5.4 — Runge-Kutta Methods

The workhorse of numerical differential equations

Learning Objectives

Understand the RK4 algorithm and why it uses four slope evaluations
Compute all four slopes ( $k_{1}$ , $k_{2}$ , $k_{3}$ , $k_{4}$ ) in a single step
Compare the accuracy of RK4 with Euler's method
Understand the error order and practical implications
Apply RK4 to systems of differential equations

Motivation

Why Runge-Kutta? The RK4 method is considered the workhorse of numerical methods in science and engineering. It provides exceptional accuracy with manageable computational cost. You will encounter RK4 in:

Engineering simulations (structural mechanics, control systems)
Physics simulations (orbital mechanics, fluid dynamics)
Biology and medicine (pharmacokinetics, population dynamics)
Climate modeling and weather prediction

From Euler to Runge-Kutta

Euler's method uses a single slope evaluation at the beginning of each step, giving $O (h)$ accuracy. Runge-Kutta methods use multiple slope evaluations at strategic points to achieve higher-order accuracy. The 4th-order method (RK4) uses four slope evaluations to achieve $O (h^{4})$ accuracy.

The RK4 Formulas

Definition: Fourth-Order Runge-Kutta (RK4)

Given $y' = f (x, y)$ with initial condition $y (x_{0}) = y_{0}$ and step size $h$ , the RK4 step is:

k_1 = f(x_n, y_n)$$ $$k_2 = f\!\left(x_n + \tfrac{h}{2},\; y_n + \tfrac{h}{2}k_1\right)$$ $$k_3 = f\!\left(x_n + \tfrac{h}{2},\; y_n + \tfrac{h}{2}k_2\right)$$ $$k_4 = f(x_n + h,\; y_n + h\,k_3)$$ $$y_{n+1} = y_n + \tfrac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4)

Intuition Behind RK4

$k_{1}$ : Slope at the start of the interval: $f (x_{n}, y_{n})$

$k_{2}$ : Slope at the midpoint using $k_{1}$ to estimate the midpoint value

$k_{3}$ : Slope at the midpoint using $k_{2}$ (a refined estimate)

$k_{4}$ : Slope at the end of the interval using $k_{3}$

The final update uses a weighted average of these four slopes:

\text{weight} = \frac{1}{6}(1 + 2 + 2 + 1)

This weighting (Simpson's rule-like) gives extraordinary accuracy by sampling the slope in a strategic pattern.

Error Analysis

Theorem: RK4 Error

Local Truncation Error: $O (h^{5})$

Global Error: $O (h^{4})$

RK4 is a fourth-order method. To reduce error by a factor of 16, halve the step size $h$ .

Comparison of Methods

Quick Reference: Comparing Numerical Methods

Method	Order	$f$ Evaluations per Step	Typical Use
Euler	$O (h)$	1	Quick sketches, educational purposes
Improved Euler (RK2)	$O (h^{2})$	2	Better accuracy, simple code
RK4	$O (h^{4})$	4	Standard default for production code

Step-by-Step RK4 Algorithm

Set up: Define the differential equation $y' = f (x, y)$ , initial condition $(x_{0}, y_{0})$ , and step size $h$ .
Compute $k_{1}$ : $k_{1} = f (x_{n}, y_{n})$
Compute $k_{2}$ : $k_{2} = f (x_{n} + h / 2, y_{n} + (h / 2) k_{1})$
Compute $k_{3}$ : $k_{3} = f (x_{n} + h / 2, y_{n} + (h / 2) k_{2})$
Compute $k_{4}$ : $k_{4} = f (x_{n} + h, y_{n} + h k_{3})$
Update: $y_{n + 1} = y_{n} + (h / 6) (k_{1} + 2 k_{2} + 2 k_{3} + k_{4})$
Advance: $x_{n + 1} = x_{n} + h$
Repeat: Continue until you reach the desired endpoint.

Worked Examples

Problem: Solve $y' = y$ , $y (0) = 1$ , with $h = 0.1$ . Find $y (0.1)$ .

Solution:

Setup: $f (x, y) = y$ , $x_{0} = 0$ , $y_{0} = 1$ , $h = 0.1$

Step	Calculation	Value
$k_{1}$	$f (0, 1) = 1$	1
$k_{2}$	$f (0.05, 1 + 0.05) = f (0.05, 1.05) = 1.05$	1.05
$k_{3}$	$f (0.05, 1 + 0.05 (1.05)) = 1.0525$	1.0525
$k_{4}$	$f (0.1, 1 + 0.1 (1.0525)) = 1.10525$	1.10525

RK4 Update:

y_1 = 1 + \frac{0.1}{6}(1 + 2(1.05) + 2(1.0525) + 1.10525)$$ $$y_1 = 1 + \frac{0.1}{6}(6.31075) = 1 + 0.10517083... = 1.10517083...

Exact solution: $y (x) = e^{x}$ , so $y (0.1) = e^{0.1} = 1.10517092...$

Error: $| 1.10517083 - 1.10517092 | \approx 8.5 \times 10^{- 8}$ (microscopic!)

Compare with Euler: Euler gives $y_1 = 1 + 0.1(1) = 1.1$ with error $\approx 0.005$ — about 59,000 times larger!

Problem: Solve $y' = x + y$ , $y (0) = 1$ , with $h = 0.2$ . Find $y (1)$ .

Solution:

We compute 5 steps, each with the full RK4 formula. Here's a summary table:

$n$	$x_{n}$	$y_{n}$	$k_{1}$	$k_{2}$	$k_{3}$	$k_{4}$
0	0.0	1.0000	1.0000	1.2000	1.2200	1.4440
1	0.2	1.2428	1.4428	1.6871	1.7115	1.9851
2	0.4	1.5836	1.9836	2.2820	2.3118	2.6460
3	0.6	2.0442	2.6442	3.0086	3.0451	3.4532
4	0.8	2.6510	3.4510	3.8961	3.9407	4.4392
5	1.0	3.4365	—	—	—	—

RK4 Result: $y (1) \approx 3.4365$

Exact solution: $y = 2 e^{x} - x - 1$ , so $y (1) = 2 e^{1} - 1 - 1 = 2 e - 2 \approx 3.4366$

Error: $| 3.4365 - 3.4366 | \approx 0.0001$ (excellent accuracy for 5 steps with $h = 0.2$ )

Problem: Solve $y' = - 2 x y$ , $y (0) = 1$ , with $h = 0.25$ . Find $y (1)$ .

Solution:

This problem has an exact solution: $y (x) = e^{- x^{2}}$ , so $y (1) = e^{- 1} \approx 0.36788$

Using RK4 with only 4 steps ( $h = 0.25$ ):

Step	$x$	RK4 $y$	Exact $y$	Error
0	0.0	1.00000	1.00000	0.00000
1	0.25	0.93941	0.93941	0.00000
2	0.50	0.77848	0.77880	0.00032
3	0.75	0.55383	0.55432	0.00049
4	1.0	0.36779	0.36788	0.00009

Result: RK4 with just 4 steps gives $y (1) \approx 0.36779$ , accurate to 4 decimal places! The error is only $0.00009$ .

Problem: Solve $y' = y$ , $y (0) = 1$ to $x = 1$ with $h = 0.5$ . Compare three methods.

Solution:

Exact solution: $y (1) = e \approx 2.71828$

Method	$y (1)$	Error	Functions Called
Euler	2.2500	0.4683	2
Improved Euler (RK2)	2.6406	0.0777	4
RK4	2.71735	0.00093	8
Exact	2.71828	—	—

Observation: RK4 with 8 function evaluations gives error $0.00093$ , while Euler with 2 evaluations gives error $0.4683$ — RK4 is 500 times more accurate for the same computational work!

Problem: Solve the system:

x' = -y, \quad y' = x, \quad x(0) = 1, \quad y(0) = 0

with $h = 0.1$ . Compute one step.

Solution:

For systems, RK4 uses vector slopes. Let $f_1 (x, y) = - y$ and $f_2 (x, y) = x$ .

At step 0: $x_{0} = 1$ , $y_{0} = 0$

Slope	$k$ (for $x$ )	$l$ (for $y$ )
$k_{1}, l_{1}$	$- 0 = 0$	$1$
$k_{2}, l_{2}$	$- (0 + 0.05) = - 0.05$	$1$
$k_{3}, l_{3}$	$- (0 + 0.05) = - 0.05$	$1$
$k_{4}, l_{4}$	$- (0 + 0.1) = - 0.1$	$1$

RK4 Updates:

x_1 = 1 + \frac{0.1}{6}(0 + 2(-0.05) + 2(-0.05) + (-0.1)) = 1 - 0.00833... \approx 0.99167$$ $$y_1 = 0 + \frac{0.1}{6}(1 + 2(1) + 2(1) + 1) = 0.1

Result: $(x_{1}, y_{1}) \approx (0.99167, 0.1)$

Exact solution: $x (t) = \cos t$ , $y (t) = \sin t$ (unit circle)

At $t = 0.1$ : $(0.99500, 0.09983)$

Error: Tiny! RK4 perfectly captures circular motion.

Practice Problems

Test your understanding with these multiple-choice questions. Click a choice to check your answer.

Progress:

0 / 6 correct

Question 1: Computing k₁

For $y' = x 2 + y$ , $y (0) = 1$ , what is $k_{1}$ ?

Question 2: Understanding k₂

In RK4, $k_{2}$ is evaluated at which point?

Question 3: The RK4 Weight

The RK4 update formula uses the weight $h / 6$ . What is the coefficient of $k_{2}$ in this formula?

Question 4: Error Order of RK4

What is the global error order of the RK4 method?

Question 5: Comparing Methods

To reduce the global error by a factor of 256, by what factor should you decrease $h$ in RK4?

Question 6: RK4 Applications

Which of the following is not a common application of RK4?

Quick Reference Card

RK4 Formula at a Glance

\begin{align} k_1 &= f(x_n, y_n)\\ k_2 &= f\!\left(x_n + \tfrac{h}{2}, y_n + \tfrac{h}{2}k_1\right)\\ k_3 &= f\!\left(x_n + \tfrac{h}{2}, y_n + \tfrac{h}{2}k_2\right)\\ k_4 &= f(x_n + h, y_n + hk_3)\\ y_{n+1} &= y_n + \tfrac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \end{align}

Key Facts

Order: Fourth-order method, $O (h^{4})$ global error
Function calls: 4 evaluations of $f$ per step
Local truncation error: $O (h^{5})$
Halve $h$ : Error is divided by $2^4 = 16$
Stability: Good stability for most problems; region depends on the problem
Best used when: You need high accuracy without adaptive step control