PERSAMAAN DIFERENSIAL - kuliah.ftsl.itb.ac.id · lebih variabel bebas. •Dibedakan menurut: •Tipe (ordiner/biasa atau parsial) •Orde (ditentukan oleh turunan tertinggi yang ada)

PERSAMAAN DIFERENSIAL

(DIFFERENTIAL EQUATION)

M E T O D E E U L E R

M E T O D E R U N G E - K U T T A


• Persamaan paling penting dalam bidang rekayasa,

paling bisa menjelaskan apa yang terjadi dalam

sistem fisik.

• Menghitung jarak terhadap waktu dengan

kecepatan tertentu, 50 misalnya.

50dt

dx

Rate equations


• Solusinya, secara analitik dengan integral,

• C adalah konstanta integrasi

• Artinya, solusi analitis tersebut terdiri dari banyak

‘alternatif’

• C hanya bisa dicari jika mengetahui nilai x dan t.

Sehingga, untuk contoh di atas, jika x(0) = (x saat

t=0) = 0, maka C = 0

dtdx 50 Ctx 50

KLASIFIKASI PERSAMAAN DIFERENSIAL

Persamaan yang mengandung turunan dari satu

atau lebih variabel tak bebas, terhadap satu atau

lebih variabel bebas.

• Dibedakan menurut:

• Tipe (ordiner/biasa atau parsial)

• Orde (ditentukan oleh turunan tertinggi yang ada)

• Liniarity (linier atau non-linier)

SOLUSI PERSAMAAN DIFERENSIAL

• Secara analitik, mencari solusi persamaan

diferensial adalah dengan mencari fungsi integral

nya.

• Contoh, untuk fungsi pertumbuhan secara

eksponensial, persamaan umum:

kPdt

dP

Rate equations

But what you really want to know is…

the sizes of the boxes (or state variables) and how they change through time

That is, you want to know:

the state equations

There are two basic ways of finding the state equations for the state variables based on your known rate equations:

1) Analytical integration

2) Numerical integration

Suatu kultur bakteria tumbuh dengan kecepatan

yang proporsional dengan jumlah bakteria yang

ada pada setiap waktu. Diketahui bahwa jumlah

bakteri bertambah menjadi dua kali lipat setiap 5

jam. Jika kultur tersebut berjumlah satu unit pada

saat t = 0, berapa kira-kira jumlah bakteri setelah

satu jam?

• Jumlah bakteri menjadi dua kali lipat setiap 5 jam, maka k = (ln 2)/5

• Jika P0 = 1 unit, maka setelah satu jam…

SOLUSI PERSAMAAN DIFERENSIAL

kPdt

dP

dtkP

dPt

t

P

P

1

0

1

0

)(ln 0

0

ttCkP

P

ktePtP 0)(

)(1)1()1)(

5)2(ln

(eP

1487.1

Rate equation State equation (dsolve in Maple)

The Analytical Solution of the Rate

Equation is the State Equation

THERE ARE VERY FEW MODELS IN ECOLOGY THAT CAN BE SOLVED

ANALYTICALLY.

SOLUSI NUMERIK

• Numerical integration

• Eulers

• Runge-Kutta

Numerical integration makes use of this relationship:

Which you’ve seen before…

Relationship between continuous and discrete time models

*You used this relationship in Lab 1 to program the

logistic rate equation in Visual Basic:

1 where,11

tt

K

NrNNN t

ttt

tdt

dyyy ttt

, known

Fundamental Approach of Numerical Integration

y = f(t), unknown

t, specified

y

t

yt, known

dt

dy

yt+t, estimated

tdt

dyyy ttt

yt+t,

unknown

Euler’s Method: yt+ t ≈ yt + dy/dt t

1 where,1

tt

K

NrNNN t

tttt

dtdN

Calculate dN/dt*1

at Nt

Add it to Nt to

estimate Nt+ t

Nt+ t becomes the new Nt

Calculte dN/dt * 1 at new Nt

Use dN/dt to estimate next Nt+ t

Repeat these steps to estimate the state

function over your desired time length

(here 30 years)

Nt/K with time, lambda = 1.7, time step = 1

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0 10 20 30 40 50

time (years)

Nt/K

EXAMPLE OF NUMERICAL INTEGRATION

dy

dty y 6 007 2.

Analytical solution to dy/dt

Y0 = 10

t = 0.5

point to

estimate

y

Euler’s Method: yt+ t ≈ yt + dy/dt t

yt = 10

m1 = dy/dt at yt

m1 = 6*10-.007*(10)2

y = m1*t

yest= yt + y

t = 0.5

y

estimated y(t+ t)

analytical y(t+ t)

dy

dty y 6 007 2.

20

RUNGE-KUTTA

METHODS

MOTIVATION

• We seek accurate methods to solve ODEs that do

not require calculating high order derivatives.

• The approach is to use a formula involving

unknown coefficients then determine these

coefficients to match as many terms of the Taylor

series expansion.

21

SECOND ORDER RUNGE-KUTTA METHOD

22

possible. as accurate as is that such

,,,

:Problem

) ,(

),(

1

21

22111

12

1

i

ii

ii

ii

y

wwFind

KwKwyy

KyhxfhK

yxfhK

TAYLOR SERIES IN ONE VARIABLE

23

hxandxbetweenisxwhere

xfn

hxf

i

hhxf

f(x)n

nn

i

n

i

i

)()!1(

)(!

)(

of expansion SeriesTaylor order The

)1(1

)(

0

th

Approximation Error

DERIVATION OF 2ND ORDER RUNGE-KUTTA METHODS – 1 OF 5

24

)(),('2

),(

:as writtenis which

)(2

),( :ODE solve toUsed

ExpansionSeriesTaylor Order Second

32

1

3

2

22

1

hOyxfh

yxfhyy

hOdx

ydh

dx

dyhyy

yxfdx

dy

iiiiii

ii


25

)(2

),(),(

:

),(),(),(

),('

ationdifferenti rule-chainby obtained is ),(' where

32

1 hOh

yxfy

f

x

fhyxfyy

ngSubstituti

yxfy

f

x

f

dx

dy

y

yxf

x

yxfyxf

yxf

iiiiii

TAYLOR SERIES IN TWO VARIABLES

26

),( and ),( between joining line theon is),(

),()!1(

1 ),(

!

1

...

2!2

1

),(),(

1

0

2

2

22

2

22

kyhxyxyx

errorionapproximat

yxfy

kx

hn

yxfy

kx

hi

yx

fhk

y

fk

x

fh

y

fk

x

fhyxfkyhxf

nn

i

i


27

) ,(),(

:ngSubstituti

) ,(

),(

thatsuch ,,,:Problem

1211

22111

12

1

21

Kyhxfhwyxfhwyy

KwKwyy

KyhxfhK

yxfhK

wwFind

iiiiii

ii

ii

ii


28

...),( ),( )(

... ),( )(

...),( ),(

:

...),() ,(

22

22211

12211

1211

11

iiiiii

iiii

iiiiii

iiii

yxfy

fhw

x

fhwyxfhwwyy

y

fK

x

fhhwyxfhwwyy

y

fK

x

fhyxfhwyxfhwyy

ngSubstituti

y

fK

x

fhyxfKyhxf


29

2

1 ,1 :solution possible One

solutions infinite unknowns 4 withequations 3

2

1 and ,

2

1 , 1

:equations threefollowing theobtain we terms,Matching

)(2

),(),(

...),( ),( )(

:for expansions twoderived We

21

2221

32

1

22

22211

1

ww

wwww

hOh

yxfy

f

x

fhyxfyy

yxfy

fhw

x

fhwyxfhwwyy

y

iiiiii

iiiiii

i

2ND ORDER RUNGE-KUTTA METHODS

30

2

1 and ,

2

1 ,1

: thatsuch ,,, Choose

) ,(

),(

2221

21

22111

12

1

wwww

ww

KwKwyy

KyhxfhK

yxfhK

ii

ii

ii

ALTERNATIVE FORM

22111

12

1

) , (

),(

KuttaeOrder Rung Second

KwKwyy

KyhxfhK

yxfhK

ii

ii

ii

31

22111

12

1

) , (

),(

FormeAlternativ

kwkwhyy

khyhxfk

yxfk

ii

ii

ii

CHOOSING , , W1 AND W2

32

Corrector Singlea with' is This

),(),(22

1

),(

),(

:becomes methodKutta -eOrder Rung Second

2

1 ,1 then,1 choosing example,For

011211

12

1

21

s Method Heun

yxfyxfh

yKKyy

KyhxfhK

yxfhK

ww

iiiiiii

ii

ii

CHOOSING , , W1 AND W2

33

Method Midpoint theis This

)2

,2

(

)2

,2

(

),(

:becomes methodKutta -eOrder Rung Second

1 ,0 ,2

1 then

2

1 Choosing

121

12

1

21

Ky

hxfhyKyy

Ky

hxfhK

yxfhK

ww

iiiii

ii

ii

2ND ORDER RUNGE-KUTTA METHODS ALTERNATIVE FORMULAS

34

211

1i2

1

2

1

2

11

) ,(

),(

)0(select mulas Kutta ForeOrder Rung Second

KKyy

KyhxfhK

yxfhK

ii

i

ii

2

11 ,

2

1, :number nonzeroany Pick

1 ,2

1,

2

1

12

2122

ww

wwww


EXAMPLE

CISE301_Topic8L4&5

35

8269.32/)1662.018.0(4

2/)1()01.01(

1662.0))01.()18.0(1(01.0

),(

18.0)1(01.0)4 ,1(

:1STEP

1 ,01.0 ,4)1(,1)(

RK2using (1.02) find tosystem following theSolve

21

30

20

1002

30

20001

32

KKxx

tx

KxhtfhK

txxtfhK

hxtxtx

x


EXAMPLE

36

6662.3)1546.01668.0(2

18269.3

2

1)01.1()01.001.1(

1546.0))01.()1668.0(1(01.0

),(

1668.0)1(01.0)8269.3,01.1(

2 STEP

21

31

21

1112

31

21111

KKxx

tx

KxhtfhK

txxtfhK

1 RK2,Using

[1,2]for t Solution

,4)1(,)(1)( 32

xttxtx

37

2ND ORDER RUNGE-KUTTA

)( iserror global and )( iserror Local

2

) ,(

),(

corrector singlea withmethod s Heun' toEquivalent

RK2as Know 1, of valueTypical

23

211

12

1

hOhO

kkh

yy

hkyhxfk

yxfk

ii

ii

ii

38

RK2

HIGHER-ORDER RUNGE-KUTTA

39

Higher order Runge-Kutta methods are available. Derived similar to second-order Runge-Kutta. Higher order methods are more accurate but require more calculations.

3RD ORDER RUNGE-KUTTA

40

RK3

)( iserror Global and )( iserror Local

46

)2 ,(

)2

1 ,

2(

),(

RK3asKnow

34

3211

213

12

1

hOhO

kkkh

yy

hkhkyhxfk

hkyh

xfk

yxfk

ii

ii

ii

ii

4TH ORDER RUNGE-KUTTA

41

RK4


226

) ,(

)2

1 ,

2(

)2

1 ,

2(

),(

45

43211

3i4

2i3

12

1

hOhO

kkkkh

yy

hkyhxfk

hkyh

xfk

hkyh

xfk

yxfk

ii

i

i

ii

ii

HIGHER-ORDER RUNGE-KUTTA

654311

543216

415

324

213

12

1

7321232790

)7

8

7

12

7

12

7

2

7

3 ,(

)16

9

16

3 ,

4

3(

)2

1 ,

2

1(

)8

1

8

1 ,

4

1(

)4

1 ,

4

1(

),(

kkkkkh

yy

hkhkhkhkhkyhxfk

hkhkyhxfk

hkhkyhxfk

hkhkyhxfk

hkyhxfk

yxfk

ii

ii

ii

ii

ii

ii

ii

42

EXAMPLE 4TH-ORDER RUNGE-KUTTA METHOD

43

)4.0()2.0(4

2.0

5.0)0(

1 2

yandycomputetoRKUse

h

y

xydx

dy

RK4

EXAMPLE: RK4

)4.0(),2.0(4

5.0)0(,1

:Problem

2

yyfindtoRKUse

yxydx

dy

44

4TH ORDER RUNGE-KUTTA

45

RK4


226

) ,(

)2

1 ,

2(

)2

1 ,

2(

),(

45

43211

3i4

2i3

12

1

hOhO

kkkkh

yy

hkyhxfk

hkyh

xfk

hkyh

xfk

yxfk

ii

i

i

ii

ii

EXAMPLE: RK4

8293.0226

7908.12.016545.01),(

654.11.0164.01)2

1,

2

1(

64.11.015.01)2

1,

2

1(

5.1)1( ),(

432101

2003004

2002003

2001002

200001

kkkkh

yy

xyhkyhxfk

xyhkyhxfk

xyhkyhxfk

xyyxfk

)4.0(),2.0(4

5.0)0(,1

:Problem

2

yyfindtoRKUse

yxydx

dy

5.0,0

1),(

0.2

00

2

yx

xyyxf

h

46

See RK4 Formula

Ste

p 1

EXAMPLE: RK4

2141.1226

2.0

0555.2),(

9311.1)2

1,

2

1(

9182.1)2

1,

2

1(

1.7893 ),(

432112

3114

2113

1112

111

kkkkyy

hkyhxfk

hkyhxfk

hkyhxfk

yxfk

)4.0(),2.0(4

5.0)0(,1

:Problem

2

yyfindtoRKUse

yxydx

dy

8293.0,2.0

1),(

0.2

11

2

yx

xyyxf

h

47

Ste

p 2

EXAMPLE: RK4

)4.0(),2.0(4

5.0)0(,1

:Problem

2

yyfindtoRKUse

yxydx

dy

xi yi

0.0 0.5

0.2 0.8293

0.4 1.2141

48

Summary of the solution

SUMMARY

• Runge Kutta methods generate an accurate

solution without the need to calculate high order

derivatives.

• Second order RK have local truncation error of

order O(h3) and global truncation error of order

O(h2).

• Higher order RK have better local and global

truncation errors.

• N function evaluations are needed in the Nth order

RK method.

49

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PERSAMAAN DIFERENSIAL - kuliah.ftsl.itb.ac.id · lebih variabel bebas. •Dibedakan menurut: •Tipe (ordiner/biasa atau parsial) •Orde (ditentukan oleh turunan tertinggi yang ada)