computational fluid dynamics, second edition
... Flows vii 10 8 10 8 11 2 11 5 11 5 11 8 11 9 12 0 12 1 12 1 12 3 12 3 12 9 13 0 13 0 13 2 13 4 13 6 13 8 13 8 13 9 14 0 14 1 14 2 14 2 14 5 14 8 15 0 16 3 16 5 16 6 16 7 16 9 17 5 17 8 17 8 17 9 18 0 18 0 18 3 18 5 18 7 18 8 19 1 viii CONTENTS ... 592 597 600 6 01 606 607 607 608 609 609 610 611 613 614 615 615 617 617 CONTENTS 19 .1. 1 Control Function Methods 19 .1. 1 .1 Basic Theory 19 .1. 1.2 Weight Functions in One Dimension 19 .1. 1.3 Weight ... node of element and local node of element ⎤ ⎡ ⎤ ⎡ (1) (1) ⎡ ⎤ K 11 K 12 K 11 K12 K13 1 ⎥ ⎢ ⎥ ⎢ (1) (2) K␣ = ⎣ K 21 K22 K23 ⎦ = ⎢ K 21 K (1) + K 11 K(2) ⎥ = ⎣ 1 1 ⎦ 22 ⎣ 12 ⎦ h 1 (2) (2) K 31 K32...
Ngày tải lên: 01/04/2014, 10:45
... )d 1 d⇠2 ]( 1 + i 1 , ⇠2 )d 1 d⇠2 c eix1 1 G0 ( 1 , ⇠2 )d 1 d⇠2 RR ix ⇠ c c eix1 1 ( 1 , ⇠2 )G0 ( 1 , ⇠2 )d 1 d⇠2 + e 1 ( 1 , ⇠2 )G0 ( 1 , ⇠2 )d 1 d⇠2 RR ix ⇠ c + e 1 (1 )( 1 , ⇠2 )G ( 1 , ... (t +1) : < : (1 + CN k (1+ t) CN k t ·(t +1) 1 CN |x|2 ) N +2 t +1 p 21 t (1 + k t · (1+ t) |x| |x|2 ) N +2 t +1 t |x|2 |x| t |x| t If t 1, |w ⇤ (t + 1) k (1 + |x|2 ) t +1 N |, |wt ⇤ (t + 1) 31 k (1 ... t | R1 If |x|2 + t, ZZ sin (x 1 ) R || 1+ ⇠02 1 e t eix·⇠ |⇠| R = e t R1 ei|x| 1 d 1 1 R 02 1+ ⇠2 t Ce t (1 + |x|2 N |x|2 ) (1 + ) 1+ t 1+ t Ce t (1 + If |x|2 ZZ d⇠2 ( substitute ⇠2 1 = ⇠2...
Ngày tải lên: 10/09/2015, 09:06
... = b 11 b 21 b 31 b 41 b12 b22 b32 b42 x12 x22 x32 x42 x13 x23 x33 x43 b13 b23 b33 b43 0 0 y 11 y 21 y 31 y 41 0 0 y12 y22 y32 y42 y13 y23 y33 y43 y14 y24 ... C1 · C 1 ·x= b A · C1 · C2 · C 1 · C 1 · x = b (A · C1 · C2 · C3 · · ·) · · · C 1 · C 1 · C 1 ·x= b (1) · · · C 1 · C 1 · C 1 · x = b (2 .1. 7) Sample page from NUMERICAL RECIPES IN C: THE ART OF ... a 11 a 21 a 31 a 41 38 Chapter Solution of Linear Algebraic Equations Pivoting Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92...
Ngày tải lên: 15/12/2013, 04:15
Đề tài " Ergodicity of the 2D Navier-Stokes equations with degenerate stochastic forcing " pdf
... { (1, 0), ( 1, 0), (1, 1) , ( 1, 1) } satisfies the assumptions of Theorem 2 .1 Therefore, (2 .1) with noise given by QW (t, x) = W1 (t) sin x1 + W2 (t) cos x1 + W3 (t) sin(x1 + x2 ) +W4 (t) cos(x1 ... of the viscosity ν > Example 2.6 Take Z0 = { (1, 0), ( 1, 0), (0, 1) , (0, 1) } whose elements are of length Therefore, (2 .1) with noise given by (2.6) QW (t, x) = W1 (t) sin x1 + W2 (t) cos x1 ... Proposition 4 .13 Define Cn = ρn +1 10 , ρn 10 ρn p 10 21 ERGODICITY OF THE 2D NAVIER-STOKES EQUATIONS with the convention that Cn = if ρn = Note that since ρ0 = 1, one has ρN 10 = N 1 Cn We begin...
Ngày tải lên: 29/03/2014, 07:20
Báo cáo hóa học: " Global behavior of 1D compressible isentropic Navier-Stokes equations with a non-autonomous external force" docx
... Problems 2 011 , 2 011 :43 http://www.boundaryvalueproblems.com/content/2 011 /1/ 43 Page 11 of 19 Proof Differentiating (1. 9) with respect to x, exploiting (1. 8), we have df dx 2 = −γ (γ − 1) ρ γ −2 ρx ... http://www.boundaryvalueproblems.com/content/2 011 /1/ 43 Page of 19 Hence, 1 t (ρ θ )2n dx ≤ C1(T) + C1(T) x (ρ γ )2n dxds x (2 :18 ) Using the Gronwall inequality to (2 .18 ), we obtain (2 .13 ) The proof of (2 .14 ) can be found ... Problems 2 011 , 2 011 :43 http://www.boundaryvalueproblems.com/content/2 011 /1/ 43 Page of 19 The last term on the right-hand side of (2. 21) can be estimated as follows, using (1. 8), conditions (1. 10) and...
Ngày tải lên: 20/06/2014, 22:20
Báo cáo hóa học: " Research Article Regularity Criterion for Weak Solutions to the Navier-Stokes Equations in Terms of the Gradient of the Pressure" docx
... them into 1. 6 , we find Using 1. 11 for p that θ 1 θ /2n Lr u 1 θ 1/ 2 1/ 22 ··· 1/ 2n BMO θ 1 θ /2n Lr u 1 1/ 2n 1 θ BMO C u Lq ≤C u C u u r/q Lr u 1. 20 1 r/q BMO , which proves 1. 18 Remark 1. 5 From ... Remark 1. 4 Inequality 1. 11 plays an important role in our proof Chen and Zhu 17 extended 1. 11 to the following inequality: u Lq ≤C u r/q Lr u 1 r/q BMO , ≤ r < q < ∞, 1. 18 and used 1. 18 to obtain 1. 12 ... < q < ∞, 1. 18 and used 1. 18 to obtain 1. 12 Kozono and Wadade 18 give another proof of 1. 18 Here, we give an elementary and short proof of 1. 18 by 1. 11 For given ≤ r < q < ∞, there exists a...
Ngày tải lên: 22/06/2014, 02:20
Module 2: Overview of System Development
... XP The Microsoft strategy is to offer a broad range of Windows-based, embedded operating-system solutions to meet the diverse needs of customers Ultimately, the design requirements of your device ... installed Included as part of the Smart Device Extensions, is a set of pre-built device profiles An embedded device profile contains information necessary to build specific types of applications that ... agreement from Microsoft, the furnishing of this document does not give you any license to these patents, trademarks, copyrights, or other intellectual property 20 01 Microsoft Corporation All...
Ngày tải lên: 18/10/2013, 17:15
Tài liệu Solution of Linear Algebraic Equations part 1 docx
... RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University Press.Programs Copyright (C) 19 88 -19 92 by Numerical Recipes Software Permission ... right-hand sides, the b’s, are changed (§2 .1 §2 .10 ) • Calculation of the matrix A 1 which is the matrix inverse of a square matrix A, i.e., A · A 1 = A 1 · A = 1, where is the identity matrix (all ... ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University Press.Programs Copyright (C) 19 88 -19 92 by Numerical Recipes Software Permission is granted for internet...
Ngày tải lên: 15/12/2013, 04:15
Tài liệu Solution of Linear Algebraic Equations part 11 ppt
... multiplication of two × matrices, a 11 a 21 a12 a22 · b 11 b 21 b12 b22 = c 11 c 21 c12 c22 (2 .11 .1) Eight, right? Here they are written explicitly: c 11 = a 11 × b 11 + a12 × b 21 c12 = a 11 × b12 + a12 × b22 c 21 = ... of formulas was, in fact, discovered by Strassen [1] The formulas are: Q1 ≡ (a 11 + a22 ) × (b 11 + b22 ) Q2 ≡ (a 21 + a22 ) × b 11 Q3 ≡ a 11 × (b12 − b22 ) Q4 ≡ a22 × (−b 11 + b 21 ) Q5 ≡ (a 11 + a12 ... ≡ (−a 11 + a 21 ) × (b 11 + b12 ) Q7 ≡ (a12 − a22 ) × (b 21 + b22 ) (2 .11 .3) Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92...
Ngày tải lên: 15/12/2013, 04:15
Tài liệu Root Finding and Nonlinear Sets of Equations part 2 docx
... zbrac"); f1=(*func)(*x1); f2=(*func)(*x2); for (j =1; j
Ngày tải lên: 15/12/2013, 04:15
Tài liệu Solution of Linear Algebraic Equations part 3 pdf
... case of a × matrix A, for example, equation (2.3 .1) would look like this: 11 α 21 α 31 α 41 α22 α32 α42 0 α33 α43 11 · 0 α44 12 β22 0 13 β23 β33 14 β24 β34 β44 a 11 a = 21 ... xi = βij xj yi − i = N − 1, N − 2, , βii j=i +1 Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University ... 13 β23 β33 14 β24 β34 β44 a 11 a = 21 a 31 a 41 a12 a22 a32 a42 a13 a23 a33 a43 a14 a24 a34 a44 (2.3.2) We can use a decomposition such as (2.3 .1) to solve the linear set A · x = (L · U) ·...
Ngày tải lên: 24/12/2013, 12:16
Tài liệu Solution of Linear Algebraic Equations part 4 docx
... (equation 2.3 .11 ) are not stored at all.] In brief, Crout’s method fills in the combined matrix of α’s and β’s, 11 α 21 α 31 α 41 12 β22 α32 α42 13 β23 β33 α43 14 β24 β34 β44 (2.3 .14 ) by ... set of equations to be solved is u1 r1 b c1 · · · u2 r2 a2 b c2 · · · ··· · · · · = · · · (2.4 .1) uN 1 rN 1 · · · aN 1 bN 1 cN 1 ··· ... i = j of equation (2.3 .12 ) and i = j +1 N sum=a[i][j]; of equation (2.3 .13 ) for (k =1; k
Ngày tải lên: 24/12/2013, 12:16
Tài liệu Solution of Linear Algebraic Equations part 5 docx
... elements are in a [1 n][m1 +1] Subdiagonal elements are in a[j n] [1 m1] (with j > appropriate to the number of elements on each subdiagonal) Superdiagonal elements are in a [1 j ][m1+2 m1+m2 +1] with j < ... RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University Press.Programs Copyright (C) 19 88 -19 92 by Numerical Recipes Software Permission ... appropriate to the number of elements on each superdiagonal { unsigned long i,j,k,tmploop; for (i =1; i
Ngày tải lên: 24/12/2013, 12:16
Tài liệu Solution of Linear Algebraic Equations part 12 pdf
... and 2.7.25): R1 = Inverse(a 11 ) R2 = a 21 × R1 R3 = R1 × a12 R4 = a 21 × R3 R5 = R4 − a22 R6 = Inverse(R5 ) c12 = R3 × R6 c 21 = R6 × R2 R7 = R3 × c 21 c 11 = R1 − R7 c22 = −R6 (2 .11 .6) Sample page ... also obtained a surprising result for matrix inversion [1] Suppose that the matrices a 11 a 21 a12 a22 and c 11 c 21 c12 c22 (2 .11 .5) are inverses of each other Then the c’s can be obtained from the ... 2 .11 Is Matrix Inversion an N Process? 10 3 in terms of which c 11 = Q1 + Q4 − Q5 + Q7 c 21 = Q2 + Q4 (2 .11 .4) c12 = Q3 + Q5 What’s the use of this? There is one fewer...
Ngày tải lên: 24/12/2013, 12:16
Tài liệu Solution of Linear Algebraic Equations part 6 pptx
... manipulation: A 1 = A 1 · (B 1 · B0 ) = (A 1 · B 1 ) · B0 = (B0 · A) 1 · B0 0 = (1 − R) 1 · B0 = (1 + R + R2 + R3 + · · ·) · B0 (2.5.7) We can define the nth partial sum of the last expression by Bn ≡ (1 + ... x [1 n] of the linear set of equations A · X = B The matrix a [1 n] [1 n], and the vectors b [1 n] and x [1 n] are input, as is the dimension n Also input is alud [1 n] [1 n], the LU decomposition of ... 1 n≤N (2.6.2) Vjk Vjn = δkn 1 k≤N 1 n≤N (2.6.3) i =1 N j =1 Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University...
Ngày tải lên: 24/12/2013, 12:16
Tài liệu Solution of Linear Algebraic Equations part 7 docx
... · W · VT ) · (V · W 1 · UT · b) − b + b = (U · W · W 1 · UT − 1) · b + b = U · (W · W 1 − 1) · UT · b + UT · b (2.6 .10 ) = (W · W 1 − 1) · UT · b + UT · b Now, (W · W 1 − 1) is a diagonal matrix ... RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University Press.Programs Copyright (C) 19 88 -19 92 by Numerical Recipes Software Permission ... RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University Press.Programs Copyright (C) 19 88 -19 92 by Numerical Recipes Software Permission...
Ngày tải lên: 24/12/2013, 12:16
Tài liệu Solution of Linear Algebraic Equations part 8 docx
... v) 1 , and is derived briefly as follows: (A + u ⊗ v) 1 = (1 + A 1 · u ⊗ v) 1 · A 1 = (1 − A 1 · u ⊗ v + A 1 · u ⊗ v · A 1 · u ⊗ v − ) · A 1 = A 1 − A 1 · u ⊗ v · A 1 (1 − λ + λ2 − ) = A 1 ... U · VT ) 1 = A 1 − A 1 · U · (1 + VT · A 1 · U) 1 · VT · A 1 (2.7 .12 ) Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge ... = −(S 1 · R) · (P − Q · S 1 · R) 1 (2.7.24) S = S 1 + (S 1 · R) · (P − Q · S 1 · R) 1 · (Q · S 1 ) or else by the equivalent formulas P = P 1 + (P 1 · Q) · (S − R · P 1 · Q) 1 · (R · P 1 ) Q...
Ngày tải lên: 24/12/2013, 12:16
Tài liệu Solution of Linear Algebraic Equations part 9 docx
... (i = 1, , M ) (2.8 .10 ) j =1 (M ) M (M ) Ri−j xj = yi i = 1, , M (2.8 .11 ) j =1 becomes M (M +1) Ri−j xj (M +1) + Ri−(M +1) xM +1 = yi i = 1, , M + (2.8 .12 ) i = 1, , M (2.8 .13 ) j =1 By ... +1) = Gj (M +1) = Hj Gj Hj (M ) − GM +1 HM +1 j (M +1) (M ) − HM +1 GM +1 j (M +1) (M ) (M ) (2.8.25) Now, starting with the initial values (1) x1 = y1 /R0 (1) G1 = R 1 /R0 (1) H1 = R1 /R0 (2.8.26) ... GM +1 j = xj we can substitute the previous order (M +1) − xj (M +1) (2.8 .18 ) xM +1 The result of this operation is (M +1) xM +1 = (M ) M − yM +1 j =1 RM +1 j xj (M ) M RM +1 j GM +1 j − R0 j=1...
Ngày tải lên: 21/01/2014, 18:20
Tài liệu Solution of Linear Algebraic Equations part 10 docx
... (i = 1, , M ) (2.8 .10 ) j =1 (M ) M (M ) Ri−j xj = yi i = 1, , M (2.8 .11 ) j =1 becomes M (M +1) Ri−j xj (M +1) + Ri−(M +1) xM +1 = yi i = 1, , M + (2.8 .12 ) i = 1, , M (2.8 .13 ) j =1 By ... +1) = Gj (M +1) = Hj Gj Hj (M ) − GM +1 HM +1 j (M +1) (M ) − HM +1 GM +1 j (M +1) (M ) (M ) (2.8.25) Now, starting with the initial values (1) x1 = y1 /R0 (1) G1 = R 1 /R0 (1) H1 = R1 /R0 (2.8.26) ... GM +1 j = xj we can substitute the previous order (M +1) − xj (M +1) (2.8 .18 ) xM +1 The result of this operation is (M +1) xM +1 = (M ) M − yM +1 j =1 RM +1 j xj (M ) M RM +1 j GM +1 j − R0 j=1...
Ngày tải lên: 21/01/2014, 18:20
Tài liệu Numerical Solution of Stochastic Differential Equations with Jumps in Finance pdf
... §8.3 → §8.4 → §8.5 → §8.6 → §8.8 ↓ 11 .1 → 11 .2 → 11 .3 → 11 .4 → 11 .5 ↓ 12 .1 → 12 .2 ↓ 13 .1 → 13 .2 → 13 .3 → 13 .4 → 13 .7 ↓ Chapter 14 ↓ 16 .1 XVI Suggestions for the Reader (ii) Engineers, ... 1. 1 → 1. 2 → 1. 3 → 1. 4 → 1. 5 → 1. 6 → 1. 7 → 1. 8 ↓ §5 .1 → §5.2 → §5.3 → §5.4 ↓ §6 .1 → §6.2 → §6.3 ↓ §7 .1 → §7.2 → §7.3 → §7.4 ↓ §8 .1 → §8.2 → §8.3 → §8.4 → §8.5 → §8.6 → §8.8 ↓ 11 .1 → 11 .2 ... Chapter ↓ §6 .1 → §6.2 → §6.3 ↓ §7 .1 → §7.2 → §7.3 → §7.4 ↓ §8 .1 → §8.2 → §8.3 → §8.4 → §8.5 → §8.6 → §8.8 → §8.9 ↓ Chapter 11 ↓ 12 .1 → 12 .2 ↓ 13 .1 → 13 .2 → 13 .3 → 13 .4 → 13 .7 ↓ Chapter 14 ↓ Chapter...
Ngày tải lên: 19/02/2014, 22:20