Exercises — Runge-Kutta 4th order (RK4) — derivation

$k_3=hf\left(x_n+\frac{h}{2},y_n+\frac{k_2}{2}\right)$. Toh yeh midpoint $x_n+\frac{h}{2}$ par sample hota hai, aur yeh $k_2$ reuse karta hai (ek refined midpoint slope). Figure s01 dekho: dono amber arrows ($k_2$ aur $k_3$) midpoint par ek hi vertical dashed line share karte hain; sirf estimated $y$ ki height alag hoti hai.

$\frac{1}{6}$ hona chahiye, na ki $\frac{h}{6}$. Hamare convention mein har $k_i$ mein $h$ ka ek factor pehle se hai (kyunki $k_i=hf$). $\frac{h}{6}$ likhne se $h$ se do baar multiply ho jaata, jo $h^2$ step deta — galat dimensions. $y_{n+1}=y_n+\frac{1}{6}(k_1+2k_2+2k_3+k_4).$ $\frac{h}{6}$ wali form tabhi correct hai jab doosre convention mein $k_i$ ek pure slope ho (koi $h$ nahi).

Yahan $f(x,y)=x+y$, $x_0=0$, $y_0=1$, $h=0.2$.

• $k_1=0.2f(0,1)=0.2(0+1)=0.2$ — left edge par slope.
• $k_2=0.2f(0.1,1+0.1)=0.2f(0.1,1.1)=0.2(0.1+1.1)=0.24$ — midpoint, $\frac{k_1}{2}=0.1$ use karke.
• $k_3=0.2f(0.1,1+0.12)=0.2f(0.1,1.12)=0.2(0.1+1.12)=0.244$ — refined midpoint, $\frac{k_2}{2}=0.12$ use karke.
• $k_4=0.2f(0.2,1+0.244)=0.2f(0.2,1.244)=0.2(0.2+1.244)=0.2888$ — right edge, poora $k_3$ use karke. $y_1=1+\frac{1}{6}(0.2+2(0.24)+2(0.244)+0.2888)=1+\frac{1.4568}{6}=\mathrm{1.2428.}$ Exact: $y=2e^x-x-1\Rightarrow y(0.2)=2e^{0.2}-1.2=1.242806$. Error $\approx 6\times 10^{-6}$.

$f(x,y)=-2y$, $y_0=3$, $h=0.5$.

• $k_1=0.5(-2\cdot 3)=-3$.
• $k_2=0.5(-2(3+\frac{-3}{2}))=0.5(-2\cdot 1.5)=-1.5$.
• $k_3=0.5(-2(3+\frac{-1.5}{2}))=0.5(-2\cdot 2.25)=-2.25$.
• $k_4=0.5(-2(3+(-2.25)))=0.5(-2\cdot 0.75)=-0.75$. $y_1=3+\frac{1}{6}(-3+2(-1.5)+2(-2.25)+(-0.75))=3+\frac{-11.25}{6}=\mathrm{1.125.}$ Bracket check karte hain: $-3+2(-1.5)+2(-2.25)+(-0.75)=-3-3-4.5-0.75=-11.25$, aur $3-\frac{11.25}{6}=3-1.875=1.125$. Exact: $y=3e^{-2x}\Rightarrow y(0.5)=3e^{-1}=1.103638$. Yahan step $h=0.5$ ek fast-decaying equation ke liye bada hai, isliye error ($\approx 0.021$) dikh raha hai — ek reminder ki "4th order" chhote $h$ ke baare mein ek statement hai.

Jab $f$ sirf $x$ par depend karta hai, toh $y$-arguments irrelevant hain, isliye: $k_1=hf(x_n),k_2=k_3=hf\left(x_n+\frac{h}{2}\right),k_4=hf(x_n+h).$ Update mein substitute karke: $y_{n+1}-y_n=\frac{1}{6}(hf(x_n)+2\cdot hf(x_n+\frac{h}{2})+2\cdot hf(x_n+\frac{h}{2})+hf(x_n+h)).$ $=\frac{h}{6}(f(x_n)+4f\left(x_n+\frac{h}{2}\right)+f(x_n+h)).$ Yeh last line bilkul Simpson's rule $\frac{h}{6}(f_0+4f_{mid}+f_1)$ hai $[x_n,x_n+h]$ par integral ke liye. Toh RK4 hai hi Simpson's rule, $y$-dependence allow karne ke liye generalize kiya gaya. Isliye weights $1,2,2,1$ hain: dono midpoint samples Simpson ke single "$4$" mein merge ho jaate hain.

Fixed length $L=b-a$ cross karne ke liye tum $N=\frac{L}{h}$ steps lete ho — yaani $N\propto \frac{1}{h}$.

"Har step $\sim C{h}^5$ add karta hai" ka matlab. Ek single step par method ka answer true curve se (kuch constant) $\times h^5$ ki bounded amount se differ karta hai. Uss constant ko $C$ kaho: yeh package karta hai solution kitna curvy hai (roughly, interval par uski 5th derivative ka size) aur $\frac{1}{120}$ ka fixed numerical factor jo RK4 chhod jaata hai. $C$ $h$ ke chhote hone par shrink nahi karta — yeh problem par depend karta hai, step par nahi — isliye ek step ke error ki poori $h$-dependence $h^5$ mein hai. Isliye hum likhte hain "local error $=C{h}^5$".

Errors add kyun hote hain (error propagation). Calculation ko ek relay ki tarah socho: har step previous step ke (pehle se thode galat) answer se start hota hai aur apni nayi $C{h}^5$ galti add karta hai. Toh har step par do cheezein hoti hain — (i) ek nayi $\sim C{h}^5$ error commit hoti hai, aur (ii) pehle se maujood errors aage carry hote hain. Ek well-behaved (stable) equation ke liye carried-forward errors blow up nahi hote; woh zyada se zyada bounded factor se badhte hain. Us case mein total sirf $N$ fresh contributions ko sum karne se dominate hota hai, ek per step: $\text{global error}\sim \underset{\text{kitne steps}}{\underbrace{N}}\times \underset{\text{ek step ki error}}{\underbrace{C{h}^5}}=\frac{L}{h}\cdot C{h}^5=CL{h}^4=O(h^4).$ $h$ ki ek power steps ki badhti number se "khaai" jaati hai: jab $h$ chhota hota hai tum zyada steps lete ho, isliye zyada chhoti errors pile up hoti hain, $h$ ka ek factor cancel ho jaata hai. Dekho Local vs Global Truncation Error. Isliye RK4 ko 4th order kehte hain (global statement), chahye har step $5$th-order accurate kyun na ho.

$f(x,y)=y$, $h=0.5$.

Step 1 ($x_0=0,y_0=1$):

• $k_1=0.5(1)=0.5$
• $k_2=0.5(1+0.25)=0.625$
• $k_3=0.5(1+0.3125)=0.65625$
• $k_4=0.5(1+0.65625)=0.828125$ $y_1=1+\frac{1}{6}(0.5+2(0.625)+2(0.65625)+0.828125)=1+\frac{3.890625}{6}=\mathrm{1.6484375.}$

Step 2 ($x_1=0.5,y_1=1.6484375$):

• $k_1=0.5(1.6484375)=0.82421875$
• $k_2=0.5(1.6484375+0.412109\dots )=1.0302734\dots$
• $k_3=0.5(1.6484375+0.515137\dots )=1.0817871\dots$
• $k_4=0.5(1.6484375+1.081787\dots )=1.3651123\dots$ $y_2=1.6484375+\frac{1}{6}(0.82421875+2(1.0302734)+2(1.0817871)+1.3651123).$ Bracket sum $=6.4134277\dots$, toh $y_2=1.6484375+1.0689046\dots =2.7173421$. Exact $e=2.7182818$; error $\approx 0.00094$. Single bade step ne $2.70833$ diya (error $\approx 0.010$). Step half karne se error roughly $16=2^4$ se cut hua — yeh 4th-order method ki pehchaan hai.

Global error $\propto h^4$. $h=0.5$ se $h=0.25$ jaane par $h$ half hota hai, toh error ${\left(\frac{1}{2}\right)}^4=\frac{1}{16}$ se scale hota hai. $\text{naya error}\approx \frac{0.00094}{16}\approx 5.9\times 10^{-5}.$ Toh hum predict karte hain $y(1)\approx e-5.9\times 10^{-5}\approx 2.718222$. (Direct computation mein error $\approx 6.5\times 10^{-5}$ aata hai — $2^4$ rule kaam karta hai.)

$f(x,y)=y$, $y_0=1$ ke saath:

• $k_1=h\cdot 1=h.$
• $k_2=h\left(1+\frac{k_1}{2}\right)=h\left(1+\frac{h}{2}\right)=h+\frac{h^2}{2}.$
• $k_3=h\left(1+\frac{k_2}{2}\right)=h\left(1+\frac{h}{2}+\frac{h^2}{4}\right)=h+\frac{h^2}{2}+\frac{h^3}{4}.$
• $k_4=h\left(1+k_3\right)=h\left(1+h+\frac{h^2}{2}+\frac{h^3}{4}\right)=h+h^2+\frac{h^3}{2}+\frac{h^4}{4}.$

Ab weights $1,2,2,1$ se combine karo: $k_1+2k_2+2k_3+k_4.$ $h$ ki power se collect karo:

• $h^1$: $1+2+2+1=6$ → contributes $6h$.
• $h^2$: $2\cdot \frac{1}{2}+2\cdot \frac{1}{2}+1=1+1+1=3$ → contributes $3h^2$.
• $h^3$: $2\cdot \frac{1}{4}+1\cdot \frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1$ → contributes $h^3$.
• $h^4$: $1\cdot \frac{1}{4}=\frac{1}{4}$ → contributes $\frac{1}{4}{h}^4$.

Toh $k_1+2k_2+2k_3+k_4=6h+3h^2+h^3+\frac{1}{4}{h}^4$. $6$ se divide karo: $y_1=1+h+\frac{h^2}{2}+\frac{h^3}{6}+\frac{h^4}{24}.$ Yeh exactly $e^h=1+h+\frac{h^2}{2}+\frac{h^3}{6}+\frac{h^4}{24}+\frac{h^5}{120}+\cdots$ ke pehle paanch terms hain. Pehla term jo RK4 miss karta hai woh $\frac{h^5}{120}$ hai — confirm karta hai local error $O(h^5)$. Dekho Taylor Series Methods.

Vector RK4 bilkul scalar RK4 jaisa hai, bas componentwise apply hota hai. ${\mathbf{u}}_0=(1,0)$, $h=0.1$, $\mathbf{f}(x,y)=(y,-x)$ lo.

$k_1=h\mathbf{f}({\mathbf{u}}_0)$: $\mathbf{f}(1,0)=(0,-1)$, toh $k_1=(0,-0.1)$.

$k_2=h\mathbf{f}({\mathbf{u}}_0+\frac{k_1}{2})$: argument $=(1+0,0-0.05)=(1,-0.05)$; $\mathbf{f}=(-0.05,-1)$; toh $k_2=(-0.005,-0.1)$.

$k_3=h\mathbf{f}({\mathbf{u}}_0+\frac{k_2}{2})$: argument $=(1-0.0025,0-0.05)=(0.9975,-0.05)$; $\mathbf{f}=(-0.05,-0.9975)$; toh $k_3=(-0.005,-0.09975)$.

$k_4=h\mathbf{f}({\mathbf{u}}_0+k_3)$: argument $=(1-0.005,0-0.09975)=(0.995,-0.09975)$; $\mathbf{f}=(-0.09975,-0.995)$; toh $k_4=(-0.009975,-0.0995)$.

Componentwise combine karo, ${\mathbf{u}}_1={\mathbf{u}}_0+\frac{1}{6}(k_1+2k_2+2k_3+k_4)$:

• $x$: $1+\frac{1}{6}(0+2(-0.005)+2(-0.005)+(-0.009975))=1+\frac{-0.029975}{6}=\mathrm{0.99500417.}$
• $y$: $0+\frac{1}{6}(-0.1+2(-0.1)+2(-0.09975)+(-0.0995))=\frac{-0.59900}{6}=-\mathrm{0.09983333.}$

Toh $x(0.1)\approx 0.9950042,y(0.1)\approx -0.0998333$. Exact: $\cos 0.1=0.9950042,-\sin 0.1=-0.0998334$. Dono lagbhag $2\times 10^{-6}$ tak match karte hain — RK4 systems ko same 4th-order accuracy ke saath handle karta hai, koi nayi machinery nahi chahiye.