Runge-kutta Methods
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- Words: 598
- Pages: 5
Ryan Sullenberger 2 May 2007 ME102
Runge-Kutta Methods Fourth-Order RK Method 25.8 : Solve the following problem with the fourth-order RK method where y(0) = 4 and y`(0) = 0. Solve from x = 0 to 5 with h = 0.5. Plot the results.
DiffEQ and Initial Conditions:
d
2
dx
2
d y + 8⋅ y = 0 dx
y + 0.6⋅
d y = y1 dx
d
2
dx
2
y = −0.6
d y − 8y = −0.6y 1 − 8y 0 dx
d y(0) = 0 dx
y( 0) = 4
System of first order ODE's and IC's:
4 Y0 := 0
Y 1 D( x , Y) := −0.6⋅ Y1 − 8⋅ Y0
4th-order RK solution: N := 100
xY := rkfixed ( Y0 , 0 , 5 , N , D)
(
xY = x Y
0
Y
1
)
0
xY =
1
2
0
0
4
0
1
0.05
3.96
-1.571
2
0.1
3.844
-3.065
3
0.15
3.656
-4.454
4
0.2
3.401
-5.714
5
0.25
3.087
-6.825
6
0.3
2.721
-7.769
7
0.35
2.313
-8.532
8
0.4
1.871
-9.105
9
0.45
1.406
-9.482
10
0.5
0.926
-9.661
11
0.55
0.443
-9.644
12
0.6
-0.035
-9.439
13
0.65
-0.498
-9.054
14
0.7
-0.938
-8.502
15
0.75
-1.346
...
〈0〉 X := xY 〈1〉 S1 := xY 〈2〉 S2 := xY
Graphed solution: 10
5
S1 S2
0
−5
− 10
0
1
2
3 X
4
5
Solving the ODE by writing a program that engages fourth-order RK method: h := 0.5
N := 10
f1( z) := z
f2( y , z) := −0.6⋅ z − 8 ⋅ y
( x y z ) :=
y ←4 0
z ←0 0
x ←0 0
for i ∈ 0 .. N − 1 x
i+ 1
← (i + 1)⋅ h
( i) k12 ← f2( y , z ) i i
k11 ← f1 z
y_new ← y +
1
z_new ← z +
1
i
i
2 2
⋅ k11 ⋅ h ⋅ k12 ⋅ h
k21 ← f1( z_new) k22 ← f2( y_new , z_new) y_new ← y +
1
z_new ← z +
1
i
i
2 2
⋅ k21 ⋅ h ⋅ k22 ⋅ h
k31 ← f1( z_new) k32 ← f2( y_new , z_new) y_new ← y + k31 ⋅ h i
z_new ← z + k32 ⋅ h i
k41 ← f1( z_new) k42 ← f2( y_new , z_new) y z
i+ 1
i+ 1
←y +
1
←z +
1
i
(x y z )
i
6 6
⋅ ( k11 + 2 ⋅ k21 + 2 ⋅ k31 + k41 ) ⋅ h ⋅ ( k12 + 2 ⋅ k22 + 2 ⋅ k32 + k42 ) ⋅ h
Programs solutions: 0
x=
0
0
0
0
0
4
0
0
1
0.5
1
1.037
1
-9.289
2
1
2
-2.428
2
-3.197
3
1.5
3
-1.557
3
5.365
4
2
4
1.154
4
4.072
5
2.5
5
1.481
5
-2.333
6
3
6
-0.293
6
-3.638
7
3.5
7
-1.132
7
0.372
8
4
8
-0.185
8
2.66
9
4.5
9
0.724
9
0.656
10
5
10
0.378
10
-1.626
y=
z=
Compare to MathCAD's rkfixed solutions:
Programs solutions graphed:
10
10 y
5 y z
5
z 0
S1 S2
−5 − 10
0
1
2
3 x
4
5
0 −5 − 10
0
1
2
3
x, x, X, X
4
5
Runge-Kutta Methods Fourth-Order RK Method 25.8 : Solve the following problem with the fourth-order RK method where y(0) = 4 and y`(0) = 0. Solve from x = 0 to 5 with h = 0.5. Plot the results.
DiffEQ and Initial Conditions:
d
2
dx
2
d y + 8⋅ y = 0 dx
y + 0.6⋅
d y = y1 dx
d
2
dx
2
y = −0.6
d y − 8y = −0.6y 1 − 8y 0 dx
d y(0) = 0 dx
y( 0) = 4
System of first order ODE's and IC's:
4 Y0 := 0
Y 1 D( x , Y) := −0.6⋅ Y1 − 8⋅ Y0
4th-order RK solution: N := 100
xY := rkfixed ( Y0 , 0 , 5 , N , D)
(
xY = x Y
0
Y
1
)
0
xY =
1
2
0
0
4
0
1
0.05
3.96
-1.571
2
0.1
3.844
-3.065
3
0.15
3.656
-4.454
4
0.2
3.401
-5.714
5
0.25
3.087
-6.825
6
0.3
2.721
-7.769
7
0.35
2.313
-8.532
8
0.4
1.871
-9.105
9
0.45
1.406
-9.482
10
0.5
0.926
-9.661
11
0.55
0.443
-9.644
12
0.6
-0.035
-9.439
13
0.65
-0.498
-9.054
14
0.7
-0.938
-8.502
15
0.75
-1.346
...
〈0〉 X := xY 〈1〉 S1 := xY 〈2〉 S2 := xY
Graphed solution: 10
5
S1 S2
0
−5
− 10
0
1
2
3 X
4
5
Solving the ODE by writing a program that engages fourth-order RK method: h := 0.5
N := 10
f1( z) := z
f2( y , z) := −0.6⋅ z − 8 ⋅ y
( x y z ) :=
y ←4 0
z ←0 0
x ←0 0
for i ∈ 0 .. N − 1 x
i+ 1
← (i + 1)⋅ h
( i) k12 ← f2( y , z ) i i
k11 ← f1 z
y_new ← y +
1
z_new ← z +
1
i
i
2 2
⋅ k11 ⋅ h ⋅ k12 ⋅ h
k21 ← f1( z_new) k22 ← f2( y_new , z_new) y_new ← y +
1
z_new ← z +
1
i
i
2 2
⋅ k21 ⋅ h ⋅ k22 ⋅ h
k31 ← f1( z_new) k32 ← f2( y_new , z_new) y_new ← y + k31 ⋅ h i
z_new ← z + k32 ⋅ h i
k41 ← f1( z_new) k42 ← f2( y_new , z_new) y z
i+ 1
i+ 1
←y +
1
←z +
1
i
(x y z )
i
6 6
⋅ ( k11 + 2 ⋅ k21 + 2 ⋅ k31 + k41 ) ⋅ h ⋅ ( k12 + 2 ⋅ k22 + 2 ⋅ k32 + k42 ) ⋅ h
Programs solutions: 0
x=
0
0
0
0
0
4
0
0
1
0.5
1
1.037
1
-9.289
2
1
2
-2.428
2
-3.197
3
1.5
3
-1.557
3
5.365
4
2
4
1.154
4
4.072
5
2.5
5
1.481
5
-2.333
6
3
6
-0.293
6
-3.638
7
3.5
7
-1.132
7
0.372
8
4
8
-0.185
8
2.66
9
4.5
9
0.724
9
0.656
10
5
10
0.378
10
-1.626
y=
z=
Compare to MathCAD's rkfixed solutions:
Programs solutions graphed:
10
10 y
5 y z
5
z 0
S1 S2
−5 − 10
0
1
2
3 x
4
5
0 −5 − 10
0
1
2
3
x, x, X, X
4
5
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