Numerical Methods for Ordinary Dierential Equations Episode 2 docx

... =1, A2 n + B +2Cn2 n ,θ =2, A2 n + B + Cθ 2 (θ−1)(θ 2) θ n ,θ=1,θ =2. 32 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS 0 1 2 3 4 √ 2 1+ √ 2 2 √ 2 2+ √ 2 1 +2 √ 2 10 10 2 1 Figure 121 (ii) ... y 6 , y  4 =2y 5 + y 1 − µ(y 1 + µ −1) (y 2 2 + y 2 3 +(y 1 + µ −1) 2 ) 3 /2 − (1 − µ)(y 1 + µ) (y 2 2 + y 2 3 +(y 1 + µ) 2 ) 3 /2 , y  5 = −2y 4...

Ngày tải lên: 13/08/2014, 05:21

... examples 121 26 Hybrid Methods 122 26 0 Historical introduction 122 26 1 Pseudo Runge–Kutta methods 123 26 2 Generalized linear multistep methods 124 26 3 General linear methods 124 26 4 Numerical ... T 0 ◦ 6 .28 318530 72 3 ◦ 6 .28 426 20831 6 ◦ 6 .28 74944 421 9 ◦ 6 .29 28884880 12 ◦ 6.3004544311 15 ◦ 6.31 020 66431 18 ◦ 6. 322 1637356 21 ◦ 6.3363486630 24 ◦ 6.3 527...

Ngày tải lên: 13/08/2014, 05:21

... (301c) 148 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS Table 306(I) Members of I 2 and their symmetries I ()(1) (2) (1, 1) (1, 2) (2, 2) (1, 1, 1) (1, 1, 2) (1, 2, 2) (2, 2, 2) σ(I) 11 121 2 622 6 306 ... 3 [1, 2] 1 2 3 [1, 3] 1 3 2 [2, 1] 2 1 3 [2, 2] 2 1 3 [2, 3] 2 3 1 [3, 1] 3 1 2 [3, 2] 3 2 1 [3, 3] 3 1 2 303 The use of labelled trees W...

Ngày tải lên: 13/08/2014, 05:21

... −0.01737078 47 20 0.00898483 −0.00878393 80 40 0.00446704 −0.00441680 149 80 0.0 022 2 721 −0.0 022 14 62 240 160 0.0011 120 3 −0.00110889 480 320 0.000555 62 −0.00055484 960 640 0.00 027 771 −0.00 027 7 62 1 621 20 4 ... 1130400. 025 2×10 −10 4.4 125 40 25 6178.9889×10 −10 4.1893 80 61150 .26 26×10 −10 4.0904 160 14949.6176×10 −10 4.04 42 320 3696.5967×10 −10 4. 021 8 640 919.13 62...

Ngày tải lên: 13/08/2014, 05:21

... constants for Adams–Moulton methods k β 0 β 1 β 2 β 3 β 4 β 5 β 6 β 7 C 0 1 1 2 1 1 2 1 2 − 1 12 2 5 12 2 3 − 1 12 1 24 3 3 8 19 24 − 5 24 1 24 − 19 720 4 25 1 720 323 360 − 11 30 53 360 − 19 720 3 160 5 95 28 8 1 427 1440 − 133 24 0 24 1 720 − 173 1440 3 160 − 863 60480 6 19087 60480 27 13 25 20 − 15487 20 160 586 945 − 6737 20 160 26 3 25 20 −...

Ngày tải lên: 13/08/2014, 05:21

... c 3 −4 2( 1−c 3 )(6c 3 −1−6c 2 3 ) c 3 (1−2c 3 ) 2( 1−c 3 )(6c 3 −1−6c 2 3 ) 1−c 3 6c 3 −1−6c 2 3 6c 3 −1−6c 2 3 12c 3 (1−c 3 ) 1 12c 3 (1−c 3 ) 1 12c 3 (1−c 3 ) 6c 3 −1−6c 2 3 12c 3 (1−c 3 ) ; ( 322 e) for case II, 0 c 2 c 2 1 2 1 2 − 1 8c 2 1 8c 2 1 1 2c 2 − 1 − 1 2c 2 2 1 6 0 2 3 1 6 ; ( 322 f) for case III, 0 1 2 1 2 0 − 1 12b 3 1 12b 3 1 − 1 2 − 6b...

Ngày tải lên: 13/08/2014, 05:21

... tableau 0 1 2 c 2 1 2 a 21 1 2 c 3 1 2 a 31 1 2 a 32 . . . . . . . . . . . . 1 2 c s 1 2 a s1 1 2 a s2 ··· 1 2 a s,s−1 1 2 1 2 b 1 1 2 b 2 ··· 1 2 b s−1 1 2 b s 1 2 + 1 2 c 2 1 2 b 1 1 2 b 2 ··· 1 2 b s−1 1 2 b s 1 2 a 21 1 2 + 1 2 c 3 1 2 b 1 1 2 b 2 ··· 1 2 b s−1 1 2 b s 1 2 a 31 1 2 a 32 . . . . . . . . . . . . . ....

Ngày tải lên: 13/08/2014, 05:21

... =5,p=8), 0 1 20 −7− √ 21 120 1 15 −7+ √ 21 120 0 7− √ 21 14 1 20 343+9 √ 21 25 20 56−15 √ 21 315 343−69 √ 21 25 20 0 1 2 1 20 49+ 12 √ 12 360 8 45 49− 12 √ 12 360 0 7+ √ 21 14 1 20 343+69 √ 21 25 20 56+15 √ 21 315 343−9 √ 21 25 20 0 1 1 20 119−3 √ 21 360 13 45 119+3 √ 21 360 0 1 20 49 180 16 45 49 180 1 20 Lobatto ... =5,p=8), 0 1 20 −7− √...

Ngày tải lên: 13/08/2014, 05:21

... − 3 4 √ 2) 0 λ (2 + √ 2) λ(1 + 3 4 √ 2) λ(1 + 1 4 √ 2) 0 1 2+ 3 √ 2 4 − λ(1+ √ 2) 2 − √ 2 8λ 2 3 √ 2 4 − λ(1− √ 2) 2 + √ 2 8λ λ 2+ 3 √ 2 4 − λ(1+ √ 2) 2 − √ 2 8λ 2 3 √ 2 4 − λ(1− √ 2) 2 + √ 2 8λ λ . RUNGE–KUTTA ... find α 1 = √ 2 − 3 2 and α 2 = 1 2 (1 − √ 2) and the tableau 0 1 2 (4 − 3 √ 2) 1 2 ( √ 2 − 1) 1 1 2 (5 − 3 √ 2) 1 2 √ 2...

Ngày tải lên: 13/08/2014, 05:21

... b 4 a 42 c 2 + b 4 a 43 c 3 = 1 6 , b 4 a 43 a 32 c 2 = 1 24 , b 2 c 2 2 (1 − c 2 )+b 3 c 2 3 (1 − c 3 )+b 4 c 2 4 (1 − c 4 ) + b 3 a 32 c 2 (2c 3 − c 2 )+b 4 a 42 c 2 (2c 4 − c 2 )+b 4 a 43 c 3 (2c 4 − ... β 3 α 2 + β 11 β 11 +β 7 +2 6 +2 4 +β 3 + 4 3 β 2 + 1 15 12 5 α 12 + β 2 α 3 + β 4 α 2 + β 12 β 12 +β 8 +β 6 +β 4 + 1 2 β 3 + 2 3 β...

Ngày tải lên: 13/08/2014, 05:21