8/19/2019 Komputasi Sistem Fisis 2

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Homework 2

Name : Nadya Amalia

Student ID : 20213042

Subject : Komputasi Sistem Fisis (FI5005)

Lecturer : Dr.rer.nat. Linus Ampang Pasasa

Use Runge-Kutta fourth order method for solving a system of first order Ordinary Differential

Equation.

ݕଵ  (ݔ) ̇ ଶݕ (ݔ) ̇

൨ =    ଵݕ0.5−4 − ଶݕ0.3  − ଵݕ0.1

൨ =  ( , (ݔ

 (0) = ݕ

ଵ(0)

ଶ(0)൨ = ቂݕ4

6ቃwith ℎ = 0.5, on the interval [0,10].

Solution

In vector format we write the Runge-Kutta fourth order method as

ାଵ +  = ଵ + ଵ) + ଶ2 + ଷ2  ,(ସ = 0,1,2,…

where

ଵ =  ଵଵଶଵ൨, ଶ  = ଵଶଶଶ൨, ଷ  =  ଵଷଶଷ൨, ସ  = 

ଵସଶସ൨

ଵ =  ℎ (ݔ, (ݕଵ), (ݕଶ)),  = 1,2.

ଶ =  ℎ ൬ݔ + ℎ2

, (+ ଵ)ݕ

1

2ଵଵ, (ݕଶ) +

1

2ଶଵ൰ ,  = 1,2.

ଷ =  ℎ ൬ݔ + ℎ2

, (+ ଵ)ݕ

1

2ଵଶ, (ݕଶ) +

1

2ଶଶ൰ ,  = 1,2.

ସ =  ℎ (ݔ + ℎ, (ݕଵ) + ଵଷ, (ݕଶ) + ଶଷ),  = 1,2.

In explicit form, we write the method as

(ଵ)ାଵݕ  = + ଵ)ݕ) 

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6( ଵଵ + 2 ଵଶ + 2 ଵଷ + ଵସ)

(ଶ)ାଵݕ  = + ଶ)ݕ) 

1

6( ଶଵ + 2 ଶଶ + 2 ଶଷ + ଶସ)

For = , we have , =  ))  = , ))  = ଵଵ  =  ℎ ଵ(ݔ, (ݕଵ), (ݕଶ)) = (0.5)(−0.5)(4) =  −1.000000

ଶଵ  =  ℎ ଶ(ݔ, (ݕଵ), (ݕଶ)) = (0.5)[4 − (0.3)(6) − (0.1)(4)] = 0.900000

ଵଶ  =  ℎ ଵ ൬ݔ + ℎ2

,   (+ ଵ)ݕ

1

2ଵଵ,   + ଶ)ݕ)

1

2ଶଵ൰

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= (0.5)(−0.5) ቂ4 + ଵଶ (−1.000000)ቃ =  −0.875000

ଶଶ  =  ℎ ଶ ൬ݔ + ℎ2

,  (+ ଵ)ݕ

1

2ଵଵ, + ଶ)ݕ) 

1

2ଶଵ൰

= (0.5)

ቄ4 − (0.3)

ቂ6 +

ଵ

ଶ(0.900000)

ቃ− (0.1)

ቂ4 +

ଵ

ଶ(−1.000000)

ቃቅ = 0.857500

ଵଷ  =  ℎ ଵ ൬ݔ + ℎ2

+ ଵ)ݕ) ,1

2ଵଶ, (ݕଶ) +

1

2ଶଶ൰

= (0.5)(−0.5) ቂ4 + ଵଶ (−0.875000)ቃ =  −0.890625

ଶଷ  =  ℎ ଶ ൬ݔ + ℎ2

, (+ ଵ)ݕ

1

2ଵଶ, (ݕଶ) +

1

2ଶଶ൰

= (0.5) ቄ4 − (0.3) ቂ6 + ଵଶ (0.857500)ቃ− (0.1) ቂ4 +ଵଶ (−0.875000)ቃቅ = 0.857563

ଵସ  =  ℎ ଵ(ݔ + ℎ, (ݕଵ) + ଵଷ, (ݕଶ) + ଶଷ)

= (0.5)(−0.5)[4 + (−0.890625)] =  −0.777344

ଶସ  =  ℎ ଶ(ݔ + ℎ, (ݕଵ) + ଵଷ, (ݕଶ) + ଶଷ)= (0.5){4 − (0.3)(6 + 0.857563) − (0.1)[4 + (−0.890625)]} = 0.815897

= ଵ)ଵݕ) = (ଵ(0.5ݕ + ଵ)ݕ) 1

6( ଵଵ + 2 ଵଶ + 2 ଵଷ + ଵସ)

= 4 +1

6[(−1.000000) + (2)(−0.875000) + (2)(−0.890625) + (−0.777344)]

= 3.115234

= ଶ)ଵݕ) = (ଶ(0.5ݕ + ଶ)ݕ) 1

6 ( ଶଵ + 2 ଶଶ + 2 ଶଷ + ଶସ)

= 6 +1

6[(0.900000) + (2)(0.857500) + (2)(0.857563) + (0.815897)]

= 6.857670

For = , we have , = . ))  = ., ))  = .ૡૠૠଵଵ  =  ℎ ଵ(ݔଵ, (ݕଵ)ଵ, (ݕଶ)ଵ) = (0.5)(−0.5)(3.115234) =  −0.778809

ଶଵ  =  ℎ ଶ(ݔଵ, (ݕଵ)ଵ, (ݕଶ)ଵ) = (0.5)[4 − (0.3)(6.857670) − (0.1)(3.115234)] = 0.815588

ଵଶ  =  ℎ ଵ ൬ݔଵ +ℎ

2 ,   + ଵ)ଵݕ)1

2 ଵଵ,   + ଶ)ଵݕ)1

2 ଶଵ൰= (0.5)(−0.5) ቂ3.115234 + ଵଶ (−0.778809)ቃ =  −0.681458

ଶଶ  =  ℎ ଶ ൬ݔଵ + ℎ2

,   + ଵ)ଵݕ)1

2ଵଵ,   + ଶ)ଵݕ)

1

2ଶଵ൰

= (0.5) ൜4 − (0.3) 6.857670 + 12

(0.815588)൨− (0.1) 3.115234 + 12

(−0.778809)൨ൠ= 0.773889

ଵଷ =  ℎ

ଵ൬ݔ

ଵ +

ℎ

2

ݕ) ,

ଵ)

ଵ +

1

2 ଵଶݕ) ,

ଶ)

ଵ +

1

2 ଶଶ൰

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= (0.5)(−0.5) 3.115234 + 12

(−0.681458)൨ =  −0.693626

ଶଷ  =  ℎ ଶ ൬ݔଵ + ℎ2

, (+ ଵ)ଵݕ

1

2ଵଶ, (ݕଶ)ଵ +

1

2ଶଶ൰

= (0.5) ቄ4 − (0.3) ቂ6.857670 +ଵଶ (0.773889)ቃ− (0.1) ቂ3.115234 +

ଵଶ (−0.681458)ቃቅ

= 0.774583

ଵସ  =  ℎ ଵ(ݔଵ + ℎ, (ݕଵ)ଵ + ଵଷ, (ݕଶ)ଵ + ଶଷ)= (0.5)(−0.5)[3.115234 + (−0.693626)] =  −0.605402

ଶସ  =  ℎ ଶ(ݔଵ + ℎ, (ݕଵ)ଵ + ଵଷ, (ݕଶ)ଵ + ଶଷ)= (0.5){4 − (0.3)(6.857670 + 0.774583) − (0.1)[3.115234 + (−0.693626)]} = 0.734082

= ଵ)ଶݕ) = (ଵ(1.0ݕ + ଵ)ଵݕ) 1

6( ଵଵ + 2 ଵଶ + 2 ଵଷ + ଵସ)

= 3.115234 + 16

[(−0.778809) + (2)(−0.681458) + (2)(−0.693626) + (−0.605402)]

= 2.426171

+ ଶ)ଵݕ) = ଶ)ଶݕ) = (ଶ(1.0ݕ1

6( ଶଵ + 2 ଶଶ + 2 ଶଷ + ଶସ)

= 6.857670 +1

6[(0.815588) + (2)(0.773889) + (2)(0.774583) + (0.734082)]

= 7.632106

Solution obtained with a computer program for the fourth-order RK method. The plot below

represents solutions for (ݔ)ଵݕ and (ݔ)ଶݕ for ݔ = 0 until ݔ = 10, with ℎ  = 0.5.

0,000000

2,000000

4,000000

6,000000

8,000000

10,000000

12,000000

14,000000

0,0 2,0 4,0 6,0 8,0 10,0 12,0

y1

y2