Chapter 2 Solving Linear Equation - Griffith University · PDF fileChapter 2 Solving Linear Equation ... (2.10) Dd d d d d d i i n n == ... • Find a mathematical model or way of

MEE3017 Computer Modeling Techniques in Engineering

Chapter 2 Solving Linear Equation • A linear equation represents the linear dependence of a quantity φ on a set of

variables x1 through xn and a set of constant coefficients α1 through αn; its form is φααα =+++ nnxxx ...2211 • If we replace φ by a singly subscripted quantity fi and the coefficients αj by a doubly

subscripted quantity αij , we may write a system of n such equations in the form

(2.1)

nnnnjnjnn

ininjijii

nnjj

nnjj

fxxxx

fxxxx

fxxxx

fxxxx

=+++++

=+++++

=+++++

=+++++

αααα

αααα

αααα

αααα

LL

LL

LL

LL

21211

22111

222222121

111212111

::::

::::

• The solution of such systems for the quantities xj, when the coefficients aij and the

values fi are given, pervades engineering applications and computational methods. For example, systems of linear algebraic equations are used directly in mathematical models for electrical, structural, and pipe networks, and in some computational methods for fitting curves to data. In other cases, such as in finite difference and finite element solutions of partial differential equations, they represent approximations of mathematical equations, which cannot be solved by analytical methods. 2.1 Fundamentals of linear algebra 2.1.1 Notation and Definitions • A matrix is a rectangular array of quantities.

An example of a matrix is

(2.2)

==

34333231

24232221

14131211

][aaaaaaaaaaaa

aA ij

Chapter 2 – Page 1


• In Eq. (2.2), the bold-faced A is the symbol for the matrix, the literal representation on the far right shows the arrangement of the elements, and the form [aij] is a shorter abstraction of the literal form.

• The matrix has no particular meaning until we associate the elements with a given concept.

• A matrix with m rows and n columns has dimension (m × n) and is referred to as an (m × n) matrix. A column vector c and a row vector r are shown in Eq.(2.3)

(2.3) [ ]4321

3

2

1

][;][ rrrrrrccc

cc ji ==

==

2.1.2 Operations • Addition and subtraction for conformable matrices A and B are summarized by

][][)()()( ijijij

nmnmnmbacCBA ±===±

××× (2.4)

• Matrix multiplication A ⋅ B is defined only if the number of columns in A is equal to

the number of rows in B.

(2.5a) ][)()()( ijnmnrrm

pPBA ==⋅×××

(2.5b) rjirjiji

r

kkjikij babababap +++=×= ∑

=L2211

1)(

• Matrix multiplication is both distributive and associative. • If A,B, and C are matrices with appropriate dimensions to satisfy the addition/

subtraction conditions of Eq. (2.4) and the multiplication conditions of Eq.(2.5a), the distributive and associative properties are expressed in Eq.(2.6) and Eq.(2.7), respectively, by

ACABACBCABACBA ⋅+⋅=⋅+⋅+⋅=+⋅ )(;)( (2.6)

CBACBACBA ⋅⋅=⋅⋅=⋅⋅ )()( (2.7) • Multiplication of a matrix A by a scalar q to form a product matrix S equal to qA is

also a defined operation. The product in this case is given by



][][][)()( ijijij

nmnmqaaqsSAq ====

×× (2.8)

2.1.3 Square Matrices • A square matrix is one that has the same number of rows as the number of columns;

for example, a matrix with dimension (n × n). Special types of square matrices are described as follows.

• A diagonal matrix D equal to [dij] satisfies the condition

jifordij ≠= 0 (2.9) • We may replace [dij] by [di], implicitly it in the correct context so that it is not

confused with the vector notation of Eq. (2.3), and express the diagonal matrix D in the form

(2.10) D d

dd

d

dd

i i

n

n

= =

−

[ ]

1

2

1

0

0

O

O

• The identity matrix I is a diagonal matrix in which every diagonal element has a unit

value; it is the matrix of the scalar unit value. The zero matrix 0 is one whose elements are all zero; it may be either a square matrix or a general rectangular matrix.

• Multiplications involving the general diagonal matrix D, the identity matrix I, and the zero matrix 0 are summarized as follows; each of D, I and 0 is assumed to be an (n x n) matrix.

[ ] [ ]iii

nnnncduucD ===⋅

××× )1()1()(

(2.11a)

[ ] [ ]jjjnnnn

drvvDr ===⋅××× )1()()1(

(2.11b)

[ ] [ ]ijiijnnnnnn

adpPAD ===⋅××× )()()(

(2.11c)

[ ] [ ]ijjijnnnnnn

adqQDA ===⋅××× )()()(

(2.11d)

ccInnnn )1()1()( ×××

=⋅ (2.12a)



rIrnnnn )1()()1( ×××

=⋅ (2.12b)

IAAAInnnnnnnnnn )()()()()( ×××××

⋅==⋅ (2.12c)

00)1()1()( ×××

=⋅nnnn

c (2.13a)

00)1()()1( nnnn

r×××

=⋅ (2.13b)

000)()()()()( nnnnnnnnnn

AA×××××

⋅==⋅ (2.13c)

• The inverse of a square matrix A is denoted by A-1 and is itself a square matrix with

the same dimension as A such that

IAAAA =⋅=⋅ −− 11 (2.14)

The analogous scalar relation is ( ). αα α α− −= =1 1 1 • Two other types of square matrices that will be useful at a later stage are the lower

triangular and upper triangular matrices. The lower triangular form L permits nonzero elements only on and below the diagonal; it may be represented by

(2.15) L ij

n n nn

= =

[ ]l

l

l l

M O

M

M O

M

l l L L L L l

11

21 22

1 2

0

• The upper triangular form U permits nonzero elements only on and above the

diagonal; it may be represented by

(2.16) U u

u u uu u

u

ij

n

n

nn

= =

[ ]

11 12 1

22 2

0

L L L L

O M

M

O M

M



2.1.4 The Determinant of a Square Matrix The determinant of an (n × n) square matrix A is written asAand is defined by either of

∑=

=n

jijijCaA

1)( for any one value of i (2.17a)

or

∑=

=n

iijijCaA

1)( for any one value of j (2.17b)

in which Cij is known as the cofactor of the element aij. • The cofactor Cij of a (n x n) square matrix is obtained by first removing row i and

column j to form a ((n-1) x (n-1)) matrix and then by performing the operation (2.18) removed)j column and i row withA of nt(determinaC ji

ij ×−= +)1( 2.1.5 The Matrix Equation for Linear Algebraic Systems • The concept of matrix multiplication expressed in Eq.(2.5a) and (2.5b) allows us to

represent the system of linear algebraic equations given by Eq.(2.1) in the form

(2.19a)

=

⋅

n

i

2

1

n

2

1

nnnj2n1n

inij2i1i

n2j22221

n1j11211

f

f

ff

x

xx

aaaa

aaaa

aaaaaaaa

M

M

M

M

M

LLL

MM

LLL

MM

LLL

LLL

or, more compactly, in the form (2.19b) fxA =⋅ Here, A is the coefficient matrix formed by coefficients of the linear system, the original right-hand sides of the system are now the components of the column vector f. and the original unknown quantities are now the components of the column vector x. • A unique solution for x requires A to be square matrix, and it requires the system of

equations to be linearly independent. A coefficient matrix with a zero determinant is singular, a unique solution for x requires a nonsingular matrix.



• The solution of Eq.(2.19b) may be derived with the help of the associative law from Eq. (2.7), the concept of the inverse from Eq.(2.14), and operations with the identity matrix from Eq. (2.12a) as follows:

fAxAAxAAxIx ⋅=⋅⋅=⋅⋅=⋅= −−− 111 )()( (2.20) Among the additional tools used in methods for solving matrix equations are row and column exchanges. • A row exchange in the coefficient matrix A of Eq. (2.19b) is simply a position

exchange of two equations. The sequence of the components of x is preserved, but those of the right-hand side vector f must be exchanged to correspond to the new positions. A column exchange in the coefficient matrix does not alter the sequence of equations but it requires a corresponding exchange in the components of x so that the components are multiplied by the appropriate coefficients. Illustrations of these concepts are as follows.

• Exchange of rows I and k:

(2.21a)

=⋅

M

M

M

M

L

M

L

M

k

i

k

i

f

fx

a

a

1

1

• Exchange of columns j and k:

(2.21b) fx

x

aa

aa

k

j

nknj

kj

=

⋅

M

LLL

MM

LLL 11

2.1.6 Computational Techniques for Basic Operations The procedure of numerical computation: • Find a mathematical model or way of presenting the problem. • Choose a numerical method or calculation formula. • Set up the computation procedure. • Draw a program flowchart. • Write a program. • Run the computation. • Check the computation results.



Example 1: Calculation of Inverse Matrix 1) Find a mathematical model or way of presenting the problem. Problem: Find the inverse matrix from 2 by 2 matrix. The notation of a mathematical model is

=−

2221

12111

aaaa

A

1−A is the inverse matrix A.

2) Choose a suitable numerical method or formula. Using a formula of inverse matrix.

=

−

−−

=−

2221

1211

2221

1211

21122211

1 1bbbb

aaaa

aaaaA

3) Set up the computation procedure

Step 1 Data input Step 2 Compute the matrix Step 3 Error control Step 4 Compute the inverse matrix Step 5 Output the result



4) Draw a program flowchart = ≠

22211211 ,,, aaaa

21122211 aaaad −←

dabdabdab

dab

///

/

1122

2121

1212

2211

←−←−←

←

22211211 ,,, bbbb

Matrix is not Non-singular

END

RESULT, OUTPUT

d: 0

DATA INPUT

START

5) Write a program

Numerical computation program includes four parts: a. Define the variables and type b. Data input c. Computation d. Data output



Program example in C

/* < PROGRAM> */ # include <stdio.h> # include <stdlib.h> main( ) int i, j; double a[2][2]; /* 22× INPUT MATRIX */ double b[2][2]; /* 2 2× OUTPUT MATRIX */ double d; /*** STEP 1 DATA INPUT***/ printf (“ 2 INVERSE MATRIX /n *”); 2× for (i=0; i<2; i++) for (j=0; j<2; j++) printf (“a%d%d=”, i+1, j+1); scanf (“%1f”, &a[I][j]); /***STEP 2 CACULATE MATRIX***/ d=a[0][0]*a[1][1]-a[0][1]*a[1][0]; /***STEP3 ERROR CONTROL***/ if (d==0. 0) printf(“IT IS NON-SING ULAR MATRIX/n”); /***STEP 4 CALCULATE INVERSE MATRIX***/ b[0][0]=a[1][1]/d; b[0][1]=-a[0][1]/d; b[1][0]=-a[1][0]/d; b[1][1]=a[0][0]/d; /***STEP 5 RESULT OUTPUT***/ printf (“<SOLUTION>/n”); printf(“B=/n”); for (i=0; i<2; i++) for (j=0; j<2; j++) printf(“ % 1f”, b[i][j]); printf (“/n”);

Data input section

Computing section

Data output section

Variables type



6) Compute the problem

=<

====

×

B

aaaa

3221

22

22

21

12

11

-3.000000 2.000000 2.0 -1.000000

7) Check the computation results Compare with other method such as MATLAB functions. * Tip A good program for numerical computation should have following characteristics: 1) short computation time 2) accurate result 3) short program steps 4) easy to read 5) easy to move to other computers Example : Compare with MATLAB matrix inverse function by using following matrix:

=

3221

A

find ?1 =−A using MATLAB program and MATLAB inverse function.



2.1.7 Introduction to Numerical Computation Linear Equations. Preliminaries Resistive Network.

R1

VVVV

R

R

R

R

R

21

105

104

103

102

101

2

1

35

34

33

32

31

==

Ω×=

Ω×=

Ω×=

Ω×=

Ω×=

V1 R2 R3 R4 V2 i2 i3

i1

R5

Ohm’s law: V=iR (2.22)

Kirchhoff’s laws:

0=∑ I (2.23)

∑ = 0U (2.24) The implication of Kirchhoff’s and Ohm’s law are that the currents i1, i2, and i3 must satisfy the following relations:

235432314

13323212

3422421

)()(

0)(

ViRRRiRiRViRiRRiR

iRiRiRRR i

=+++−−=−++−

=−−++

mnmnmm

nn

nn

bxaxaxa

bxaxaxabxaxaxa

=+⋅⋅⋅++

=+⋅⋅⋅++=+⋅⋅⋅++

2............

211

22222121

11212111

(2.25)



The coefficients and are given real or complex numbers. Eq. (2,25) is a system if m equations with n variables, or “unknowns”, Our job is to determine their values.

ija ib.,...,, 21 nxxx

Eq. (2.25) can be written compactly in matrix form as (2.26) bAx = where ( )ijaA = is an m×n matrix of coefficients, and ( )ibb = and ( )jxx = are column vectors of dimension m and n, respectively. EXAMPLE 2 In the rectangular region shown in Figure 2.2(a), the electric potential is zero on the boundaries. The charge distribution, however, is uniform and given by

286 m×.02ε=vp

Solve Poisson’s equation to determine the potential distribution in the rectangular region. Solution To determine the potential distribution in the rectangular region, we use Poisson’s equation.

20

2

2

2

22 −=

ερ

−=∂

Φ∂+

∂Φ∂

=Φ∇ v

yx

with zero potential Φ on the boundaries. 0=

Figure 2.2 Geometry of the 6 rectangular region and the 28m× mh 2= mesh.



By establishing the rectangular grid shown in Figure 2.2(b), we realize that we have six nodes and, hence, six unknown potentials for which to solve. Replacing ∇ by its finite difference representation, we obtain

Φ2

02)4(1,1,1,,1,12 =+Φ−Φ+Φ+Φ+Φ −+−+ jijijijijih

It should be noted that although the although the mesh size was not explicitly used in solving Laplace’s equation in the previous example, h is included as a part of the matrix formation in solving Poisson’s equation. In SI system of units, h should be in meters. By applying the preceding difference equation at the various nodes in Figure 2.2(b), we obtain the following matrix equation:

−−

−−

−−

411000140100104110011401001041000114

=

ΦΦΦΦΦΦ

6

5

4

3

2

1

−−−−−−

888888

Instead of solving the resulting six equations, we may note some symmetry considerations in Figure 2.2(b). It is clear that

6521 Φ=Φ=Φ=Φ and that 43 Φ=Φ Taking these symmetry considerations into account, the number of equations reduces to two, and we obtain the following solution:

,56.41 =Φ 72.53 =Φ



2.2 Direct Methods for Linear Systems 2.2.1 Gaussian Elimination Example of Elimination Solve the system of Eqs.

(2.27) 39521744

132

321

321

321

=++=++

=++

xxxxxx

xxx

By subtracting the multiple 2 of the first equation from the second equation and the first equation from the third equation, the first “derived system” is obtained:

(2.28) 26412

132

32

32

321

=+−=+

=++

xxxxxxx

The first equation, after being multiplied by 2, becomes 2 (2.29) 624 321 =++ xxx Subtract this from the middle equation in Eq.(2.27) to see that result is 21)67()24()44( 321 −=−+−+− xxx The same rationale yields the third equation in the first derived system. Continue the computation by subtracting the multiple 2 of the second equation of the first derived system from the third equation to obtain the second derived system:

44

12132

3

32

321

=−=+

=++

xxxxxx

The derivation of this upper triangular system of equation is called the forward elimination process. From the third equation of the the system above, we immediately calculate

144

3 ==x

Then the second equation, with 1 substituted for x3, implies that



12

112 −=

−−=x

Finally, from the first equation, with their numerical values replacing x2 and x3, we see that

21

2)1(31

1 −=−−−

=x

and thereby obtain the solution, which 21

1 −=x , 12 −=x , 13 =x .

The computation of the unknowns from the upper triangular system is known as back substitution. Let us assume that in Eq.(2.25) the matrix of coefficients is nonsingular; that is, assume m = n , and that a solution exists and is unique. Then Eq.(2.25) can be rewritten as

(2.30)

1,211

1,22222121

1,11212111

2............

+

+

+

=+⋅⋅⋅++

=+⋅⋅⋅++

=+⋅⋅⋅++

nnnnnnn

nnn

nnn

axaxaxa

axaxaxaaxaxaxa

where, for later notation convenience, we have defined )1(1, niba ini ≥≤=+ .

Assume that , and subtract the multiplying 0≠na11

1

aai of the first equation from the ith

equation for i = 2, ⋅ ⋅ ⋅, n. The coefficient of x1 in the ith equation then becomes 0, and we thereby obtain the first derived system which has the form

)1(1,

)1(2

)1(2

)1(1,2

)1(22

)1(22

1,11212111

+

+

+

=+⋅⋅⋅++

=+⋅⋅⋅++=+⋅⋅⋅++

nnnnnn

nnn

nnn

axaxa

axaxaaxaxaxa

MMM (2.31)

where iji

ijij aaa

aa11

1)1( −= (2 ≤ i ≤ n, 2 ≤ j ≤ n+1) (2.32)

Assume neat that , and subtract the multiple of the second equation of (2.31) from the i

0)1(22 ≠a )1(

22)1(

2 / aaith equation (i=3, ⋅ ⋅ ⋅ , n). We thereby obtain the second derived system:



)2(1,

)2(3

)2(3

)2(1,3

)2(33

)2(33

)1(1,2

)1(23

)1(232

)1(22

1,11313212111

+

+

+

+

=+++

=++=+++=++++

nnnnnn

nnn

nnn

nnn

axaxa

axaxaaxaxaxaaxaxaxaxa

L

MMM

L

L

L

(2.33)

By repeating this process until the (n-1)st derived system has been constructed, we obtain

)1(1,

)1(

)2(1,3

)2(33

)2(33

)1(1,2

)1(23

)1(232

)1(22

1,11313212111

−+

−

+

+

+

=+

=++=+++=++++

nnnn

nnn

nnn

nnn

nnn

axa

axaxaaxaxaxaaxaxaxaxa

MMM

L

L

L

(2.34)

where the relation for obtaining the coefficients of the kth derived system from the coefficients of preceding system has the general form

)1()1(

)1(1)( −

−

−− −= k

kjkkk

kikk

ijk

ij aaa

aa (i = k+1, ⋅ ⋅ ⋅ , n; j = k+1, ⋅ ⋅ ⋅ , n+1) (2.35)

In Eq. (2.35), k ranges from 1 to n-1. The process is started by assigning (i = 1, ⋅ ⋅ ⋅ , n; j = 1, ⋅ ⋅ ⋅ , n+1) ijij aa =)0(

By inspection of Eq.(2.34), we see that the coefficient matrix of the (n-1)st derived system is in upper triangular form. The remaining step in solving this system is easy. The value of xn can be obtained from the final equation of Eq.(2.34)since nonsingularity of A implies that necessarily

. Specifically, 0)1( ≠−nnna

)1(

)1(1,

−

−+= n

nn

nnn

n aa

x

and the (backward) recursive formula for obtaining the values of the unknowns xk in terms of the previously calculated values xj (j > k) is

−= ∑+=

−−+−

n

kjj

kkj

knkk

kkk xaa

ax

1

)1()1(1,)1(

1 (k = n−1, ⋅ ⋅ ⋅ , 1) (2.36)

Gaussian elimination: Formula (2.36) is called Back Substitution. Process (2.26-2.35) is referred to as Forward Elimination.



The coefficients of the successive derived systems associated with Example 2.1 are given in Table 2.1.

TABLE 2.1 Derived System in Forward Elimination Original System A(0)

952744312

311

First Derived System A(1)

6412312

31

1−

Second Derived System A(2)

412312

41

1−

2.2.2 Gaussian Elimination with Pivoting Because of the effect of propagated rounding errors, native Gaussain elimination is sometimes unsatisfactory. Example 2.3 Assume that in the absence of round off error, the last two equations of the (n − 2)nd derived system are

3210

1

1

=+=+

−

−

nn

nn

xxxx

where the zero coefficients is obtained from previous calculations. The solution is

. As we know, resultants of arithmetic operations usually have round off errors: Therefore, the computed derived system of equations is actually

11 == −nn xx

321

1

1

=+=+

−

−

nn

nn

xxxxε

Since 0≠ε , in the computation of the (n − 1)st derived system the native elimination process is continued with the nonzero element ε, and the last two equations of the final derived system are



εε

ε23)21(

11

−=−

=+−

n

nn

x

xx

Back substitution applied to these equations results in the values

ε

ε

ε

nn

n

xx

x

−=

−

−=

−

1

21

23

1

For ε ≈ zero, ε2

−3 and ε2

− 1 ≈ε2− , xn ≈1 which is the correct value. But xn-1 will have

considerable error because xn, being very close to 1, will induce subtractive cancellation in numerator. The method of back substitution (2.36) implies that the values of all the other unknowns xn-2, ⋅ ⋅ ⋅ ,x1 obtained by the use of the erroneous value of xn-1 are also suspect. The difficulty in the system discussed above is not due simple to ε being small but rather to its being small relative to other coefficients in the same column. The element used for elimination is termed a )1( −k

kka pivot. The subtractive cancellation difficulty just observed can be remedied if we choose as the kth pivot element the coordinate having the largest magnitude among all , k ≤ i ≤ n, k ≤ j ≤ n. The

element a is then put in the diagonal position by interchanging rows i* and k and columns j* and k. That is, equations i* and k, and unknowns x

)1(**−kji

)1

a

**−kj

)1( −kkka

(i

j* and xk are interchanged. This operation is called pivoting for maximal size or, more simply, maximal pivoting, see Fig 2.2. We need to find the maximum of (n-k+1)2 numbers before computing the kth derived system. At the expense of some increase in round off error propagation, a popular alternative procedure is to perform partial pivoting, where the maximal element is chosen from only the kth column. That is, we choose the kth pivot element, to be any coordinate a that

maximizes

)1(*

−kki

)1( −kika , i ≥ k. Then the kth and i*th rows are interchanged to put the pivot

element in the diagonal position. Fig.2.2 illustrates the partial and maximal pivoting strategies.



)1()1(

)1()1(

)2(,1

)2(1,1

)1(2

)1(22

111

−−

−−

−−

−−−

knn

k

kkn

kkk

knk

kkk

n

n

aa

aaaa

aaaa

L

MM

L

L

MMO

L

L

nk

Search this portion Of kth column for pivot

Partial Pivoting

)1()1(

)1()1(

)2(,1

)2(1,1

)1(2

)1(22

111

−−

−−

−−

−−−

knn

knk

kkn

kkk

knk

kkk

n

n

aa

aaaa

aaaa

L

MM

L

L

MMO

L

L

Search this portion Of matrix for pivot

Fig.2.2 Partial and Maximal Pivoting.

2.3 Iterative Methods for Linear Systems 2.3.1 Jacobi Iteration Consider again (2.1?) with m = n. Assume that A is nonsingular and the rows have been exchanged, as necessary, so that the diagonal elements are nonzero. Eq. (2.1?) can then be rewritten so that the ith equation is explicit for xi:

( )

( )

( )nnnnnnnn

n

nn

nn

bxaxaxaa

x

bxaxaxaa

x

bxaxaxaa

x

−+++−=

−+++−=

−+++−=

−− 11,2211

2232312122

2

1131321211

1

1

1

1

L

M

L

L

(2.45)



Assume that an initial approximation of the solution has been given, or simply choose an arbitrary vector. Let denote this initial approximation. ( T

nxxX )0()0(1

)0( ,,L= )

)

Substitute it into the right-hand side of Eq.(2.45) and evaluate. The elements of the resulting vector give the next approximations of the unknowns. Let vector denote these approximations, and then substitute the new vector X

( TnxxX )1()1(

1)1( ,,L=

( )Tnx )2()2 ,,L

(1), into the right side of Eq.(2.45) to get a further approximation, , and so on. The general step is given by xX (

1)2( =

( )

( )

( )nk

nnnk

nk

nnn

kn

knn

kkk

knn

kkk

bxaxaxaa

x

bxaxaxaa

x

bxaxaxaa

x

−+++−=

−+++−=

−+++−=

−−+

+

+

)(11,

)(22

)(11

)1(

2)(

2)(

323)(

12122

)1(2

1)(

1)(

313)(

21211

)1(1

1

1

1

L

M

L

L

(2.46)

Scheme (2.46) is called the Jacobi Iteration Method. It can be proven that under certain conditions, for k → ∞, the sequence of vector x(k) converges to the exact solution of equations (2.1). One such condition is each diagonal element of the matrix of coefficient satisfy the condition:

∑≠=

⟩n

ijj

ijii aa1

, i = 1, … , n. (2.47)

If this condition is satisfied, then A is said to be diagonally dominant. The criteria for stopping the iteration process are usually either: 1. The number of iterations has exceeded some predetermined maximum K, or 2. The difference between successive values of all xi’s are less than some predetermined

tolerance, ε. One iteration stop ⇒ n division, n2 multiplication and n2 additions (or subtractions.) 2.3.2 Gauss Seidal Iteration Consider again the recursive Jacobi iteration scheme (2.46). Observe that in calculating the "new” value of , the previous value of is used on the right-hand side although the “new” value, is already known.

)1(2

+kx )(1

kx)1(

1+kx



Similarly, for obtaining the new value , the “odd” values and are used, although the new, and presumably more accurate values and of these variables are already available. A modification typically (but not always) giving faster convergence can be devised if in the calculation of (2 ≤ i ≤ n), the updated new values , … , are used in phase of the earlier values , … , in (2.46). This modification results in the Gauss Seidal iteration method, which is defined by the recursive scheme.

)1(3

+kx )(1

kx

)(1

kx

)1(2

+kx)1

)(1

kix −

)1(1

+kx (2

+kx

)1( +kix

)1(1

+kx )1(1+

−k

ix

( )

( )

( )nk

nnnk

nk

nnn

kn

knn

kkk

knn

kkk

bxaxaxaa

x

bxaxaxaa

x

bxaxaxaa

x

−+++−=

−+++−=

−+++−=

+−−

+++

++

+

)1(11,

)1(22

)1(11

)1(

2)(

2)(

323)1(

12122

)1(2

1)(

1)(

313)(

21211

)1(1

1

1

1

L

M

L

L

(2.48)

Iterative methods are particularly popular for solution of band systems arising in numerical method for partial differential equations. For such equations, in many cases the width k is relatively wide-often proportional to

- but the band itself has relatively few nonzero entries. Let p denote a bound to the effort required by elimination for solution of banded systems is proportional to k

2/1n2n, and

for iterative methods, proportional to 2pnM, where M is the number of iterations necessary for acceptable accuracy. Thus iterative methods are preferable if

2pnM < k2n, that is

pn

pkM ~2

2

<

Unfortunately, it is typically difficult to asses M until computations have already begun.



Table 2.x Solution of Linear Equations. Computation Methods Matrix Condition Characteristics

Direct Methods

*Gaussian Elimination *Gauss-Jordan Elimination *Cholesky’s Method *Lu Decomposition Method

Square Matrix Square Matrix Symmetric Matrix Symmetric Matrix

For linear systems of small or moderate size, either Gaussian Elimination or Lu Decomposition is effective and efficient. (Size < zero)

Iterative Methods

*Jacobi Iteration *Gauss-Seidal Method *Successive over- Relaxation (SOR Method)

Diagonally Dominant.

∑≠=

>n

ijj

ijii aa1

,

i = 1, … , n.

For large and high-order linear equations, for example, in solving differential equation, Iterative Methods are attractive. Fast convergence.



EXAMPLE 2 **********(will be deleted)*************** In the rectangular region shown in Figure 2.2(a), the electric potential is zero on the boundaries. The charge distribution, however, is uniform and given by

286 m×.02ε=vp

Solve Poisson’s equation to determine the potential distribution in the rectangular region. Solution To determine the potential distribution in the rectangular region, we use Poisson’s equation.

20

2

2

2

22 −=

ερ

−=∂

Φ∂+

∂Φ∂

=Φ∇ v

yx

with zero potential Φ on the boundaries. 0=

Figure 2.2 Geometry of the rectangular region and the mesh. 286 m× mh 2= By establishing the rectangular grid shown in Figure 2.2(b), we realize that we have six nodes and, hence, six unknown potentials for which to solve. Replacing ∇ by its finite difference representation, we obtain

Φ2

02)4(1,1,1,,1,12 =+Φ−Φ+Φ+Φ+Φ −+−+ jijijijijih



It should be noted that although the although the mesh size was not explicitly used in solving Laplace’s equation in the previous example, h is included as a part of the matrix formation in solving Poisson’s equation. In SI system of units, h should be in meters. By applying the preceding difference equation at the various nodes in Figure 2.2(b), we obtain the following matrix equation:

−−

−−

−−

411000140100104110011401001041000114

=

ΦΦΦΦΦΦ

6

5

4

3

2

1

−−−−−−

888888

Instead of solving the resulting six equations, we may note some symmetry considerations in Figure 2.2(b). It is clear that

6521 Φ=Φ=Φ=Φ and that 43 Φ=Φ Taking these symmetry considerations into account, the number of equations reduces to two, and we obtain the following solution:

,56.41 =Φ Φ 72.53 = To improve the accuracy of the potential distribution, finer mesh such as the one shown in Figure 2.2 is required. Because of the large number of nodes in this case, symmetry should be used, and a solution for only one-quarter of the rectangular geometry is desired. The application of the difference equation at nodes 1, 2, 4, and 5 should proceed routinely, whereas special care should be exercised at the boundary nodes 3, 6, 9, 10, and 11, and also at the corner node 12. For example, applying the difference equation at node 6 yields

02)4(165932 =+Φ−Φ+Φ+Φ+Φ ah

Or

02)42(165932 =+Φ−Φ+Φ+Φ

h



In equation )4(10432122

2

2

22 Φ−Φ+Φ+Φ+Φ=

∂Φ∂

+∂

Φ∂=Φ

hyx

5Φ

∇ , symmetry was used to

complete the five-point star difference equation. Specifically the potential at node a to the right of 6 was taken equal to . Similarly at the corner node 12, we obtain

02)4(1129112 =+Φ−Φ+Φ+Φ+Φ cbh

Because of symmetry, and bΦ=Φ11 cΦ=Φ9 , hence,

02)422(1129112 =+Φ−Φ+Φ

h

Fig. 2.3 The finer mesh solution and symmetry consideration of example 2. The matrix equation for the twelve nodes shown in Figure 2.3 is then



−−

−−

−−

−−

−−

−−

4221412

1421421

114111141

142111411

1141142

1141114

ΦΦΦΦΦΦΦΦΦΦΦΦ

12

11

10

9

8

7

6

5

4

3

2

1

=

−−−−−−−−−−−−

222222222222

The “2” coefficient in the coefficient matrix (to the left) of equation appears whenever symmetry consideration is used at boundary and corner nodes. It should be noted that the

coefficient matrix in equation is the same for both Laplace’s and Poissons equations. The constant vector on the right-hand side of equation, however, depends on the charge distribution within and the potential at the boundaries of the region of interest. Furthermore, if instead of a uniform charge distribution we have a given charge distribution

1212×

),( yxvρ , the constants vector on the right-hand side of equation should reflect the value of ),( yxvρ calculated at each node. Solution of equation gives

65.696.582.334.669.566.332.579.412.335.305.3,04.2

121110

987

654

321

=Φ=Φ=Φ=Φ=Φ=Φ=Φ=Φ=Φ=Φ=Φ=Φ


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