Runge-Kutta 4th order (RK4) — derivation

Solving $\dfrac{dy}{dx}=f(x,y)$ , $y(x_0)=y_0$ , without knowing the closed-form solution.

HOW: deriving the general RK4 skeleton

A 4-stage explicit RK method has the form $y_{n+1} = y_n + (a_1 k_1 + a_2 k_2 + a_3 k_3 + a_4 k_4)$ where each stage uses the previous ones:

\begin{aligned} k_1 &= h\,f(x_n,\; y_n)\\ k_2 &= h\,f(x_n + c_2 h,\; y_n + b_{21}k_1)\\ k_3 &= h\,f(x_n + c_3 h,\; y_n + b_{31}k_1 + b_{32}k_2)\\ k_4 &= h\,f(x_n + c_4 h,\; y_n + b_{41}k_1 + b_{42}k_2 + b_{43}k_3) \end{aligned}

Why this shape? Each $k_i$ is an estimate of $h\cdot(\text{slope})$ at a sample point. We tune the constants $a_i, c_i, b_{ij}$ so the whole thing agrees with the Taylor series of the true solution up to $h^4$ .

Step 1 — Taylor expansion of the true solution (the target)

Since $y'=f$ , differentiate repeatedly (chain rule, remember $y$ depends on $x$ ): $y' = f,\qquad y'' = f_x + f_y f,\qquad y''' = f_{xx}+2f_{xy}f+f_{yy}f^2 + f_y(f_x+f_y f).$

Why this step? These are the exact quantities the true solution carries. Our method must reproduce them.

So the exact step is

y(x_n+h)=y_n + h f + \tfrac{h^2}{2}(f_x+f_y f) + \tfrac{h^3}{6}y''' + \tfrac{h^4}{24}y'''' + O(h^5).

Step 2 — Taylor expand each $k_i$ and match

Expanding $k_2,k_3,k_4$ as 2-variable Taylor series and forcing the weighted sum to equal the exact expansion above produces a system of order conditions. The classic solution (there are infinitely many; this is the elegant symmetric one) is:

$c_2=c_3=\tfrac12,\quad c_4=1,\qquad a_1=a_4=\tfrac16,\quad a_2=a_3=\tfrac13,$ and $b_{21}=b_{32}=\tfrac12$ , $b_{43}=1$ , all others zero.

Why these numbers? The weights $\left(\tfrac16,\tfrac13,\tfrac13,\tfrac16\right)$ are exactly Simpson's rule weights — because when $f$ depends only on $x$ , RK4 reduces to Simpson's $\tfrac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4)$ -style integration, which is $O(h^5)$ accurate. RK4 is Simpson's rule generalized to $y$ -dependence!

Step 3 — The classical RK4 formulas

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

Worked Example 1

Solve $y'=x+y$ , $y(0)=1$ , find $y(0.1)$ with $h=0.1$ .

$k_1 = h f(0,1) = 0.1(0+1)=0.1$ — Why? slope at start.
$k_2 = 0.1\,f(0.05,\,1.05)=0.1(0.05+1.05)=0.11$ — Why? midpoint using half of $k_1$ .
$k_3 = 0.1\,f(0.05,\,1.055)=0.1(0.05+1.055)=0.1105$ — Why? refined midpoint using half of $k_2$ .
$k_4 = 0.1\,f(0.1,\,1.1105)=0.1(0.1+1.1105)=0.12105$ — Why? endpoint using full $k_3$ .

$y_1 = 1 + \tfrac16(0.1 + 2(0.11)+2(0.1105)+0.12105)=1.11034.$

Exact solution $y=2e^x-x-1 \Rightarrow y(0.1)=1.110342$ . Error $\approx 10^{-6}$ ! Why so good? Fourth-order accuracy.

Worked Example 2 (Forecast-then-Verify)

Solve $y'=y$ , $y(0)=1$ , one step $h=1$ . Forecast: true value is $e=2.71828$ ; Euler gives $2$ . Predict RK4 lands very close to $e$ .

$k_1 = 1\cdot 1 = 1$
$k_2 = 1\cdot(1+0.5)=1.5$
$k_3 = 1\cdot(1+0.75)=1.75$
$k_4 = 1\cdot(1+1.75)=2.75$ $y_1 = 1 + \tfrac16(1+3+3.5+2.75)=1+\tfrac{10.25}{6}=2.70833.$

Verify: $2.70833$ vs $2.71828$ , error $\approx 0.01$ even with a huge step $h=1$ . Notice $y_1 = 1+1+\tfrac12+\tfrac16+\tfrac1{24}$ = the first five terms of $e^x$ 's series — direct proof RK4 matches Taylor to $h^4$ .

Common Mistakes

Recall Feynman: explain to a 12-year-old

You want to walk across a hill in the fog and land at the right height. Euler checks the ground slope only where he starts and walks straight — he overshoots. RK4 is smarter: he peeks at the slope at the start, then guesses the middle, checks it, re-checks the middle, then checks the far end — and takes a cleverly weighted average of all four peeks (the two middle peeks count twice). That average almost perfectly matches the real curved path. So even with big steps, he lands almost exactly on target.

Active Recall

What IVP does RK4 solve?

y'=f(x,y)

with

y(x_0)=y_0

, marching one step

h

at a time.

Write the four RK4 slope formulas (with

k_i=hf

k_1=hf(x_n,y_n)

;

k_2=hf(x_n+\frac h2,y_n+\frac{k_1}2)

;

k_3=hf(x_n+\frac h2,y_n+\frac{k_2}2)

;

k_4=hf(x_n+h,y_n+k_3)

What is the RK4 update formula?

y_{n+1}=y_n+\frac16(k_1+2k_2+2k_3+k_4)

What are the local and global error orders?

Local

O(h^5)

, global

O(h^4)

Why do

k_2,k_3

get weight 2?

They estimate the midpoint slope (best represents the interval); matches Simpson's

1,2,2,1

weighting.

Where is each

k_i

sampled?

k_1

left edge,

k_2,k_3

midpoint,

k_4

right edge.

What does RK4 reduce to when

f=f(x)

only?

Simpson's rule for the integral

\int f\,dx

How does RK4 avoid computing derivatives of

f

It replaces derivatives with extra function evaluations at interior points, matching the Taylor series.

Why is global error one order lower than local?

1/h

steps accumulate the per-step

O(h^5)

error into

O(h^4)

Connections

Euler's Method — the 1st-order ancestor (single left-edge slope).
Modified Euler / Heun's Method — the 2nd-order sibling (average of edges).
Simpson's Rule — source of the $1,2,2,1$ weights.
Taylor Series Methods — what RK4 mimics without derivatives.
Local vs Global Truncation Error — why 4th order globally.
Adaptive Step Size (RKF45) — RK4 with error control.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, RK4 ka core idea bahut simple hai. Jab humein $y'=f(x,y)$ solve karna hota hai lekin exact formula nahi milta, tab hum step-by-step aage badhte hain. Euler method sirf shuru ki slope dekh kar seedha chal deta hai — isliye galat direction mein overshoot ho jata hai, kyunki slope to raste mein badalti rehti hai. RK4 smart hai: wo char jagah slope check karta hai — start pe ( $k_1$ ), do baar beech mein ( $k_2, k_3$ ), aur end pe ( $k_4$ ).

Phir in char slopes ka ek weighted average leta hai: formula $y_{n+1}=y_n+\frac16(k_1+2k_2+2k_3+k_4)$ . Notice karo ki beech wali slopes ( $k_2,k_3$ ) ko do-do baar count kiya — kyunki midpoint interval ka sabse accha representative hota hai. Ye weights $1,2,2,1$ actually Simpson's rule se aate hain! Isliye RK4 ko "Simpson's rule for ODEs" bhi kehte hain.

Kyun itna accurate? Kyunki ye weighted combination true solution ke Taylor series ko $h^4$ tak exactly match karta hai — bina koi derivative calculate kiye. Derivatives ki jagah bas extra function evaluations use hote hain. Result: local error $O(h^5)$ , global error $O(h^4)$ . Matlab step chhota karoge to error bahut tezi se girega. Exam mein RK4 favourite isliye hai kyunki bade step pe bhi answer exact ke kareeb aata hai — jaise humne dekha $y'=y$ mein $h=1$ pe bhi $e$ ke bahut paas nikla.

Yaad rakho: $k_1$ left, $k_2$ - $k_3$ middle, $k_4$ right, aur middles ko double weight — bas yehi RK4 ki poori kahani hai.

Runge-Kutta 4th order (RK4) — derivation

The Big Picture

WHAT are we trying to do?

HOW: deriving the general RK4 skeleton

Step 1 — Taylor expansion of the true solution (the target)

Step 2 — Taylor expand each $k_i$ and match

Step 3 — The classical RK4 formulas

Worked Example 1

Worked Example 2 (Forecast-then-Verify)

Common Mistakes

Active Recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Numerical Methods

Connections