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Addison Wesley Longman, Inc." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 302 "All rights reserved. \+ No part of this publication may be reproduced, stored in a retrieval s ystem, or transmitted, in any form or by any means, electronic, mechan ical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 15 "Initializations" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "with(linalg):\nwith(plots) :\nread(`pvac.txt`):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "Relaxation - the Concept" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 249 "T he idea of relaxation has been attributed to the British engineer, Ric hard Southwell, [1], although some claim that Gauss was familiar with \+ the concept. Conceived in the \"by-hand\" era before computers, the i dea underlies the more modern method of " }{TEXT 347 10 "successive" } {TEXT -1 1 " " }{TEXT 348 14 "overrelaxation" }{TEXT -1 62 " (SOR). O ur exposition will be in terms of the linear system " }{TEXT 349 1 "A " }{TEXT 260 1 "x" }{TEXT -1 3 " = " }{TEXT 261 1 "y" }{TEXT -1 22 " f or which the matrix " }{TEXT 350 1 "A" }{TEXT -1 36 " is the strictly \+ diagonally dominant" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "A := matrix(3,3,[-12,-1,1,1,13,7,3,4,8]); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "and for which the vector " }{TEXT 262 1 "y" }{TEXT -1 3 " is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 25 "y := vector([-11,48,35]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Then, th e solution is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "xe := linsolve(A,y);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "the system \+ variables are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "vars := [seq(x[k],k=1..3)];" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "and \+ the equations of the system are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "for k from 1 to 3 do\nq||k : = evalm(A &* vars)[k] = y[k];\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "The residual \+ vector for this system would be " }{TEXT 263 1 "R" }{TEXT -1 3 " = " } {TEXT 351 1 "A" }{TEXT 264 1 "x" }{XPPEDIT 18 0 "``-``" "6#,&%!G\"\"\" F$!\"\"" }{TEXT 265 1 "y" }{TEXT -1 44 ", which we make into a functio n in Maple via" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 46 "R := map(unapply,evalm(A&*vars - y),op(vars)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Clearly, " }{TEXT 266 1 "R" }{TEXT -1 18 " would be z ero if " }{TEXT 256 1 "x" }{TEXT -1 3 " = " }{TEXT 257 1 "x" } {XPPEDIT 18 0 "``[e]" "6#&%!G6#%\"eG" }{TEXT -1 18 ". Indeed, we find " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "R(op(convert(xe,list)));" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "If we start an iteration process at " }{TEXT 258 1 "x" }{XPPEDIT 18 0 "``[0]" "6# &%!G6#\"\"!" }{TEXT -1 3 " = " }{TEXT 259 1 "0" }{TEXT -1 41 ", the re sidual vector would initially be " }{TEXT 267 1 "R" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "``-``" "6#,&%!G\"\"\"F$!\"\"" }{TEXT 268 1 "y" }{TEXT -1 101 ", in which the largest absolute value is 48, occurring in the \+ second equation. In fact, we see that " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "R(0,0,0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "If we could make the largest residual zero, that is, if we coul d " }{TEXT 269 34 "relax the largest residual to zero" }{TEXT -1 80 ", we should be led to a more accurate solution. Our strategy, then, is to keep " }{XPPEDIT 18 0 "x[1]=x[3]" "6#/&%\"xG6#\"\"\"&F%6#\"\"$" } {TEXT -1 17 " = 0, and adjust " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"# " }{TEXT -1 33 " to make the second component of " }{TEXT 270 1 "R" } {TEXT -1 160 " become zero. We would be aided in this task if we solv ed each equation for the variable with the coefficient of greatest mag nitude in that equation, obtaining" }}{PARA 0 "" 0 "" {TEXT -1 1 " " } }{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[1]=(-x[2]+x[3]+11) /12" "6#/&%\"xG6#\"\"\"*&,(&F%6#\"\"#!\"\"&F%6#\"\"$F'\"#6F'F'\"#7F-" }}{PARA 259 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[2]=(-x[1]-7*x[3]+ 48)/13" "6#/&%\"xG6#\"\"#*&,(&F%6#\"\"\"!\"\"*&\"\"(F,&F%6#\"\"$F,F-\" #[F,F,\"#8F-" }}{PARA 260 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[3]= (-3*x[1]-4*x[2]+35)/8" "6#/&%\"xG6#\"\"$*&,(*&F'\"\"\"&F%6#F+F+!\"\"*& \"\"%F+&F%6#\"\"#F+F.\"#NF+F+\"\")F." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "which we implement as functions in Mapl e via" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "for k from 1 to 3 do\nX[k] := unapply(solve(q||k,x[k] ),op(remove(has,vars,k)));\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "So far, the astute rea der will recognize the fixed-point iteration formula " }{TEXT 271 1 "x " }{TEXT -1 3 " = " }{TEXT 272 1 "g" }{TEXT -1 1 "(" }{TEXT 273 1 "x" }{TEXT -1 58 "). However, the actual iteration we do will be differen t." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The value of " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT -1 38 " whi ch drives the second component of " }{TEXT 274 1 "R" }{TEXT -1 11 " to zero is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "X[2](0,0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "If " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT -1 42 " is set equal to this value while keep ing " }{XPPEDIT 18 0 "x[1]=x[3]" "6#/&%\"xG6#\"\"\"&F%6#\"\"$" }{TEXT -1 58 " = 0, then the next member of the sequence of iterates is " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 1 " " } {TEXT 275 1 "x" }{XPPEDIT 18 0 "``[1]" "6#&%!G6#\"\"\"" }{TEXT -1 7 " \+ = (0, " }{XPPEDIT 18 0 "x[2](0,0)" "6#-&%\"xG6#\"\"#6$\"\"!F)" }{TEXT -1 7 ", 0) = " }{XPPEDIT 18 0 "``(0,48/13,0)" "6#-%!G6%\"\"!*&\"#[\"\" \"\"#8!\"\"F&" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "which we represent in Maple as the list" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z := [0,X [2](0,0),0];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Since the vector " }{TEXT 277 1 "x" } {TEXT -1 34 " has changed, the residual vector " }{TEXT 276 1 "R" } {TEXT -1 15 " will change to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "`R` = evalf(R(op(z)));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "Now, the third component of the residual vector has the l argest magnitude, so " }{XPPEDIT 18 0 "x[3]" "6#&%\"xG6#\"\"$" }{TEXT -1 91 " must be varied to bring this component of the residual to zero . The appropriate value of " }{XPPEDIT 18 0 "x[3]" "6#&%\"xG6#\"\"$" }{TEXT -1 15 " is computed by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "new := X[3](z[1],z[2]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "so " }{TEXT 278 1 "x" }{XPPEDIT 18 0 "``[2]" "6#&%!G6#\"\" #" }{TEXT -1 31 ", the next iterate, is given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "z := evalf( [z[1],z[2],new]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "The residual generated by " } {TEXT 279 1 "x" }{XPPEDIT 18 0 "``[2]" "6#&%!G6#\"\"#" }{TEXT -1 8 " i s then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "`R` = R(op(z));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "The second co mponent of " }{TEXT 281 1 "R" }{TEXT -1 31 " has the largest magnitude , so " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT -1 86 " is adjus ted to bring this component of the residual to zero. This is done by \+ giving " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT -1 10 " the va lue" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "new := X[2](z[1],z[3]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "so " }{TEXT 280 1 "x" }{XPPEDIT 18 0 "``[3]" "6#&%!G6#\"\"$" }{TEXT -1 30 ", the n ext iterate is given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z := [z[1],new,z[3]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "The residual generated by " }{TEXT 282 1 "x" }{XPPEDIT 18 0 "`` [3]" "6#&%!G6#\"\"$" }{TEXT -1 8 " is then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "`R` = R(op(z));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 70 "The first component of the residual now has the largest magnitude, so " }{XPPEDIT 18 0 "x[1]" "6#&%\"xG6#\"\"\"" }{TEXT -1 45 " must be adjusted to bring this component of " }{TEXT 283 1 "R" } {TEXT -1 37 " to zero. This is done by assigning " }{XPPEDIT 18 0 "x[ 1]" "6#&%\"xG6#\"\"\"" }{TEXT -1 10 " the value" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "new := X[1] (z[2],z[3]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "so " }{TEXT 284 1 "x" }{XPPEDIT 18 0 "``[4 ]" "6#&%!G6#\"\"%" }{TEXT -1 31 ", the next iterate, is given by" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z := [new,z[2],z[3]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "The residual generated by \+ " }{TEXT 285 1 "x" }{XPPEDIT 18 0 "``[4]" "6#&%!G6#\"\"%" }{TEXT -1 8 " is then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 15 "`R` = R(op(z));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "The component of " }{TEXT 286 1 "R" }{TEXT -1 50 " with the largest absolute value \+ is the third, so " }{XPPEDIT 18 0 "x[3]" "6#&%\"xG6#\"\"$" }{TEXT -1 94 " must be adjusted to bring this component of the residual to zero. This is done by assigning " }{XPPEDIT 18 0 "x[3]" "6#&%\"xG6#\"\"$" }{TEXT -1 10 " the value" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "new := X[3](z[1],z[2]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "so " }{TEXT 287 1 "x" }{XPPEDIT 18 0 "``[5]" "6#&%!G6#\"\"&" } {TEXT -1 31 ", the next iterate, is given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z := [z[1],z[2],n ew];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "The residual generated by " }{TEXT 288 1 "x" } {XPPEDIT 18 0 "``[5]" "6#&%!G6#\"\"&" }{TEXT -1 8 " is then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "` R` = R(op(z));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "The second component of " }{TEXT 289 1 "R" }{TEXT -1 35 " now has the largest magnitude, so " } {XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT -1 101 " must be adjuste d to bring this component of the residual vector to zero. This is don e by assigning " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT -1 10 " the value" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "new := X[2](z[1],z[3]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "so that " } {TEXT 290 1 "x" }{XPPEDIT 18 0 "``[6]" "6#&%!G6#\"\"'" }{TEXT -1 31 ", the next iterate, is given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z := [z[1],new,z[3]];" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "The residual generated by " }{TEXT 291 1 "x" }{XPPEDIT 18 0 "``[6]" "6#&%!G6#\"\"'" }{TEXT -1 8 " is then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "`R` = R(op( z));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "in which the third component has the largest mag nitude. We therefore drive this third component of " }{TEXT 292 1 "R " }{TEXT -1 38 " to zero by assigning to the variable " }{XPPEDIT 18 0 "x[3]" "6#&%\"xG6#\"\"$" }{TEXT -1 10 " the value" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "new := X[ 3](z[1],z[2]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "giving us as the next iterate the \+ vector " }{TEXT 293 1 "x" }{XPPEDIT 18 0 "``[7]" "6#&%!G6#\"\"(" } {TEXT -1 25 ", represented in Maple as" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z := [z[1],z[2],new];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 33 "The residual vector generated by " }{TEXT 294 1 "x" } {XPPEDIT 18 0 "``[7]" "6#&%!G6#\"\"(" }{TEXT -1 8 " is then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "` R` = R(op(z));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "The second component of R now has \+ the largest magnitude, so it is driven to zero by assigning to " } {XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT -1 10 " the value" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "new := X[2](z[1],z[3]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "This gives " }{TEXT 295 1 "x" }{XPPEDIT 18 0 "``[8]" "6#&%!G6#\"\")" }{TEXT -1 22 ", the next \+ iterate, as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "z:=[z[1],new,z[3]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "and the itera tion continues until it converges to the exact solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "print( xe);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 516 "Although it can become tedious, selecting the co mponent with the largest magnitude in the residual vector is something which can be done \"by-hand\" for a system of modest size. Clearly, \+ having to pick, repeatedly, the largest number from a list of 10,000 w ould be highly susceptible to error. Moreover, having a computer make the many comparisons required to select the largest magnitude is not \+ efficient, and this version of the relaxation algorithm is not a viabl e method for solving large problems on a computer." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 159 "However, the underlying \+ concept does motivate the method of successive overrelaxation, the SOR technique, which is capable of being implemented on a computer. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 25 "Successive Overrelaxat ion" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 178 "Southwell discovered that often, when driving a component of t he residual to zero, the other components would move in the same direc tion. For example, the residual generated by " }{TEXT 297 1 "x" } {XPPEDIT 18 0 "``[1]" "6#&%!G6#\"\"\"" }{TEXT -1 3 " is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 262 "" 0 "" {TEXT 296 1 "R" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "MATRIX([[7.3], [0], [-20]])" "6#-%'MATRIXG6#7%7#$\" #t!\"\"7#\"\"!7#,$\"#?F*" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 60 "The third component is driven to zero by increasin g it from " }{XPPEDIT 18 0 "-20" "6#,$\"#?!\"\"" }{TEXT -1 34 " to 0. \+ The residual generated by " }{TEXT 352 1 "x" }{XPPEDIT 18 0 "``[2]" " 6#&%!G6#\"\"#" }{TEXT -1 3 " is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 263 "" 0 "" {TEXT 298 1 "R" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "MA TRIX([[9.8], [17.7], [0]])" "6#-%'MATRIXG6#7%7#$\"#)*!\"\"7#$\"$x\"F*7 #\"\"!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "Driving the third component of the residual upwards from " }{XPPEDIT 18 0 "-20" "6#,$\"#?!\"\"" }{TEXT -1 110 " to 0 also drove the first c omponent upwards from 7.3 to 9.8, and the second component upwards fro m 0 to 17.7." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "This led Southwell to \"overcompensate\" or " }{TEXT 299 9 "ove rrelax" }{TEXT -1 170 ", pushing the residual, not to zero, but beyond zero. All that is needed is a more efficient way of implementing the calculations on a computer. Formulating the system " }{TEXT 353 1 "A " }{TEXT 300 1 "x" }{TEXT -1 3 " = " }{TEXT 301 1 "y" }{TEXT -1 78 " a s if for performing Gauss-Seidel iteration is the basis for this effic iency." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Assuming that the " }{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 32 "th equ ation has been solved for " }{XPPEDIT 18 0 "x[k]" "6#&%\"xG6#%\"kG" } {TEXT -1 40 ", we write the linear system in the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 264 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[ k](m+1)=x[k](m)+1/a[kk" "6#/-&%\"xG6#%\"kG6#,&%\"mG\"\"\"F,F,,&-&F&6#F (6#F+F,*&F,F,&%\"aG6#%#kkG!\"\"F," }{TEXT -1 1 " " }{XPPEDIT 18 0 "``( y[k]-Sum(a[ks]*x[s](m),s=1..n))" "6#-%!G6#,&&%\"yG6#%\"kG\"\"\"-%$SumG 6$*&&%\"aG6#%#ksGF+-&%\"xG6#%\"sG6#%\"mGF+/F8;F+%\"nG!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 "as in the Jacobi \+ method. Then, recognizing that the expression in parentheses is actua lly the " }{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 36 "th component of t he residual vector " }{TEXT 302 1 "R" }{TEXT -1 10 ", we write" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 265 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "x[k](m+1)=x[k](m)+1/a[kk]" "6#/-&%\"xG6#%\"kG6#,&%\"mG \"\"\"F,F,,&-&F&6#F(6#F+F,*&F,F,&%\"aG6#%#kkG!\"\"F," }{TEXT -1 1 " " }{TEXT 303 1 "R" }{XPPEDIT 18 0 "``[m]" "6#&%!G6#%\"mG" }{TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "By in troducing " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 53 ", the re laxation factor, this equation is modified to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 266 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[k]( m+1)=x[k](m)+omega/a[kk]" "6#/-&%\"xG6#%\"kG6#,&%\"mG\"\"\"F,F,,&-&F&6 #F(6#F+F,*&%&omegaGF,&%\"aG6#%#kkG!\"\"F," }{TEXT -1 1 " " }{TEXT 304 1 "R" }{XPPEDIT 18 0 "``[m]" "6#&%!G6#%\"mG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "Finally, write \+ the general equation in the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 267 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[k](m+1)=(1-omega)*x [k](m)+omega/a[kk]" "6#/-&%\"xG6#%\"kG6#,&%\"mG\"\"\"F,F,,&*&,&F,F,%&o megaG!\"\"F,-&F&6#F(6#F+F,F,*&F0F,&%\"aG6#%#kkGF1F," }{TEXT -1 1 " " } {XPPEDIT 18 0 "``(y[k]-Sum(a[ks]*x[s](m+1),s=1..k-1)-Sum(a[ks]*x[s](m) ,s=k+1..n))" "6#-%!G6#,(&%\"yG6#%\"kG\"\"\"-%$SumG6$*&&%\"aG6#%#ksGF+- &%\"xG6#%\"sG6#,&%\"mGF+F+F+F+/F8;F+,&F*F+F+!\"\"F?-F-6$*&&F16#F3F+-&F 66#F86#F;F+/F8;,&F*F+F+F+%\"nGF?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 84 "which is now consonant with the Gauss-Sei del method, except for the introduction of " }{XPPEDIT 18 0 "omega" "6 #%&omegaG" }{TEXT -1 38 ", the relaxation factor. Clearly, if " } {XPPEDIT 18 0 "omega=1" "6#/%&omegaG\"\"\"" }{TEXT -1 62 ", the SOR te chnique is precisely the Gauss-Seidel method. If " }{TEXT 354 1 "A" } {TEXT -1 45 " is a symmetric positive definite matrix and " }{XPPEDIT 18 0 "0 " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 " " {TEXT -1 14 "Implementation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "Implementing the SOR method for s mall systems is more conveniently done in matrix form. To accomplish \+ this, write the linear system " }{TEXT 355 1 "A" }{TEXT 305 1 "x" } {TEXT -1 3 " = " }{TEXT 306 1 "y" }{TEXT -1 12 " in the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 268 "" 0 "" {TEXT 307 1 "y" }{TEXT -1 1 " " }{XPPEDIT 18 0 "``-``" "6#,&%!G\"\"\"F$!\"\"" }{TEXT -1 3 " A " }{TEXT 308 1 "x" }{TEXT -1 3 " = " }{TEXT 309 1 "0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "and multiply though by " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 8 ", giving" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 269 "" 0 "" {TEXT 312 1 "0" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 2 " (" }{TEXT 310 1 "y" }{TEXT -1 1 " " }{XPPEDIT 18 0 "``-``" "6#,&%!G\"\"\"F$!\"\"" } {TEXT -1 3 " A " }{TEXT 311 2 "x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 6 "Write " }{TEXT 356 1 "A" }{TEXT -1 3 " = \+ " }{TEXT 357 1 "L" }{TEXT -1 3 " + " }{TEXT 358 1 "D" }{TEXT -1 3 " + \+ " }{TEXT 359 1 "U" }{TEXT -1 8 ", where " }{TEXT 360 1 "L" }{TEXT -1 41 " is zero on and above the main diagonal, " }{TEXT 361 1 "U" } {TEXT -1 45 " is zero on and below the main diagonal, and " }{TEXT 362 1 "D" }{TEXT -1 50 " is zero off the main diagonal. In addition, \+ add " }{TEXT 363 1 "D" }{TEXT -1 1 " " }{TEXT 317 1 "x" }{TEXT -1 48 " to each side of the equation, thereby obtaining" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 270 "" 0 "" {TEXT 364 1 "D" }{TEXT -1 1 " " } {TEXT 315 1 "x" }{TEXT -1 4 " = " }{TEXT 365 1 "D" }{TEXT -1 1 " " } {TEXT 316 1 "x" }{TEXT -1 2 " " }{XPPEDIT 18 0 "-omega" "6#,$%&omegaG !\"\"" }{TEXT -1 1 "(" }{TEXT 366 1 "L" }{TEXT -1 3 " + " }{TEXT 367 1 "D" }{TEXT -1 3 " + " }{TEXT 368 1 "U" }{TEXT -1 2 ") " }{TEXT 313 1 "x" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 1 " " }{TEXT 314 1 "y" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "Bring the term " }{XPPEDIT 18 0 "-omega" "6#,$%&omegaG !\"\"" }{TEXT 369 1 "L" }{TEXT -1 1 " " }{TEXT 318 1 "x" }{TEXT -1 20 " to the left, giving" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 271 "" 0 "" {TEXT -1 1 "(" }{TEXT 370 1 "D" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT 371 1 "L" }{TEXT -1 2 ") " }{TEXT 319 1 "x " }{TEXT -1 4 " = [" }{XPPEDIT 18 0 "``(1-omega)" "6#-%!G6#,&\"\"\"F'% &omegaG!\"\"" }{TEXT 372 1 "D" }{TEXT -1 1 " " }{XPPEDIT 18 0 "-omega " "6#,$%&omegaG!\"\"" }{TEXT 373 1 "U" }{TEXT -1 2 "] " }{TEXT 320 1 " x" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 1 " " }{TEXT 321 1 "y" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "Let the vector " }{TEXT 322 1 "x" }{TEXT -1 46 " on the l eft side be the updated one, namely, " }{TEXT 323 1 "x" }{XPPEDIT 18 0 "``(m+1)" "6#-%!G6#,&%\"mG\"\"\"F(F(" }{TEXT -1 20 ", while keeping \+ the " }{TEXT 324 1 "x" }{TEXT -1 45 " on the right side as the \"old\" one, namely, " }{TEXT 325 1 "x" }{XPPEDIT 18 0 "``(m)" "6#-%!G6#%\"mG " }{TEXT -1 13 ". This gives" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 272 "" 0 "" {TEXT -1 1 "(" }{TEXT 374 1 "D" }{TEXT -1 3 " + " } {XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT 375 1 "L" }{TEXT -1 2 ") " } {TEXT 326 1 "x" }{XPPEDIT 18 0 "``(m+1)" "6#-%!G6#,&%\"mG\"\"\"F(F(" } {TEXT -1 4 " = [" }{XPPEDIT 18 0 "``(1-omega)" "6#-%!G6#,&\"\"\"F'%&om egaG!\"\"" }{TEXT 376 1 "D" }{TEXT -1 1 " " }{XPPEDIT 18 0 "-omega" "6 #,$%&omegaG!\"\"" }{TEXT 377 1 "U" }{TEXT -1 2 "] " }{TEXT 327 1 "x" } {XPPEDIT 18 0 "``(m)" "6#-%!G6#%\"mG" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 1 " " }{TEXT 328 1 "y" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Solve for " }{TEXT 329 1 "x" }{XPPEDIT 18 0 "``(m+1)" "6#-%!G6#,&%\"mG\"\"\"F(F(" }{TEXT -1 10 " to obtain" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 273 "" 0 " " {TEXT 330 1 "x" }{XPPEDIT 18 0 "``(m+1)" "6#-%!G6#,&%\"mG\"\"\"F(F( " }{TEXT -1 4 " = (" }{TEXT 378 1 "D" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 2 "L)" }{XPPEDIT 18 0 "``^(-1)" "6#)% !G,$\"\"\"!\"\"" }{TEXT -1 1 "[" }{XPPEDIT 18 0 "``(1-omega)" "6#-%!G6 #,&\"\"\"F'%&omegaG!\"\"" }{TEXT 379 1 "D" }{TEXT -1 1 " " }{XPPEDIT 18 0 "-omega" "6#,$%&omegaG!\"\"" }{TEXT 380 1 "U" }{TEXT -1 2 "] " } {TEXT 331 1 "x" }{XPPEDIT 18 0 "``(m)" "6#-%!G6#%\"mG" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 1 "(" }{TEXT 381 1 "D " }{TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT 382 1 " L" }{TEXT -1 1 ")" }{XPPEDIT 18 0 "``^(-1)" "6#)%!G,$\"\"\"!\"\"" } {TEXT 332 1 "y" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "which has the fixed-point form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 274 "" 0 "" {TEXT 333 1 "x" }{XPPEDIT 18 0 "``(m+1)" " 6#-%!G6#,&%\"mG\"\"\"F(F(" }{TEXT -1 3 " = " }{TEXT 383 1 "B" }{TEXT -1 1 " " }{TEXT 334 1 "x" }{XPPEDIT 18 0 "``(m)" "6#-%!G6#%\"mG" } {TEXT -1 3 " + " }{TEXT 335 1 "C" }{TEXT -1 3 " = " }{TEXT 340 1 "g" } {TEXT -1 1 "(" }{TEXT 341 1 "x" }{XPPEDIT 18 0 "``(m)" "6#-%!G6#%\"mG " }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }}{PARA 275 " " 0 "" {TEXT -1 1 " " }{TEXT 403 1 "B" }{TEXT -1 4 " = (" }{TEXT 400 1 "D" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 2 "L)" }{XPPEDIT 18 0 "``^(-1)" "6#)%!G,$\"\"\"!\"\"" }{TEXT -1 1 "[" }{XPPEDIT 18 0 "``(1-omega)" "6#-%!G6#,&\"\"\"F'%&omegaG!\"\"" }{TEXT 401 1 "D" }{TEXT -1 1 " " }{XPPEDIT 18 0 "-omega" "6#,$%&omegaG!\"\"" }{TEXT 402 1 "U" }{TEXT -1 2 "] " }}{PARA 0 "" 0 "" {TEXT -1 4 "and " }}{PARA 276 "" 0 "" {TEXT -1 1 " " }{TEXT 407 1 "C" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 1 "(" }{TEXT 405 1 "D" } {TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT 406 1 "L" }{TEXT -1 1 ")" }{XPPEDIT 18 0 "``^(-1)" "6#)%!G,$\"\"\"!\"\"" }{TEXT 404 2 "y " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 13 "Example 39 .12" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "Consider the linear system " }{TEXT 408 1 "A" }{TEXT 409 1 "x" }{TEXT -1 3 " = " }{TEXT 410 1 "y" }{TEXT -1 11 " for which " }{TEXT 411 1 "A" }{TEXT -1 44 " is the symmetric, positive definite matrix " }}{PARA 0 "" 0 "" {TEXT -1 1 "\010" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "A := matrix(3,3,[8,-5,-7,-5,14,1,-7,1,8]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "and " }{TEXT 412 1 "y" }{TEXT -1 14 " is the vector" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "y := vector([-23,26,19]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "We see that " }{TEXT 413 1 "A" }{TEXT -1 63 " is symmetric. That it is positive definite a lso, we see from " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 25 "definite(A,positive_def);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "o r from the actual eigenvalues" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eigenvals(evalf(op(A)));" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "which are clearly all positive." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "the exact solution is " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "xe := linsolve(A,y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Decompose " }{TEXT 384 1 " A" }{TEXT -1 4 " as " }{TEXT 385 1 "L" }{TEXT -1 3 " + " }{TEXT 386 1 "D" }{TEXT -1 3 " + " }{TEXT 387 1 "U" }{TEXT -1 8 ", where " }{TEXT 388 1 "L" }{TEXT -1 2 ", " }{TEXT 389 1 "D" }{TEXT -1 6 ", and " } {TEXT 390 1 "U" }{TEXT -1 13 " are given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 145 "L := matrix(3,3, (i,j) -> if i>j then A[i,j] else 0 fi);\nU := matrix(3,3, (i,j) -> if i " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Notice that we have used " }{TEXT 391 1 "d" }{TEXT -1 13 " in place of " }{TEXT 392 1 "D" }{TEXT -1 62 ", which in Maple is res erved for the differentiation operator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 "The matrix " }{TEXT 393 1 "B" } {TEXT -1 4 " = (" }{TEXT 394 1 "D" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "o mega" "6#%&omegaG" }{TEXT 395 1 "L" }{TEXT -1 1 ")" }{XPPEDIT 18 0 "`` ^(-1)" "6#)%!G,$\"\"\"!\"\"" }{TEXT -1 1 "[" }{XPPEDIT 18 0 "``(1-omeg a)" "6#-%!G6#,&\"\"\"F'%&omegaG!\"\"" }{TEXT 396 1 "D" }{TEXT -1 1 " \+ " }{XPPEDIT 18 0 "-omega" "6#,$%&omegaG!\"\"" }{TEXT 397 1 "U" }{TEXT -1 9 "] is then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 68 "B := map(simplify,evalm(inverse(d+omega*L)&*(( 1-omega)*d-omega*U)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "and the vector " }{TEXT 336 1 " C" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 1 " (" }{TEXT 398 1 "D" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "omega" "6#%&omeg aG" }{TEXT -1 2 "L)" }{XPPEDIT 18 0 "``^(-1)" "6#)%!G,$\"\"\"!\"\"" } {TEXT 337 1 "y" }{TEXT -1 8 " is then" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "C := evalm(omega*invers e(d+omega*L)&*y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "The eigenvalues of " }{TEXT 399 1 "B" }{TEXT -1 70 " determine the rate of convergence of the SOR iterat ion. In terms of " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 10 " , they are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 18 "q := eigenvals(B);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Since we wish to pick " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 152 " so that the largest magnitude of the eigenvalues is as small as possible, we \+ begin with Figure 39.5, a graph of the three eigenvalues as a function of " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 1 "." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 191 "p1 \+ := plot(map(abs,[q]),omega=0..2,color=[black,red,green], thickness=[3, 1,1], labels=[``,``]):\np2 := textplot([1.75,-.07,`w`], font=[SYMBOL,1 2]):\ndisplay([p1,p2],xtickmarks=5, ytickmarks=5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "The optimal value of " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 67 " appears to be approximately 1.55, determined from the graph. For " } {XPPEDIT 18 0 "omega=1.55" "6#/%&omegaG$\"$b\"!\"#" }{TEXT -1 12 " the matrix " }{TEXT 414 1 "B" }{TEXT -1 8 " becomes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "B1 := subs( omega=1.55,op(B));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "and the vector " }{TEXT 338 1 "C" }{TEXT -1 8 " becomes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "C1 := subs(omega=1.55,op(C));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Starting with the initial vector " }{XPPEDIT 18 0 "[10,10 ,10]^T" "6#)7%\"#5F%F%%\"TG" }{TEXT -1 111 ", we obtain the first ten \+ iterates and the norms of the differences between the iterate and the \+ exact solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 102 "x := evalf(vector([10,10,10]));\nfor k from 1 to 10 do\nx := evalm(B1&*x+C1);\nprint(k,x,norm(x-xe));\nod:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "To verify that the iteration converges more slowly for ot her values of " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 9 ", we \+ try " }{XPPEDIT 18 0 "omega=1.75" "6#/%&omegaG$\"$v\"!\"#" }{TEXT -1 11 " and obtain" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 29 "B2 := subs(omega=1.75,op(B));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 " for the matrix " }{TEXT 419 1 "B" }{TEXT -1 5 ", and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "C 2 := subs(omega=1.75,op(C));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "for the vector " }{TEXT 339 1 "C" }{TEXT -1 43 ". Again iterating from the initial vector " } {XPPEDIT 18 0 "[10,10,10]^T" "6#)7%\"#5F%F%%\"TG" }{TEXT -1 9 ", we fi nd" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "x := evalf(vector([10,10,10]));\nfor k from 1 to 10 \+ do\nx := evalm(B2&*x+C2);\nprint(k,x,norm(x-xe));\nod:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "With " }{XPPEDIT 18 0 "omega=1.75" "6#/%&omegaG$\"$v\"!\"#" } {TEXT -1 70 ", the tenth iterate is farther away from the exact soluti on than with " }{XPPEDIT 18 0 "omega=1.55" "6#/%&omegaG$\"$b\"!\"#" } {TEXT -1 34 ". We conclude that our choice of " }{XPPEDIT 18 0 "omega " "6#%&omegaG" }{TEXT -1 73 " based on the graph of the magnitudes of \+ the eigenvalues was appropriate." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 3 "" 0 " " {TEXT -1 20 "Equivalence of Forms" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "When " }{XPPEDIT 18 0 "omega=1" "6 #/%&omegaG\"\"\"" }{TEXT -1 151 ", the SOR method reduces to the Gauss -Seidel method, an equivalence we will now verify for the system in Ex ample 39.12 whose Gauss-Seldel equations are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 277 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[1]^(m+1) \+ = ``(5/8)*x[2]^m+``(7/8)*x[3]^m-23/8" "6#/)&%\"xG6#\"\"\",&%\"mGF(F(F( ,(*&-%!G6#*&\"\"&F(\"\")!\"\"F()&F&6#\"\"#F*F(F(*&-F.6#*&\"\"(F(F2F3F( )&F&6#\"\"$F*F(F(*&\"#BF(F2F3F3" }{TEXT -1 1 " " }}{PARA 278 "" 0 "" {TEXT -1 4 " " }{XPPEDIT 18 0 "x[2]^(m+1)=``(5/14)*x[1]^m-``(1/14)* x[3]^m+13/7" "6#/)&%\"xG6#\"\"#,&%\"mG\"\"\"F+F+,(*&-%!G6#*&\"\"&F+\"# 9!\"\"F+)&F&6#F+F*F+F+*&-F/6#*&F+F+F3F4F+)&F&6#\"\"$F*F+F4*&\"#8F+\"\" (F4F+" }{TEXT -1 1 " " }}{PARA 279 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[3]^(m+1) = ``(7/8)*x[1]^m-``(1/8)*x[2]^m+19/8" "6#/)&%\"xG6#\" \"$,&%\"mG\"\"\"F+F+,(*&-%!G6#*&\"\"(F+\"\")!\"\"F+)&F&6#F+F*F+F+*&-F/ 6#*&F+F+F3F4F+)&F&6#\"\"#F*F+F4*&\"#>F+F3F4F+" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "When " } {XPPEDIT 18 0 "omega=1" "6#/%&omegaG\"\"\"" }{TEXT -1 13 ", the matrix " }{TEXT 415 1 "B" }{TEXT -1 8 " becomes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "BB := subs(omega=1,o p(B));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "and the vector " }{TEXT 342 1 "C" }{TEXT -1 8 " becomes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "CC := subs(omega=1,op(C));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "Consequent ly, the SOR equations in the fixed-point iteration " }{TEXT 343 1 "x" }{XPPEDIT 18 0 "``(m+1)" "6#-%!G6#,&%\"mG\"\"\"F(F(" }{TEXT -1 3 " = \+ " }{TEXT 416 1 "B" }{TEXT -1 1 " " }{TEXT 344 1 "x" }{XPPEDIT 18 0 "`` (m)" "6#-%!G6#%\"mG" }{TEXT -1 3 " + " }{TEXT 345 1 "C" }{TEXT -1 4 " \+ are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 280 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "x[1]^(m+1)=``(5/8)*x[2]^m+``(7/8)*x[3]^m-23/8" "6#/) &%\"xG6#\"\"\",&%\"mGF(F(F(,(*&-%!G6#*&\"\"&F(\"\")!\"\"F()&F&6#\"\"#F *F(F(*&-F.6#*&\"\"(F(F2F3F()&F&6#\"\"$F*F(F(*&\"#BF(F2F3F3" }{TEXT -1 1 " " }}{PARA 281 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "x[2 ]^(m+1)=``(25/112)*x[2]^m+``(27/112)*x[3]^m+93/112" "6#/)&%\"xG6#\"\"# ,&%\"mG\"\"\"F+F+,(*&-%!G6#*&\"#DF+\"$7\"!\"\"F+)&F&6#F(F*F+F+*&-F/6#* &\"#FF+F3F4F+)&F&6#\"\"$F*F+F+*&\"#$*F+F3F4F+" }{TEXT -1 1 " " }} {PARA 282 "" 0 "" {TEXT -1 9 " " }{XPPEDIT 18 0 "x[3]^(m+1)=`` (465/896)*x[2]^m+``(659/896)*x[3]^m-219/896" "6#/)&%\"xG6#\"\"$,&%\"mG \"\"\"F+F+,(*&-%!G6#*&\"$l%F+\"$'*)!\"\"F+)&F&6#\"\"#F*F+F+*&-F/6#*&\" $f'F+F3F4F+)&F&6#F(F*F+F+*&\"$>#F+F3F4F4" }{TEXT -1 1 " " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 183 "and we want to sh ow these equations are indeed the Gauss-Seidel equations. Clearly, th e first equation in each formulation is the same. In the second Gauss -Seidel equation, replace " }{XPPEDIT 18 0 "x[1]^(m+1)" "6#)&%\"xG6#\" \"\",&%\"mGF'F'F'" }{TEXT -1 26 " from the first, obtaining" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 283 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[2]^(m+1)=``(25/112)*x[2]^m+``(27/112)*x[3]^m+93/112" "6#/)&%\" xG6#\"\"#,&%\"mG\"\"\"F+F+,(*&-%!G6#*&\"#DF+\"$7\"!\"\"F+)&F&6#F(F*F+F +*&-F/6#*&\"#FF+F3F4F+)&F&6#\"\"$F*F+F+*&\"#$*F+F3F4F+" }{TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "the s econd equation in the SOR formulation. In the third Gauss-Seidel equa tion, replace " }{XPPEDIT 18 0 "x[2]^(m+1)" "6#)&%\"xG6#\"\"#,&%\"mG\" \"\"F*F*" }{TEXT -1 52 " from the second equation. This introduces an other " }{XPPEDIT 18 0 "x[1]^(m+1)" "6#)&%\"xG6#\"\"\",&%\"mGF'F'F'" } {TEXT -1 15 ". Now replace " }{TEXT 417 3 "all" }{TEXT -1 1 " " } {XPPEDIT 18 0 "x[1]^(m+1)" "6#)&%\"xG6#\"\"\",&%\"mGF'F'F'" }{TEXT -1 48 " from the first equation, and the result will be" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 284 "" 0 "" {TEXT -1 3 " " }{XPPEDIT 18 0 "x [3]^(m+1)=``(465/896)*x[2]^m+``(659/896)*x[3]^m-219/896" "6#/)&%\"xG6# \"\"$,&%\"mG\"\"\"F+F+,(*&-%!G6#*&\"$l%F+\"$'*)!\"\"F+)&F&6#\"\"#F*F+F +*&-F/6#*&\"$f'F+F3F4F+)&F&6#F(F*F+F+*&\"$>#F+F3F4F4" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 142 "the th ird equation of the SOR formulation. Thus, we have verified that the \+ matrix form of the SOR iteration actually uses the updated values " } {TEXT 418 1 "x" }{XPPEDIT 18 0 "``[m+1]" "6#&%!G6#,&%\"mG\"\"\"F(F(" } {TEXT -1 51 " in the same way that Gauss-Seidel iteration does.." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "To execut e these substitutions in Maple, write the Gauss-Seidel equations as" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "unassign('x','X');\nfor k from 1 to 3 do\nX[k] := solve(evalm(A \+ &* vars-y)[k], x[k]);\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "Substitute for " } {XPPEDIT 18 0 "x[1]" "6#&%\"xG6#\"\"\"" }{TEXT -1 83 " in the second e quation, using its value from the first equation, thereby obtaining" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "X2 := subs(x[1]=X[1],X[2]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "This is the s econd equation in the SOR formulation." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 77 "Finally, in the third equation of the Gauss-Seidel formulation the variables " }{XPPEDIT 18 0 "x[1]" "6#&% \"xG6#\"\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\" \"#" }{TEXT -1 28 " both appear. The variable " }{XPPEDIT 18 0 "x[2] " "6#&%\"xG6#\"\"#" }{TEXT -1 78 " must be updated with the most recen t version of this variable. The variable " }{XPPEDIT 18 0 "x[1]" "6#& %\"xG6#\"\"\"" }{TEXT -1 95 " must likewise be replaced with its updat ed version, but that update contains the old value of " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT -1 28 ". Thus, the replacement of \+ " }{XPPEDIT 18 0 "x[1]" "6#&%\"xG6#\"\"\"" }{TEXT -1 39 " must be done after the replacement of " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" } {TEXT -1 11 ", lest the " }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" } {TEXT -1 4 " in " }{XPPEDIT 18 0 "x[1]" "6#&%\"xG6#\"\"\"" }{TEXT -1 46 " be updated twice. With this due care, we get" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "subs([x[1]= X[1], x[2]=X2], X[3]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 141 "on the righthand side of the t hird equation, and we have verified that the matrix form of the SOR it eration actually uses the updated values " }{TEXT 346 1 "x" }{XPPEDIT 18 0 "``(m+1)" "6#-%!G6#,&%\"mG\"\"\"F(F(" }{TEXT -1 50 " in the same \+ way that Gauss-Seidel iteration does." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 " " 0 "" {TEXT -1 10 "References" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 99 "[1] Richard V. Southwell, Relaxation Meth ods in Engineering Science, Oxford University Press, 1940." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "[2] James M. Orte ga, Numerical Analysis, A Second Course, Academic Press, 1992." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "1" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }