Iterative Methods

Chapter 2

Iterative Methods

2.1 Introduction

In this section, we will consider three diﬀerent iterative methods for solving a sets of equations. First, we consider a series of examples to illustrate iterative methods. To construct an iterative method, we try and re- arrange the system of equations such that we generate a sequence. 2

2.1.1 Simple Iteration Example

1 Example 2.1.1: Let us consider the equation

x y=e−x f(x)= x + e− 2=0 . (2.1) − α 2 When solving an equation such as (2.1) for α y=2−x

(k) where f(α)=0,0 <α< 2, we can generate a sequence x ∞ from some initial value (guess) { }k=0 x(0) by re-writing the equation as x x =2 e− , − k (k+1) x( ) (0) i.e. by computing x =2 e− from some x . If the series converges, it will converge to the − solution. For example, let us consider x(0) = 1 and x(0) = 1: −

21 k x(k) x(k) 0 1.0 -1.0 1 1.63212 -0.71828 2 1.80449 -0.05091 3 1.83544 0.947776 4 1.84046 1.61240 5 1.84126 1.80059 6 1.84138 1.83480 7 1.84140 1.84124 8 1.84141 1.84138 9 ......

In this example, both sequences appear to converge to a value close to the root α =1.84141 where 0 <α< 2. Hence, we have constructed a simple algorithm for solving an equation and it appears to be a robust iterative method.

However, (2.1) has two solutions: a positive root at 1.84141 and a negative root at -1.14619. Why do we only ﬁnd one root?

(k+1) (k) If f(x)= 0 has a solution x = α then x = g(x ) will converge to α, provided g′(α) < 1 | | and x(0) is suitably chosen.

The condition g′(α) < 1 is a necessary condition. | | In the above example, x x g(x)=2 e− and g′(x)= e− , − and

g′(x) < 1 if x> 0 . | |

So this method can be used to ﬁnd the positive root of (2.1). However, it will never converge to the negative root. Hence, this kind of approach will not always converge to a solution.

2.1.2 Linear Systems

Let us adopt the same approach for a linear system.

Example 2.1.2:

22 Consider the following set of linear equations:

10x1 + x2 = 12

x1 + 10x2 = 21

Let us re-write these equations as

x = (12 x )/10 1 − 2 x = (21 x )/10 . 2 − 1

Thus, we can use the following:

x(k+1) =1.2 x(k)/10 1 − 2 x(k+1) =2.1 x(k)/10 , 2 − 1

(k) (k) (k) T (0) to generate a sequence of vectors x = (x1 , x2 ) from some starting vector, x .

If 0 x(0) = , 0 then 0 1.2 0.99 1.002 x(0) = , x(1) = , x(2) = , x(3) = , ... 0 2.1 1.98 2.001 where 1 x(k) as k , → 2 → ∞ which is indeed the correct answer. So we have generated a convergent sequence.

Let us consider the above set of linear equations again. Possibly the more obvious rearrangement was

x = 21 10x 1 − 2 x = 12 10x . 2 − 1

Thus, we can generate a sequence using:

x(k+1) = 21 10x(k) 1 − 2 x(k+1) = 12 10x(k) , 2 − 1

If, we again use 0 x(0) = , 0 then 0 21 99 1011 x(0) = , x(1) = , x(2) = − , x(3) = , ... 0 12 198 1992 23 − Clearly, this sequence is not converging! Why?

Example 2.1.3:

Let us consider the above example (2.1.2) again. Can we ﬁnd a method that allows the system to converge more quickly?

(1) (0) Let us look at the computation more carefully. In the ﬁrst step x1 is computed from x2 and in (1) (0) the second step we compute x2 from x1 .

(1) (0) It seems more natural, from a computational point of view, to use x1 rather then x1 in the second step. i.e. to use the latest available value. In eﬀect, we want to compute the following:

x(k+1) =1.2 x(k)/10 1 − 2 x(k+1) =2.1 x(k+1)/10 , 2 − 1 which gives, 0 1.2 1.002 1 x(0) = , x(1) = , x(2) = , 0 1.98 1.9998 → 2 1 which converges to 2 much more rapidly! In the following sections, we will consider, in general terms, iterative methods for solving a system Ax = b. First, though we introduce some important results about a sequence of vectors

2.2 Sequences of Vectors

2.2.1 The Limit of a Sequence

x(k) ∞ Let k=0 be a sequence in a Vector Space V . How do we know if this sequence has a limit? First observe that x = y ; x = y. i.e. two distinct objects in a Vector Space can have the k k k k same size. However, from rule 1 for norms (1.1) we know that if x y = 0, then x y. k − k ≡ So if lim x(k) x =0 k k − k →∞ then lim x(k) = x k →∞ The vector x is the limit of the sequence.

24 2.2.2 Convergence of a Sequence

(k) Suppose the sequence x ∞ converges to x, where { }k=0 x(k+1) = Bx(k) + c .

If x(k) x for k , then x satisﬁes the equation: → → ∞ x = Bx + c , and so we have x(k+1) x = B(x(k) x) , − − and thus, taking norms, x(k+1) x B x(k) x . k − k≤k kk − k If B < 1 then k k x(k+1) x < x(k) x , k − k k − k i.e. we have a monotonically decreasing sequence, or, in other words, the error in the approximations decreases.

Say we start from an initial guess x(1) x = B(x(0) x). Then − − x(2) x = B(x(1) x) − − = B B(x(0) x) − = B2(x(0) x) , − and so on, to give x(k) x = Bk(x(0) x) . − − Taking norms, and using rule 5 (1.9) for sub-ordinate matrix norms

(k) k (0) k 1 (0) k 2 2 (0) k (0) x x B (x x) B − B (x x) B − B (x x) B (x x) . k − k≤k kk − k≤k kk kk − k≤k kk k k − k≤···≤k k k − k If B < 1, then B k 0 as k and hence, x(k) x as k . k k k k → → ∞ → → ∞ Recall that ρ(B) B ( 1.5) so a necessary condition for convergence is ρ(B) < 1. Furthermore, ≤k k § it is possible to show that if ρ(B) < 1 , then B < 1 . k k and if ρ(B) > 1 , then B > 1 , k k although we do not prove these results in this course.

Hence, ρ(B) < 1 is not only a necessary condition, but it is also suﬃcient condition. 25 2.2.3 Spectral radius and rate of convergence

In numerical analysis, to compare diﬀerent methods for solving systems of equations we are interested in determining the rate of convergence of the method. As we will see below the spectral radius is a measure of the rate of convergence.

Consider the situation where BN N has N linearly independent eigenvectors. As before we have ×

x(k+1) x = B(x(k) x) , − − or substituting in for v(k) = x(k) x, we have −

v(k+1) = Bv(k) .

(0) N Now write v = i=1 αiei where ei are the eigenvectors (with associated eigenvalues λi) of B, then P

N N N (1) v = B αiei = αiBei = αiλiei , i=1 ! i=1 i=1 X X X N N N (2) 2 v = B αiλiei = αiλiBei = αiλi ei , i=1 ! i=1 i=1 X X X continuing this sequence gives N (k) k v = αiλi ei . i=1 X Now suppose λ > λ (i =2,...,N), then | 1| | i| N (k) k k v = α1λ1 e1 + αiλi ei i=2 X N k k λi = λ α e1 + α ei . 1 1 i λ " i=2 1 # X

Given that λi/λ1 < 1, for large k, (k) k v α λ e1 . ≃ 1 1 Hence, the error associated with x(k), the kth vector in the sequence, is given by v(k) which varies as the kth power of the largest eigenvalue. In other words, it varies as the kth power of the spectral radius ρ(B) (= λ ). So the spectral radius is a good indication of the rate of convergence. | 1|

2.2.4 Gerschgorin’s Theorem

The above result means that if we know the magnitude of the largest vector of the iteration matrix we can estimate the rate of convergence of a system of equations for a particular method. However, this

26 requires the magnitudes of all eigenvalues to be known, which would probably have to be determined numerically.

The Gerschgorin Theorem is a surprisingly simple result concerning eigenvalues that allows us to put bounds on the size of the eigenvalues of a matrix without actually ﬁnding the eigenvalues themselves.

The equation Ae = λe, where (λ, e) are an eigenvalue, eigenvector pair of the matrix A, can be written in component notation as

N N aij ej = aiiei + aij ej = λei . j=1 j=1 X Xj=i 6 Rearranging implies N e (a λ)= a e , i ii − − ij j j=1 Xj=i 6 and thus, N e a λ a e . | i| | ii − |≤ | ij | | j | j=1 Xj=i 6 Suppose the component of eigenvector e with the largest absolute value is e , such that e e | l| | l| ≥ | j | for j (note e = 0 for all j). Then from above j 6 N N e a λ a e a e | l| | ll − |≤ | lj | | j|≤ | lj | | l| j=1 j=1 Xj=l Xj=l 6 6 so,dividing by el gives | | N a λ a . | ll − |≤ | lj | j=1 Xj=l 6 Each eigenvalue lies inside a circle with centre a and radius N a with j = l. ll j=1 | lj | 6 P However, we don’t know l without ﬁnding λ and e.

But we can say that the union of all such circles must contain all the eigenvalues. This is Gerschgorin’s Theorem.

Example 2.5.1: Determine the bounds on the eigenvalues for the matrix

2 1 0 0 −  1 2 1 0  A = − − .    0 1 2 1  − −     0 0 1 2   −    27 Gerschgorin’s Theorem implies that the union of all circles

N a λ a . | ll − |≤ | lj | j=1 Xj=l 6 must contain all eigenvalues.

For l = 1 and 4 we get the relation 2 λ 1. | − |≤ For l =2 and 3 we get 2 λ 2. | − |≤ The matrix is symmetric - the eigenvalues are real so Gerschgorin’s Theorem implies

0 λ 4 . ≤ ≤

The eigenvalues of A are

λ1 =3.618, λ2 =2.618, λ3 =1.382, and λ4 =0.382. 0 1 2 3 4 hence, the largest eigenvalue is indeed less than 4.

2.3 The Jacobi Iterative Method

The Jacobi Iterative Method follows the iterative method shown in Example 2.1.2.

Consider the linear system

Ax = b, AN N = [aij ] , xN = [xi] , bN = [bi] . ×

Let us try to isolate xi. The ith equation looks like

N aij xj = bi . j=1 X Assuming a = 0 for all i, we can re-write this as ii 6 N a x = b a x , ii i i − ij j j=1 Xj=i 6 so,

N 1 xi = bi aij xj  a − ii j=1  Xj=i   6  giving the recurrence relation  

N (k+1) 1 (k) x = bi aij x  , (2.2) i a − j ii j=1  Xj=i   6   28  for each xi (i =1,...,N). This is known as the Jacobi Iterative Method.

In matrix form, we have

A = D L U , (2.3) − − where D is a diagonal matrix with elements aii,

L is a strictly lower triangular matrix, L = [lij ] such that

a , i>j − ij lij =  0, i j ,  ≤  and U is a strictly upper triangular matrix, U = [uij ] such that

a , i

Dividing each equation by aii is equivalent to writing

1 1 x = D− (L + U)x + D− b

1 where the elements of D− are 1/aii, so we have pre-multiplied by the inverse of D. Hence, the matrix form of the iterative method (2.2), known as the Jacobi Iteration Method is

(k+1) 1 (k) 1 x = D− (L + U)x + D− b . (2.4)

1 The matrix BJ = D− (L + U) is called the iteration matrix for the Jacobi Iteration method.

2.3.1 Convergence of the Jacobi Iteration Method

From 2.2.2 recall that an iterative method of the form x(k+1) = Bx(k) + c will converge provided § B < 1 and that a necessary and suﬃcient condition for this is to be true is ρ(B) < 1. k k 1 Thus, for the Jacobi method, we require B = D− (L + U) < 1 for convergence and, hence, k J k k k ρ(BJ ) < 1.

29 Example 2.3.1: Let us return once more to Example 2.1.2 and recast it in the form of the Jacobi iterative method. The linear system we wish to solve is

10 1 x1 12 Ax = = = b .       1 10 x2 21       The first thing we need to do is find D and L + U where A = D L U : − − 10 1 10 0 0 1 A = D = and L + U = −  1 10 ⇒  0 10  1 0  −       1 0 1/10 hence, D− (L + U)= BJ = − .  1/10 0  − Now choosing the matrix norm sub-ordinate to the infinity norm we find 1 BJ = < 1 . k k∞ 10

Alternatively we can consider the spectral radius of BJ. The eigenvalues of BJ are given by

λ2 1/100 = 0 − and so 1 ρ(BJ)= , 10 which in this case is equal to BJ . k k So if x is the limit to our sequence then 1 x(k+1) x x(k) x . k − k∞ ≤ 10 k − k∞ In Example 2.1.2. we had 0 1 x(0) = and x = , 0 2 so x(0) x = 2 and k − k∞ 1 x(1) x 2=0.2 . k − k∞ ≤ 10 × Remember, 1.2 x(1) = , 2.1 so, 0.2 x(1) x = . − 0.1 and indeed, x(1) x 0.2 . k − k∞ ≤ Since the size of ρ(BJ) is an indication of the rate of convergence we see here that this system converges at a rate of ρ(BJ)=0.1. The smaller the spectral radius the more rapid the convergence. So is it possible to modify this method to make it faster? 30 2.4 The Gauss-Seidel Iterative Method

To produce a faster iterative method we amend the Jacobi Method to make use of the new values as they become available (e.g. as in Example 2.2.2.).

Expanding out the Jacobi Method (2.4) we have

(k+1) 1 (k) 1 x = D− (L + U)x + D− b

1 (k) 1 (k) 1 = D− Lx + D− Ux + D− b .

1 1 (k) Here D− L is a lower triangular matrix so the ith row of D− Lx contains the values

(k) (k) (k) (k) x1 , x2 , x3 ,...,xi 1 . −

(components up to, but not including the diagonal).

1 Likewise, D− U is an upper triangular matrix so the ith row contains

(k) (k) (k) xi+1, xi+2,...,xN .

(k+1) If we compute the xi ’s in the order of increasing i (i.e. from the top of the vector to the bottom) then when computing (k+1) xi , we have available (k+1) (k+1) (k+1) x1 , x2 ,...,xi 1 . −

Hence, a more eﬃcient version of the Jacobi Method is to compute (in the order of increasing i)

(k+1) 1 (k+1) 1 (k) 1 x = D− Lx + D− Ux + D− b .

This is equivalent to ﬁnding x(k+1) from

1 (k+1) 1 (k) 1 (I D− L)x = D− Ux + D− b , − or, (k+1) 1 1 1 (k) 1 1 1 x = (I D− L)− D− Ux + (I D− L)− D− b . − − This is known as the Gauss-Seidel Iterative Method.

The iteration matrix becomes

1 1 1 B = (I D− L)− D− U GS − 1 1 = [D(I D− L)]− U − 1 = (D L)− U . − 31 The way of deriving the Gauss-Seidel method formally is as follows:

A = D L U , − − so Ax = b becomes (D L)x = Ux + b , − and hence, 1 1 x = (D L)− Ux + (D L)− b , − − generating the recurrence relation

(k+1) 1 (k) 1 x = (D L)− Ux + (D L)− b . (2.5) − −

1 The iteration matrix for the Gauss-Seidel method is given by B = (D L)− U. Thus, for GS − convergence (from 2.2.2) we require that § 1 B = (D L)− U < 1 . k GSk k − k

Example 2.4.1: Again we reconsider the linear system used in Examples (2.1.2, 2.1.3 & 2.3.1) and recast it in the form of the Gauss-Seidel Method:

10 1 A = ,  1 10   and since A = D L U, we have − − 10 0 0 1 D L = and U = − . −  1 10 0 0      Then

1 1/10 0 1 1/10 0 0 1 (D L)− = , so (D L)− U = − , −  1/100 1/10 −  1/100 1/10 0 0  − −       and thus the Gauss-Seidel iteration matrix is

1 0 1/10 BGS = (D L)− U = − . − 0 1/100   Clearly, the norm of the iteration matrix is

1 1 BGS = (D L)− U = < 1 , k k k − k∞ 10 and hence, the method will converge for this example.

Let us look at the eigenvalues to get a feel for the rate of convergence. The eigenvalues are given by

λ 1/10 det − − =0 ,  0 1/100 λ −  32  or, 1 λ λ =0 , − 100 so we have 1 λ =0 or λ = , 100 and hence, 1 1 ρ(B )= ρ (D L)− U = . GS − 100 Observe that in this example even though B = B we have ρ(B ) = [ρ(B )]2 (cf Example k GSk k J k GS J 2.3.1), implying that Gauss-Seidel converges twice as fast as Jacobi.

2.5 The Successive Over Relaxation Iterative Method

The third iterative method we will consider is a method which accelerates the Gauss-Seidel method.

Consider the system Ax = b, with A = D L U as before. When trying to solve Ax = b, we − − obtain an approximate solution x(k) of the true solution x. The quantity r(k) = b Ax(k) is called − a residual and it is a measure of the accuracy of x(k). Clearly, we would like to make the residual r(k) to be as small as possible for each approximate solution x(k).

(k) (k+1) (k+1) Now remember, when calculating xi , the components x1 ,...,xi 1 are already known. So in − the Gauss-Seidel iterative method for the most recent approximation, the residual vector is given by

r(k) = b Dx(k) + Lx(k+1) + Ux(k) . −

Ultimately, we wish to make x x(k) as small as possible. However, as we don’t know x yet, we − instead consider x(k+1) x(k) as a measure for x x(k). We now wish to calculate x(k+1) such that − −

D(x(k+1) x(k))= ω(b Dx(k) + Lx(k+1) + Ux(k)) , − − where ω is called the relaxation parameter. Re-arranging, we get

(D ωL)x(k+1) = ((1 ω)D + ωU)x(k) + ωb , − − and hence, the recurrence relation is given by

(k+1) 1 (k) 1 x = (D ωL)− ((1 ω)D + ωU)x + (D ωL)− ωb . (2.6) − − −

The process of reducing residuals at each stage is called “Successive Relaxation”. If0 <ω< 1, the iterative method is known as a “Successive Under Relaxation” and they can be used to obtain convergence when the Gauss-Seidel scheme is not convergent. For choices of ω> 1 the scheme

33 is a “Successive Over-Relaxation” and is used to accelerate convergent Gauss-Seidel iterations. Note, ω = 1 is simply the Gauss-Seidel Iterative Method.

The iteration matrix for the S.O.R. method - Successive Over-Relaxation with ω > 1 is given by 1 BSOR = (D ωL)− [(1 ω)D + ωU] . − −

The iteration matrix BSOR can be derived by splitting A in the following way:

1 1 A = D L U = D 1 + D L U, ω> 0 . − − − ω ω − − Thus Ax = b can be written as

1 1 D L x = 1 D + U x + b ω − − − ω (D ωL)x = ((1 ω)D + ωU) x + ωb , − − so, 1 BSOR = (D ωL)− [(1 ω)D + ωU] − −

The aim is to choose ω such that the rate of convergence is maximised, that is the spectral radius,

ρ(BSOR(ω)), is minimised. How do we ﬁnd the value of ω that does this? There is no complete answer for general N N systems, but it is known that if, for each, 1 i N, a = 0 then × ≤ ≤ ii 6

ρ(BSOR) 1 ω . ≥ | − | This means that for convergence we must have 0 <ω< 2.

Example 2.6.1: We return once more to the linear system considered throughout this chapter in Examples (2.1.1, 2.1.2, 2.3.1 & 2.4.1) and recast it here in terms of the SOR iterative method. Recall, 10 1 A = ,  1 10 and A = D L U such that   − − 10 0 0 1 10(1 ω) ω (1 ω)D + ωU = (1 ω) + ω − = − − , − −  0 10 0 0   0 10(1 ω) −       and 10 0 0 0 10 0 (D ωL)= ω = −  0 10 −  1 0  ω 10 − Now      

1 1/10 0 (D ωL)− = , −  ω/100 1/10 − 34  thus the iteration matrix

1 1 ω ω/10 BSOR = (D ωL)− [(1 ω)D + ωU]= − − . ω(1 ω) ω2 − −  − +1 ω − 10 100 −   The eigenvalues of this matrix are given by ω2 ω2(1 ω) [(1 ω) λ] +1 ω λ − =0 , − − 100 − − − 100 ω2 ω2 ω2(1 ω) λ2 λ (1 ω)+ +1 ω + (1 ω) +1 ω − =0 , − − 100 − − 100 − − 100 ω2 λ2 λ 2(1 ω)+ + (1 ω)2 =0 . − − 100 − Solving this quadratic for λ gives

1 ω2 ω2 ω4 1/2 λ = 2(1 ω)+ 4(1 ω)2 + 4(1 ω) + 4(1 ω)2 2 − 100 ± − − 100 104 − − ! ω2 1 ω2 ω4 1/2 = (1 ω)+ 4(1 ω) + − 200 ± 2 − 100 104 ω2 ω ω2 1/2 = (1 ω)+ 4(1 ω)+ . − 200 ± 20 − 100 1 When ω = 1 (the Gauss-Seidel Method), one root is 0 and the other is 100 . Changing ω changes these roots. Suppose we select ω such that ω2 4(1 ω)+ =0 , − 100 so there are equal roots to the equation. Then this implies, ω2 = 2(1 ω) and λ = ω 1 . 200 − − − The smallest value of ω (ω > 1) producing equal roots is

ω =1.002512579 , which is not very diﬀerent (ω 1) to Gauss-Seidel! ≈ However, the spectral radius of the SOR iteration matrix is just

ρ(BSOR)=0.002512579 compared with ρ(BGS)=0.01.

ρ(B) is very sensitive to ω. If you can ‘hit’ the right value, the improvement in speed of convergence of the iteration method is signiﬁcant.

Although this example is only a 2 2 matrix, the comments apply in general. For a larger set of × equations, convergence of Gauss-Seidel can be slow and SOR with an optimum value of ω (if it can be found) can be a major improvement. 35 2.6 Convergence of the SOR Method for Consistently Or- dered Matrices

In general, it is not easy to ﬁnd an appropriate ω for the SOR method and so an ω is usually chosen which lies in the range 1 <ω< 2 and leads to a spectral radius, ρ(BSOR) which is as small as reasonably possible. However, there are a set of matrices for which it is relatively easy to ﬁnd the optimum ω.

Consider the linear system Ax = b and let A = D L U. If the eigenvalues of − −

1 1 1 αD− L + D− U , α =0 , α 6 are independent of α, then the matrix is said to be Consistently Ordered, and the optimum ω for the SOR iterative method is 2 w = . 1+ 1 ρ2(B ) − J p Explanation

First, we note that for such a matrix, consistently ordered (eigenvalues are the same for all α) implies that the eigenvalues of 1 1 1 αD− L + D− U α 1 1 are the same as those for D− L + D− U = BJ , the Jacobi iterative matrix (i.e. put α = 1).

Now consider the eigenvalues of BSOR. They satisfy the polynomial

det(B λI)=0 SOR − or 1 det (D ωL)− ((1 ω)D + ωU) λI =0 , − − − and hence,

1 det(D ωL)− det [(1 ω)D + ωU λ(D ωL)]=0 , − − − − =0 | 6 {z } so λ satisfy det [(1 ω λ)D + ωU + λωL]=0 . − − Since ω = 0, the non-zero eigenvalues satisfy 6 (1 ω λ) 1 det − − D + U + √λL ω√λ =0 , √ √ ω λ λ 36 and thus, 1 1 1 (λ + ω 1) det √λD− L + D− U − I =0 . √ − √ λ ω λ When consistently ordered, the eigenvalues of

1 1 1 √λD− L + D− U √λ 1 are the same as those of D− (L + U)= BJ .

Let the eigenvalues of BJ be µ, then the non-zero eigenvalues of BSOR satisfy λ + ω 1 µ = − . ω√λ If we put ω = 1 (i.e. recover Gauss-Seidel), then µ = λ , or λ = µ2. (Recall Example 2.4.1 where √λ this result was also found).

For ω = 0, we have 6 µ2ω2λ = λ2 +2λ(ω 1)+(ω 1)2 , − − or, λ2 + λ(2ω 2 µ2ω2) + (ω 1)2 =0 . − − −

The eigenvalues λ of BSOR are then given by µ2ω2 1 λ = (ω 1)+ 4(ω 1)2 4(ω 1)µ2ω2 4(ω 1)2 + µ4ω4 − − 2 ± 2 − − − − − µ2ω2 p µ4ω4 =1 ω + (1 ω)µ2ω2 + − 2 ± r − 4 µ2ω2 µ2ω2 =1 ω + µω (1 ω)+ . − 2 ± r − 4 For each µ2 there are 2 values of λ, these may be real or complex. If complex (note ω > 1), then

λλ = (ω 1)2 or λ = ω 1 . − | | − Hence, ρ(B )= ω 1 . SOR −

For the fastest convergence we require ρ(BSOR) to be as small as possible. It can be shown that the best outcome is to make the roots of BSOR equal when µ = ρ(BJ ), i.e. when µ is largest. This implies µ2ω2 ω +1=0 . 4 − Solving for ω yields 1 1 µ2 ω = ± 2 − , pµ /2 2 1 (1 µ2) = 2 − − , µ 1 1 µ2 ! ∓ − 2 = p . 1 1 µ2 ∓ − 37 p We are looking for the smallest value of ω and so we take the positive root of the above equation.

Hence, with µ = ρ(BJ ), the best possible choice for ω is

2 ω = . 1+ 1 (ρ(B ))2 − J p Example 2.6.1: We again return to Example (2.2.1, 2.2.2, 2.3.1 & 2.4.1) and show that it is a consistently ordered matrix and determine the minimum ω, and hence, the fastest rate of convergence for the SOR method.

As before we have 10 1 A = ,  1 10 then   1 1 1 1 0 10α αD− L + D− U = − , α  α 0  − 10 and the eigenvalues are given by  

1 α 1 λ2 =0 so λ2 = , − −10α −10 100 and hence, the matrix is consistently ordered.

Then by applying the above formulae and recalling that the eigenvalues of BJ are µ =1/10 (Example 2.3.1) we have

2 ω = 1+ 1 (ρ(B ))2 − J 2 = p , 1+ 1 (1/100) − = 1.0025126 . p

This is essentially the same value as we found in Example 2.5.1.

Thus the fastest rate of convergence for this particular system is

ρ(BSOR)=0.0025126 , as shown in Example 2.5.1