ode numerik

8/13/2019 Ode Numerik

1/19

Numeri cal Di ff erentiation

DEFINITION

fdf

d

lim f f( )

( ) ( ) ( )x

x

x h

x h x

h0

Analytical solution employing this definition is demonstrated as follows

Example : What is y x( ) atx=agiven that f(x x) 3 ?Solution : Following the above definition;

y x hy x h y x

h

ha h a

h

h

a a h ah h a

h

ha h ah h

h

h a ah h

a

( )( ) ( )

( )

lim

lim

lim

lim

lim

0

0

0

3 3

03 3

0 3 3

3

3 3

3 2 2 3 3

2 2 3

2 2

2

Numerical solution of this problem is basically based on simplification on the limit term in such that value

of h h rather than h 0 . It is logical, therefore, the smaller one choose the value for hthe closerthe result to the true value (i.e. obtained by analytical solution). Accordingly, one way to express numerical

solution is as follows;

f

df

d

f f( )

( ) ( ) ( )x

x

x

x h x

h h x

or

fdf

d

f f( )

( ) ( ) ( )x

x

x

x x x

x


2/19

TAYLOR SERIES

More flexible way to formulate derivative term numerically is based on Taylor series expansion. Taylors

theorem states that:

f f d fd

d fd

d fd

d f

d

1 2 3

4

( )!

( )!

( )!

( )!

( )

!

( ).....

x x x x x xx

x xx

x xx

x x

x

0

0

0

1

01

2

02

3

03

40

4

0 1 2 3

4

[1a]

or

f f d f

d

d f

d

d f

d

d f

d

1 2 3

4

( )!

( )!

( )

!

( )

!

( )

!

( ).....

x x x

x x x

x

x x

x

x x

x

x x

x

0

0

0

10

1

20

2

30

3

40

4

0 1 2 3

4

[1b]

The above two series provide a mean to predict value of functionf(x)atx0+xwhen values off(x0),x0, and

xare given.

For partial derivative:

f ff f

f f

f f f

( , ) ( , )!

( , )

!

( , )

!( , )

!( , )

!

( , )

!

( , )

!

( , )

x x y y x yx x y

x

y x y

y

x x yx

y x yy

x x y

x

y x y

y

x x y

0 0 0 0

1 10 01

1 10 0

1

2 20 02

2 20 02

3 30 03

3 30 03

4 40 0

1 1

2 2

3 3 4

x 4 .....

[2]

x0

x

x0+x

f(x0) (x0+x)


3/19

Example : What is the error inyof equation y x x 2 3 atx=1.1 if it is predicted using Taylorsseries ?

Solution : By Taylors series expansion

y y

y x y x y x y x

x x x x

( . ) ( . )

( ) . ( ).

( ).

( )

.. .

( )

. . ( ) . ( )

. . .

.

11 1 01

0101

2

01

6

2 01 6 101

212

01

612

2 1 1 01 6 1 1 0 005 12 1 0 000167 12

3 0 7 0 06 0 002004

3762004

2 3

3 22 3

3 2

Analytical solution ofyatx=1.1 is y 2 11 11 3 7623. . .


4/19

DERIVATIVE NUMERICAL FORMULATION BASED ON TAYLOR SERIES

A. First derivative

Forward difference

f fdf

d truncation( ) ( )

( )x x x x

x

x0 0

0

so that

df

d=

f ftruncation

( ) ( ) ( )x

x

x x x

x

0 0 0

or

df

d

f f( ) ( ) ( )x

x

x x x

x

0 0 0

Backward difference

f fdf

d truncation( ) ( )( )

x x x xx

x0 0 0

df

d

f f( ) ( ) ( )x

x

x x x

x

0 0 0

Central difference

f fdf

d truncation( ) ( )

( )x x x x

x

x0 0

0

f fdf

d truncation( ) ( )

( )x x x x

x

x0 0

0

f f dfd

( ) ( ) ( )x x x x x xx

0 002

so that

df

d

f f( ) ( ) ( )x

x

x x x x

x

0 0 0

2


5/19

B. Second derivative

Central difference

f fdf

d

d f

d

2

truncation( ) ( )( ) ( )

x x x xx

x

x x

x0 0

02

0

2

f f dfd

d f

d

2truncation( ) ( )

( ) ( )x x x x xx

x x

x0 0

0 2 022

2

0

22

000d

)(fd)(f2)(f)(f

x

xxxxxxx

so that

2

000

2

0

2 )(f+)(f2)(f

d

)(fd

x

xxxxx

x

x

C. High accuracy first derivative

f fdf

d

d f

d

2

truncation( ) ( )( ) ( )

x x x xx

x

x x

x0 0

02

0

2

or

df

d=

f f d f

d

2

truncation

( ) ( ) ( ) ( )x

x

x x x

x

x x

x

0 0 0 0

2

Substituting the second derivative term the result obtained in (B) above:

dfd

= f f f f f

f f f f f

- f f f

truncation

truncation

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) (

xx

x x xx

x x x x x x

x

x x x

x

x x x x x

x

x x x x

0 0 0 0 0 02

0 0 0 0 0

0 0

22

2

2

4 3

0

2

x

x

) truncation

or

df

d

- f f f ( ) ( ) ( ) ( )x

x

x x x x x

x

0 0 0 04 3

2


6/19

Numeri cal I ntegration

DEFINITION

f x dx f u xii

n

( ) ( )

limx 0

1

Point uiis a point in x-axis choosen in such that the product off(ui).xrepresenting the shaded area and this

area should be regarded as the itharea which can be related to pointxi.

Alternative procedure to obtain solution of integral are:

1. Analitical approach;a) Employing definition of integralb) Deriving general form of an equation based on definition of integral to obtain Integration Rule or

Formula. For example based on derivation of antiderivative of polinomial, it can be shown that;

ax dxa

bx

b b

11

2. Numerical approach : most often the procedure is based on Interpolatory Numerical Integrationmethod

xix0=a xn=b

f(x)

x

ui(xi,x)

(ui)

(xi)


7/19

PREVIEW ON ANALITICAL APPROACH

Example:

Find x dxa

b

2 using definition of integral and verify the result using antiderivative formula at a=2

and b=3Solution:

By definition:

f x dx f u xii

n

( ) ( )

limx 0

1

When x 0 then f u f xi i( ) ( ) , so

f x dx f x xii

n

( ) ( )

limx 0

1

So that

x dx x xii

n2 2

1

limx 0 ( )

The figure above shows thatxi= a + i.xin which i=1,2,3,...nin such thatx0=a,xn=band x= (a-b)/n.

Subtituting these, one will get;

x dxx

a i x x

x a x a i x i x

x a

b a

na i

b a

ni

b a

n

i

n

i

n

i

n

2 2

1

2 2 2 3

1

2

1

2

2

3

0

0 2

0 2

lim

lim

lim

( )

( )

Note that the term xhas been eliminated from the equation so that thr term of x 0 becomesmeaningless. On the otherhand, when the length of xdecreases, the value of nwill increase. This means

xix0=a xn=b

f(x)x


8/19

that nwill be infinity as xbecomes infinitesimal. Based on this fact the above expression can be written as

the following:

x dxn

a b a

na i

b a

ni

b a

n

n a b a

n n a b a

ni

nb a

ni

i

n

i

n

i

n

i

n

2 2

2

2

3

1

2

1

2

1

3

2

1

2

2

lim

lim lim lim

Now, the first term becomes;

limn a b a

na b a n

i

n

2

1

2,

And since i = (n(n+1))/2 and i2= (n(n+1)(2n+1))/6, the second and the third terms become;

lim lim

lim

n a

b a

ni

n a

b a

n

n n

n a b

a b

na b

a b

na

a

n

a b a b a

i

n

2 2

1

2

2 2

2

2

1

2

22

22

33

2 2 3

( )

and

lim lim

lim

nb a

ni n

b a

n

n n n

n

b b

n

b

na b

a b

n

a b

na b

a

n

a b

n

a a

n

a

n

i

n

32

1

3

3 3 3

2

22 2

2

22 2

2

3 3 3

2

1 2 1

6

3 2 6

3

2 2

3

2 2 3 2 6

( ) ( )

1

3

1

3

3 2 2 3b a b a b a

Putting all the terms back together will result in the following:

xa

dx a b a a b a b a b a b a b a

b a a a b b

b

2 2 2 2 3 3 2 2 3

2 2

2 1

3

1

3

1

3

( ) ( )

Given those a=2 and b=3;

xa

dx b a a a b b

b

2 2 2

2 2

1

3

1

33 2 2 2 3 3

6 3333

.

Alternativelly, using power antiderivative formula, one may get;


9/19

xa

dx x

b

2 3

2

3

2 2

1

3

1

33

1

32

6 3333

( ) ( )

.


10/19

NUMERICAL APPROACH

f x dx g x dx( ) ( )

Whereg(x) is a continues function that:

1. Satisfyf(x) atxi, in which i=1, 2, ..., n.2. Relativelly easy to integrate.

Polynomials form is commonly used asg(x). Procedure:

1. Construct polynomials that pass through the given points. For example; use Lagrange interpolatingpolynomials.

2. Integrate the obtained polynomials

LAGRANGE INTERPOLATING POLYNOMIALS

g x f x x xx xi

j

i jjj i

n

i

n

( ) ( )

00

So that, for example, if n=2 then :

g xx x x x

x x x xf x

x x x x

x x x xf x

x x x x

x x x xf xo( ) ( ) ( ) ( )

1 2

0 1 0 2

0 2

1 0 1 21

0 1

2 0 2 12

And to constructg(x) that pass through pointsf(1)=0,f(2)=3,f(3)=16 then;

g x

x x x x x x

x x

( )

2 3

1 2 1 30

1 3

2 1 2 33

1 2

3 1 3 216

5 12 72


11/19

NUMERICAL INTEGRATION EMPLOYING LAGRANGE POLYNOMIALS WITH n=2

f x dx

x x x x

x x x x f x

x x x x

x x x x f x

x x x x

x x x x f x dx

f x

hx x x x x x x dx

f x

hx x x x

a

b

ox

x

x

x

( )

( ) ( ) ( )

( )

( )

1 2

0 1 0 2

0 2

1 0 1 21

0 1

2 0 2 12

02

21 2 1 2

12

20

0

2

0

2

2

x x x dx

f x

h

x x x x x x x dx

hf x f x f x

x

x

x

x

2 0 2

2

2

20 1 0 1

0 1 2

0

2

0

2

2

34

( )

( ) ( ) ( )

NUMERICAL INTEGRATION EMPLOYING LAGRANGE POLYNOMIALS WITH n=1

f x dx

x x

x xf x

x x

x xf x dx

f x

hx x dx

f x

hx x dx

f x

hx x x

f x

hx x x

f x

a

b

o

x

x

o

x

x

x

x

o

x

x

x

x

o

( )

( ) ( )

( ) ( )

( ) ( )

( )

1

0 1

0

1 0

1

11

0

21

1 20

0

1

0

1

0

1

0

1

0

11

2

1

2

hx x x x x

f x

hx x x x x

f x

h x x x x f x

h x x x xo

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

12

12

02

0 11

12

0 1 02

02

12 0 1 02 1 02 0 1 12

( )

( ) ( )

f x

hx x

f x

hx x

hf x f x

o

o o

( )( )

( )( )

( ) ( )

1

2

1

2

2

1 02 1

1 02


12/19

N Points Rules Name Formula

1 2 Trapezoidal( )

( ) ( )b a

f x f x

0 12

2 3 Simpsons 1/3( )

( ) ( ) ( )b a

f x f x f x

0 1 246

3 4 Simpsons 3/8( )

( ) ( ) ( ) ( )b a

f x f x f x f x

0 1 2 33 3

8

4 5 Booles( )

( ) ( ) ( ) ( ) ( )b a

f x f x f x f x f x

7 32 12 32 790

0 1 2 3 4

5 6( )

( ) ( ) ( ) ( ) ( ) ( )b a

f x f x f x f x f x f x

19 75 50 50 75 19288

0 1 2 3 4 5

OTHER APPROACH

1. Minimizing Trapezoidal rule error Ricradson Extrapolation, Romberg Integration

II I

j k

kj k j k

k,, ,

4

4 1

11 1 1

1

2. Area Substitution Gauss Quadrature

Two-points I f f

1

3

1

3.

Three Points see pp 517-519


13/19

Ordinary dif ferential Equation

GENERAL FORM OF ODE

Speed (v) change of falling parachute with respect to time:

dv

dt

g c

m

v

In whichg,m,c are gravity acceleration, mass, and drag constant respectivelly. If these three variables are

regarded as constant parameter, this equation can be express mathematically as:

dv

dtv f v ( )

Note that the above equation state that: first derivative of vis function of vitself. This patern is the basic

form of ODE..

NUMERICAL FORMULATION OF ODEBASED ON GEOMETRIC INTERPRETATION.

Euler Method

Keep in mind thaty=f(x,y) is an ODE buty=f(x) is not. Graphicaly,y=f(x,y) can be plotted as shown in

figure 1. Recall that numerical formulation for first derivative using forward differnce is:

y x y x x y x

x

y x x y x y x y x

i

i i

i i i i

( ) ( ) ( )

( ) ( ) ( , )

Since, according to definition of ODE, y x y f x yi i i i( , ) ( , ) then:

y x x y x f x y xi i i i( ) ( ) ( , ) Plot of the last form is sown in figure 2.

y y

x x

y=f(x) y=f(x)

y(xiyi)=f(x,y)

Figure 1 Figure 2xixi

y(xi) y(xi+x)

x

xi+x

y(xiyi)=f(x,y)


14/19

Example on Euler Method

Example: Ify is function ofx, plot numerically the first derivative ofx2.y+x.y3=x+ygiven thaty(2)=3.

Use x=0.1.

NOTE: - at least one initial condition should be known. In this case y(2)=3.

- in practical field implementation value of x is up to the modelerSolution:

x y x y x y2 3 First derivative of this function is:

2 3 12 3 2x y x y y x y y y After solving fory we get

y

x y y

x y x

1 2

3 1

3

2 2

or written in complete notation:

y x y x y x y x

x y x x

( , ) ( ) ( )

( )

1 2

3 1

3

2 2

First derivative numerical solution by Euler method:

y x x y x y x y x

y x x y x y x

x y x xx

i i i i

i

i i i

i i i

( ) ( ) ( , )

( ) ( ) ( )

( )

1 2

3 1

3

2 2

Therefore

y

y y y y

y

y y y y

y

( )

( . ) ( ) ( ) ( )

( ).

.

.

( . ) ( . ) . ( . ) ( . )

. ( . ) ..

. . . .

2 3

2 1 2 1 2 2 2 2

2 3 2 2 10 1

3 1 2 2 3 3

2 3 3 2 101

2 933

2 2 2 1 1 2 2 1 2 1 2 1

2 1 3 2 1 2 1 101

2 933 1 2 2 1 2 933 2

3

2 2

3

2 2

3

2 2

933

2 1 3 2 933 2 1 10 1

2870

2 3 2 809

3

2 2. . ..

.

( . ) .

y


15/19

IMPROVEMENT ON EULER METHOD (PERDICTOR-CORRECTOR METHOD)

A. Heuns Method

Lets denote y x yi i( ) andy x x yi i x( ) , then formulation of:

Predictor: y y f x y xi x i i i 0 ( , )

Corrector: y yf x y f x x y

xi x ii i i i x

10

2

( , ) ( , )

B. Modif ied Euler M ethod

Predictor: y y f x yx

i x i i i 1

2

0

2

( , )

Corrector:

y y f x x y xi x i i i x

1 12

012

( , )

Note:

xi+ x means the value of xiadded by a length as much as x yi+x means the value ofyatx=xi+ x aproximated by the formula yi+ x.k means the value ofyatx=xiadded by a length as much as the product of

xand k.

General form isyi+x=yi+ .xwhere = average slope. Therefore, the three methods presented so far areonly different in the way they predict the average slope.

(xi,yi)

y=f(xi,yi)

x

x.f(xi,yi)(xi,yi)

y=f(xi,yi)

x

x.f(xi,yi)

y=f(xix,yi+x)

(xi,yi)

y=f(xi,yi)

x

x.f(xi,yi)

y=f(xix,yi+x)

Euler Heuns Modified Euler


16/19

C. Runge-Ku tta Method

General form of numerical ODE solution

y y x y x xi x i i i ( , , )

Where(x ,y , x)=a k a k a k

k f x y

k f x p x y q x k

k f x p x y q x k q x k

k f x p x y q x k q x k q x k

k f x p x y

i i n n

i i

i i

i i

i i

n i n

1 1 2 2

1

2 1 11 1

3 2 21 1 22 2

4 3 31 1 32 2 33 3

1

( , )

( , )

( , )

( , )

( ,

i n n n n n

n i n i n j jj

n

q x k q x k q x k

k f x p x y x q k

1 1 1 1 2 2 1 1 1

1 11

1

, , ,

,

)

,

or

For example, derivation of the second order RK (n=2), in which:

y y a k a k x

k f x y

k f x p x y q k x

i x i

i i

i i

( )

( , )

( , )

1 1 2 2

1

2 1 11 1

Substituting the k1and k2it will becomey y a k a k x

y a k x a k x

y a x f x y a x f x p x y q k x

i x i

i

i i i i i

( )

( , ) ( , )

1 1 2 2

1 1 2 2

1 2 1 11 1

[1]

Taylor series expansion for a function with two dependent varibles:

g x r y s g x y rg x y

xs

x y

y( , ) ( , )

( , ) ( , )

Employing this expansion for the form off(xi+p1x,yi+q11k1x) in eqn. [1]:

f x p x y q k x

f x y p x f x yx

q k x f x yx

i i

i ii i i i

( , )

( , ) ( , ) ( , )

1 11 1

1 11 1

[2]


17/19

substituting back [2] into [1]:

y y a x f x y a x f x p x y q k x

y a x f x y a x f x y p xf x y

xq k x

f x y

y

y a a x f x y a p xf x

i x i i i i i

i i i i ii i i i

i i i

1 2 1 11 1

1 2 1 11 1

1 2 2 1 2

( , ) ( , )

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

( ) ( , )(

i i i i

y

xa q k x

f x y

y

, ) ( , )

2 11 12

Since k1=f(xi,yi) and after algebra manipulation one will get:

y y a a x f x y

a p xf x y

x

a q x f x yf x y

y

i x i i i

i i

i ii i

( ) ( , )

( , )

( , )( , )

1 2

2 12

2 112

[3]

Recall that based on Taylor series expansion:

y y x x y x dydx

x d ydx

i x i ( ) ........2 2

22

Since dy/dx=f(x,y), then

y y x f x yx df x y

dx

y x f x yx f x y

x

f x y

y

dy x

dx

i x i

i

( , )( , )

........

( , )( , ) ( , ) ( )

........

2

2

2

2

Or

y y x f x y

xf x y

x

x f x yf x y

y

i x i

1

1

2

1

2

2

2

( , )

( , )

( , )( , )

[4]

Comparing [3] and [4] we conclude that

a a

a p

a q

1 2

2 112

2 1112

1

[5]

Solving system of equation [5] and choosing a2to be then a1= andp1=q11=1, then

y y k k xi x i

1

2

1

21 2


18/19

1. Secod Oder:

y y k k x

k f x y

k f x x y x k

i x i

i ii i

1

2

1

21 2

12 1

Where:

( , )

( , )

2. Third Order

y y k k k x

k f x y

k f x x y x k

k f x x y x k x k

i x i

i i

i i

i i

1

64

2

1 2 3

1

212

12 1

3 1 2

Where:

( , )

( , )

( , )

3. Fourth Order

y y k k k k x

k f x y

k f x x y x k

k f x x y x k

k f x x y x k

i x i

i i

i i

i i

i i

1

62 21 2 3 4

1

212

12 1

312

12 2

4 3

Where:

( , )

( , )

( , )

( , )

4. Fifth Order

y y k k k k k x

k f x y

k f x x y x k

k f x x y x k x k

k f x x y x k x k

k f x x y

i x i

i i

i i

i i

i i

i i

1

907 32 12 32 71 3 4 5 6

1

214

14 1

314

18 1

18 2

412

12 2 3

534

Where:

( , )

( , )

( , )

( , )

( , 3

16 19

16 4

637 1

27 2

127 3

127 4

87 5

x k x k

k f x x y x k x k x k x k x k i i

)

( , )


19/19

ode numerik

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