2 solution of navier stokes system of equations using fvm fdm

... Processing 20 .4 .2 Solution of Navier- Stokes System of Equations with Multithreading 20 .5 Summary References xiii 617 617 619 621 622 622 623 627 627 628 628 630 634 639 640 644 645 650 6 52 6 52 654 ... 7.3 .2 Three-Dimensional FVM Equations 7.4 FVM- FDV Formulation 7.5 Example Problems 7.6 Summary References 193 195 195 196 197 197 197 197 20 4 20 4 20 5 20 7 20 7 20 8 21 3 21 4 21 8 21 8 21 9 21 9 22 3 22 5 22 7 ... with FDM equations at nodes and given by u3 − 2u2 + u1 − 2u2 = f2 (1/3 )2 u4 − 2u3 + u2 − 2u3 = f3 (1/3 )2 Thus we obtain 9(u3 − 2u2 + 0) − 2u2 = − 38 9(u3 − − 2u3 + u2 ) − 2u3 = − 32 or 20 9...

Ngày tải lên: 01/04/2014, 10:45

... Introduction M2 = G(x1 , x2 z2 , t s)· ·@ 0 GS (0, z2 , y1 , y2 , s) 21 GS (0, z2 , y1 , y2 , s) 21 ⇤ (z2 ,s) ⇤ (z2 ,s) O1 GS (0, z2 , y1 , y2 , s) 22 O2 GS (0, z2 , y1 , y2 , s) 22 ⇤ (z2 ,s) ⇤ (z2 ,s) ... p2 + y2O(x,t) k p2 e k RR |y |2 Ct t 4⇡t (t+1) y2O(x,t) (t+1) k e p2 t k |y |2 Ct y2O(x,t) re |y |2 Ct ·(y x) p2 t dy dy 2e |y |2 2Ct ·(y x) (y1 x1 )2 (y2 x2 )2 (y1 x1 )2 (y2 x2 )2 RR dy (y1 x1 )2 ... (0, z2 , y1 , y2 , s) 23 O2 GS (0, z2 , y1 , y2 , s) 23 (z2 ,s) ⇤ O1 ⇤ O2 (z2 ,s) A @x1 G(x1 , x2 z2 , t s) M3 = 0 B C · B GS (0, z2 , y1 , y2 , s) GS (0, z2 , y1 , y2 , s) GS (0, z2 , y1 , y2 ,...

Ngày tải lên: 10/09/2015, 09:06

... b11  b21  b31 b41   b 12  b 22  b 32 b 42    x 12  x 22  x 32 x 42  x13  x23  x33 x43  b13  b23  b33 b43  0 0 y11  y21 y31 y41 0 0 y 12 y 22 y 32 y 42 y13 y23 y33 y43  y14 y24  y34 ... equation that looks like this (in the case of a single right-hand side vector):       x1 b1 a11 a 12 a13 a14  a 22 a23 a24   x2   b2  (2. 2.1) ·  =    0 a33 a34 x3 b3 0 a44 x4 ... 37 2. 1 Gauss-Jordan Elimination Elimination on Column-Augmented Matrices  Consider the linear matrix equation      a 12 a 22 a 32 a 42 a13 a23 a33 a43 a14 x11 a24   x21  · a34...

Ngày tải lên: 15/12/2013, 04:15

... basis of H We write Z2 \ {(0, 0)} = Z2 ∪ Z2 , + − where Z2 = (k1 , k2 ) ∈ Z2 | k2 > ∪ (k1 , 0) ∈ Z2 | k1 > , + Z2 = (k1 , k2 ) ∈ Z2 | − k ∈ Z2 , − + ERGODICITY OF THE 2D NAVIER- STOKES EQUATIONS ... Ergodicity of the 2D Navier- Stokes equations with random forcing, Comm Math Phys 22 4 (20 01), 65–81 [BKL 02] ——— , Exponential mixing of the 2D stochastic Navier- Stokes dynamics, Comm Math Phys 23 0 (20 02) , ... sin x2 , and cos x2 Example 2. 7 Take Z0 = { (2, 0), ( 2, 0), (2, 2) , ( 2, 2) }, which corresponds to case of Theorem 2. 3 with G generated by (0, 2) and (2, 0) In ˜ this case, H is the set of functions...

Ngày tải lên: 29/03/2014, 07:20

... (2. 10) yields t u2n dx + n(2n − 1) u2n 2 ρ 1+θ u2 dxds x 0 t t f 2n dxds + (2n − 1) ≤ C1 + 0 u2n dxds 0 t u2n 2 ρ 2 −1−θ dxds +n(2n − 1) (2: 11) t ≤ C1 (T) + n(2n − 1) 0 t t (2 −1−θ )n n − 2n ... dx + 2n(2n − 1) u2n 2 ρ 1+θ u2 dxds x 2n 0 t 0 (2: 10) u2n 2 ρ γ ux dxds + 2n u2n dx + 2n(2n − 1) = t f u2n−1 dxds 0 Applying the Young inequality and condition f(r(x, ·), ·) Î L2n([0,T],L2n[0,1]) ... | |2 + ||utx | |2 + ||uttx | |2 + || H H H ≤ C2 (T) ||ux | |2 + ||ρx | |2 + ||utx | |2 + ||uttx | |2 H H H ∂2f || ∂x∂t (4 :27 ) +||frr | |2 + ||fr | |2 ||ux | |2 + ||frt | |2 L Inserting (4 .27 ) into (4 .26 ),...

Ngày tải lên: 20/06/2014, 22:20

... |∇u |2 |u |2 dx |∇|u |2 |2 dx − ∇π · |u |2 u dx ≤ ∇π ≤ C ∇π ≤ C u∇u ≤ 1 /2 L2 1 /2 L2 1 /2 BMO ∇π ∇π |u · ∇u |2 L C ∇π u L4 L4 u 1 /2 BMO L4 u 2. 2 L4 2/ 3 BMO u , L4 which yields u L4 ≤ u0 T L4 exp C ∇π 2/ 3 ... regularity of weak solutions of the Navier- Stokes system, ” Journal of Differential Equations, vol 62, no 2, pp 186 21 2, 1986 M Struwe, “On partial regularity results for the Navier- Stokes equations, ” ... “L3,∞ -solutions of Navier- Stokes equations and e backward uniqueness,” Russian Mathematical Surveys, vol 58, no 2, pp 21 1 25 0, 20 03 H Beir˜ o da Veiga, “A new regularity class for the Navier- Stokes...

Ngày tải lên: 22/06/2014, 02:20

... Windows 20 00/XP Platform Builder Windows 20 00/XP Target Designer Pentium class x86 Same as XP The Microsoft strategy is to offer a broad range of Windows-based, embedded operating -system solutions ... with a minimum of rewriting 22 Module 2: Overview of System Development The Managed Application Model (continued) Topic Objective NET Compact Framework (CF) To introduce the features of the NET ... trademarks of their respective owners Module 2: Overview of System Development iii Instructor Notes Presentation: 60 Minutes This module provides students with an overview of the system development...

Ngày tải lên: 18/10/2013, 17:15

... positive-definite ( 2. 9), tridiagonal ( 2. 4), band diagonal ( 2. 4), Toeplitz ( 2. 8), Vandermonde ( 2. 8), sparse ( 2. 7) • Strassen’s “fast matrix inversion” ( 2. 11) 36 Chapter Solution of Linear Algebraic Equations ... set of equations to be solved can be written as the N ×N set of equations (AT · A) · x = (AT · b) (2. 0.4) where AT denotes the transpose of the matrix A Equations (2. 0.4) are called the normal equations ... nullspace of the matrix A) The task of finding the solution space of A involves • Singular value decomposition of a matrix A This subject is treated in 2. 6 In the opposite case there are more equations...

Ngày tải lên: 15/12/2013, 04:15

... b11 b21 b 12 b 22 = c11 c21 c 12 c 22 (2. 11.1) Eight, right? Here they are written explicitly: c11 = a11 × b11 + a 12 × b21 c 12 = a11 × b 12 + a 12 × b 22 c21 = a21 × b11 + a 22 × b21 (2. 11 .2) c 22 = a21 ... + a 22 ) × (b11 + b 22 ) Q2 ≡ (a21 + a 22 ) × b11 Q3 ≡ a11 × (b 12 − b 22 ) Q4 ≡ a 22 × (−b11 + b21 ) Q5 ≡ (a11 + a 12 ) × b 22 Q6 ≡ (−a11 + a21 ) × (b11 + b 12 ) Q7 ≡ (a 12 − a 22 ) × (b21 + b 22 ) (2. 11.3) ... Computations, 2nd ed (Baltimore: Johns Hopkins University Press), §§5 .2, 5.3, 12. 6 [2] 2. 11 Is Matrix Inversion an N Process? We close this chapter with a little entertainment, a bit of algorithmic...

Ngày tải lên: 15/12/2013, 04:15

... 1-800-8 72- 7 423 (North America only),or send email to trade@cup.cam.ac.uk (outside North America) a x2 x3 b x1 d c f e 351 9.1 Bracketing and Bisection 3 52 Chapter Root Finding and Nonlinear Sets of Equations ... == *x2) nrerror("Bad initial range in zbrac"); f1=(*func)(*x1); f2=(*func)(*x2); for (j=1;j

Ngày tải lên: 15/12/2013, 04:15

... case of a × matrix A, for example, equation (2. 3.1) would look like this:       α11  21 α31 α41 22 α 32 α 42 0 α33 α43 β11   · 0 α44 β 12 22 0 β13 23 β33 β14 24  β34 β44 a11 a =  21 ...  21 a31 a41 a 12 a 22 a 32 a 42 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) · x = L · (U · x) = b (2. 3.3) by first ... aii i (2. 2.3) (2. 2.4) j=i+1 The procedure defined by equation (2. 2.4) is called backsubstitution The combination of Gaussian elimination and backsubstitution yields a solution to the set of equations...

Ngày tải lên: 24/12/2013, 12:16

... (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,  β11  21  α31 α41 β 12 22 α 32 α 42 β13 23 β33 α43  β14 24   β34 β44 (2. 3.14) ... αi2 β2j + · · · + αij βjj = aij (2. 3.8) (2. 3.9) (2. 3.10) Equations (2. 3.8)– (2. 3.10) total N equations for the N + N unknown α’s and β’s (the diagonal being represented twice) Since the number of ... = 1, 2, 3, , N these two procedures: First, for i = 1, 2, , j, use (2. 3.8), (2. 3.9), and (2. 3.11) to solve for βij , namely i−1 βij = aij − αik βkj (2. 3. 12) k=1 (When i = in 2. 3. 12 the...

Ngày tải lên: 24/12/2013, 12:16

... δb (2. 5 .2) Subtracting (2. 5.1) from (2. 5 .2) gives A · δx = δb (2. 5.3) Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -43108-5) Copyright (C) 1988-19 92 by ... improved solution x 2. 5 Iterative Improvement of a Solution to Linear Equations Obviously it is not easy to obtain greater precision for the solution of a linear set than the precision of your ... routine whose analog in 2. 3 is lubksb 55 2. 5 Iterative Improvement of a Solution to Linear Equations A δx x+ b δx δb A−1 Figure 2. 5.1 Iterative improvement of the solution to A · x = b The...

Ngày tải lên: 24/12/2013, 12:16

... inverses of each other Then the c’s can be obtained from the a’s by the following operations (compare equations 2. 7 .22 and 2. 7 .25 ): R1 = Inverse(a11 ) R2 = a21 × R1 R3 = R1 × a 12 R4 = a21 × R3 ... R4 − a 22 R6 = Inverse(R5 ) c 12 = R3 × R6 c21 = R6 × R2 R7 = R3 × c21 c11 = R1 − R7 c 22 = −R6 (2. 11.6) Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -43108-5) ... to N log2 With this “fast” matrix multiplication, Strassen also obtained a surprising result for matrix inversion [1] Suppose that the matrices a11 a21 a 12 a 22 and c11 c21 c 12 c 22 (2. 11.5) are...

Ngày tải lên: 24/12/2013, 12:16

... 56 Chapter Solution of Linear Algebraic Equations But (2. 5 .2) can also be solved, trivially, for δb Substituting this into (2. 5.3) gives A · δx = A · (x + δx) − b (2. 5.4) #include "nrutil.h" ... roundoff error, then after one application of equation (2. 5.10) (that is, going from x0 ≡ B0 · b to x1 ) the first neglected term, of order R2 , will be smaller than the roundoff error Equation (2. 5.10), ... matrix Define the residual matrix R of B0 as 58 Chapter Solution of Linear Algebraic Equations We can define the norm of a matrix as the largest amplification of length that it is able to induce...

Ngày tải lên: 24/12/2013, 12:16

... solutions of A⋅x = d solutions of A ⋅ x = c′ SVD solution of A ⋅ x = c range of A d c′ c SVD solution of A⋅x = d (b) Figure 2. 6.1 (a) A nonsingular matrix A maps a vector space into one of the ... of the elements wj From (2. 6.1) it now follows immediately that the inverse of A is 62 Chapter Solution of Linear Algebraic Equations If we want to single out one particular member of this solution- set ... same permutation of the columns of U, elements of W, and columns of V (or rows of VT ), or (ii) forming linear combinations of any columns of U and V whose corresponding elements of W happen to...

Ngày tải lên: 24/12/2013, 12:16

... j,jl,jm,jp,ju,k,m,n2,noff,inc,iv; float v; n2=ija[1]; Linear size of matrix plus for (j=1;j

Ngày tải lên: 24/12/2013, 12:16

... equation (2. 9 .2) in components, one readily obtains the analogs of equations (2. 3. 12) – (2. 3.13), 1 /2 i−1 Lii = aii − L2 ik (2. 9.4) k=1 and Lji = Lii i−1 aij − Lik Ljk k=1 j = i + 1, i + 2, , N (2. 9.5) ... ) (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 ) we can recurse away At each stage M we use equations (2. 8 .23 ) and (2. 8 .24 ) to find ... (2. 8.4) k=1 But (2. 8.4) says that Ajk is exactly the inverse of the matrix of components xk−1 , which i appears in (2. 8 .2) , with the subscript as the column index Therefore the solution of (2. 8 .2) ...

Ngày tải lên: 21/01/2014, 18:20

... equation (2. 9 .2) in components, one readily obtains the analogs of equations (2. 3. 12) – (2. 3.13), 1 /2 i−1 Lii = aii − L2 ik (2. 9.4) k=1 and Lji = Lii i−1 aij − Lik Ljk k=1 j = i + 1, i + 2, , N (2. 9.5) ... (2. 8.4) k=1 But (2. 8.4) says that Ajk is exactly the inverse of the matrix of components xk−1 , which i appears in (2. 8 .2) , with the subscript as the column index Therefore the solution of (2. 8 .2) ... ) (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 ) we can recurse away At each stage M we use equations (2. 8 .23 ) and (2. 8 .24 ) to find...

Ngày tải lên: 21/01/2014, 18:20

... (20 02) , Borodin & Salminen (20 02) , J¨ckel a (20 02) , Joshi (20 03), Sch¨nbucher (20 03), Shreve (20 03a, 20 03b), Cont & o Tankov (20 04), Glasserman (20 04), Higham (20 04), Milstein & Tretjakov (20 04), ... 23 3 23 3 24 5 25 2 26 6 27 1 Regular Strong Taylor Approximations with Jumps 27 3 6.1 Discrete-Time Approximation 27 3 6 .2 Strong Order 1.0 Taylor ... (1997), Duffie, Pan & Singleton (20 00), Kou (20 02) , Sch¨nbucher (20 03), o Glasserman & Kou (20 03), Cont & Tankov (20 04) and Geman & Roncoroni (20 06) The areas of application of SDEs with jumps go far...

Ngày tải lên: 19/02/2014, 22:20