Transcript PPT

Euler Method
Civil Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
7/12/2016
http://numericalmethods.eng.usf.edu
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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
A polluted lake with an initial concentration of a bacteria is 107
parts/m3, while the acceptable level is only 5×106 parts/m3. The
concentration of the bacteria will reduce as fresh water enters the
lake. The differential equation that governs the concentration C of
the pollutant as a function of time (in weeks) is given by
dC
 0.06C  0, C (0)  107
dt
Find the concentration of the pollutant after 7 weeks. Take a step
size of 3.5 weeks.
dC
dt
 0.06C
f t, C   0.06C
Ci 1  Ci  f t i , Ci h
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Solution
Step 1:
For i  0, t0  0, C0  107
C1  C0  f t0 , C0 h


 107  f 0, 107 3.5
  
  6 10 3.5
 107   0.06 107 3.5
 107
5
 7.9 106 parts / m3
C1 is the approximate concentration of bacteria at
t  t1  t0  h  0  3.5  3.5 weeks
C 3.5  C1  7.9 106 parts/m 3
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Solution Cont
Step 2:
For i  1, t1  3.5, C1  7.9 106
C2  C1  f t1 , C1 h


  0.067.9 10 3.5
  4.74 10 3.5
 7.9 106  f 3.5, 7.9 106 3.5
 7.9 106
 7.9 106
6
5
 6.241106 parts/m 3
C2 is the approximate concentration of bacteria at
t  t2  t1  h  3.5  3.5  7 weeks
C 7  C2  6.241106 parts/m 3
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Solution Cont
The exact solution of the ordinary differential equation is
given by the solution of a non-linear equation as
C (t )  1 107 e
 3t 


 50 
The solution to this nonlinear equation at t=7 weeks is
C (7)  6.5705 106 parts/m 3
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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 Concentration of bacteria after 7 weeks as a
function of step size
Step size
h
7
3.5
1.75
0.875
0.4375
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C7
5.8×106
6.241×106
6.4164×106
6.4959×106
6.5337×106
Et
770470
329470
154060
74652
36763
|t | %
11.726
5.0144
2.3447
1.1362
0.55952
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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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