SE301: Numerical Methods Topic 8 Ordinary Differential Equations (ODEs) Lecture 28-36

CISE301_Topic8L4&5 1

SE301: Numerical Methods

Topic 8 Ordinary Differential

Equations (ODEs)Lecture 28-36

KFUPM(Term 101)Section 04

Read 25.1-25.4, 26-2, 27-1


Outline of Topic 8 Lesson 1: Introduction to ODEs Lesson 2: Taylor series methods Lesson 3: Midpoint and Heun’s method Lessons 4-5: Runge-Kutta methods Lesson 6: Solving systems of ODEs Lesson 7: Multiple step Methods Lesson 8-9: Boundary value Problems


Lecture 31Lesson 4: Runge-Kutta

Methods


Learning Objectives of Lesson 4 To understand the motivation for using

Runge Kutta method and the basic idea used in deriving them.

To Familiarize with Taylor series for functions of two variables.

Use Runge Kutta of order 2 to solve ODEs.


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.


Second Order Runge-Kutta Method

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 Two Variables

The Taylor Series discussed in Chapter 4 is extended to the 2-independent

variable case.This is used to prove RK formula.


Taylor Series in One Taylor Series in One VariableVariable

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 OrderRunge-Kutta Methods – 1 of 5

)(),('2

),(

:as writtenis which

)(2

),( :ODE solve toUsed

ExpansionSeriesTaylor Order Second

32

1

32

22

1

hOyxfh

yxfhyy

hOdx

ydh

dx

dyhyy

yxfdx

dy

iiiiii

ii



)(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 Taylor Series in Two VariablesVariables

),( 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



) ,(),(

:ngSubstituti

) ,(

),(

thatsuch ,,,:Problem

1211

22111

12

1

21

Kyhxfhwyxfhwyy

KwKwyy

KyhxfhK

yxfhK

wwFind

iiiiii

ii

ii

ii



...),( ),( )(

... ),( )(

...),( ),(

:

...),() ,(

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



1 ,1 :solution possible One

solutions infinite unknowns 4 withequations 32

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


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

) , (

),(

FormeAlternativ

kwkwhyy

khyhxfk

yxfk

ii

ii

ii

22111

12

1

) , (

),(

KuttaeOrder Rung Second

KwKwyy

KyhxfhK

yxfhK

ii

ii

ii


Choosing , , w1 and w2

Corrector Singlea with' is This

),(),(22

1

),(

),(

:becomes methodKutta -eOrder Rung Second2

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

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 MethodsAlternative Formulas

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


Second order Runge-Kutta Method Example

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


Second order Runge-Kutta Method Example

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


Lecture 32Lesson 5: Applications of Runge-Kutta Methods to Solve First Order ODEs

Using Runge-Kutta methods of differentorders to solve first order ODEs


2nd Order Runge-Kutta

RK2

)( 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


Higher-Order Runge-Kutta

Higher order Runge-Kutta methods are available.

Derived similar to second-order Runge-Kutta.

Higher order methods are more accurate butrequire more calculations.


3rd Order Runge-Kutta

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

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


Example4th-Order Runge-Kutta Method

)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


4th Order Runge-Kutta

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

5.0,0

1),(

0.2

00

2

yx

xyyxf

h

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

See RK4 Formula


Example: RK4

8293.0,2.0

1),(

0.2

11

2

yx

xyyxf

h

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


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

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.


Remaining Lessons in Topic 8Lesson 6:Solving Systems of high order ODE

Lesson 7:Multi-step methods

Lessons 8-9:Methods to solve Boundary Value Problems

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SE301: Numerical Methods Topic 8 Ordinary Differential Equations (ODEs) Lecture 28-36

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