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Runge–Kutta Methods

Intuition

One slope sample per step limits you to first order. So sample the slope at several cleverly chosen points inside the step and take a weighted average — each extra, well-placed sample cancels another term of the Taylor error. That escalation is the Runge–Kutta family; RK4, with four samples for fourth order, is the classic sweet spot and the general-purpose workhorse of scientific computing.

The family

\[y_{n+1} = y_n + \Delta t\sum_{i=1}^{s} b_i k_i, \qquad k_i = f(\text{strategic point } i)\]
Method Evals Order Note
RK1 1 1 = forward Euler
RK2 (midpoint) / Heun 2 2 first "free lunch"
RK3 3 3 good cost/accuracy compromise
RK4 4 4 the classic

RK4

\[k_1 = h f(t_n, y_n) \qquad k_2 = h f\!\left(t_n + \tfrac h2,\ y_n + \tfrac{k_1}2\right)\]
\[k_3 = h f\!\left(t_n + \tfrac h2,\ y_n + \tfrac{k_2}2\right) \qquad k_4 = h f(t_n + h,\ y_n + k_3)\]
\[\boxed{\;y_{n+1} = y_n + \frac{k_1 + 2k_2 + 2k_3 + k_4}{6}\;}\]

Weights \(\frac16(1,2,2,1)\) — Simpson's rule in disguise: two midpoint samples count double.

def RK4_step(f, t, y, h):
    k1 = f(t, y)
    k2 = f(t + 0.5*h, y + 0.5*h*k1)
    k3 = f(t + 0.5*h, y + 0.5*h*k2)
    k4 = f(t + h, y + h*k3)
    return y + h*(k1 + 2*k2 + 2*k3 + k4) / 6

(RK3 uses \(k_3 = f(t+h,\ y + h(2k_2 - k_1))\) and weights \(\frac16(1,4,1)\); see the equation page for both.)

Order vs conservation

RK4's global error \(\mathcal{O}(\Delta t^4)\) makes it spectacularly accurate over short-to-medium horizons — on the cross-field benchmark it is visually indistinguishable from the exact cycloid. But RK methods are not symplectic: on Hamiltonian systems their tiny per-step energy error drifts monotonically. Rule of thumb: RK4 for accuracy over finite times, leapfrog for fidelity over long times.

Common mistakes

  • Halving \(h\) when you meant to halve error. Halving \(h\) cuts RK4 error by 16× — check the convergence page before over-resolving.
  • Counting steps, not evaluations. RK4 costs 4 evaluations per step; at equal cost, compare RK4 with step \(h\) against Euler with step \(h/4\).
  • Reusing \(k_1\) across steps incorrectly — each stage belongs to its own step (unless you implement FSAL variants deliberately).

Knowledge graph position

Prerequisites: Forward Euler, Backward Euler. Leads to: production ODE solving; contrast with leapfrog.

Quiz

Q1 (computational). RK4 with \(N = 100\) steps gives error \(10^{-4}\). Estimate the error with \(N = 200\).

Answer

\(\mathcal{O}(\Delta t^4)\): doubling \(N\) divides error by \(2^4 = 16\) → \(\approx 6\times10^{-6}\).

Q2 (conceptual). Why do the RK4 weights sum to 1 (after the \(\frac16\) factor)?

Answer

Consistency: for $f = $ const the step must reduce to \(y + h f\) exactly; \(\frac{1+2+2+1}{6} = 1\) guarantees it.

Q3 (MCQ). At equal computational cost, which usually wins on a smooth, non-Hamiltonian problem?

  • (a) forward Euler (b) Heun (c) RK4 (d) they tie
Answer

(c). Higher order wins whenever the solution is smooth: the error constant is similar but the power of \(\Delta t\) is far better.