Runge 2 nd Order Method - IISER Punepgoel/RungeKutta.pdfRunge 2 nd Order Method Major: All...
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Runge 2nd Order Method
Major: All Engineering Majors
4/13/2010 http://numericalmethods.eng.usf.edu 1
Major: All Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
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Runge-Kutta 2nd Order Method
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Heun’s Method
y
yi+1, predicted
( )hkyhxfSlope ii 1, ++=
( )ii yxfSlope ,=
Heun’s method
2
11 =a
1=p
Here a2=1/2 is chosen
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xxi xi+1
yi
Figure 1 Runge-Kutta 2nd order method (Heun’s method)
( ) ( )[ ]iiii yxfhkyhxfSlopeAverage ,,2
1 1 +++=
11 =p
111 =q
resulting in
hkkyy ii
++=+ 211
2
1
2
1
where
( )ii yxfk ,1 =
( )hkyhxfk ii 12 , ++=
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Runge-Kutta 2nd Order Method
Runge Kutta 2nd order method is given by
For0)0(),,( yyyxf
dx
dy==
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Runge Kutta 2nd order method is given by
( )hkakayy ii 22111 ++=+
where
( )ii yxfk ,1 =
( )hkqyhpxfk ii 11112 , ++=
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Midpoint MethodHere 12 =a is chosen, giving
01 =a
2
11 =p
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2
111 =q
resulting in
hkyy ii 21 +=+
where
( )ii yxfk ,1 =
++= hkyhxfk ii 12
2
1,
2
1
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Ralston’s MethodHere
3
22 =a is chosen, giving
3
11 =a
4
31 =p
http://numericalmethods.eng.usf.edu6
4
4
311 =q
resulting in
hkkyy ii
++=+ 211
3
2
3
1
where
( )ii yxfk ,1 =
++= hkyhxfk ii 12
4
3,
4
3
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How to write Ordinary Differential Equation
How does one write a first order differential equation in the form of
( )yxfdx
dy,=
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Example
( ) 50,3.12 ==+− yey
dx
dy x
is rewritten as
( ) 50,23.1 =−=− yye
dx
dy x
In this case
( ) yeyxfx 23.1, −=
−
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Example
A ball at 1200K is allowed to cool down in air at an ambient temperature
of 300K. Assuming heat is lost only due to radiation, the differential
equation for the temperature of the ball is given by
( ) ( ) Kd
12000,1081102067.2 8412=θ×−θ×−=
θ −
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( ) ( ) Kdt
12000,1081102067.2 =θ×−θ×−=
Find the temperature at 480=t seconds using Heun’s method. Assume a step size of
240=h seconds.
( )8412 1081102067.2 ×−θ×−=θ −
dt
d
( ) ( )8412 1081102067.2, ×−θ×−=θ−
tf
hkkii
++=+ 211
2
1
2
1θθ
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SolutionStep 1: Kti 1200)0(,0,0 00 ==== θθ
( )
( )
( )10811200102067.2
1200,0
,
8412
01
×−×−=
=
=
−
f
tfk oθ ( )
( )( )
( )09.106,240
2405579.41200,2400
, 1002
=
−++=
++=
f
f
hkhtfk θ
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( )5579.4
10811200102067.2 8412
−=
×−×−=− ( )
( )017595.0
108109.106102067.2
09.106,240
8412
=
×−×−=
=
−
f
( ) ( )
( )
K
hkk
16.655
2402702.21200
240017595.02
15579.4
2
11200
2
1
2
12101
=
−+=
+−+=
++= θθ
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Solution Cont
Step 2: Khtti 16.655,2402400,1 101 ==+=+== θ
( )
( )
( )108116.655102067.2
16.655,240
,
8412
111
×−×−=
=
=
−
f
tfk θ ( )
( )( )
( )87.561,480
24038869.016.655,240240
, 1112
=
−++=
++=
f
f
hkhtfk θ
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( )38869.0
108116.655102067.2 8412
−=
×−×−=− ( )
( )20206.0
108187.561102067.2
87.561,480
8412
−=
×−×−=
=
−
f
( ) ( )
( )
K
hkk
27.584
24029538.016.655
24020206.02
138869.0
2
116.655
2
1
2
12112
=
−+=
−+−+=
++= θθ
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Solution Cont
The exact solution of the ordinary differential equation is given by thesolution of a non-linear equation as
300−θ
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( ) 9282.21022067.00033333.0tan8519.1300
300ln92593.0 31
−×−=−+
− −− tθθ
θ
The solution to this nonlinear equation at t=480 seconds is
K57.647)480( =θ
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Comparison with exact results
800
1200
Tem
pera
ture
,θ(K
)
Exact h=120
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Figure 2. Heun’s method results for different step sizes
-400
0
400
0 100 200 300 400 500
Time, t(sec)
Tem
pera
ture
,
h=240
h=480
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Effect of step size
Table 1. Temperature at 480 seconds as a function of step size, h
Step size, h θ(480) Et |єt|%
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480
240
120
60
30
−393.87
584.27
651.35
649.91
648.21
1041.4
63.304
−3.7762
−2.3406
−0.63219
160.82
9.7756
0.58313
0.36145
0.097625
K57.647)480( =θ (exact)
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Effects of step size on Heun’s Method
400
600
800T
em
pera
ture
,θ(4
80)
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Figure 3. Effect of step size in Heun’s method
-400
-200
0
200
0 100 200 300 400 500
Step size, h
Tem
pera
ture
,
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Step size,
h
θ(480)
Euler Heun Midpoint Ralston
Comparison of Euler and Runge-Kutta 2nd Order Methods
Table 2. Comparison of Euler and the Runge-Kutta methods
Euler Heun Midpoint Ralston
480
240
120
60
30
−987.84
110.32
546.77
614.97
632.77
−393.87
584.27
651.35
649.91
648.21
1208.4
976.87
690.20
654.85
649.02
449.78
690.01
667.71
652.25
648.61
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K57.647)480( =θ (exact)
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Comparison of Euler and Runge-Kutta 2nd Order Methods
Table 2. Comparison of Euler and the Runge-Kutta methods
Step size, %t∈
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hEuler Heun Midpoint Ralston
480
240
120
60
30
252.54
82.964
15.566
5.0352
2.2864
160.82
9.7756
0.58313
0.36145
0.097625
86.612
50.851
6.5823
1.1239
0.22353
30.544
6.5537
3.1092
0.72299
0.15940
K57.647)480( =θ (exact)
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Comparison of Euler and Runge-Kutta 2nd Order Methods
900
1000
1100
1200Tem
pera
ture
, Ralston
Midpoint
θ(K
)
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Figure 4. Comparison of Euler and Runge Kutta 2nd order methods with exact results.
500
600
700
800
900
0 100 200 300 400 500 600
Time, t (sec)
Tem
pera
ture
,
Analytical
Ralston
Euler
Heun
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http://numericalmethods.eng.usf.edu 19
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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 and MAPLE, blogs, related physical problems, please visit
http://numericalmethods.eng.usf.edu/topics/runge_kutta_2nd_method.html
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THE ENDTHE END
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