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Replace the value of boundary points, u¹ = 2 U3,2 14.2 2 2 U23 2 |U3.3 [u4,3 where we have an updated u value we use its current value: 6.8+7.2+ 2.8 +5.2 U3,3 2 U4,3 月 = = = = 5.5 4 7.7 +5.5+ 5.225 +4.225 4 8.7 +5.6625+ 9.4+ 6.8875 5.5+ 8.4+ 4.225 +8.9 = 5.6625 「U2,2 5.5 2 5.6625 U3,2 5.3125 2 想你 |U4.2 = 6.7563 2 U2.3 7.0516 2 U3.3 7.616 2 = 6.7563 4 5.6625+ 6.7563+ 6.8875 +8.9 5.3125 +7.0516+ 9.2 +8.9 4 「U2.2 U3,2 1 = 5.3125 14.2 1 U23 1 = 7.616 1 1 = 7.0516 = LU 4.3 Now, we should compute infinity norm for vector u?-u²: 3.5 2.8 5.225 5.2 4.225 6.8875 and

Replace the value of boundary points, u¹ = 2 U3,2 14.2 2 2 U23 2 |U3.3 [u4,3 where we have an updated u value we use its current value: 6.8+7.2+ 2.8 +5.2 U3,3 2 U4,3 月 = = = = 5.5 4 7.7 +5.5+ 5.225 +4.225 4 8.7 +5.6625+ 9.4+ 6.8875 5.5+ 8.4+ 4.225 +8.9 = 5.6625 「U2,2 5.5 2 5.6625 U3,2 5.3125 2 想你 |U4.2 = 6.7563 2 U2.3 7.0516 2 U3.3 7.616 2 = 6.7563 4 5.6625+ 6.7563+ 6.8875 +8.9 5.3125 +7.0516+ 9.2 +8.9 4 「U2.2 U3,2 1 = 5.3125 14.2 1 U23 1 = 7.616 1 1 = 7.0516 = LU 4.3 Now, we should compute infinity norm for vector u?-u²: 3.5 2.8 5.225 5.2 4.225 6.8875 and

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Replace the value of boundary points, u¹ = u²2,2 = 2 U3,2 U4,2² U2,3²2 ૫૩ 32 6.8 +7.2 + 2.8 +5.2 4 U3,3 LU4,3 where we have an updated u value we use its current value: = = = 2 U4,3² = 7.7 +5.5+ 5.225 +4.225 4 8.7 +5.6625+ 9.4+ 6.8875 4 5.5+ 8.4+ 4.225 +8.9 [22,2 13,2 2 4 5.6625 +6.7563+ 6.8875 +8.9 5.3125 +7.0516+ 9.2 + 8.9 4 2 14,2 2 12,3 2 = 5.5 U3,3 4 2 5.5 5.6625 5.3125 = 5.6625 = 6.7563 7.0516 7.616 [212,2 = 6.7563 13,2 1 = 5.3125 U4,2 1 U2,3 1 = 7.616 1 = 7.0516 = U4,3 Now, we should compute infinity norm for vector u² − u²¹: 3.5 2.8 5.225 5.2 4.225 6.8875 and
Transcribed Image Text:Replace the value of boundary points, u¹ = u²2,2 = 2 U3,2 U4,2² U2,3²2 ૫૩ 32 6.8 +7.2 + 2.8 +5.2 4 U3,3 LU4,3 where we have an updated u value we use its current value: = = = 2 U4,3² = 7.7 +5.5+ 5.225 +4.225 4 8.7 +5.6625+ 9.4+ 6.8875 4 5.5+ 8.4+ 4.225 +8.9 [22,2 13,2 2 4 5.6625 +6.7563+ 6.8875 +8.9 5.3125 +7.0516+ 9.2 + 8.9 4 2 14,2 2 12,3 2 = 5.5 U3,3 4 2 5.5 5.6625 5.3125 = 5.6625 = 6.7563 7.0516 7.616 [212,2 = 6.7563 13,2 1 = 5.3125 U4,2 1 U2,3 1 = 7.616 1 = 7.0516 = U4,3 Now, we should compute infinity norm for vector u² − u²¹: 3.5 2.8 5.225 5.2 4.225 6.8875 and
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