Overview

By extending the Taylor expansion of the function f up to order 5, a reasoning similar to that used for second-order Runge–Kutta methods leads to a nonlinear system of eight equations containing ten unknowns (see Scheid, ref. [32]). The final result is the classical fourth-order Runge–Kutta method, which is a highly useful and widely applied tool.

Runge–Kutta Order 4 Algorithm

Given: step size h, initial condition (t₀, y₀), and a maximum number of iterations N.

For 0 ≤ n ≤ N:

k₁ = h f(tₙ, yₙ)

k₂ = h f(tₙ + h/2 , yₙ + k₁/2)

k₃ = h f(tₙ + h/2 , yₙ + k₂/2)

k₄ = h f(tₙ + h , yₙ + k₃)

yₙ₊₁ = yₙ + (1/6)(k₁ + 2k₂ + 2k₃ + k₄)

tₙ₊₁ = tₙ + h
        

Write tₙ₊₁ and yₙ₊₁.

Remark

The Runge–Kutta method of order 4 is very frequently used because of its high accuracy, which is clearly illustrated in the following example.

Example

Consider again the differential equation:

y′(t) = −y(t) + t + 1 ,    y(0) = 1

We evaluate the constants kᵢ. For the first iteration (h = 0.1):

k₁ = 0.1 f(0, 1) = 0.1(−1 + 1) = 0
k₂ = 0.1 f(0.05, 1) = 0.1(−1 + 1.05) = 0.005
k₃ = 0.1 f(0.05, 1.0025) = 0.1(−1.0025 + 1.05) = 0.00475
k₄ = 0.1 f(0.1, 1.00475) = 0.1(−1.00475 + 1.1) = 0.009525

y₁ = 1 + (1/6)(0 + 2·0.005 + 2·0.00475 + 0.009525)
   = 1.0048375
        

Second iteration:

k₁ = 0.1 f(0.1, 1.0048375) = 0.00951625
k₂ = 0.1 f(0.15, 1.009595625) = 0.014040438
k₃ = 0.1 f(0.15, 1.011857719) = 0.0138142281
k₄ = 0.1 f(0.2, 1.018651728) = 0.0181348272

y₂ = 1.0048375 + (1/6)(k₁ + 2k₂ + 2k₃ + k₄)
   = 1.0187309014
        

The following table compares the numerical and exact solutions, and gives the absolute error.

The error is on the order of 10⁻⁶, which compares very favorably with lower-order methods. A slight increase in error over successive iterations is also observed, showing once again the propagation of error from one step to the next.