Numerical Methods for Nonlinear Variable Problems Episode 2 pot

... of Sec. 2. 5 of this chapter. We assume that Q is a polygonal domain of U 2 (see Remark 3.8 for the nonpolygonal case), and we consider a triangulation 2T h of Q satisfying (2. 21)- (2. 23). Then ... (2. 10) T-OonQ. (2. 11) Before proving that (2. 10), (2. 11) imply u e H 2 (Q), we shall recall a classical lemma (also very useful in the analysis of fourth-order problems)...

Ngày tải lên: 12/08/2014, 16:21

... strongly in H S (Q), for every s < 2 (cf. Necas [1])), so that u e H 2 (fi) with ||Au|| L2(n) < ||/||L2(Q). (2. 20) • 2. 5. Finite element approximations of (2. 1) Henceforth we shall ... U 2 . Consider a "classical" triangulation 2T h of Q, i.e. 2T h is a finite set of triangles T such that TcQ VTef», U T = Q, (2. 21) ^0^ = 0 V T u T 2 e ^ and T x...

Ngày tải lên: 27/06/2014, 18:20

... /m i/O < i/O < 1, x 2 ) - «i,0)j «(1, 0); Xi ; x 2 = 0 = (, <1, I/O < x 2 < 1, i/O < x, (4 .29 7) x (4 .29 7) 2 (4 .29 7) 3 (4 .29 7) 4 (4 .29 7) 5 (4 .29 7) 6 (K\u • n) in the above ... the Solution of Elliptic Problems for Partial Differential Operators 379 we indeed have dv 1 {v A (b 2 - c 2 ) + v B (c 2 - a 2 ) + vda 2 - b 2 )}, dx t...

Ngày tải lên: 12/08/2014, 16:21

... Variational Problems 321 1. Introduction 321 2. A Family of Linear Variational Problems 321 3. Internal Approximation of Problem (P) 326 4. Application to the Solution of Elliptic Problems for Partial ... \\v\\ 2 - U\\ \\v\\ - \\L\\ \\v\\ + n _ / [p (imi + IILII) I2V (u\\ + \\L\\y W2 2 V/8/ 2^ ' l ' ' Hence J(u)->+oo as ||«|| -+ +oo. (4. 12)...

Ngày tải lên: 12/08/2014, 16:21

... (5 .21 ) and if u e (H 2 (Q)) 2 , p e H^ty/U, we have \\Ph ~ PWmnyut ^ C7i(||u|| (H 2( n 2 + \\p\\ H Hn)l«l ( 5 - 33 ) ||(u h + ViA,) - u|| (L2(S2) )2 < C/i 2 (|ju|| (H2(n) )2 ... e (H 3 (Q)) 2 , p e H 2 (Q)/R, we have K - u|| (ffl(n)) 2 < C7i 2 (||u|| (H 3 (n)) 2 + ||p|| H 2 (n)/R ), (5 .26 ) \\Ph ~ PWHHCWM ^ C/i(||u|| (H3(n)) 2 + WPWHHO)!...

Ngày tải lên: 12/08/2014, 16:21

... dx y dx 2 8 2 u 8 2 v d 2 u 3 2 v\} ox 1 ox 2 dx 2 ox\)) 'dh^d 2 ^ dx 2 dx 2 ' dx\ dx = ioAuAv + (1 - ff) Jo I. \ 1 ox 2 ox l ox 2 L(v) = \ fvdx, feL 2 (Q). (4.336) The ... M k . If M i2 of Step 2 does not belong to M v M h then M n , M i2 will not coincide, and we can use the M i2 defined in Step 2 (see Fig. 4 .2 (where M v...

Ngày tải lên: 12/08/2014, 16:21

... 22 1, 22 4, 24 4, 24 5, 24 6, 24 8, 25 3, 25 4, 25 9, 26 6, 26 7, 26 8, 26 9, 27 0, 27 1, 27 3, 308, 310, 3 12, 314, 315, 318, 438, 444, 450 Perrier, P. 195, 21 1, 21 4, 21 9, 22 1, 24 4, 24 5, 24 6, 24 8, 25 3, 25 4, 25 9, ... 24 1 ,24 2 convex 134 Korn 21 4 ,23 0 ,23 1 ,24 0 lifting 21 3 NACA 00 12 225 , 23 0, 23 1, 23 2, 23 3, 23 4 ; , 23 5, 23 6,...

Ngày tải lên: 12/08/2014, 16:21

... =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 + y...

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

... 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