Transcript Euler Method for Solving Ordinary Differential Equations

Numerical Methods
Golden Section Search Method Example
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Example
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The cross-sectional area A of a gutter with equal base and
edge length of 2 is given by (trapezoidal area):
Max. f ( )  A  4 sin  (1  cos )  4 sin   2 sin(2 )
Find the angle  which maximizes the cross-sectional area
of the gutter. Using an initial interval of [0,  ] find the
2
solution after 2 iterations.
  0.05
Convergence achieved if “ interval length ” is within
5
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Solution
The function to be maximized is f ( )  4 sin  (1  cos )
Iteration 1: Given the values for the boundaries of
xL  0 and xu   / 2we can calculate the initial intermediate
points as follows:
5 1
5 1
( xu  x L )  0 
(1.5708)  0.97080 f (0.97080)  5.1654
2
2
5 1
5 1
x 2  xu 
( xu  x L )  1.5708
(1.5708)  0.60000 f (0.60000)  4.1227
2
2
x1  x L 
f2
f1
X1=?
XL
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X2
X1
XL=X2
Xu
X2=X1
Xu
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Solution Cont
x1  x L 
5 1
5 1
( xu  x L )  0.60000
(1.5708 0.60000)  1.2000
2
2
To check the stopping criteria the difference between xu
and x L is calculated to be
xu  xL  1.5708 0.60000 0.97080
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Solution Cont
Iteration 2
x L  0.60000
xu  1.5708
f ( x1 )  f ( x2 )
x1  1.2000
f (1.2000)  5.0791
x 2  0.97080
f (0.97080)  5.1654
x L  0.60000
xu  1.2000
x1  0.97080
x 2  xu 
XL
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X2
X 1 Xu
5 1
5 1
( xu  x L )  1.2000
(1.2000 0.6000)  0.82918
2
2
xu  x L
 1.2000 0.6000  0.9000
2
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Theoretical Solution and
Convergence
Iteration
1
2
3
4
5
6
7
8
9
xl
0.0000
0.6002
0.6002
0.8295
0.9712
0.9712
1.0253
1.0253
1.0253
xu
1.5714
1.5714
1.2005
1.2005
1.2005
1.1129
1.1129
1.0794
1.0588
x1
0.9712
1.2005
0.9712
1.0588
1.1129
1.0588
1.0794
1.0588
1.0460
x2
0.6002
0.9712
0.8295
0.9712
1.0588
1.0253
1.0588
1.0460
1.0381
f(x1)
5.1657
5.0784
5.1657
5.1955
5.1740
5.1955
5.1908
5.1955
5.1961
xu  x L 1.0253 1.0588

 1.0420
2
2
f(x2)
4.1238
5.1657
4.9426
5.1657
5.1955
5.1937
5.1955
5.1961
5.1957

1.5714
0.9712
0.6002
0.3710
0.2293
0.1417
0.0876
0.0541
0.0334
f (1.0420)  5.1960
The theoretically optimal solution to the problem
happens at exactly 60 degrees which is 1.0472 radians
and gives a maximum cross-sectional area of 5.1962.
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THE END
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Acknowledgement
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This material is based upon work supported by the National Science
Foundation under Grant # 0717624. Any opinions, findings, and
conclusions or recommendations expressed in this material are those of
the author(s) and do not necessarily reflect the views of the National
Science Foundation.
The End - Really