Advanced Engineering Advanced Engineering MathematicsMathematics

A.EmamzadehA.Emamzadeh

Spring Spring 20092009

Syllabus# Topic Reference Duration (weeks)

1 Numerical solution of ODE [1] Chapter 5 & 11 2

2 Numerical solution of PDE [1] Chapter 12 2

3 Matrix Algebra [1] Chapters 6, 7 and 10 2

4 Integral equations [2] Chapter 16 1

5 Fourier series and Integrals [3] Chapter 1 1

6 Partial differential equations [3] Chapters 2 to 6 2

7 Special functions [4] Appendix B 1

8 Complex functions [5] Chapter 2 to 10 2

9 Monte Carlo method [2] Chapter 19 1

10 Integral transforms [4] Chapter 13 1

Total 15

Reference texts

Reference texts

[1] Faires & Burden Numerical Methods 3rd ed. ,Brooks/Cole 2003

[2] FrobergIntroductions to Numerical Analysis, Addison-Wesley

[3] Powers Boundary Value Problems

[4] Pipes & HarvillApplied Mathematics for Engineers and Physicists, McGraw-Hill

[5] Churchill Complex variable and applications

Ordinary Differential Equations

1) Introduction

2) Initial Value Problems (IVP)

3) Boundary Value Problems (BVP)

1) Introduction

1- Sources of errors1.1- Initial data error1.2- Round off error1.3- Truncation error

2- Type of problems2.1- well – behaved2.2- ill – conditioned

3- Stability

4- Big Oh

2) Methods of solving IVP

1- Introduction

2- Taylor series method

3- Single step methods

4- Multi step methods

5- Extrapolation methods

3) Methods of solving BVP

1- Shooting methods

2- Finite difference methods

3- Variational methods

4- Eigen Value Problems (EVP)

Types of problems in IVP

Single first order equation System of n first order equations Single n th order equation

Single first order equation

, with initial condition (IC),),( yxfy

00 )( yxy

To find y at x1 to a specified accuracy.

System of n first order equations

),,...,,(

),...,,(

1

111

nnn

n

yyxfy

yyxfy

With initial conditions

00

1001

)(

)(

nn yxy

yxy

To find y1,…yn at x1 to a specified accuracy.

Where tn1 y,,yY

and tn1 f,,fF

In matrix form

00 Y)Y(x

Y)F(x,Y

Single n th order equation

),,,,,,( )1()( nn yyyyxfy with initial conditions

00)1(

100 )(,,)( nn yxyyxy

This equation with the given initial conditions can be transformed into a system of n first order equations as follows,

Let nn yyyyyy )1(

21 then

nn yyyxfy ,,,, 21

and

001001 )(,)( nn yxyyxy

The whole system can be written in the matrix form

00 )(

),(

YxY

YxFY

Where tnt

n fyyyFyyY ,,,,,, 321 and

Finally we are dealing with the form

00 )(

),(

yxy

yxfy

as a single first order equation or

as a system of n first order equations

Taylor series method

00 )(

),(

yxy

yxfy

To find y at x1 correct to md

Let x1 – x0 = h then the Taylor expansion of y about

x = x0 is

yh

yh

yhyhxyxy!3!2

3

0

2

0001

0

0

0

0

000

0

),(

y

ff

x

f

yf

y

ff

x

f

xy

y

ff

x

fy

yxfy

y givenishere

20

10

3 2

y

y

yxyGiven

:Example

onsoand

Find y(h) correct to 2d where h=0.1, 0.5,1 and 2.

14)0(32

12)0(2

5)0(21

2)0(

)5()5(

)4(2)4(

yyyyyy

yyyyy

yyyy

yhavewe

5432

60

7

2

1

6

521 hhhhhhy

Taking number of terms n=6

When h=0.1 y(0.1)=.8107845000

h=0.5 y(0.5)=.3265625000

h=1 y(1)=.4500000000

h=2 y(2)=3.400000000

Taking number of terms n=7

y(0.1)= .8107846111

y(0.5)= .3282986111

y(1)= .5611111111

y(2)= 10.51111111

Taking n=10

y(0.1)= .8107846019

y(0.5)= .3276448999

y(1.0)= .4951609347

y(2.0)= 9.887477949

Compare the accuracy

Single step methodsDef.:

Local truncation error

Global truncation error

Order of a method

First order method (Euler's method)

Statement of the problem

00

,

yxy

yxfy

given

To find y at x=b correct to md

Let b-x0=h then

200000 , hyxfhyyhyby o

Here O(h2) is the local truncation error

In the standard form

,1,0,1

11

nyxfhk

kyy

nn

nn where

Second order methods

errortruncationglobaltheisand

where1.

2

12

1

221

2,

2

,

hO

ky

hxfhk

yxfhk

hOkyy

nn

nn

nn

g.t.e.beforeas

where2.

2

12

1

2211

,

,2

1

2

1

hO

kyhxfhk

yxfhk

hOkkyy

nn

nn

nn

A third order method

123

12

1

33211

2,

2,

2

,

,2,146

1

kkyhxfhk

ky

hxfhk

yxfhk

nhOkkkyy

nn

nn

nn

nn

where

A 4th order method (runge – kutta)

34

23

12

1

443211

,

2,

2

2,

2

,

226

1

kyhxfhk

ky

hxfhk

ky

hxfhk

yxfhk

hOkkkkyy

nn

nn

nn

nn

nn

where

Example:

20,10

2sin

yy

yyxy

Find y(0.1) and y’(0.1) using h=0.1 and a 2nd order method.

20

10

2sin

p

y

pyxp

py thenLet

Take k for the slope of y and l for the slope of p then

101002102

000101

2101

2101

2sin

2sin

2

12

1

lpkyhxhllphk

pyxhlphk

llpp

kkyy

where

Now calculate [Note the argument of sine must the in radiance].

Modified Euler's method

g.t.e.theiswhere

2

2111

1

,,2

,

hO

hOyxfyxfh

yy

yxfhyy

pnnnnn

cn

nnnpn

error.truncationlocaltheiswhere

2

21 ,2

hO

hOyxhfyy nnnn 1

Mid - Point rule

How to control the truncation error1) By lowering the step size and repeat the

whole calculations: Runge-Kutta method

2) By considering the 1st neglected term in the Taylor expansion:

Merson’s method

A 2nd and third order method

Fehlberge’s method

A single-step method with error estimator Merson’s method

5431

431

5

314

213

12

1

55411

89230

1

22

32

,

83

8,

2

66,

3

3,

3

,

46

1

kkkkE

kkk

yhxhfk

kky

hxhfk

kky

hxhfk

hy

hxhfk

yxhfk

hOkkkyy

nn

nn

nn

nn

nn

nn

where

Fehlberg method

543216

43215

3214

213

12

1

54311

654311

~

40

11

4104

1859

2565

35442

27

8,

2

4104

845

513

36808

216

439,

2197

7296

2197

7200

2197

1932,

13

12

32

9

32

3,

8

3

4

1,

4

,

5

1

4104

2197

2565

1408

216

2555

2

50

9

56430

28561

12825

6656

135

16

kkkkkyh

xhfk

kkkkyhxhfk

kkkyh

xhfk

kkyh

xhfk

kyh

xhfk

yxhfk

kkkkyy

kkkkkyy

nn

nn

nn

nn

nn

nn

nn

nn

where

Multi step methods Predictor formulas Corrector formulas

Predictor formulas

543211

43211

3211

211

11

1

1440

475

1440

2877

1440

7298

1440

9982

1140

7923

1140

4277

720

251

720

1274

720

2616

720

2774

720

1901

24

9

24

37

24

59

24

55

12

5

12

16

12

23

2

1

2

3

nnnnnnnn

nnnnnnn

nnnnnn

nnnnn

nnnn

nnn

ffffffhyy

fffffhyy

ffffhyy

fffhyy

ffhyy

fhyy

Corrector formulas

432111

32111

2111

111

11

1440

27

1440

173

1440

482

1440

798

1440

1427

1440

475

720

19

720

106

720

264

720

646

720

251

24

1

24

5

24

19

24

9

12

1

12

8

12

5

2

1

2

1

nnnnnnnn

nnnnnnn

nnnnnn

nnnnn

nnnn

ffffffhyy

fffffhyy

ffffhyy

fffhyy

ffhyy

Extrapolation methods1.Introduction:Consider calculating as the area of a unit circle

(Fig.1), by calculation of the area of n sided polygon inscribed in the circle.

n

1

Fig. 1nA0Let be the area of n sided polygon.

n

nnA n

n 2sin

2sin

20

Using Taylor expansion of Sin x

66

44

22

753

0 !7

2

!5

2

!3

22

2

n

a

n

a

n

a

nnnn

nAn

Where a2, a4, a6, … are constants, do not depend on n.

Now if we double n then

6

64

42

220 64164 n

a

n

a

n

aA n

To form a linear combination of

in such away that in the linear combination the first term be and the second term vanishes, that is, for some and

nn AA 200 and

88

66

44

1200 n

b

n

b

n

bAAA n

saynn

Where b4, b6, b8, … are constants, do not depend on n.

04

1 andThen

The implies that 34

31 and

To go further we can repeat the same procedure by doubling the sides again and hence.

8

86

64

421 2566416 n

b

n

b

n

bA n

Again to form a linear combination of

in such away that in the linear combination the first term be p and the second term vanishes, that is for some and .

nn AA 211 and

,88

66

22

11 n

c

n

cAAA n

saynn

Where c6, c8, … are constants, do not depend on n.

,88

66

22

11 n

c

n

cAAA n

saynn

1,2,nfor

andgeneralIn

andthatimpliesthis

andThen

14

4

14

1

15

16

15

1

,016

1

n

n

n

The result of above operations can be presented in an extrapolation table as follows.

nnnn

nnn

nn

n

AAAA

AAA

AA

A

322

41

80

22

140

120

0 31

34

151

1516

Note that truncation error of the columns respectively are

on.soand,1

,1

,1

642

nO

nO

nO

631

6364

Numerically1. Starting with a triangle (the least accuracy),

we have

Are these coefficients ( and ) general?

30A

60A

120A

240A

480A

1.299038

2.598077

3.000001

3.105829

3.132629

3.03109

3.133975

3.141105

3.141563

3.140834

3.141581

3.141593

3.141593

3.141593

3.141593

40A

80A

160A

320A

640A

2

2.828427

3.061468

3.121446

3.136549

3.10457

3.139148

3.141439

3.141584

3.141453

3.141591

3.141593

3.141593

3.141593

3.141593

50A

100A

200A

400A

800A

2. 377642

2.938927

3.09017

3.12869

3.138364

3.126022

3.140584

3.3014153

3.141489

3.141555

3.141593

3.141593

3.141593

3.141594

3.141594

1600A 3.14078

63.14159

33.14159

33.14159

33.14159

33.14159

3

100A

200A

400A

800A

1600A

2. 938926

3.09017

3.12869

3.138364

3.140786

3.140584

3.14153

3.141589

3.141593

3.141593

3.141593

3.141593

3.141593

3.141593

3.141593

3200A 3.14139

13.14159

33.14159

33.14159

33.14159

33.14159

3

1. yes, whenever the accuracy parameter is changed by factor of 2. and the terms of the Taylor expansion are alternatively zero.

2. No otherwise

Extrapolation in differentiation Starting with central difference formula for

the first and second derivative

)2(2

)1(2

22

2

hOh

hxfxfhxfxf

hOh

hxfhxfxf

Denote the RHS of (1) as and the RHS of (2) as hF0hS0

Similar extrapolation table looks like

hhh

hh

h

FFF

FF

F

22

14

0

12

0

0 31

34

151

1516

The truncation error of the columns in order are

.onsoandand642 , hOhOhO

Similarly for the second derivative

Extrapolation in integrationConsider the trapezoidal rule

1

100

20

22

n

iin

h

hb

a

fffh

T

hOTdxxf

where

hhh

hh

h

TTT

TT

T

22

14

0

12

0

0 31

34

151

1516

Extrapolation in IVPSuppose y'=f(x,y), y(x0)=y0 to find y(b)

correct to md, with extrapolation method.

Start with Mid-Point rule

20011 ,2 hOyxhfyy

Either y-1is known previously or it should be

calculated by, say, Euler's method and corrected by modified Euler's method to have the same accuracy as O(h2)

therefore with h<0

Ec

E

yxfyxfh

yy

yxhfyy

110001

0001

,,2

,

and

If we denote then in extrapolation notationhYy 01 by

hhh

hh

h

YYY

YY

Y

22

14

0

12

0

0 31

34

151

1516

And so on.

Compare this method with single step methods and multi step methods.

ExampleConsider y'=y , y(0)=1 . To find y(1)

correct to 3d .

start with h=1 , y0=1 , x0=0

21

021

11

10

11

8

21

8

131

8

131

8

5

8

5

2

11

4

11

2

1

2

11

2

1

5.222

12

101

2

11,0

Yyy

yy

h

Y

yy

cE

cE

with

To continue calculate and and extrapolate.

6666.26

16

2

7

6

5

8

21

3

4

2

5

3

13

4

3

1 21

01

01

1

YYY

41

0Y 81

0Y

Shooting methodsStatement of a boundary value problem

byay

yyxfy

,

,,Given

Find y for a<x<b correct to md.

Shooting method changes the BVP into an IVP by assuming a value for y'(a) , say S, then using a root finder to find the

root of y(b,S)-=0 up to a desired accuracy.

1. Secant method

2. Newton’s method

Let us start with secant method as the root finder

Step 1. Guess S0=y'(a)Step 2. Solve the IVP

,,,,, 0Sayayyyxfy

to find y(b, S0) .

Is y(b, S0)=? if yes then stop, otherwise continue

001

001

,

,

SbyifSayS

SbyifSayS otherwiseGuess

Step 4. Solve the IVP

,,,,, 1Sayayyyxfy

to find y(b, S1).

Step 3.

Is y(b, S1)=If yes then stop, otherwise continue.

Step 5. Find the next Sn+1 ,using secant method.

,2,1

,,

,

1

11

n

SbySby

SSSbySS

nn

nnnnn

Step 6. Solve the IVP

1

1

,

,,,,,

n

n

Sby

Sayayyyxfy

findto

Step 7. Test for convergence

stopthenyesifIs TolSby n 1,

Step 8. Go to step 5.

Example 1:Given

31,32

1

22

yy

yx

y [note that this is a linear differential equation]

To find y for 12

1 x correct to 3d.

0001.3,1

27500.36250.3

21

36250.3

2

1

6250.3,12

1

7500.3,11

2

2

01

00

Sy

S

SyS

SyS

x y

0.5 3

0.6 2.8667

0.7 2.8287

0.8 2.8501

0.9 2.9112

1 3.0001

Example 2:

Given

82,41

21

yy

y

y

xyy [this is a nonlinear differential equation]

To find y for 1<x<2 correct to 3d.

,2,1

,2,2

8,2

08,2

1

11

n

SySy

SSSySS

SySF

nn

nnnnn

000000.8,2333333.1

9902356.7,23331706.1

9688716.7,2328125.1

2580645.8,2375.1

111111.7,216667.1

4.6,21

16,22

66

55

44

33

22

11

00

SyS

SyS

SyS

SyS

SyS

SyS

SyS

x y

1 4

1.1 4.1450

1.2 4.3165

1.9 7.0793

2 7.9994

In case of Newton's method as the root finder the steps will be as follows

Step 1. Guess S0=y'(a)Step 2. Solve the system

1

0

,,

0

S

ayS

ay

Say

ay

S

y

y

f

S

y

y

f

S

y

yyxfy

(1)

By any IVP method find y(b, S0), y' (b, S0) and

00 ,,, SbS

ySb

S

y

Step 3. Find a new Sn using Newton’s method

,1,0,

,1

n

Sbsy

SbySS

n

nnn

Step 4. Solve the system(1) with new Sn=y'(a)

Step 5. Test for convergence, i.e.

stopthenyesifTolSby n ,

Step 6. Go to step 3.

Example: Find y for 1<x<2, given

thenanddenoteuslet

system(1)solvetonow

guess

US

yY

S

y

S

yy

y

y

xyy

2

82,41

21

0

if

11

01

1

41

412

21

0

2

2

U

Y

Sp

y

Uy

p

xY

y

pU

UY

y

p

xpp

py then

Solve by 4th order Runge-Kutta with h=0.01 (say) then

66666667.1

24,216,2

1

00

S

SS

ySy

impliesthis

and

Next five iterations yields

000000.6,2000000.8,233333333.1

00018279.6,200012208.8,2333353689.1

04715063.6,203137248.8,233854166.1

82666667.6,25333333.8,2416666667.1

6666667.10,26666667.10,266666667.1

555

444

333

222

111

SS

ySyS

SS

ySyS

SS

ySyS

SS

ySyS

SS

ySyS

Finite difference methodsFinite difference methodsStatement of the problem:Given the BVP

byay

yyxfy

,

,,

Find y for a<x<b correct to md.

There are two cases:

1. Linear equation, in general

byay

xryxqyxpy

,

2. Non linear equation

byay

yyxfy

,

,,

The above two cases are called BVPs with separated boundary conditions.

BVP with general linear boundary values are of the form

54321

54321

,,

BbyBbyBayBayB

AbyAbyAayAayA

yyxfy

BVP with general boundary conditions are of the form

0,,,,

0,,,,

,,

2

1

bybyayaybg

bybyayayag

yyxfy

The idea of finite difference methods is to discretise the equation by dividing the interval [a,b] into n equal divisions, i.e.

hn

ab say

bxniihxxax ni ,1,,1, 00 then

The BVP becomes

n

iiii

yyax

yyxfy

,,

,,

00

Now we replace the derivatives by an approximate value of finite difference such as:

42112

32112

211

11

4642

22

22

h

yyyyyy

h

yyyyy

h

yyyy

h

yyy

iiiiii

iiiii

iiii

iii

Central-difference expressions with error of order h2

Central-difference expressions with error of order h4

4321123

3321123

22112

2112

6

12395639128

81313812

16301612

88

h

yyyyyyyy

h

yyyyyyy

h

yyyyyy

h

yyyyy

iiiiiiii

iiiiiii

iiiiii

iiiii

Forward-difference expressions with error of order h

41234

3123

2112

1

464

33

2

h

yyyyyy

h

yyyyy

h

yyyy

h

yyy

iiiiii

iiiii

iiii

iii

4

12345

31234

2123

12

31426241122

51824143

2542

34

h

yyyyyyy

h

yyyyyy

h

yyyyy

h

yyyy

iiiiii

i

iiiiii

iiiii

iiii

Forward-difference expressions with error of order h2

Backward-difference expressions with error of order h

44321

3321

211

1

464

33

2

h

yyyyyy

h

yyyyy

h

yyyy

h

yyy

iiiiii

iiiii

iiii

iii

454321

34321

2321

21

21124261432

31424185

4522

43

h

yyyyyyy

h

yyyyyy

h

yyyyy

h

yyyy

iiiiiii

iiiiii

iiiii

iiii

Backward-difference expressions with error of order h2

In linear case we have, using the central difference forms with truncation error of O(h2),

1,,1,2

2

0

112

11

niyy

xryxqh

yyxp

h

yyy

n

iiiii

iiii

After some simplifications

1,,1,0

11

niyy

DyCyByA

n

iiiiiii

Where

ii

ii

ii

ii

rhD

hpC

qhB

hpA

2

2

2

2

24

2

In matrix form

11

2

11

1

2

1

11

222

11

nnnnn CD

D

AD

y

y

y

BA

CBA

CB

O

O

This linear system can be solved by

1.Direct methods (gauss elimination, …)

2. Iterative methods (Jacobi's method, Gauss Seidel method, Successive Over Relaxation method, SOR).

Example:

3d.tocorrectforFind 12

1

31,32

1

22

xy

yy

yx

y

Take n=5 h=0.1 then

96.2

04.0

04.0

46.1

8.18.000

9.06.17.00

08.04.16.0

007.02.1

4

3

2

1

y

y

y

y

Using Gauss elimination method

91111111.285000000.2

8285714.2866666667.2

43

21

yy

yy

In non linear case, using central difference formulas with truncation error O(h2)

1,,1,,

2/,,2

0

112

11

niyy

hyyyxfhyyy

n

iiiiiii

Using simple iteration method, with suitable initial guess

,1,0

1,,1,

2,,

22

1

0

)1(1

)(1)(

2)(1

)1(1

1

k

niyy

h

yyyxf

hyyy

n

ki

kik

iik

ik

iki

Example:

82,41

21

yy

y

y

xyy

Find y for 1<x<2 correct to 5d.

Upon discritisation we have,

,1,0,1,,18,4

1

222

1

0

)(

)1(1

)(1

)1(1

)(1

2)(

1)1(

1)1(

kniyy

yh

yy

xh

yyhyyy

n

ki

ki

ki

i

ki

kik

ik

ik

i

Now, takeN=5 h=0.2 After k= 50 iterations and Tol=

610

1.2 4.30868851

1.4 4.74347837

1.6 5.37398310

1.8 6.34221887

ix iy

Take N=10 h=0.1

After k= iterations

1.1

1.2 4.31464627

1.3

1.4 4.75747894

1.5

1.6 5.39795103

1.7

1.8 6.37355180

1.9

ix iy

To improve the accuracy1. Either increase n smaller hLarger system of equations2. Use extrapolation,In the above example

Improved 3

14.1 y less accurate

3

44.1 y more accurate y(1.4)

4762145797.4

75747894.43

474347837.4

3

1

hO

Continue to improve accuracy.

Iterative methods for system of linear equations.To solve

1) Ax=b

Where

nnnnn

n

b

b

b

x

x

aa

aa

A

11

1

111

,,x

2)

nnnnn

nn

bxaxa

bxaxa

11

11111

3)

nibxa

nibxaxan

jijij

inini

,,1

,,1

1

11

4)

Pivoting , ScalingJacobi’s method: guess nixi ,,1)0(

,1,0

,,11

1

)()1(

k

nixab

ax

n

ijj

kjiji

ii

ki

Test for convergence at each iteration

Tolxx kj

kj )()1( For all j=1,…,n

Guess – Seidel method

,1,0,,,1

1

,,1

1

)(1

1

)1()1(

)0(

kni

xaxaba

x

nix

n

ij

kjij

i

j

kjiji

ii

ki

iGuess

Test for convergence

Successive Over Relaxation (SOR) method

,1,0,,,1

,,1

)(1

1

)1()()1(

)0(

kni

xaxaba

xx

nix

n

ij

kjij

i

j

kjiji

ii

ki

ki

i

Guess

Test for convergence

required?isor bopt .

1. No general formula for

2. depends on the form of A.

3. 0< <2

When 1< <2 we have over relaxation and when 0< <1 we have under relaxation

4. to be calculated numerically.

bb

bb

b

b

Example: solve

96.2

04.0

46.1

8.18.000

9.06.17.00

08.04.16.0

007.02.1

4

3

2

1

x

x

x

x

w 1 1.1 1.2 1.3 1.4 1.5 1.6

k 34 27 21 15 18 24 30

bw

Minimum When k=1428.1

Variational methodsVariational methods IntroductionDistance between two points

2

1

2

1

222 1

x

x

x

xdx

dx

dydydxyI

To minimize I[y] set its derivative to zero.

There are certain restrictions on each y which must pass through ,etc. 2211 ,, yxyx &

2y

1y

1x 2x

4y

3y 2y1y

E-L eqn. yyxFy

yyxFydx

d

,,,,

If F is independent of 0

y

Fy

If F is independent of Cy

F

y

F

dx

dy

0

If F is independent of

0

y

FyF

dx

d

Cy

FyFx

or

Now to solve the BVP

010

yy

xfyxqyxp

The Rayleigh - Ritz method minimizes the E-L equation

1

0

22 2 dxxuxfxuxqxuxpuI

Choose u to be a linear combination of some basis function such as few first terms of Taylor expansion or in general iicu

Where are chosen in such away to satisfy the boundary conditions

Now for minimization find

and equate to zero

These are called the normal equations which can be solved by a method of linear systems.

i 010 ii ni

c

I

i

,,1

nic

I

i

,,10

Example:Let us choose the basis to be the linear functions

10

00

1

11

11

1

1

xx

xxxh

xx

xxxh

xx

xx

x

i

iii

ii

iii

i

i

i

xi

x0

1

1ix ix1ix 1

The normal equations yield a tridiagonal linear system which can be solved by previously introduced methods.

Example:

010

422 22

yy

xyyxyxGiven

Use h=0.1 and linear approximation.Now given

0

byay

xfQyy

Then E-L equation

b

a

dxfuQuuuI 222

to be minimized.

B.C.&If2

210 xcxccu

210210

,0,0,0 cccc

I

c

I

c

Igivesthen

B.C.&If3

32

210 xcxcxccu

on.so&

givesthen 3,2,1,03,2,1,00

icic

Ii

i

Eigen Value Problems(Homogeneous BVP)

Consider the problem:

010

2

yy

yy

To find the nontrivial solution of the above system

,2,1

,2,10sin0sin

01

sin

00

sincos

kk

kkB

y

xBy

y

xBxAy

k

B.C.&

B.C.&

The Eigen Values are

and are the Eigen functions.

Let us use finite difference method

222 kk xBy kkk sin

1,,10

2

0

2211

niyy

yhyyy

n

iiii

In matrix form, let th 222

0

0

1

1

1

11

1

1

1

ny

y

t

t

t

O

O

This homogeneous linear system has nontrivial solution when

0

1

1

11

1

det

t

t

t

A

O

O

Choosing n=2 then 21h

and an approximation to the analytical value

80 2 t8696.922

1

Choosing n=4 then 41h

6274.5422

32

3726.922

0

10

11

01

23

22

21

t

t

t

and

The analytical value of

8264.88

4784.39

8696.9

23

22

21

Again for improvement there are two methods

1) To increase n and have smaller h leading to a higher degree polynomial equation and hence more round off error.

2) To use extrapolation technique.

In this case

Improved

8301.9

3726.93

48

3

13

4

3

1 21

21

21

accuratemoreaccurateless

Numerical Solution of PDE(Finite difference method)

We are going to consider

1) Heat equation

2) Steady – State equation

3) Wave equation

Consider the heat equation in the form

xftxu

ttbu

ttau

tbxax

uD

t

u

,

,

,

0,2

2

To find u(x,t) using finite difference method.

1. Explicit method

As before let nixixxax i ,,1,0, 00

and

ijji

j

utxu

jtjttt

,

,1,0,0 00

denote

The partial derivatives will be as,

)difference(central

)difference(forward

2

2

1

2

2

1

2xO

x

uuu

x

u

tOt

uu

t

u

jiijiji

ij

ijij

ij

Finally, let 2.

x

Dtr

Upon substitution in the equation,

2111 2 xtOuuuruu jiijjiijij

Stability condition is

When 2

1r

2

1r

jijiij uuu 111 2

1

The explicit method looks like

i, ji-1, j i+1, j

i, j+1

In the whole problem

iju

a bx

t

Problem:Problem: show that when the local truncation error will reduce to

The restriction on r causes higher number of operations which leads to higher round off error.

Hence we move to implicit procedures.

6

1r

42 xtO

2.Implicit methods

Let us replace by a linear combination of the known time step and the unknown (next) time step as follows,

2

2

x

u

10

11

2

2

2

2

where

ijijij x

u

x

uD

t

u

When then we have the Explicit method.

When then we have the Implicit Crank - Nicolson method

2

1

When then we have the fully implicit method (backward difference method)

Implicit Crank – Nicolson method

11111111 222 jiijjijiijjiijij uuuuuur

uu

The diagram of this method looks like

i-1j+1 ij+1 i+1j+1

i-1j ij i+1j

Rearrangement of the equation yields

,1,0,1,,2,12

12

21

2

11

11111

jni

ur

urur

b

bur

urur

jiijjiij

ijjiijji

where

This system of (n-1) equation in (n-1) unknown must be solved for each time step j . Therefore j can be omitted at each time step, that is

For each j=0,1,…

1,,2,12

12 11

ni

bur

urur

iiii

Taking into consideration the boundary conditions then in matrix form,

nnn ur

b

b

ur

b

u

u

rr

rr

r

rr

O

O

2

2

12

21

2

21

1

2

01

1

1

This system can be solved by SOR method,

For each j=0,1,…

,1,0;1,,2,1

21

21)(1

)()1(1

)()1(

kni

ur

urur

br

uu ki

ki

kii

ki

ki

k is the number of iterations

In this special case is found to be.opt

nr

rb

cos1

,11

22

Example Given

201020

1000,

0100,20

16.0,00,0

0,2002

2

xx

xxxu

ttu

Dttu

txx

uD

t

u

Find at x=4,8,12,16

and at x=2,4,…,18

,xu

As you know

Take r=1 and then and for each j

216.0

x

tr

4x 100t

00

4

6

6

4

25.000

5.025.00

05.025.0

005.02

50

4

3

2

1

uu

u

u

u

u

&thatnote

100

27273.3

09091.5

09091.5

27273.3

4

3

2

1

t

u

u

u

u

at

For the next time step and00 u 1005 u

545455.52

18182.4

18182.4

545455.2

4

3

2

1

b

b

b

b

Now at t=200

7373.55

8582.17

33188.7

1057.3

50

4

3

2

1

4

3

2

1

4

3

2

1

u

u

u

u

b

b

b

b

u

u

u

u

A

& so on.

Elliptic ProblemsConsider the Poisson equation

mjni

jkcyihax

yxgyxu

dycbxa

yxfy

u

x

uyxu

ji

,,1,0,,1,0

,,

,

,,2

2

2

22

andfor

andlet

boundariestheonwith

on

ax 0 1x 2x 3x 4x nxb

cy 0

1y2y

dym

y

x

Let us use the central differential formula for both derivatives, then

1,,1;1,,1

2

22

222211

211

2

2

11

2

11

mjni

fkhukhuuhuuk

fk

uuu

h

uuu

ijijijijjiji

ijijijijjiijji

or

The diagram looks like

ij+1

i+1jiji-1j

ij-1

Five point formula.

This linear system of (n-1)(m-1) equations and unknowns can be solved by Gauss – Seidel method or by SOR method with

mnc

cb

coscos2

1

11

22

where

Example:Example:Find , given the Poisson equation ji yxu ,

80,6022

2

2

2

yxy

u

x

u

u=0 on the boundaries.

Take a) h=k=2 b) h=k=1

For the case a) there are 6 unknowns ijui=1,2 ; j=1,2,3

2111

2221

2331

uu

uu

uu

8

6 x

y

3,2,1,2,1

84

11111

ji

uuuuu ijijjijiij

Using the symmetric property (in this particular problem) we have

56.456.4

72.572.5

56.456.4

2313

2212

2111

uu

uu

uu

For the case b) there are 35 linear equations in 35 unknowns 7,,1;5,,1? jiuij

Using the symmetric property the system will be reduced to 12 equations in 12 unknowns.

x x x

x x x

x x x

x x x

8

6 x

y

Preparing the system to be solved by SOR it looks like

b

kij

kij

kij

kji

kji

bkij

kij

ji

uuuuuuu

7,,15,,1

484

)()(1

)1(1

)(1

)1(1

)()1(

The results are as follows

647.6960.5181.3

335.6686.5657.3

319.5794.4123.3

353.3047.3042.2

Hyperbolic Problem Hyperbolic Problem (Wave equation)(Wave equation)

Two dimensional heat equation in Cartesian coordinates

(AAlternative lternative DDirection irection IImplicit methodmplicit method) ADIADI

Fourier SeriesFourier SeriesPower series play an integral part in real(& complex)

analysis.

Taylor Series in ODE

Fourier Series in PDE

Definition:Definition: Two nonzero f(x) and g(x) are said to be orthogonal on the interval with respect to the weight function (x) if their scalar product vanishes:

bxa

b

adxxgxfx 0

ExampleExample

0sinsin

2,0

1sinsin

2

0

dxmxnx

xnmmxnx

onarthogonal

areand

2l – periodic function means xflxf 2

Problem:Problem: the set of function

l

xn

l

xn sin,cos,1

are orthogonal over the interval with respect to the weight function (x)=1 .

lxl

Furthermore,

l

l

l

l

l

lldx

l

xndx

xnldxl

222 sin

2cos;2

Let f(x) be sufficiently smooth function defined on and periodic with period 2l.

Then we seek to express f(x) as an infinite linear sum of sines and cosines having the same period as f(x), namely 2l.

We assume that

lxl

10 sincos

2

1

nnn l

xnb

l

xnaaxf

Where and the coefficients and are found to be

0a na nb

l

ln

l

ln

l

l

dxl

xnxf

lb

n

dxl

xnxf

la

dxxfl

a

sin1

,2,1

cos1

10

CorollaryCorollaryWhen f(x) is periodic, piecewise smooth function, its

Fourier series converges to [f(x+)+f(x-)]/2.

Example 1. Find the Fourier series of f(x)=x 0<x<2l and is 2l – periodic

02

sin1

00cos1

21

2

0

2

0

2

00

nn

ldx

l

xnx

lb

ndxl

xnx

la

lxdxl

a

l

n

l

n

l

Therefore

1 1

sin12

1sin2

n n l

xn

nl

l

xn

n

llxf

y

x

L

2L

2L 4L 6L

Example 2.Example 2.Find the Fourier series of

and is of period 2l lxlxxf 2

1

2

2

122

222

12

2

2

22

2

22

22

22

0

6

1

,14

3

cos14

3

00sin1

014

cos1

3

21

n

n

n

l

ln

nl

ln

l

l

n

n

llllx

l

xn

n

llxf

ndxl

xnx

lb

nn

ldx

l

xnx

la

ldxx

la

hence

At

Complex FormComplex Form

Knowing that

2cos

2sin

ii

ii

ee

i

ee

and

Then the Fourier series of f(x) takes the form

lxin

n

nnlxin

n

nn

n

lixn

lixn

n

lxin

lxin

n

eiba

eibaa

i

eeb

eea

axf

11

0

1

0

222

222

l

l

lxin

n

n

lxin

n

nnn

nnn

dxexfl

c

ecxf

niba

c

niba

cac

2

1

02

02

,20

0

where

then

andLet

Fourier Sine and Cosine SeriesFourier Sine and Cosine Series1. When f(x) is an even function then

1

0

0

cos2

0,cos2

nn

l

nn

l

xna

axf

bdxl

xnxf

la

The cosine expansion of f(x)

2. When f(x) is an odd function then

1

00

sin

sin2

,0

nn

l

nn

l

xnbxf

dxl

xnxf

lbaa

The sine expansion of f(x)

Fourier IntegralFourier IntegralThe Fourier series extension to none periodic

functions leads to Fourier Integral

Consider a periodic function of period 2l. Its Fourier expansion is

xfl

.,sincos2

0

l

nxbxa

axf nnnnnl

where

Now let and set

that is

llnn

1

l

1

1

sinsincoscos1

1

n

l

l nln

l

l nln

l

l ll

duuufxduuufx

duufl

xf

If and f is absolutely integrable

then using the idea of definite integrals,

xfxf ll

lim

0sinsincoscos

1

dduuufxduuufxxf

0

sincos dxBxAxfor

duuufBduuufA

sin1

,cos1

where

Example:Example:Find the Fourier integral representation of the function

0sin1

sin2cos

1cos

1

10

11

1

1

duuufB

duuduuufA

x

xxf

if

if

dx

xf

0

sincos2

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