Transcript PPT

Euler Method
Industrial 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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Euler Method
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Euler’s Method
y
dy
 f x, y , y 0  y 0
dx
Slope

Rise
Run

y1  y0
x1  x0
 f  x0 , y 0 
y1  y0  f x0 , y0 x1  x0 
 y0  f x0 , y0 h
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True value
Φ
x0,y0
y1, Predicted
value
Step size, h
x
Figure 1 Graphical interpretation of the first step of Euler’s method
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Euler’s Method
y
yi 1  yi  f xi , yi h
True Value
h  xi 1  xi
yi+1, Predicted value
Φ
yi
h
Step size
xi
xi+1
x
Figure 2. General graphical interpretation of Euler’s method
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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
The open loop response, that is, the speed of the motor to a
voltage input of 20 V, assuming a system without damping is
20  0.02 
dw
 0.06w
dt
If the initial speed is zero w0  0 , and using Euler’s method,
what is the speed at t  0.8 s ? Assume a step size of h  0.4 s.
dw
 1000  3w
dt
f t , w  1000  3w
wi 1  wi  f ti , wi h
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Solution
Step 1: For i  0, t0  0, w0  0
w1  w0  f t0 , w0 h
 0  f 0,0 0.4
 0  1000  300.4
 0  1000  0.4
 400 rad/s
w1 is the approximate speed of the motor at
t  t1  t0  h  0  0.4  0.4 s
w0.4  w1  400 rad/s
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Solution Cont
Step 2: For i  1, t1  0.4, w2  400
w2  w1  f t1 , w1 h
 400  f 0.4, 400 0.4
 400  1000  3400 0.4
 400   200 0.4
 320 rad/s
w2 is the approximate speed of the motor at
t  t2  t1  h  0.4  0.4  0.8 s
x0.8  w2  320 rad/s
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Solution Cont
The exact solution of the ordinary differential equation is
given by
 1000   1000  3t
wt   

e
 3   3 
The solution to this nonlinear equation at t = 0.8 s is
w0.8  303.09 rad/s
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Comparison of Exact and
Numerical Solutions
Figure 3. Comparing exact and Euler’s method
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Effect of step size
Table 1 Speed of motor at 0.8 seconds as a function of
step size, h
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Step size, h
w0.8
Et
|t | %
0.8
0.4
0.2
0.1
0.05
800
320
324.8
314.18
308.58
−496.91
−16.906
−21.706
−11.023
−5.4890
163.95
5.5778
7.1615
3.6370
1.8110
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Comparison with exact results
Figure 4. Comparison of Euler’s method with exact solution for different step sizes
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Effects of step size on Euler’s
Method
Figure 5. Effect of step size in Euler’s method.
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Errors in Euler’s Method
It can be seen that Euler’s method has large errors. This can be illustrated using
Taylor series.
dy
1 d2y
1 d3y
2
3
xi 1  xi  




y i 1  y i 
x

x

x

x
 ...
i

1
i
i

1
i
2
3
dx xi , yi
2! dx x , y
3! dx x , y
i
yi 1  yi  f ( xi , yi )xi 1  xi  
i
i
i
1
1
2
3
f ' ( xi , yi )xi 1  xi   f ' ' ( xi , yi )xi 1  xi   ...
2!
3!
As you can see the first two terms of the Taylor series
yi 1  yi  f xi , yi h are the Euler’s method.
The true error in the approximation is given by
Et 
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f xi , yi  2 f xi , yi  3
h 
h  ...
2!
3!
Et  h 2
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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/euler_meth
od.html
THE END
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