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8

The Eigen Value Problem

8.1. INTRODUCTION

Computation of eigen values and the corresponding eigen vectors of a matrix is of practical importance. For example, in solid mechanics, when we consider an element in a continuum, subjected to normal and shear stresses, we usually find principal stresses which are the maximum and minimum stresses in an element.

Consider the wedge of unit thickness, subjected to normal and shear stresses. If it has to be in equilibrium, the following system of equations, written in matrix notation, must be satisfied.

LMσ

x

− σθ

xy

σ xy

σ y − σθ

OP Lcos θO = L0O

Q MNsin θPQ MN0PQ

For non-trivial solution to exist, the determinant of the coefficient matrix must be zero.

σ x − σθ

σ xy

σ xy

=0

σ y − σθ

σθ2 – σθ (σx + σy) + (σx σy – σxy2) = 0

This is a quadratic equation whose roots give two eigen values corresponding to principal stresses.

In this chapter, we will describe power method for finding largest eigen value of a matrix and inverse power method for finding smallest eigen value of a matrix. Several methods will be discussed for finding eigen values of symmetric tridiagonal matrices.

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9

Numerical Solution of

Ordinary Differential Equations

9.1. INTRODUCTION

A physical situation that concerns with the rate of change of one quantity with respect to another gives rise to a differential equation.

Consider the first order ordinary differential equation dy

= f (x, y)

...(1) dx with the initial condition

...(2) y(x0) = y0

Many analytical techniques exist for solving such equations. But these methods can be applied to solve only a selected class of differential equations.

However, a majority of differential equations appearing in physical problems cannot be solved analytically. Thus it becomes imperative to discuss their solution by numerical methods.

In numerical methods, we donot proceed in the hope of finding a relation between variables but we find the numerical values of the dependent variable for certain values of independent variable.

It must be noted that even the differential equations which are solvable by analytical methods, can be solved numerically also.

9.2. INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS

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Index

A

A general error formula 36

Acceptance region 716

Acceptance sampling 701

Accuracy of numbers 28

Adams-Moulton (or Adams-Bashforth) formula 546

Advantages of statistical quality control 701

Advantages/features of ‘C’ language 6

Aitken’s method 308

Aitken’s ∆2 method 150

Algorithmic errors 27

Alternative hypothesis 715

Analysis of data 701

Analysis of time-series 755

Analysis of variance (ANOVA) 745

Angle between two lines of regression 684

Annihilation of off-diagonal elements 465

ANOVA table 745

Applications of t-distribution 726

Approximate numbers 28

Approximation of functions 616

Area diagrams 678

Arithmetic operations with normalized floating point numbers 50

Arrays and string 15

Assignable causes 702

Assumptions for interpolation 165

Automatic error monitoring 550

Averaging operator 169

B

Backward differences 168

Bar diagrams 678

Basic structure of ‘C’ program 10

Bender-Schmidt explicit finite difference scheme

658

Bender-Schmidt 657

Bessel’s interpolation formula 253

Bisection method 63

Bolzano method 64

Boole’s rule 336

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260

NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’

Using Bessel’s formula,

RS f (0) + f (1) UV + FG u − 1IJ ∆f (0) + u (u − 1) RS ∆

2

T 2 W H 2K

T

2

f(u) =

FG

H

(u − 1) u −

+

f(.25) =

UV

W

f (− 1) + ∆2 f (0)

2

3!

IJ

K

1 u

2

∆3 f (− 1)

FG 32 + 35IJ + (.25 – .5) (3) + (.25) (.25 − 1) RS − 5 + 2 UV

H 2 K

2

T 2 W

+

(.25 − 1) (.25 − .5) (.25)

(7)

3!

= 32.9453125

Hence y25 = 32.9453125.

Example 2. Apply Bessel’s formula to find the value of f(27.4) from the table : x :

25

26

27

28

29

30 f(x) :

4.000

3.846

3.704

3.571

3.448

3.333.

Sol. Taking origin at 27 and h = 1 x = a + uh ⇒

27.4 = 27 + u × 1

∴ u = 0.4

The difference table is as follows : u

103f(u)

–2

4000

–1

3847

103 ∆f(u)

103 ∆2f(u)

103 ∆3f(u)

103 ∆4f(u)

103 ∆5f(u)

– 154

12

– 142

0

–3

3704

9

– 133

1

3571

2

3448

3

3333

4

1

–7

10

– 123

–3

–2

8

– 115

Bessel’s formula is

RS f (0) + f (1) UV + FG u − 1IJ ∆f (0) + u (u − 1) RS ∆ f (0) + ∆ f (− 1) UV

2!

2

T 2 W H 2K

W

T

F 1I

(u − 1) G u − J u

H 2 K ∆ f (− 1) + (u + 1) u (u − 1) (u − 2) RS ∆ f (− 1) + ∆ f (− 2) UV

+

3!

4!

2

W

T

F 1I

(u − 2) (u − 1) G u − J u (u + 1)

H 2K

∆ f (− 2)

+

2

2

f(u) =

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43

ERROR ANALYSIS AND ESTIMATION

Rounding off all other numbers to two decimal digits, we have 0.15, 15.45, 0.00, 8.12,

0.02, 0.64 and 0.17.

The sum S is given by

S = 305.1 + 143.3 + 0.15 + 15.45 + 0.00 + 8.12 + 0.02 + 0.64 + 0.17

= 472.59 = 472.6.

To determine absolute error, we note that first two numbers have each an absolute error of 0.05 and remaining seven numbers have an absolute error of 0.005 each.

∴ Absolute error in all 9 numbers

= 2(0.05) + 7(0.005) = 0.1 + 0.035 = 0.135 ≈ 0.14.

In addition to the above absolute error, we have to take into account rounding error which is 0.01. Hence the total absolute error in S = 0.14 + 0.01 = 0.15

Thus,

S = 472.6 ± 0.15.

Example 17. 5.5 = 2.345 and 6.1 = 2.470 correct to four significant figures. Find the relative error in taking the difference of these numbers.

1

× 0.001 = 0.0005

2

∆x1

∆x2

∆x1

0.0005

+

∴ Relative error <

=2

=2

= 0.008 .

X

X

0.125

X

Example 18. 10 = 3.162 and e ~

– 2.718 correct to three decimal places. Find the percentage error in their difference.

0.0005

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