H Ix - x0. We have seen that the Jacobi and Gauss-Seidel iterative techniques can be written x = Tx-+e and x = T,x-+e using the matrices T, = D(L+ U) and T,- (D- L)-U. If p(T,) or p(T,) is less than...

Proof this theoremH Ix - x0.<br>We have seen that the Jacobi and Gauss-Seidel iterative techniques can be written<br>x = Tx-+e and x = T,x-+e<br>using the matrices<br>T, = D(L+ U) and T,- (D- L)-U.<br>If p(T,) or p(T,) is less than 1, then the corresponding sequence (x will converge to<br>the solution x of Ax = b. For example, the Jacobi scheme has<br>` =D'(L + U)xª-» + D-'b,<br>and, if (x1, converges to x, then<br>x= D(L + U)x + D 'b.<br>This implies that<br>Dx = (L+ U)x+b and (D - L- U)x = b.<br>Since D-L-U = A, the solution x satisfies Ax = b.<br>We can now give easily verified sufficiency conditions for convergence of the Jacobi<br>and Gauss-Seidel methods. (To prove convergence for the Jacobi scheme see Exercise 14,<br>and for the Gauss-Seidel scheme see (Or2), p. 120.)<br>erle ARRoeet Mete ep det pet c ep d<br>y i Cip L em ia<br>d e ri<br>7.3 The Jacobi and Gauss-Siedel Iterative Techniques<br>459<br>Theor 7.21 If A is strictly diagonally dominant, then for any choice of x, both the Jacobi and<br>Gauss-Scidel methods give sequences (x , that converge to the unique solution of<br>Ax = b.<br>the spectral radius of the iteration<br>The relations ne rapidity of converge<br>l be seen from Corollary 7.20. The inequane<br>norm, so it follows from the statement after Theorem 7.15 on page 446 that<br>Ix - x|| p(T)ʻ[x® – x||.<br>(7.12)<br>Thus we would like to select the iterative technique with minimal p(T) < for a particular<br>system Ax = b. No general results exist to tell which of the two techniques, Jacobi or Gauss-<br>Seidel, will be most successful for an arbitrary linear system. In special cases, however, the<br>answer is known, as is demonstrated in the following theorem. The proof of this result can<br>be found in [Y), pp. 120-127.<br>Theorem 7.22 (Stein-Rosenberg)<br>If a s0, for each i +j and a, > 0, for each i= 1,2, ...,n, then one and only one of the<br>following statements holds:<br>(i) Os P(T,) < p(T) < I;<br>(i) p(T) = p(T,) = 0;<br>(ii) I< p(T) < p(T):<br>(iv) p(T) = p(T,) = 1.<br>For the special case described in Theorem 7.22, we see from part (i) that when one<br>method gives convergence, then both give convergence, and the Gauss-Seidel method con-<br>verges faster than the Jacobi method. Part (ii) indicates that when one method diverges then<br>both diverge, and the divergence is more pronounced for the Gauss-Seidel method.<br>ERCISE SET 7.3<br>1. Find the first two iterations of the Jacobi method for the following linear systems, using x<br>= 0:<br>a. 3 - + =1,<br>3 + 6r + 2y =0.<br>3 + 3x + 7x) = 4.<br>10 -<br>= 9,<br>-A+ 10r - 2, 7<br>- 23 + 10r = 6.<br>e. 10n + Sx<br>Sx, + 10 - 4xy<br>-6,<br>d.<br>4 + + +<br>6.<br>-25,<br>- - 3 + + 4<br>-6,<br>4x + 8 - xa= -11.<br>2 + + Sa - 4- AS = 6,<br>-A- Ay- + 4<br>+ S -11.<br>-6,<br>2 - + a+ 4xs = 6.<br>Find the first two iterations of the Jacobi method for the following lincar systems, using x -0<br>2.<br>a. 4x + A - xy = 5,<br>- + 3p + =-4.<br>h -2+ + a-4,<br>X-2 - a = -4,<br>2 +2x + Sx = 1.<br>N+ 2-0.<br>C.<br>4x + - + -2,<br>d.<br>-4<br>-0,<br>XI+4 - - -1.<br>-N +4r - X<br>- 5.<br>- -+ S+ 0,<br>- A+ 4<br>- 0,<br>+ 4-<br>- 4+ 4 - -2,<br>- A+ 4- 6.<br>X-+ + 3=1.<br>+ L<br>-6,<br>-<br>

Extracted text: H Ix - x0. We have seen that the Jacobi and Gauss-Seidel iterative techniques can be written x = Tx-+e and x = T,x-+e using the matrices T, = D(L+ U) and T,- (D- L)-U. If p(T,) or p(T,) is less than 1, then the corresponding sequence (x will converge to the solution x of Ax = b. For example, the Jacobi scheme has ` =D'(L + U)xª-» + D-'b, and, if (x1, converges to x, then x= D(L + U)x + D 'b. This implies that Dx = (L+ U)x+b and (D - L- U)x = b. Since D-L-U = A, the solution x satisfies Ax = b. We can now give easily verified sufficiency conditions for convergence of the Jacobi and Gauss-Seidel methods. (To prove convergence for the Jacobi scheme see Exercise 14, and for the Gauss-Seidel scheme see (Or2), p. 120.) erle ARRoeet Mete ep det pet c ep d y i Cip L em ia d e ri 7.3 The Jacobi and Gauss-Siedel Iterative Techniques 459 Theor 7.21 If A is strictly diagonally dominant, then for any choice of x, both the Jacobi and Gauss-Scidel methods give sequences (x , that converge to the unique solution of Ax = b. the spectral radius of the iteration The relations ne rapidity of converge l be seen from Corollary 7.20. The inequane norm, so it follows from the statement after Theorem 7.15 on page 446 that Ix - x|| p(T)ʻ[x® – x||. (7.12) Thus we would like to select the iterative technique with minimal p(T) < for="" a="" particular="" system="" ax="b." no="" general="" results="" exist="" to="" tell="" which="" of="" the="" two="" techniques,="" jacobi="" or="" gauss-="" seidel,="" will="" be="" most="" successful="" for="" an="" arbitrary="" linear="" system.="" in="" special="" cases,="" however,="" the="" answer="" is="" known,="" as="" is="" demonstrated="" in="" the="" following="" theorem.="" the="" proof="" of="" this="" result="" can="" be="" found="" in="" [y),="" pp.="" 120-127.="" theorem="" 7.22="" (stein-rosenberg)="" if="" a="" s0,="" for="" each="" i="" +j="" and="" a,=""> 0, for each i= 1,2, ...,n, then one and only one of the following statements holds: (i) Os P(T,) < p(t)="">< i;="" (i)="" p(t)="p(T,)" =="" 0;="" (ii)="">< p(t)="">< p(t): (iv) p(t) = p(t,) = 1. for the special case described in theorem 7.22, we see from part (i) that when one method gives convergence, then both give convergence, and the gauss-seidel method con- verges faster than the jacobi method. part (ii) indicates that when one method diverges then both diverge, and the divergence is more pronounced for the gauss-seidel method. ercise set 7.3 1. find the first two iterations of the jacobi method for the following linear systems, using x = 0: a. 3 - + =1, 3 + 6r + 2y =0. 3 + 3x + 7x) = 4. 10 - = 9, -a+ 10r - 2, 7 - 23 + 10r = 6. e. 10n + sx sx, + 10 - 4xy -6, d. 4 + + + 6. -25, - - 3 + + 4 -6, 4x + 8 - xa= -11. 2 + + sa - 4- as = 6, -a- ay- + 4 + s -11. -6, 2 - + a+ 4xs = 6. find the first two iterations of the jacobi method for the following lincar systems, using x -0 2. a. 4x + a - xy = 5, - + 3p + =-4. h -2+ + a-4, x-2 - a = -4, 2 +2x + sx = 1. n+ 2-0. c. 4x + - + -2, d. -4 -0, xi+4 - - -1. -n +4r - x - 5. - -+ s+ 0, - a+ 4 - 0, + 4- - 4+ 4 - -2, - a+ 4- 6. x-+ + 3=1. + l -6, - p(t):="" (iv)="" p(t)="p(T,)" =="" 1.="" for="" the="" special="" case="" described="" in="" theorem="" 7.22,="" we="" see="" from="" part="" (i)="" that="" when="" one="" method="" gives="" convergence,="" then="" both="" give="" convergence,="" and="" the="" gauss-seidel="" method="" con-="" verges="" faster="" than="" the="" jacobi="" method.="" part="" (ii)="" indicates="" that="" when="" one="" method="" diverges="" then="" both="" diverge,="" and="" the="" divergence="" is="" more="" pronounced="" for="" the="" gauss-seidel="" method.="" ercise="" set="" 7.3="" 1.="" find="" the="" first="" two="" iterations="" of="" the="" jacobi="" method="" for="" the="" following="" linear="" systems,="" using="" x="0:" a.="" 3="" -="" +="1," 3="" +="" 6r="" +="" 2y="0." 3="" +="" 3x="" +="" 7x)="4." 10="" -="9," -a+="" 10r="" -="" 2,="" 7="" -="" 23="" +="" 10r="6." e.="" 10n="" +="" sx="" sx,="" +="" 10="" -="" 4xy="" -6,="" d.="" 4="" +="" +="" +="" 6.="" -25,="" -="" -="" 3="" +="" +="" 4="" -6,="" 4x="" +="" 8="" -="" xa="-11." 2="" +="" +="" sa="" -="" 4-="" as="6," -a-="" ay-="" +="" 4="" +="" s="" -11.="" -6,="" 2="" -="" +="" a+="" 4xs="6." find="" the="" first="" two="" iterations="" of="" the="" jacobi="" method="" for="" the="" following="" lincar="" systems,="" using="" x="" -0="" 2.="" a.="" 4x="" +="" a="" -="" xy="5," -="" +="" 3p="" +="-4." h="" -2+="" +="" a-4,="" x-2="" -="" a="-4," 2="" +2x="" +="" sx="1." n+="" 2-0.="" c.="" 4x="" +="" -="" +="" -2,="" d.="" -4="" -0,="" xi+4="" -="" -="" -1.="" -n="" +4r="" -="" x="" -="" 5.="" -="" -+="" s+="" 0,="" -="" a+="" 4="" -="" 0,="" +="" 4-="" -="" 4+="" 4="" -="" -2,="" -="" a+="" 4-="" 6.="" x-+="" +="" 3="1." +="" l="" -6,="">
Jun 04, 2022
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