Partial Differential Equations part 2
... equation (19.1 .20 ) is r n+1 j+1 /2 − r n j+1 /2 ∆t = s n+1 /2 j+1 − s n+1 /2 j ∆x s n+1 /2 j − s n−1 /2 j ∆t = v r n j+1 /2 − r n j −1 /2 ∆x (19.1.35) If you substitute equation (19.1 .22 ) in equation ... (19.1.40) Then ξ =1−iα sin k∆x − α 2 (1 − cos k∆x)(19.1.41) so |ξ| 2 =1−α 2 (1 − α 2 )(1 − cos k∆x) 2 (19.1. 42) The stability criterion |ξ| 2 ≤ 1 is therefore α 2 ≤...
Ngày tải lên: 07/11/2013, 19:15
Partial Differential Equations part 1
... 1-800-8 72- 7 423 (North America only),or send email to trade@cup.cam.ac.uk (outside North America). Chapter 19. Partial Differential Equations 19.0 Introduction The numerical treatment of partial differential ... Recipes dealing with partial differential equations alone. (The references [1-4] provide, of course, available alternatives.) In most mathematics books, partial...
Ngày tải lên: 28/10/2013, 22:15
... the whole interval. Figure 16.1 .2 illustrates the idea. In equations, k 1 = hf(x n ,y n ) k 2 =hf x n + 1 2 h, y n + 1 2 k 1 y n+1 = y n + k 2 + O(h 3 ) (16.1 .2) As indicated in the error term, ... organization about it: k 1 = hf(x n ,y n ) k 2 =hf(x n + h 2 ,y n + k 1 2 ) k 3 =hf(x n + h 2 ,y n + k 2 2 ) k 4 =hf(x n + h, y n + k 3 ) y n+1 = y n + k 1 6 + k 2 3 +...
Ngày tải lên: 15/12/2013, 04:15
... equation (19 .2. 4) is ξ =1− 4D∆t (∆x) 2 sin 2 k∆x 2 (19 .2. 5) The requirement |ξ|≤1leads to the stability criterion 2D∆t (∆x) 2 ≤ 1(19 .2. 6) 848 Chapter 19. Partial Differential Equations Sample ... (19 .2. 19) write D j+1 /2 = 1 2 D(u n j+1 )+D(u n j ) (19 .2. 22) Implicit schemes are not as easy. The replacement (19 .2. 22) with n → n +1leaves us with a nasty se...
Ngày tải lên: 15/12/2013, 04:15
Tài liệu Partial Differential Equations part 4 ppt
... ∆t /2. In each substep, a different dimension is treated implicitly: u n+1 /2 j,l = u n j,l + 1 2 α δ 2 x u n+1 /2 j,l + δ 2 y u n j,l u n+1 j,l = u n+1 /2 j,l + 1 2 α δ 2 x u n+1 /2 j,l + δ 2 y u n+1 j,l (19.3.16) The ... us u n+1 j,l = u n j,l + 1 2 α δ 2 x u n+1 j,l + δ 2 x u n j,l + δ 2 y u n+1 j,l + δ 2 y u n j,l (19.3.13) Here α ≡ D∆t ∆ 2 ∆ ≡ ∆x...
Ngày tải lên: 15/12/2013, 04:15
Tài liệu Partial Differential Equations part 5 ppt
... then adding the three equations, we get u j 2 + T (1) · u j + u j +2 = g (1) j ∆ 2 (19.4. 32) This is an equation of the same form as (19.4 .29 ), with T (1) =21 −T 2 g (1) j =∆ 2 (g j−1 −T·g j +g j+1 ) (19.4.33) After ... Similarly, ρ jl = 1 JL J−1 m=0 L−1 n=0 ρ mn e 2 ijm/J e 2 iln/L (19.4.3) If we substitute expressions (19.4 .2) and (19.4.3) in our model problem (19.0.6...
Ngày tải lên: 15/12/2013, 04:15
Tài liệu Partial Differential Equations part 6 doc
... turns out to be ρ s 1 − π 2 2J 2 (19.5.11) The number of iterations r required to reduce the error by a factor of 10 −p is thus r 2pJ 2 ln 10 π 2 1 2 pJ 2 (19.5. 12) In other words, the number ... (19.5.31) implicitly in two half-steps: u n+1 /2 − u n ∆t /2 = − L x u n+1 /2 + L y u n ∆ 2 − ρ u n+1 − u n+1 /2 ∆t /2 = − L x u n+1 /2 + L y u n+1 ∆ 2 − ρ (19.5.35) (cf...
Ngày tải lên: 15/12/2013, 04:15
Tài liệu Partial Differential Equations part 7 doc
... (19.6. 32) gives τ h L H (u+h 2 u 2 )−L h (u+h 2 u 2 ) =L H (u)−L h (u)+h 2 [L H (u 2 )−L h (u 2 )] + ··· =(H 2 −h 2 )τ 2 +O(h 4 ) (19.6.38) For the usual case of H =2hwe therefore have τ 1 3 τ h 1 3 τ h (19.6.39) The ... execution time: 8 82 Chapter 19. Partial Differential Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING...
Ngày tải lên: 24/12/2013, 12:16
Tài liệu Integration of Ordinary Differential Equations part 1 doc
... 16.6 of this chapter treats the subject of stiff equations, relevant both to ordinary differential equations and also to partial differential equations (Chapter 19). ... involving ordinary differential equations (ODEs) can always be reduced to the study of sets of first-order differential equations. For example the second-order equation d 2 y dx 2 + q(x) dy dx = ... gene...
Ngày tải lên: 15/12/2013, 04:15
Tài liệu Solution of Linear Algebraic Equations part 2 ppt
... equation a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 · x 11 x 21 x 31 x 41 x 12 x 22 x 32 x 42 x 13 x 23 x 33 x 43 y 11 y 12 y 13 y 14 y 21 y 22 y 23 y 24 y 31 y 32 y 33 y 34 y 41 y 42 y 43 y 44 = b 11 b 21 b 31 b 41 b 12 b 22 b 32 b 42...
Ngày tải lên: 15/12/2013, 04:15