cise301 : numerical methods topic 8 ordinary differential equations (odes) lecture 28-36

CISE301_Topic8L4&5 KFUPM 1

CISE301: Numerical Methods

Topic 8 Ordinary Differential

Equations (ODEs)Lecture 28-36

KFUPM

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 (RK) method and the basic idea used in deriving them.

To get familiar with Taylor series for functions of two variables.

To use RK method 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 as possible.


Runge-Kutta Method

1

2 1

1 1 2 2

1 2

( , )( , )

( ) (

Second Order Rung

)Problem:

, , ,such that is as accurate as poss

e-Kutta (

ible

RK2

.

)K h f t xK h f t h x Kx t h x t w K w K

Find w wx(t h)


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 Variable

1( ) ( )

0

The Taylor Series expansion of ( )

( ) ( ) ( )! !

i nni n

i

f x

h hf x h f x f xi n

where x is between x and x h

Approximation Error


Taylor Series in One Variable- Another Look -

hxxx

xfdxdh

nxf

dxdh

ihxf

f(x)

hxfdxxfdhxf

dxdh

nn

i

i

iii

ii

i

and between is

)(!

1)(!1)(

ofexpansion SeriesTaylor The

)()()(

Define

1

0

)(


Definitions

2

22

2

2

22

2

1

0

),(),(2),(),(

),(),(),(

),(),(

),(

yyxfk

yxyxfkh

xyxfhyxf

yk

xh

yyxfk

xyxfhyxf

yk

xh

yxfyxfy

kx

h

xfhyxf

xh

Define

i

ii

i


Taylor Series Expansion

)0,0(

''

)0.0(

1

)0.0(

0

2

),(

4),(

(0,0)at evaluated sderivative Parial)2)(1(,

yx fkfhyxfy

kx

h

yxfy

kx

h

yxxy)f(x


Taylor Series in Two Variables

),( and ),(between joining line on the is),(

),(!

1),(!

1),(

, ofexpansion SeriesTaylor The1

0

kyhxyxyxerrorionapproximat

yxfy

kx

hn

yxfy

kx

hi

kyhxf

y)f(xnn

i

i

x x+h

y

y+k


Runge-Kutta Method

1

2 1

1 1 2 2

1 2

( , )( , )

( ) ( )Problem:

RK

, , ,such that is as accurate as possibl

2

e.

K h f t xK h f t h x Kx t h x t w K w K

Find w wx(t h)


Runge-Kutta Method

)( ),()()(

),(21 ),(

)),(,( ),()()(

...)('''6

)(''2

)(')()(

possible. as many terms asmatch to,,, :Problem

3'22

'2221

2''

21

32

21

hOffhwfhwxtfhwwtxhtx

xtfx

ht

hffhfhffhxhtf

xtfhxhtfhwxtfhwtxhtx

txhtxhthxtxhtx

wwFind

xt

xt


Runge-Kutta Method

)( ),()()(

...)('''6

)(''2

)(')()(

3'22

'2221

32

hOffhwfhwxtfhwwtxhtx

txhtxhthxtxhtx

xt

1,1,5.0,5.0solution possibleOne

5.0,5.0,1

21

2221

ww

wwww


Runge-Kutta Method

1

2 1

1 2

( , )( , )

1( ) ( )2

RK2K h f t xK h f t h x K

x t h x t K K


Runge-Kutta MethodAlternative Formula

1

2 1

1 2

( , )( , )

( ) ( )2

RK2F f t xF f t h x hF

hx t h x t F F


Runge-Kutta MethodAlternative Formula

1

2 1

1 2

( , )(

Alternative F

, )

( ) ( )2

ormF f t xF f t h x hF

hx t h x t F F

1

2 1

1 2

( , )( , )

1( ) ( )2

RK2K h f t xK h f t h x K

x t h x t K K


Runge-Kutta MethodAlternative Formulas

1

2 1

1 2

RK2 Formulas (select 0) ( , )( , )

1 1( ) ( ) 12 2

K h f t xK h f t h x K

x t h x t F F

21,

211,

number zero-nonany Picksolutionanother

5.0,5.0,1

21

2221

ww

wwww


Runge-Kutta Method

1

2 1

3 2

4 3

1 2 3 4

( , )1 1( , )2 21 1( , )2 2

( , )

Fourth Order Runge-Kutta (R

1( ) ( ) 2 26

K4)K h f t x

K h f t h x K

K h f t h x K

K h f t h x K

x t h x t K K K K


Second order Runge-Kutta Method Example

2 3

Solve the following system to find (1.02) using RK2

( ) 1 ( ) , (1) 4, 0.01

x

x t x t t x h



2 3

2 31

2 32 1

1 2

Solve the following system to find (1.02) using RK2

( ) 1 ( ) , (1) 4, 0.01

STEP1:

( , ) 0.01(1 ) 0.18

( , ) 0.01(1 ( 0.18) ( .01) ) 0.16621 1(1 0.01) (1) 4 (0.182 2

x

x t x t t x h

K h f t x x t

K h f t h x K x t

x x K K

0.1662) 3.8269



6662.3)1546.01668.0(218269.3

21)01.1()01.001.1(

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

1668.0)1(01.0),(

2 STEP

21

3212

321

KKxx

txKxhtfhK

txxtfhK


RK2 Using [1,2]for tSolution

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

xttxtx


Summary RK methods generate an accurate solution

without the need to calculate high order derivatives.

Second order RK have local truncation error of order O(h3).

Fourth order RK have local truncation error of order O(h5).

N function evaluations are needed in the Nth order RK method.


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


Learning Objectives of Lesson 5 Use Runge-Kutta methods of different

orders to solve first order ODEs.


Runge-Kutta Method

1

2 1

1 1 1 2 2

1 2

1

( , )( , )

Problem:, , ,

such that i

Second Order Runge Kutta (

s as accurate as possible.

RK2)

i i

i i

i i

i

K f x yK f x h y K hy y w K w K

Find w wy


Runge-Kutta Methods

RK2

1

2 1

1 1 2

Second Order Runge-Kutta ( , )( , )

R

2

( K2)

i i

i i

i i

K f x yK f x h y K h

hy y K K


Runge-Kutta Methods

1

2 1

3 1 2

1 2 3

(Third

, )1 1( , )2 2

( , 2 )1( ) ( ) 46

Order Runge Kutta (RK3)

i i

i i

i i

K f x y

K f x h y K h

K f x h y K h K h

y x h y x K K K

RK3


Runge-Kutta Methods

1

2 1

3 2

4 3

1 1 2 3 4

( , )1 1( , )2 21 1( , )2

Fourth Order R

2(

unge Kutta (

, )

2 2

K4

6

R )

i i

i i

i i

i i

i i

K f x y

K f x h y K h

K f x h y K h

K f x h y K hhy y K K K K

RK4


Runge-Kutta Methods

Higher order Runge-Kutta methods are available.

Higher order methods are more accurate butrequire more calculations.

Fourth order is a good choice. It offers good accuracy with a reasonable calculation effort.


Fifth Order Runge-Kutta Methods

654311

543216

415

324

213

12

1

7321232790

)78

712

712

72

73,(

)169

163,

43(

)21,

21(

)81

81,

41(

)41,

41(

),(

KKKKKhyy

hKhKhKhKhKyhxfK

hKhKyhxfK

hKhKyhxfK

hKhKyhxfK

hKyhxfK

yxfK

ii

ii

ii

ii

ii

ii

ii


Second Order Runge-Kutta Method

needed steps of# Determine

)(

),(

:Given

00

hyxy

yxfdxdy



211

12

1

00

2

),(),(

2needed steps of# Determine

)(

),(

:Given

KKhyy

hKyhxfKyxfK

formulaRK

hyxy

yxfdxdy

ii

ii

ii



2112

1112

111

01

2101

1002

001

2

),(),(

:2 Step

2

),(),(

:1 Step

KKhyy

hKyhxfKyxfK

hxx

KKhyy

hKyhxfKyxfK

211

12

1

00

2

),(),(

2needed steps of# Determine

)(

),(

:Given

KKhyy

hKyhxfKyxfK

formulaRK

hyxy

yxfdxdy

ii

ii

ii


Example 1Second Order Runge-Kutta Method

)02.1(),01.1(findto

4)1(,1

equation aldifferenti thesolve tomethod Kutta Rungeorder second theUse

32

yy

yxydxdy



)02.1(),01.1(2

4)1(,1

:Problem

32

yyfindtoRKUse

yxydxdy

4,1

1),(

0.01h

00

32

yxxyyxf



8269.3)62.1618(201.04

2

62.16))01.()18.0(1(),(

0.18)1(),(

:1 Step

2101

30

201002

30

20001

KKhyy

xyhKyhxfK

xyyxfK

)02.1(),01.1(2

4)1(,1

:Problem

32

yyfindtoRKUse

yxydxdy

4,1

1),(

0.01h

00

32

yxxyyxf



)02.1(),01.1(2

4)1(,1

:Problem

32

yyfindtoRKUse

yxydxdy

8269.3,01.11),(

0.01h

11

32

yxxyyxf



6662.3)46.1568.16(201.08269.3

2

46.15))01.()1668.0(1(),(

68.16)1(),(

:2 Step

2112

31

211112

31

21111

KKhyy

xyhKyhxfK

xyyxfK

)02.1(),01.1(2

4)1(,1

:Problem

32

yyfindtoRKUse

yxydxdy

8269.3,01.11),(

0.01h

11

32

yxxyyxf


Example 1Summary of the solution

6662.302128269.301110000.40010

...

yxi ii

)02.1(),01.1(2

4)1(,1

:Problem

32

yyfindtoRKUse

yxydxdy

Summary of the solution


Solution after 100 steps


Example 24th-Order Runge-Kutta Method

)4.0()2.0(42.0

5.0)0(

1 2

yandycomputetoRKUsehy

xydxdy

See RK4 Formula


Example 2Fourth Order Runge-Kutta Method

)4.0(),2.0(4

5.0)0(,1

:Problem

2

yyfindtoRKUse

yxydxdy


Example 2Fourth Order Runge-Kutta Method

5.0,01),(

0.2h

00

2

yxxyyxf

8293.0226

7908.12.016545.01),(

654.11.0164.01)21,

21(

64.11.015.01)21,

21(

5.11),(

:1Step

432101

2003004

2002003

2001002

200001

KKKKhyy

xyhKyhxfK

xyhKyhxfK

xyhKyhxfK

xyyxfK

)4.0(),2.0(4

5.0)0(,1

:Problem

2

yyfindtoRKUse

yxydxdy

See RK4 Formula


Runge-Kutta Methods

43211

34

23

12

1

226

),(

)21,

21(

)21,

21(

),((RK4) Kutta RungeOrder Fourth

KKKKhyy

hKyhxfK

hKyhxfK

hKyhxfK

yxfK

ii

ii

ii

ii

ii

RK4


Example 2 Fourth Order Runge-Kutta Method

8293.0,2.01),(

0.2h

00

2

yxxyyxf

2141.12262.0

0555.2),(

9311.1)21,

21(

9182.1)21,

21(

1.7893 ),(:2Step

432112

3114

2113

1112

111

KKKKyy

hKyhxfK

hKyhxfK

hKyhxfK

yxfK

)4.0(),2.0(4

5.0)0(,1

:Problem

2

yyfindtoRKUse

yxydxdy


Example 2Summary of the solution

2141.14.028293.02.015.00.00ii yxi

)4.0(),2.0(4

5.0)0(,1

:Problem

2

yyfindtoRKUse

yxydxdy

Summary of the solution


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

cise301 : numerical methods topic 8 ordinary differential equations (odes) lecture 28-36

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