MECN 3500 Inter - Bayamon Lecture 7 Numerical Methods for Engineering MECN 3500 Professor: Dr. Omar...

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LectureLecture

77Numerical Methods for EngineeringNumerical Methods for Engineering

MECN 3500 MECN 3500

Professor: Dr. Omar E. Meza CastilloProfessor: Dr. Omar E. Meza [email protected]

http://www.bc.inter.edu/facultad/omeza

Department of Mechanical EngineeringDepartment of Mechanical Engineering

Inter American University of Puerto RicoInter American University of Puerto Rico

Bayamon CampusBayamon Campus

Lecture 7Lecture 7MEC

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Tentative Lectures ScheduleTentative Lectures Schedule

TopicTopic LectureLecture

Mathematical Modeling and Engineering Problem SolvingMathematical Modeling and Engineering Problem Solving 11

Introduction to MatlabIntroduction to Matlab 22

Numerical ErrorNumerical Error 33

Root FindingRoot Finding 4-5-64-5-6

System of Linear EquationsSystem of Linear Equations 77

Least Square Curve FittingLeast Square Curve Fitting

Polynomial Interpolation Polynomial Interpolation

Numerical IntegrationNumerical Integration

Ordinary Differential Equations Ordinary Differential Equations


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Gauss EliminationGauss Elimination

Linear Algebraic EquationsLinear Algebraic Equations

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To solve linear algebraic equations using To solve linear algebraic equations using the technique the technique Gauss EliminationGauss Elimination. .

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Course ObjectivesCourse Objectives


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An equation of the form An equation of the form ax+by+c=0 ax+by+c=0 or or equivalently equivalently ax+by=-c ax+by=-c is called a linear is called a linear equation in equation in xx and and yy variables. variables.

ax+by+cz=dax+by+cz=d is a linear equation in is a linear equation in three variables, three variables, x, yx, y, and , and zz..

Thus, a linear equation in n variables isThus, a linear equation in n variables is

aa11xx11+a+a22xx22+ … +a+ … +annxxnn = b = b A solution of such an equation consists A solution of such an equation consists

of real numbers of real numbers cc11, c, c22, c, c33, … , c, … , cnn. If you . If you need to work more than one linear need to work more than one linear equations, a system of linear equations equations, a system of linear equations must be solved simultaneously.must be solved simultaneously.

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IntroductionIntroduction


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For small number of equations (n For small number of equations (n ≤ 3) ≤ 3) linear equations can be solved readily linear equations can be solved readily by simple techniques such as “by simple techniques such as “method method of eliminationof elimination.”.”

Linear algebra provides the tools to Linear algebra provides the tools to solve such systems of linear equations.solve such systems of linear equations.

Nowadays, easy access to computers Nowadays, easy access to computers makes the solution of large sets of makes the solution of large sets of linear algebraic equations possible and linear algebraic equations possible and practical.practical.

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Solving Small Number of EquationsSolving Small Number of Equations


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There are many ways to solve a system There are many ways to solve a system of linear equations:of linear equations: Graphical methodGraphical method Cramer’s ruleCramer’s rule Method of eliminationMethod of elimination Computer methodsComputer methods..

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Solving Small Number of EquationsSolving Small Number of Equations


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The Graphical MethodThe Graphical Method

For two equations:

Solve both equations for x2:

2222121

1212111

bxaxa

bxaxa

22

21

22

212

1212

11

12

112 intercept(slope)

a

bx

a

ax

xxa

bx

a

ax


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Or equate and solve for x1

12

11

22

21

12

1

22

2

12

11

22

21

22

2

12

1

1

22

2

12

11

12

11

22

21

22

21

22

21

12

11

12

112

0

aa

aa

ab

ab

aa

aa

ab

ab

x

a

b

a

bx

a

a

a

a

a

bx

a

a

a

bx

a

ax


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Ill-conditioned(Slopes are too close)

No solution Infinite solutions


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1111


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12

12

21

21

x3x3x2x

rearrange 3xx

3xx2

2x1 – x2 = 3x1 + x2 = 3

One solution


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2x1 – x2 = 3

2x1 – x2 = – 1

No solution


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6x1 – 3x2 = 92x1 – x2 = 3

Infinite solutions


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2x1 – x2 = 3

2.1x1 – x2 = 3

Ill conditioned


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on Cramer’s RuleCramer’s Rule

Compute the Compute the determinant Ddeterminant D 2 x 2 matrix2 x 2 matrix

3 x 3 matrix 3 x 3 matrix 211222112221

1211 aaaaaa

aaD

3231

222113

3331

232112

3332

232211

333231

232221

131211

aa

aaa

aa

aaa

aa

aaa

aaa

aaa

aaa

D

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To find To find xxkk for the following system for the following system

Replace kReplace kthth column of a column of as with bs with bss (i.e., (i.e., aaikik b bi i ) )

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nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

...

...

...

2211

22222121

11212111

)

)(

ijk D(a

matrix newDx


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3 x 3 matrix3 x 3 matrix

333231

232221

131211

aaa

aaa

aaa

D

33231

22221

11211

33

33331

23221

13111

22

33323

23222

13121

11

1

1

1

baa

baa

baa

DD

Dx

aba

aba

aba

DD

Dx

aab

aab

aab

DD

Dx

1818


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1919


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on Example 9.3Example 9.3

44.05.03.01.0

67.09.15.0

01.052.03.0

321

321

321

xxx

xxx

xxx

5.03.01.0

9.115.0

152.03.0

D

8.19

44.03.01.0

67.015.0

01.052.03.01

5.29

5.044.01.0

9.167.05.0

101.03.01

9.14

5.03.044.0

9.1167.0

152.001.01

33

22

11

DD

Dx

DD

Dx

DD

Dx

2020


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Eliminate x2

Subtract to get

2222121

1212111

bxaxa

bxaxa

2122221212112

1222122211122

baxaaxaa

baxaaxaa

21122211

1212112

21121122

2121221

2121221211211122

aaaa

babax

aaaa

babax

babaxaaxaa

Elimination MethodElimination Method

Not very practical for large number (> 4) of equations

2121


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on Example 9.3Example 9.3

2 x 2 matrix2 x 2 matrix

3)1(2)2(3

)18)(1()2(3

4)1(2)2(3

)2(2)18(2

2

1

x

x

22

1823

21

21

xx

xx

2222


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nnnnnn

n

n

b

b

b

x

x

x

aaa

aaa

aaa

2

1

2

1

21

22221

11211

The system can be written in a matrix format as

NAVIE GAUSS EliminationNAVIE GAUSS Elimination

Gauss elimination is the most important algorithm to solve systems of linear equations.

It involves by combining equations to eliminate unknowns.

It involves 2 phases:

1. Forward elimination phase: reduce the set of equations to an upper triangular system.

2. Back substitution: work from the last equation up.

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In the first step of the forward elimination phase, x1 is eliminated from all equations except the first one.

The coefficient of x1 in the first equation is called the pivot element.

The second step is to eliminate x2 from the third equation through the nth equation.

Do the same for all variables x3 to xn-1.

The goal is to set up upper triangular matrix

The back-substitution phase starts from the last equation up, to find the values of x1, x2, …, xn.


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Gauss Elimination Pseudocode

Forward elimination phase

Back substitution


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Use Gauss elimination to solve

Carry 6 significant figures.

Solution

1. Forward elimination: eliminate x1 from equation (2):

- (0.1/3)

5617.19293333.000333.7

3.193.071.0

261667.000666667.000333333.01.0

32

321

321

xx

xxx

xxx

NAVIE GAUSS Elimination

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Eliminate x1 from equation (3): - (0.3/3)

After eliminating x1 from equations (2) and (3), the system becomes

Eliminate x2 from equation (3):

+ (0.190000/7.00333)

After eliminating x2 from equation (3), the system becomes


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2. Back Substitution: find the value of x3 from equation (3):

Substitute the value of x3 in equation (2) to find the value of x2:

Substitute the values of x2 and x3 in equation (1) to find the value of x1:

0000.70120.10

0843.703 x

50000.200333.7

)0000.7(293333.05617.19

5617.19)0000.7(293333.000333.7

2

2

x

x

00000.33

)0000.7(2.0)50000.2(1.085.7

85.7)0000.7(2.0)50000.2(1.03

1

1

x

x


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0det AD

Pitfalls of Elimination MethodsPitfalls of Elimination Methods

Division by zero Division by zero (Partially solved by the (Partially solved by the pivoting technique) pivoting technique)

Round-off errors Round-off errors (Important when (Important when large number of equation are to be large number of equation are to be solved)solved)

Ill-conditioned systemsIll-conditioned systems: :

Small changes in coefficients result in Small changes in coefficients result in large changes in the solution.large changes in the solution.

WhenWhen

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0000.10000.10000.1

0001.20000.30003.0

21

21

xx

xx

Techniques for Improving SolutionsTechniques for Improving Solutions

Use of more significant figures Use of more significant figures (The (The simplest remedy) simplest remedy)

Pivoting Pivoting (Determine the largest (Determine the largest available coefficient in the column available coefficient in the column below the pivot element and switch below the pivot element and switch rows so that the largest element is the rows so that the largest element is the pivot element.)pivot element.)

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on Techniques for Improving SolutionsTechniques for Improving Solutions

Scaling Scaling (Divide each row by the largest (Divide each row by the largest element in that row) element in that row)

2

000,100000,1002

21

21

xx

xxWithout Scaling:

With Scaling:2

100002.0

21

21

xx

xx

With Scaling and Pivoting:

100002.0

2

21

21

xx

xx

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on MATLAB Script File: NGaussElim.mMATLAB Script File: NGaussElim.m

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3333


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>>Enter Matrix A > [3 -0.1 -0.2; 0.1 7 -0.3; 0.3 -0.2 10]A = 3.0000 -0.1000 -0.2000 0.1000 7.0000 -0.3000 0.3000 -0.2000 10.0000

>>Enter Solution Vector B > [7.85 -19.3 71.4]

B = 7.8500 -19.3000 71.4000

S = 3.0000 -2.5000 7.0000

>>Enter Matrix A > [3 -0.1 -0.2; 0.1 7 -0.3; 0.3 -0.2 10]A = 3.0000 -0.1000 -0.2000 0.1000 7.0000 -0.3000 0.3000 -0.2000 10.0000

>>Enter Solution Vector B > [7.85 -19.3 71.4]

B = 7.8500 -19.3000 71.4000

S = 3.0000 -2.5000 7.0000


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MECN 3500 Inter - Bayamon Lecture 7 Numerical Methods for Engineering MECN 3500 Professor: Dr. Omar...

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Transcript of MECN 3500 Inter - Bayamon Lecture 7 Numerical Methods for Engineering MECN 3500 Professor: Dr. Omar...