Chapter 08.04
Runge-Kutta 4th Order Method for
Ordinary Differential Equations
After reading this chapter, you should be able to
1. develop Runge-Kutta 4th order method for solving ordinary differential equations,
2. find the effect size of step size has on the solution,
3. know the formulas for other versions of the Runge-Kutta 4th order method
What is the Runge-Kutta 4th order method?
Runge-Kutta 4th order method is a numerical technique used to solve ordinary differential
equation of the form
dy
 f  x, y , y 0  y 0
dx
So only first order ordinary differential equations can be solved by using the Runge-Kutta 4th
order method. In other sections, we have discussed how Euler and Runge-Kutta methods are
used to solve higher order ordinary differential equations or coupled (simultaneous)
differential equations.
How does one write a first order differential equation in the above form?
Example 1
Rewrite
dy
 2 y  1.3e  x , y 0   5
dx
in
dy
 f ( x, y ), y (0)  y 0 form.
dx
08.04.1
08.04.2
Chapter 08.04
Solution
dy
 2 y  1.3e  x , y 0   5
dx
dy
 1.3e  x  2 y, y 0   5
dx
In this case
f x, y   1.3e  x  2 y
Example 2
Rewrite
ey
dy
 x 2 y 2  2 sin( 3 x), y 0   5
dx
in
dy
 f ( x, y ), y (0)  y 0 form.
dx
Solution
dy
 x 2 y 2  2 sin( 3 x), y 0   5
dx
dy 2 sin( 3x)  x 2 y 2

, y 0  5
dx
ey
In this case
2 sin( 3x)  x 2 y 2
f  x, y  
ey
The Runge-Kutta 4th order method is based on the following
yi 1  yi  a1k1  a2 k 2  a3 k3  a4 k 4 h
ey
where knowing the value of y  yi at xi , we can find the value of y  yi 1 at xi 1 , and
h  xi 1  xi
Equation (1) is equated to the first five terms of Taylor series
3
dy
1 d2y
1 d3y
2




xi 1  xi 
yi 1  yi 
x

x

x

x

xi , yi
i 1
i
i 1
i
2 xi , yi
3 xi , yi
dx
2! dx
3! dx
4
1d y
xi 1  xi 4

4 xi , yi
4! dx
dy
 f  x, y  and xi 1  xi  h
Knowing that
dx
1
1
1
y i 1  y i  f  xi , y i h  f ' xi , y i h 2  f '' xi , y i h 3  f ''' xi , y i h 4
2!
3!
4!
Based on equating Equation (2) and Equation (3), one of the popular solutions used is
1
y i 1  y i  k1  2k 2  2k 3  k 4 h
6
(1)
(2)
(3)
(4)
Runge-Kutta 4th Order Method
08.04.3
k1  f xi , yi 
(5a)
1
1


k2  f  xi  h, yi  k1h 
2
2 

(5b)
1
1


k 3  f  x i  h, y i  k 2 h 
2
2


k 4  f xi  h, yi  k 3 h 
(5c)
(5d)
Example 3
A ball at 1200 K is allowed to cool down in air at an ambient temperature of 300 K.
Assuming heat is lost only due to radiation, the differential equation for the temperature of
the ball is given by
d
 2.2067 10 12  4  81108 ,  0   1200 K
dt
where  is in K and t in seconds. Find the temperature at t  480 seconds using RungeKutta 4th order method. Assume a step size of h  240 seconds.
Solution
d
 2.2067  10 12  4  81  10 8
dt
f t ,   2.2067  10 12  4  81 108
1
 i 1   i  k1  2k 2  2k 3  k 4 h
6
For i  0 , t 0  0 ,  0  1200K



k1  f t0 ,0 

 f 0,1200
 2.2067  10 12 1200 4  81 108 
 4.5579
1
1


k 2  f  t0  h,  0  k1h 
2
2


1
1


 f  0  240,1200   4.5579  240 
2
2


 f 120,653.05
 2.2067  10 12 653.05 4  81 108 
 0.38347
1
1


k 3  f  t 0  h,  0  k 2 h 
2
2


1
1


 f  0  240,1200   0.38347   240 
2
2


 f 120,1154.0
08.04.4
Chapter 08.04

 2.2067 1012 1154.04  81108
 3.8954
k4  f t0  h, 0  k3h 

 f 0  240,1200   3.894 240
 f 240,265.10
 2.2067 1012 265.104  81108 
 0.0069750
1
 1   0  ( k1  2 k 2  2 k 3  k 4 ) h
6
1
 1200   4.5579  2 0.38347   2 3.8954   0.069750240
6
 1200   2.1848  240
 675.65 K
1 is the approximate temperature at
t  t1
 t0  h
 0  240
 240
1   240
 675.65 K
For i  1, t1  240,1  675.65 K
k1  f t1 ,1 
 f 240,675.65
 2.2067  10 12 675.65 4  81 108 
 0.44199
1
1


k 2  f  t1  h, 1  k1h 
2
2


1
1


 f  240  240,675.65   0.44199240 
2
2


 f 360,622.61
 2.2067  1012 622.614  81 108 
 0.31372
1
1


k3  f  t1  h, 1  k 2 h 
2
2


1
1


 f  240  240 ,675.65   0.31372  240 
2
2


 f 360,638.00
 2.2067  10 12 638.00 4  81  108 
Runge-Kutta 4th Order Method
08.04.5
 0.34775
k4  f t1  h,1  k3h
 f 240  240,675.65   0.34775 240
 f 480,592.19
 2.2067  10 12 592.19 4  81 108 
 0.25351
1
 2   1  ( k1  2 k 2  2 k 3  k 4 ) h
6
1
 675.65   0.44199  2 0.31372  2 0.34775   0.25351  240
6
1
 675.65   2.0184   240
6
 594.91K
 2 is the approximate temperature at
t  t2
 t1  h
 240  240
 480
2   480
 594.91 K
Figure 1 compares the exact solution with the numerical solution using the Runge-Kutta 4th
order method with different step sizes.
Temperature, θ(K)
1600
1200
h=120
800
Exact
h=240
400
h=480
0
0
200
400
600
-400
Time,t(sec)
Figure 1 Comparison of Runge-Kutta 4th order method
with exact solution for different step sizes.
08.04.6
Chapter 08.04
Table 1 and Figure 2 show the effect of step size on the value of the calculated temperature at
t  480 seconds.
Table 1 Value of temperature at time, t  480 s for different step sizes
Step size, h
480
240
120
60
30
 480
Et
-90.278 737.85
594.91 52.660
646.16 1.4122
647.54 0.033626
647.57 0.00086900
| t | %
113.94
8.1319
0.21807
0.0051926
0.00013419
Temperature, θ(480)
800
600
400
200
0
0
100
200
300
400
500
-200
Step size, h
Figure 2 Effect of step size in Runge-Kutta 4th order method.
In Figure 3, we are comparing the exact results with Euler’s method (Runge-Kutta 1st order
method), Heun’s method (Runge-Kutta 2nd order method), and Runge-Kutta 4th order
method.
The formula described in this chapter was developed by Runge. This formula is same as
Simpson’s 1/3 rule, if f x, y  were only a function of x . There are other versions of the 4th
order method just like there are several versions of the second order methods. The formula
developed by Kutta is
1
yi 1  yi  k1  3k 2  3k 3  k 4 h
(6)
8
where
k1  f xi , yi 
(7a)
1
1


k 2  f  xi  h, yi  hk1 
3
3


2
1


k 3  f  xi  h, yi  hk1  hk 2 
3
3


k 4  f xi  h, yi  hk1  hk 2  hk3 
(7b)
(7c)
(7d)
Runge-Kutta 4th Order Method
08.04.7
This formula is the same as the Simpson’s 3/8 rule, if f x, y  is only a function of x .
Temperature, θ(K)
1400
1200
4th order
1000
800
Exact
600
Heun
400
Euler
200
0
0
100
200
300
400
500
Time, t(sec)
Figure 3 Comparison of Runge-Kutta methods of 1st (Euler), 2nd, and 4th order.
ORDINARY DIFFERENTIAL EQUATIONS
Topic
Runge-Kutta 4th order method
Summary
Textbook notes on the Runge-Kutta 4th order method for
solving ordinary differential equations.
Major
General Engineering
Authors
Autar Kaw
Last Revised February 6, 2016
Web Site
http://numericalmethods.eng.usf.edu