 1 reduce ho jaata hai.
• Taylor Series Methods — woh target series jo Ex 7 reproduce karta hai.
• Local vs Global Truncation Error — $16\times$ rule in Ex 6.
• Adaptive Step Size (RKF45) — G jaise cells ke liye $h$ automatically kaise choose karein.