Rac 2015 1

MSMS33200200 Numerical AnalysisNumerical AnalysisMSMS33200200 Numerical AnalysisNumerical AnalysisPart 2Part 2

Dr. Ir. Rachman SetiawanEngineering Design Centre( h l b )(Mechanical Design Lab.)

Tel. 2500979E-mail: rachmans@edc.ms.itb.ac.id

8. Curve Fitting8. Curve Fitting8.1 Introduction

Purpose: To represent a function based on knowledge its behaviour at certain discrete points.pThere are two approaches:

Interpolation produces a function that matches the given data exactly

Regression produces an approximate function for the given data

Application: Trend Analysis

57

Trend AnalysisHypothesis testing

Source of Data:Experimental dataDiscrete numerical solution

Rac 2015 2

58

Interpolation (polynomial) Regression (linear)

Comparison between Interpolation & Regression

Interpolation Regression The data is very precise Curve fits each points exactly Example: relationship generates

from exact solution/calculation

The data contains error or noise Curve does not necessarily fit

each points exactly but represents the general trend of the data

59

Example: experimental result

Rac 2015 3

8.2 Interpolation8.2.1 Newtons Divided-Difference

Making use of a Polynomial form to fit a curve Polynomial g y yinterpolation:

It requires (n + 1) datapoints for n-th order polynomial

60

While Polynomial is the form of fitting, one of the methods is Newtons Divided-difference.

The simplest method of Newtons DD is by using Linear form Linear Interpolation

001

0101

01

01

0

01

xxxx

xfxfxfxf

xxxfxf

xxxfxf

Finite divided differenceapproximation of the first derivative

61

Datapoints: [x0, f(x0)], [x1, f(x1)]

Rac 2015 4

General form of Newtons Interpolating Polynomials

110010

xfbxxxxxxbxxbbxf nnn

011

0122

011

00

,,,,

,,,

xfxfwhere

xxxxfb

xxxfbxxfb

xfb

ji

nnn

62

0

01111011

,,,,,,,,,,

,

xxxxxfxxxfxxxxf

xxxfxf

xxf

n

nnnnn

ji

jiji

Recursive nature of Newtons divided differences:

010120010 ,,,

nxxxxxxxxxxf

xxxxxxxfxxxxfxfxf

63

110011 ,,,, nnn xxxxxxxxxxf

Rac 2015 5

Algorithm

f( )f(xi) = yi

ji

jiji xx

xfxfxxf

,

S d t

First stage

64

-Second stage-yint2 f2(x)-Error

Effects on higher order (Ex 18.3): Ln 2 = 0.6931472 (true value)

order Data points required

f(x) t (%)

1st (linear) 2 0.4621 33.3

2nd (quadratic) 3 0.5658 18.4

3rd (cubic) 4 0.6288 9.3

65

Rac 2015 6

8.2.3 Lagrange Form

nj

i

n

iin xxxx

xLxfxLxf ;

Rationale: Each term of Li(xi) is 1 at x = xi and zero at all other

datapoints takes the value of f(xi) at data point xi. Therefore, the curve passes through each data points A

h t i ti f I t l ti

ijj jii xx00

66

characteristic of Interpolation

Example: Lagrange Interpolating polynomial orde-3

3

xfLxfLxfLxfLxfxLxf

323

2

13

1

03

02

32

3

12

1

02

0

131

3

21

2

01

00

30

3

20

2

10

13

332211000

3

xfxxxx

xxxx

xxxxxf

xxxx

xxxx

xxxx

xfxxxx

xxxx

xxxxxf

xxxx

xxxx

xxxxxf

xfLxfLxfLxfLxfxLxfi

ii

67

Rac 2015 7

8.2.4 Interpolating Polynomial Form

nnn xaxaxaaxf ...2210

Steps: Build system of linear equations in form of matrix problem Solve using ordinary Gauss/Gauss-Jordan/etc..

nnf 210

nn xfxaxaxaa ... 00202010

n xfaxxx 00

20

2001

68

nnnnnn

nn

xfxaxaxaa

xfxaxaxaa

...

...

2210

112

12110

nnn

nnn

n

xf

xfxf

a

aa

xxx

xxx

2

1

2

1

2

12

11

1111

8.2.5 Inverse Interpolation

Evenly-spacedx = ??

1429.01667.02000.02500.03333.05000.00000.17654321

xfx

x = ??

69

Mostly unevenly-spacedSample Problem!!: Find x, to give f(x) = 0.3

Rac 2015 8

8.2.5 Inverse Interpolation Obvious method:

Switching x f(x) f(x) x ?? oscillation in resultSwitching x f(x), f(x) x ?? oscillation in result Alternative method:

Interpolating polynomial, then root of polynomial Steps:

Choose order of polynomials Set up linear algebraic system Find coefficients: a0, a1, ... , an

70

Solve for known f(x), to find x using root finding algorithm:

baaxaxaxaaxaxaxaabn

n

nn

002

210

2210

' :where0...'

...

Root finding problem

8.2.6. Spline Interpolation

Segmen IISegmen III Principles:

Segmen ISegmen II

... 1. At first and end points, the value of functions = datapoints

2 At interior nots the value

11

2

nnm yxf

yxf

001

71

2. At interior nots, the value of adjacent functions must be equal

2

11 ijij xfxf ij xf jth Segmentith data

Rac 2015 9

8.2.6. Spline Interpolation

Principles:Segmen IISegmen III

3. At interior nots, slope of adjacent functions must be equal

4. Assume, the first point has d d i ti

11 ijij xfdxdxf

dxd

Segmen ISegmen II

...

4

3

72

zero second derivative:3

ij xf jth Segmentith data

0022

xfdxd I

8.2.6. Spline InterpolationExample: Quadratic Splines (Text Book EX. 18.9):

Quadratic:

Datapoints: (3.0,2.5); (4.5,1.0); (7.0,2.5);(9.0,0.5)

Principles:

jijijij cxbxaxf 2

5.239 111 cba 0.14525.20 111 cba1 2

73

5.0981 111 cba

5.27495.2749

0.14525.20

333

222

222

cbacba

cba1 2

301414

099

3322

2211

baba

baba4 01 a

Rac 2015 10

8.2.6. Spline InterpolationExample: Quadratic Splines (Text Book EX. 18.9):

In Matrix form:

2

1

1

1

...

...

c

acba

Solve for: a1, b1, c1, a2, ..., c3

74

3c

0.90.73.916.246.1

0.75.446.1876.664.05.40.35.5

2

2

xxxxfxxxxfxxxf

III

II

I

Other Advanced Topics: 8.2.5 Matrix Elimination coefficients* 8.2.6 Piece-wise Spline*p

Self evaluation Newtons DD:

...,,

,,,,,,,,,,

012

0

01111011

xxxfxx

xxxfxxxfxxxxfn

nnnnn

75

Lagrange:

...

;

2

00

xf

xxxx

xLxfxLxfn

ijj ji

ji

n

iiin

Rac 2015 11

8.3 Regression8.3.1 Review on Statistics

Consider a population of xi containing n members.p p i g Mean/average:

Standard deviation:

n

iiyn

y1

1

1

2

yys

n

ii s

OVC y

In a normalised form Coefficient of variance

76

1

atau1

222

1

nn

yys

ns

iiy

iy y

OVC

77

Normal distribution

Rac 2015 12

If the population is composed of a number of measurements, or numerical calculation: Mean Accuracy (the closeness to the true value) Standard deviation Precision (the closeness among

other members) Regression makes use of this principle since, often, the

population is scattered due to random error (from calculation or measurement)

Like in interpolation, the representing curve can be of the form of linear, quadratic of general polynomial.

78

The big issue here is to choose the criteria for the best fit

8.3.2 The Best-fit Criterion Supposed n data yi, to be fitted linearly (a0 + a1xi), so that containing

error/residual, e that is defined by:

The proposed criteria for Best fit : Minimise the sum of errors

Minimise the sum of the absolute values of error

Minimise the maximum of the error (Minimax)

iii xaaye 10

iemin

iemin

79

Minimise the maximum of the error (Minimax)

Minimise the sum of the squares of errors

2min ie

iemaxmin

Rac 2015 13

Alternatives of The Best-fit Criterion:

iemin

iemin

80

iemaxmin

The best proposed criterion is chosen to be the Least Squares Widely used in even more sophisticated regressions and will be used for the entire discussion

2mod,, elfitimeasuredir yyS

ii xaay 10

81

iiy 10

Rac 2015 14

8.3.3 Quantification of Error of the Fitting The error Sum of squares (SSE) Sr

2

Other form: normalisation of SSE R-square (r2):

This represents the relative discrepancy between squares

t

rt

SSSr 2

2iir xfyS

2yyS it

82

p p y qof errors and the spread of the original data, yi .

By taking square root Correlation coefficient (r). For a perfect fit r = 1, for total non-correlated r = 0.

8.3.4 Least-Squares Linear Regression Minimisation first derivatives = 0 For a0 : For a1 :0 1

Solving for the two equations:

ii

ii

iir

iir

xanay

xaay

xaayaS

xaayS

10

10

100

210

0

02

iiii

iiii

iiir

iir

xyxaxa

xaxaxy

xxaayaS

xaayS

210

210

101

210

0

02

83

g q

Example 17.1

xaya

xxnyxyxn

aii

iiii

10

221

Rac 2015 15

Use common sense to fit data, not necessarily linear. It could be parabolic, exponential, logarithmic, etc

Linearization if necessary !

84

Linearization: Know the basic relationship ! Modify the equation into linear relationshipy q p

xaay

xbayeay xb

10

11

'lnln1

85

Rac 2015 16

* Favourite Text Book Problems:17.6, compare the results with that of Ms Excel17 13 compare the results with that of Ms Excel17.13, compare the results with that of Ms Excel

86

8.3.5 Polynomial Regression (Least Squares) The minimisation of Sum of squares of errors (Least

Squares): 22q )

That is by differentiate Sr wrt each a0, a1, am:

Solving for the system of equations a0, a1, am

22210 mimiiir xaxaxaayS

0,,0;010

m

rrr

aS

aS

aS

m yaxxxn 2

87

imi

ii

ii

i

mm

imi

mi

mi

miiii

miiii

iii

yx

yxyx

y

a

aaa

xxxx

xxxxxxxxxxxn

22

1

0

221

2432

132

Rac 2015 17

8.3.6 n-Dimensional Space Regression So far, we have 1 variable to regulate 1 function:

Example: x f (x)Example: x f (x) Consider a problem with more than 1 variables to regulate

1 response function Application: A phenomenon that is regulated by more than

one input variables. The simplest form is Linear relationship having 2 input

variables (Multiple Linear Regression):

88

So, the line in the two-dimensional space becomes flat-plane in the three-dimensional space.

22110 xaxaay

89

Two-dimensional linear regression

Rac 2015 18

The problem solving remains the same Express the sum of the squares of errors, Sr Differentiate Sr wrt each coefficients, a0, a1, and a2.r , 0, 1, 2 Build a system of equations in matrix form Solve it

ii

ii

i

iiii

iiii

ii

yxyx

y

aaa

xxxxxxxx

xxn

2

1

2

1

0

22212

21211

21

90

8.3.7 Advanced Topics Non-linear regression Fourier Approximation Fourier Approximation Response Surface modelling Metamodelling/Neural Network

91

Rac 2015 19

9. Numerical Integration9. Numerical Integration9.1 Introduction

Integration is summation with infinitesimally small division Numerical integration is approximation of the integration g pp g

with finite difference

nnn

b

an

b

a

xaxaaxf

dxxfdxxfI

10

92

9. Numerical Integration9. Numerical Integration9.1 Introduction

Basic Approach Trapezoidal Concept

There are two approaches in refining the approximate the integral: I th b f Increase the number of

sections Use higher-order terms

93

Rac 2015 20

9. Numerical Integration9. Numerical Integration 9.1 Introduction

94

Multiple-section approach Higher-order-term approach

9. Numerical Integration9. Numerical Integration9.2 Trapezoidal Rule (Linear)9.2.1 Basic Rule

1

1

curvetheunderareathegconsiderin

afbfabI

axab

afbfafxf

dxxfdxxfIb

a

b

a

1

1

95

3"121

:degree) (secondError 2

abfE

ff

t

Rac 2015 21

9. Numerical Integration9. Numerical Integration9.2.2Multiple Section

nab 1 1

11

11

111

11 1

Error (summation of error at all sections, still 2nd

n

ii xfxfxfn

abI1

0 221

1

11 1

1

Derivation Eq. 21.8 & 21.9

96

degree):

it fn

abE "121 3

9. Numerical Integration9. Numerical IntegrationProgramming realisation

FUNCTION Trapm (h,n,f)sum = f0

start

DO I = 1, n-1sum = sum + 2* fi

ENDDOsum = sum + fnTrapm = h*sum/2

END Trapm

97

end

Rac 2015 22

9. Numerical Integration9. Numerical Integration9.3 Simpsons Rule 1/3 (Quadratic)9.3.1 Basic Rule

bb

210

2

20

2

46

polynomialLagrangeorderSecond;

xfxfxfabI

xfbxax

dxxfdxxfIb

a

b

a

1

4

1

But, third-order accuracy (see Box. 21.3)

98

)4(5

5)4(

2880

2901

fab

abfEt

9.3.2Multiple Aplication

nn

xfxfxfxfabI21

24

nj

ji

i xfxfxfxfnI

,...6,4,2,...5,3,10 243

41

4

4

4 4 1

11 1

1

11

11

5

)4(1

abfE

99

290

fEt

Rac 2015 23

9.4 Simpsons 3/8 Rule9.4.1 Basic Rule

bb

Comparison from 1/3 rule: 3/8 third order approximation whilst 1/3 only second

3210

20

3

338

;

xfxfxfxfabI

bxax

dxxfdxxfIaa

100

3/8 third order approximation, whilst 1/3 only second order but with the same level of accuracy

1/3 rule requires 3 datapoints, whilst 3/8 4 datapoints For multiple application, 1/3 rule is used for no. of points:

3, 5, 7, 9, , whilst 3/8 is used for no. of points: 4, 7, 10, 13 ,

9.5 Composite methodIn order to adapt to the number of

segments, integration can be g , gformed as a composite of various previously described methods, i.e. Simpsons 1/3, and Simpsons 3/8 rules

101

Rac 2015 24

9.6 Higher Order

102

9.7 Open Integrals

103

Rac 2015 25

*Text Book Problems: 21.1 and 21.3 21 9 21.9 21.18

104

10. Integration of Equations10. Integration of Equations10.1 Introduction

Chapter 9 discusses methods to calculate the integral from a number of discrete datapoints, e.g. experimental resultsp , g p

Basically, the methods discussed previously is still applicable. While not having datapoints, we have equation, so solve the equation for any input data, xi (Newton-Cotes)

However, different strategies are preferable! Three approaches are available:

Recursive Roombergs integration

105

Recursive Roomberg s integration Gauss-Quadrature Gauss-Legendre

Rac 2015 26

10. Integration of Equations10. Integration of Equations Roombergs Integration

Provides a way to combine two applications of the Trapezoidal rule with lower level of accuracy to compute a p y pthird estimate with higher accuracy

Richardsons Extrapolation

lmml

m IIhhhII

11

2

41212 2/14 hOhhforhIhII

106

812

6

1212

8/631

6364

4/151

1516

2/33

hOhhforIII

hOhhforIII

hOhhforhIhII

lm

lmlm

10. Integration of Equations10. Integration of Equations Roombergs Integration

Richardson Extrapolation: Example 22.1: I = integral estimate with 1 segment and Il = integral Im integral estimate with 1 segment, and Il integral

estimate with 2 segment, can be combined to reach the third estimate:

%6.16

367467.11728.0310688.1

34

31

34

t

lm III

In general form:

107

144

11,1,1

1

,

kkjkj

k

kj

III j : more or less accurate integralk : level of integration

Rac 2015 27

10. Integration of Equations10. Integration of Equations10.2 Gauss Quadrature

Consider a linear approximation of numerical integration (Basic g (Trapezoidal). Observe the error!

Now, by modifying the intersection of the area and the curve, the approximation could be more accurate!

a b

108

ba? ?

10. Integration of Equations10. Integration of Equations10.2.1 Gauss-Legendre (2-point)

Systematic implementation of 2-point gauss quadrature to estimate third order integralg

First, normalised the range of integration:a to b

-1 to 1

109

Rac 2015 28

10. Integration of Equations10. Integration of Equations The method can approach up to cubic integrals:

1

0 1xy

y

xfcxfcI 4 unknowns (c0, c1, x0, x1)4 equations

33

22

xyxy

1100 xfcxfcI

0

21

1

110

1

110

dxxbfcafc

dxbfcafc

4 equations

...5773505.03

11

0

10

x

cc

110

032

1

1

310

1

1

210

1

dxxbfcafc

dxxbfcafc ...5773505.03

13

1 x

10. Integration of Equations10. Integration of Equations Step-by-step Application

The integral variable, x, is transformed to new variable, xd:

Consequently, the integral equation is transformed to using the above transformation relationship, to become:

d

d

dxabdx

xababx

2

2

b

1

111

Then, solve the integral using Gauss-Quadrature with known xo, x1, co, c1:

dddda

dxxfIdxxfI

1

1100 xfcxfcI dd

Rac 2015 29

10. Integration of Equations10. Integration of Equations10.2.2 Gauss-Legendre (Higher points)

nn xfcxfcxfcI 1100 With 2*(n+1) coefficients, it could solve system of

polynomial equations for the order up to {2*(n+1)-1}, to give c0, c1, cn (weights), and x0, x1, xn (function arguments)

Hence, accurate for polynomial with order up to {2*(n+1)-1}.

nn1100

112

10. Integration of Equations10. Integration of Equations10.2.2 Gauss-Legendre (Higher points)

113

Rac 2015 30

10. Integration of Equations10. Integration of Equations 10.3 Improper Integrals

114

10. Integration of Equations10. Integration of Equations 10.3 Improper Integrals

Transform infinite integration limit to finite limit (x 1/t)

Set the datapoints at sufficiently close to infinity Use Open integral method, e.g. Midpoint method (Tab.

21.4):

,11/1

/12 badtt

ft

dxxfa

b

b

a

115

12

1 23

23

0

n

n

ii

x

x

xfxfxfhdxxfn

Rac 2015 31

1111. . Numerical Numerical DifferentiDifferentiationation

Use Finite-divided-difference to estimate f(x), f(x) f (x), ...

Alternative Methods: Forward FDD: accuracy O(h) Fig. 23.1 Backward FDD: accuracy O(h) Fig. 23.2 Centered FDD: accuracy O(h2) Fig. 23.3

Another alternative method to increase accuracy Richardson extrapolation

116

lmml

m DDhhhDD

11

2

1122. Differential Equations. Differential Equations12.1 Introduction

Differential equations that will be solved are in form of:

dy

Recall the approximation by Taylor series:

The numerical solution for the diff. equation solves for yi at

yxfdxdy ,

hyxfyy

hdxdyyy

iiii

ii

,1

1

117

q yieach desired input xi

It requires one initial condition, x0. The method is then called Eulers method (linear)

Rac 2015 32

1122. Differential Equations. Differential Equations Type of Problems to be solved:

We know the slope at any point (dy/dx) Solve for each position y Solve for each position, y

Example: Falling parachutist Free vibration

118

1122. Differential Equations. Differential Equations Problems:

Truncation, due to discretization of x in predicting y, consisting of 2 parts:g p Local truncation error Propagated truncation error

...!3'''

!2''' 321 h

xfhxfhxfxfxf iiiii

(truncated)

119

Round-off, caused by limited numbers of significant digits that can be retained by computer

Rac 2015 33

1122. Differential Equations. Differential Equations Improvement (smaller step size)

120

1122. Differential Equations. Differential Equations Improvement (higher order)

Midpoint method

121

Rac 2015 34

1122. Differential Equations. Differential Equations12.2 Higher order methods

Heuns method: is used to extrapolate a Predictor yxfy ' is used to extrapolate a Predictor

The slope at the predictor is calculated and then used to estimate a better general slope, as a Corrector to the previous slope (average)

iii yxfy ,

hyxfyy iiii ,0 1

''

,' 0 111

iii yxfyhyyy '0

122

The next predicted result is, then: Or:

2''' 1 ii yyy

hyyy ii 11

hyxfyxfyy

hyxfyy

iiiiii

iiii

2,,:Corrector

,:Predictor0

111

01

1122. Differential Equations. Differential Equations Improvement (higher order)

Heuns method

123

Rac 2015 35

1122. Differential Equations. Differential Equations12.3 Runge-Kutta

Runge-Kutta methods are generalisation as well as extension of the previous two methods (Heuns dan Mid-p (point)

The following general form, hence, can be used:

n represents the order of approximation: second-order, thi d d t

nn

iiii

kakakahhyxyy

2211

1 ,,

124

third-order, etc

hkqhkqhkqyhpxfk

hkqhkqyhpxfkhkqyhpxfk

yxfk

nnnnninin

ii

ii

ii

11,122,111,11

22212123

11112

1

,,

,,

,

1122. Differential Equations. Differential Equations Imagine, as in Heun and Midpoint, the algorithm contains

two important steps: Calculation of p and q. Calculation of k

Before actually calculating the next y Heun and Midpoint belong to second order Runge-Kutta

(RK2), with certain p, q, k. Description Eq. 25.36 to 25.38

125

Rac 2015 36

1122. Differential Equations. Differential Equations Various types of Runge-Kutta:

Second order (RK2):hkkyy 11

hkyhxfk

yxfk

hkkyy

ii

ii

ii

12

1

211

,,

22

hkyhxfk

yxfkhkyy

ii

ii

ii

12

1

21

21,

21

,

(Heun)

(Mid-point)

126

yf ii 12 2,

2

hkyhxfk

yxfk

hkkyy

ii

ii

ii

12

1

211

43,

43

,32

31

(Ralston)

1122. Differential Equations. Differential Equations Third order (RK3):

yxfk

hkkkyy

ii

ii

1

3211

,

461

hkkkyy ii 3211 46

1

Fourth order (RK4):

hkhkyhxfk

hkyhxfk

yxfk

ii

ii

ii

213

12

1

2,21,

21

,

yxfk ii1 , hkkkkyy ii 43211 221

6

127

hkyhxfk

hkyhxfk

hkyhxfk

ii

ii

ii

34

23

12

,21,

21

21,

21

hkkkkyy ii 43211 226

Rac 2015 37

1122. Differential Equations. Differential Equations Graphical Description of Slope Modification in R-K

128

Heun (R-K 2) R-K 4

1122. Differential Equations. Differential Equations Comparison

129

Rac 2015 38

1122. Differential Equations. Differential Equations Systems of ODE

nyyyxfddy ,...,,, 2111

nnn

n

n

yyyxfdxdy

yyyxfdxdydx

,...,,,

,...,,,

21

2122

211

0000

130

Initial condition:

Selected: step size,

002

01

0 ,...,,, nyyyx

h

BoundaryBoundary--value Problemvalue Problem Heat transfer: from hotter surface, T1, to cooler surface T2,

in one-dimensional medium (rod)

Known: Th, and Tc.

Thot

0'22

xTThdx

Tda

Tcool

131

Find: temperature distribution at the medium

Rac 2015 39

BoundaryBoundary--value Problemvalue Problem Solution Strategy#1: Shooting method

xTThTd 0'2

Steps:

a

a

a

TThdxdzz

dxdT

TThdxdT

dxd

xTThdx

';

'

02One 2nd order ODEbecomesTwo 1st ODEs (system of ODEs)

132

Assume z(0) calculate to find Tn if Tn known Tc, Assume z(0), Interpolate to find appropriate z(0) use for overall

solution, T (x). Example 27.1

BoundaryBoundary--value Problemvalue Problem Strategy #2: Finite difference

0'2

xTThTd

Grid model:

0'2

0

211

2

ia

iii

a

TThx

TTT

xTThdx

133

T0=Th T1 T2 Tn = Tc

x

Rac 2015 40

BoundaryBoundary--value Problemvalue Problem For each step, build system of equation:

1121012 0'20'2 h TThTTTTThTTT

12 2112 21

22123

1212

'20'2

0'2

00

nannc

nannn

a

aa

TThx

TTTTThx

TTT

TThx

TTT

TThx

TThx

Th = Hot temp(known)Th = Cold temp(known)

134

nnnn

n

b

bb

T

TT

a

aaaaa

2

1

1

1

0

2212

11211 Matrix of Linear Algebraic EquationSolve for Ts (temperatureDistribution)

BoundaryBoundary--value Problemsvalue Problems Buckling

Analytical Approach

135

0222

2

2

2

2

ypdx

ydEIPy

dxyd

EIM

dxyd

Rac 2015 41

BoundaryBoundary--value Problemsvalue Problems Analytical approach: Eigenvalue

Numerical approach: Finite difference: Numerical approach: Finite difference:

positionatcolumntheofdeflectionare

02

0

22

11

22

2

ii

iiii

xy

ypx

yyy

ypdx

yd

136

above) Procedure Problems ValueBoundary (see

position at column theofdeflection are

ii xy

Eigenvalue ProblemsEigenvalue Problems General Form

0 XIA Possible solutions:

{X} = 0, i.e. all x-s zero trival (useless) [[A]-[I]] = {0}, i.e. det [[A]-[I]] = 0 non-trivial, with

usefull {X}

Hence, the latter solution is sought after:d t [[A] [I]] 0 li i l f ( t f l i l

137

det [[A]-[I]] = 0 polinomial of (roots of polynomial search)

Solve for i using methods of Muller or Bairstow, or even simpler methods as in Roots of equation problems.

Rac 2015 42

Eigenvalue ProblemsEigenvalue Problems Alternate method: Power Method

0

XXA

XIA

Procedure:

)for search (iterative

1

ii XAX

XXA

The result: the largest Eigenvalue.In order to obtain the smallest Eigenvalue,[A] [A]-1, with the result of (Eigenvalue)-1.

1. Trial of {X}, and solve for [A]{X}2. Find and {Xnew}3. Feed {Xnew} for [A]{Xnew}4. Repeat the process until error is small enough

138

Eigenvalue ProblemsEigenvalue Problems Application #1: Mohrs Circle State of stress

Find Principal Stresses:

zzyzx

yzyyx

xzxyx

ij

02

1

xzxyx

139

stresses) (principal,,0

0

321

322

13

2

2

IIIzzyzx

yzyyx

Root finding of Third order Polynomial Equation Problem

Rac 2015 43

Eigenvalue ProblemsEigenvalue Problems Application #2: MDOF Vibration

0F

02

02

2122

2111

xxkxm

xxkxm

140

First Approach: Mixed numerical-analytical Trial solution:

tAx ii sin

Eigenvalue ProblemsEigenvalue Problems

tAx

tAx

ii

ii

sin

sin2

0sin2sinsin

0202

0sinsin2sin

02

212

22

2122

212

11

212

11

2111

tAtAktAmxxkxm

AAkAmtAtAktAm

xxkxm

141

02 21222 AAkAm

00

2

2

2

1

2

22

1

2

1

AA

mk

mk

mk

mk

Eigenvalue Problem

Rac 2015 44

Eigenvalue ProblemsEigenvalue Problems Solution:

Natural frequencies: 1 and 2 Mode Shape: A1 and A2 for each modep 1 2

142

Partial Differential EquationsPartial Differential Equations General Form

0222

DuCuBuA

Classification

022

D

yC

yxB

xA

143

Rac 2015 45

Math Function:

Partial Differential EquationsPartial Differential Equations Elliptic Equations

Example:

Name : Laplace equationProblem: Find temperature

022

2

2

yT

xT

144

Problem: Find temperature distribution over 2D areawith known boundaryvalues

Partial Differential EquationsPartial Differential Equations Elliptic Equations

Numerical Approach 022

2

2

yT

xT

yx

04:hence , grid, squareset weif

022

,1,1,,1,1

21,,1,

2,1,,1

bTa

TTTTTyx

yTTT

xTTT

jijijijiji

jijijijijiji

145

...

T

bTaTemp. Distribution over area

Matrix of Linear Algebraic Equations!

Rac 2015 46

Math Function:

Partial Differential EquationsPartial Differential Equations Parabolic Equations

Example:

Name: Heat conduction equationP bl T t di t ib ti

2

2

'xTk

tT

146

Problem: Temperature distribution over a rod length atdifferent time

Partial Differential EquationsPartial Differential Equations Parabolic Equations

Numerical Approach

2

2

'xTk

tT

lilililili

li

li

li

li

li

TTTTTxtk

xTTTk

tTT

111

2

211

1

2

' :parameterconstant set the

2'

147

T0=Th T1 T2 Tn = Tc

x

Rac 2015 47

Math Function:

Partial Differential EquationsPartial Differential Equations Hyperbolic Equations

Example:

Wave equation

2

2

22

2 1ty

cxy

148