Runge-Kutta 4th Order Method for Ordinary Differential ...
[Pages:15]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 dx
=
f (x, y), y(0) =
y0
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 dx
=
f (x, y),
y(0) =
y0
form.
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
e y dy + x2 y 2 = 2sin(3x), y(0) = 5
dx
in
dy dx
=
f (x, y),
y(0) =
y0
form.
Solution
e y dy + x2 y 2 = 2sin(3x), y(0) = 5
dx
dy = 2sin(3x) - x2 y 2 , y(0) = 5
dx
ey
In this case
f (x, y) = 2sin(3x) - x2 y 2
ey
The Runge-Kutta 4th order method is based on the following
( ) yi+1 = yi + a1k1 + a2k2 + a3k3 + a4k4 h
(1)
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
( ) ( ) ( ) yi+1
=
yi
+
dy dx
xi , yi
xi+1 - xi
+
1 2!
d2y dx 2
xi , yi
xi+1 - xi
2
+
1 3!
d3y dx3
xi , yi
xi+1 - xi
3
(2)
( ) +
1 4!
d4y dx 4
xi , yi
xi+1 - xi
4
Knowing that
dy dx
=
f (x, y)
and
xi+1
- xi
=h
yi+1
=
yi
+
f
(xi ,
yi )h +
1 2!
f
' (xi ,
yi )h2
+
1 3!
f
( ) '' xi , yi h3
+
1 4!
f
(''' xi ,
yi )h4
(3)
Based on equating Equation (2) and Equation (3), one of the popular solutions used is
yi+1
=
yi
+
1 6
(k1
+
2k2
+
2k3
+
k4
)h
(4)
Runge-Kutta 4th Order Method
k1 = f (xi , yi )
k2
=
f
xi
+
1 2
h,
yi
+
1 2
k1h
k3
=
f
xi
+
1 2 h, yi
+
1 2
k2
h
k4 = f (xi + h, yi + k3h)
08.04.3
(5a)
(5b) (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?108
dt
( ) f (t, ) = -2.2067 ?10-12 4 - 81?108
i+1
=
i
+
1 6
(k1
+
2k2
+
2k3
+
k4
)h
For i = 0 , t0 = 0 , 0 = 1200K
k1 = f (t0 ,0 ) = f (0,1200)
( ) = -2.2067 ?10-12 12004 - 81?108
= -4.5579
k2
=
f
t0
+
1 2
h,
0
+
1 2
k1h
= f 0 + 1 (240),1200 + 1 (- 4.5579)? 240
2
2
= f (120,653.05)
( ) = -2.2067 ?10-12 653.054 - 81?108
= -0.38347
k3
=
f
t
0
+
1 2
h,
0
+
1 2
k
2
h
= f 0 + 1 (240),1200 + 1 (- 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
=
0
+
1 6
(k1
+
2k2
+
2k3
+
k4 )h
= 1200 + 1 (- 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 ?10-12 675.654 - 81?108
= -0.44199
k2
=
f
t1
+
1 2
h,1
+
1 2
k1h
= f 240 + 1 (240),675.65 + 1 (- 0.44199)240
2
2
= f (360,622.61)
( ) = -2.2067 ?10-12 622.614 - 81?108
= -0.31372
k3
=
f
t1
+
1 2
h,1
+
1 2
k
2h
= f 240 + 1 (240),675.65 + 1 (- 0.31372)? 240
2
2
= f (360,638.00)
( ) = -2.2067 ?10-12 638.004 - 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 ?10-12 592.194 - 81?108
= -0.25351
2
= 1
+
1 6
(k1
+ 2k2
+ 2k3
+
k4 )h
= 675.65 + 1 (- 0.44199 + 2(- 0.31372) + 2(- 0.34775) + (- 0.25351))? 240
6
= 675.65 + 1 (- 2.0184)? 240
6
= 594.91K
2 is the approximate temperature at t = t2 = t1 + h = 240 + 240 = 480
2 = (480)
594.91K
Figure 1 compares the exact solution with the numerical solution using the Runge-Kutta 4th order method with different step sizes.
1600
Temperature, (K)
1200 800 400
h=120
Exact h=240
0 0
-400
h=480
200
400
600
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)
-90.278 594.91 646.16 647.54 647.57
E t
737.85 52.660 1.4122 0.033626 0.00086900
|t | %
113.94 8.1319 0.21807 0.0051926 0.00013419
800
Temperature, (480)
600
400
200
0 0
-200
100 200 300 400 500 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
yi+1
=
yi
+
1 8
(k1
+
3k 2
+
3k3
+
k4
)h
(6)
where
k1 = f (xi , yi )
(7a)
k2
=
f
xi
+
1 3
h,
yi
+
1 3
hk1
(7b)
k3
=
f
xi
+
2 3 h, yi
-
1 3
hk1
+
hk
2
(7c)
k4 = f (xi + h, yi + hk1 - hk2 + hk3 )
(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, (K)
1400 1200 1000
800 600 400 200
0
0
4th order Exact
Heun Euler
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 October 13, 2010
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