Transcript PPT

Runge 2nd Order Method
Computer Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
7/12/2016
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Runge-Kutta 2nd Order Method
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Runge-Kutta 2nd Order Method
For
dy
 f ( x, y ), y (0)  y0
dx
Runge Kutta 2nd order method is given by
yi 1  yi  a1k1  a2 k2 h
where
k1  f xi , yi 
k2  f xi  p1h, yi  q11k1h 
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Heun’s Method
Heun’s method
Slope  f xi  h, yi  k1h 
y
Here a2=1/2 is chosen
1
a1 
2
p1  1
Slope  f xi , yi 
q11  1
resulting in
1 
1
yi 1  yi   k1  k2 h
2 
2
where
k1  f xi , yi 
yi+1, predicted
Average Slope 
yi
xi
1
 f xi  h, yi  k1h   f xi , yi 
2
xi+1
x
Figure 1 Runge-Kutta 2nd order method (Heun’s method)
k 2  f xi  h, yi  k1h 
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Midpoint Method
Here a2  1 is chosen, giving
a1  0
p1 
1
2
q11 
1
2
resulting in
yi 1  yi  k2h
where
k1  f xi , yi 
1
1


k 2  f  xi  h, yi  k1h 
2
2


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Ralston’s Method
Here a 2  2 is chosen, giving
3
1
3
3
p1 
4
3
q11 
4
resulting in
a1 
2 
1
yi 1  yi   k1  k 2 h
3 
3
where
k1  f xi , yi 
3
3


k 2  f  xi  h, yi  k1h 
4
4


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How to write Ordinary Differential
Equation
How does one write a first order differential equation in the form of
dy
 f  x, y 
dx
Example
dy
 2 y  1.3e  x , y 0   5
dx
is rewritten as
dy
 1.3e  x  2 y, y 0   5
dx
In this case
f x, y   1.3e  x  2 y
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Example
A rectifier-based power supply requires a capacitor to temporarily
store power when the rectified waveform from the AC source drops
below the target voltage. To properly size this capacitor a first-order
ordinary differential equation must be solved. For a particular power
supply, with a capacitor of 150 μF, the ordinary differential equation to
be solved is

 18 cos(120 (t ))  2  v(t ) 
dv(t )
1

dt
150  10 6

 0.1  max 



0.04

,0 


Find voltage across the capacitor at t= 0.00004s. Use step size
h=0.00002

 18 cos(120 (t ))  2  v 
dv
1


dt

 0.1  max 
150  10 6 


8
,0 



 18 cos(120 (t ))  2  v 




0
.
1

max
,
0



0
.
04





1 
1
 v i   k1  k 2  h
2 
2
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f t , v  
vi 1
0.04
1
150  10 6
Solution
Step 1: i  0, t0  0, v0  v(0)  0
k1  f t0 , vo   f 0,0 
1
150 10 6


 18 cos(120 (0))  2  (0) 

  2.6660 106

0
.
1

max
,
0



0.04






k 2  f t0  h, v0  k1h   f 0  0.00002,0  2.6660 106 0.00002  f 0.00002, 53.32
1

150 10 6

 18 cos(120 (0.00002))  2  (53.32) 
,0   666.67
 0.1  max 
0.04





1 
1
1
1

v1  v0   k1  k 2 h  0   2.6660 106   666.67 0.00002
2 
2
2
2

 0  1.3327 106 0.00002  26.653V

9

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Solution Cont
Step 2: i  1, t1  t0  h  0  0.00002  0.00002 v1  26.653 V
k1  f t1 , v1   f 0.00002, 26.653
1

150 10 6

 18 cos(120 (0.00002))  2  (26.653) 
,0   666.67
 0.1  max 
0.04



k 2 f t1  h, v1  k1h   f 0.00002  0.00002, 26.653   666.67 0.00002   f 0.00004, 26.640

1
150 10 6

 18 cos120 0.00004   2  26.640 


0
.
1

max
,0   666.67


0.04



1 
1
1
1

v2  v1   k1  k 2 h  26.653    666.67    666.67 0.00002
2 
2
2
2

 26.653   666.67 0.00002  26.647V
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Solution Continued
The solution to this nonlinear equation at t=0.00004 seconds is
v(0.00004)  15.974V
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Comparison with exact results
Figure 2. Heun’s method results for different step sizes
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Effect of step size
Table 1. Effect of step size for Heun’s method
Step size, h
0.00004
0.00002
0.00001
0.000005
0.0000025
v0.00004
53.307
26.640
15.980
15.918
15.970
−37.333
−10.666
−0.0056605
0.055825
0.0044682
v0.00004  15.974V
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|t | %
Et
233.71
65.771
0.035436
0.34947
0.027974
(exact)
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Effects of step size on Heun’s
Method
Figure 3. Effect of step size in Heun’s method
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Comparison of Euler and RungeKutta 2nd Order Methods
Table 2. Comparison of Euler and the Runge-Kutta methods
Step size,
h
0.00004
0.00002
0.00001
0.000005
0.0000025
v0.00004
Euler
Heun
Midpoint
Ralston
106.64
53.307
26.640
15.996
15.993
53.307
26.640
15.980
15.918
15.970
−0.026667
−0.026667
11.642
15.917
15.968
35.529
17.751
15.363
15.917
15.968
v0.00004  15.974V
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(exact)
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Comparison of Euler and RungeKutta 2nd Order Methods
Table 2. Comparison of Euler and the Runge-Kutta methods
Step size,
h
0.00004
0.00002
0.00001
0.000005
0.0000025
t %
Euler
Heun
567.59
233.71
66.771
0.13146
0.11268
233.71
65.269
0.031301
0.35683
0.037561
v(0.00004)  15.974V
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Midpoint
Ralston
100.17
122.47
100.17
11.152
27.101
3.8009
0.33187 0.33187
0.012523 0.012523
(exact)
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Comparison of Euler and RungeKutta 2nd Order Methods
Figure 4. Comparison of Euler and Runge Kutta 2nd order
methods with exact results.
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/runge_kutt
a_2nd_method.html
THE END
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