3 time integration methods for tvd schemes

... 284 286 298 298 30 2 30 3 30 5 30 6 30 8 30 8 30 9 30 9 30 9 31 5 32 0 32 4 32 7 32 7 33 2 33 4 33 5 33 7 33 7 34 0 34 2 34 2 34 3 34 6 34 6 34 7 34 7 34 8 35 3 35 5 35 5 35 8 36 2 36 5 36 6 36 7 36 8 36 9 37 6 37 7 37 7 37 8 x CONTENTS ... References 38 0 38 0 38 1 38 4 39 1 39 6 39 9 39 9 39 9 402 402 404 12 Incompressible Viscous Flows via Finite Element Methods 12.1 Primitive Variable Methods 12.1.1 Mixed Methods 12.1.2 Penalty Methods 12.1 .3 ... ) = (1)(1)(−1) dx x=0 ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ −1 2 .33 3 −1.888 u1 ⎣−1.888 4.666 −1. 833 ⎦ ⎣u2 ⎦ = ⎣1. 833 ⎦ + ⎣ ⎦ 0.666 u3 −1. 833 2 .33 3 ⎡ ⎤ ⎡ ⎤ u1 ⎣u2 ⎦ = ⎣ ⎦ −1 u3 ∗ = (1) du Note that, although dx (0) =...

Ngày tải lên: 01/04/2014, 10:45

... (G( (3. 23) 3) )) 1, from Theorem 3. 12 to 3. 14, every term will be dominated by |W(x, t)|; if t  1, |F b (G 3) F (L(⇠, t) (⇠, t))|  Cp t and other terms are bounded by some constant number 3. 3 ... d⇠2 " " e B 4 By Lemma 3. 3, E(|⇠|2 ) cos (|⇠|A(|⇠|2 )t) is analytic for ⇠ R2 for ⇠  " Therefore the integrand inside the above integration is correspondingly analytic for ⇠ C2 With Cauchy’s ... Then for the short-wave component F (G· )(x, t), we have the following structure: Theorem 3. 13 For sufficient large R, if |x|  Mt, there exists some 35 Chapter Fundamental Solution constant C, for...

Ngày tải lên: 10/09/2015, 09:06

... b21  b31 b41   b12  b22  b32 b42    x12  x22  x32 x42  x 13  x 23  x 33 x 43  b 13  b 23  b 33 b 43  0 0 y11  y21 y31 y41 0 0 y12 y22 y32 y42 y 13 y 23 y 33 y 43  y14 y24  y34 y44 ... 37 2.1 Gauss-Jordan Elimination Elimination on Column-Augmented Matrices  Consider the linear matrix equation      a12 a22 a32 a42 a 13 a 23 a 33 a 43 a14 x11 a24   x21  · a34 x31 a44 ... vector):       x1 b1 a11 a12 a 13 a14  a22 a 23 a24   x2   b2  (2.2.1) ·  =    0 a 33 a34 x3 b3 0 a44 x4 b4 Here the primes signify that the a’s and b’s not have their original numerical...

Ngày tải lên: 15/12/2013, 04:15

... Phys 2 13 (2000), 291 33 0 [KS01] ——— , A coupling approach to randomly forced nonlinear PDE’s I, Comm Math Phys 221 (2001), 35 1 36 6 [Mat98] J C Mattingly The stochastically forced Navier-Stokes ... tn in (3. 9), we then get µ − ν dn ≤ − α for every n > N , and therefore µ − ν TV ≤ − α by 2 Corollary 3. 5, thus leading to a contradiction As an immediate corollary, we have Corollary 3. 17 If ... convenient for the modes which are directly forced In the case when all of the modes are forced, the choice ζt = (1 − t/T )J0,t ξ for t ∈ [0, T ] produces the well-known Bismut-Elworthy-Li formula...

Ngày tải lên: 29/03/2014, 07:20

... inequality to (2.18), we obtain (2. 13) The proof of (2.14) can be found in [3] , please refer to Lemma 2 .3 in [3] for detail Lemma 2 .3 Under the assumptions in Theorem 2.1, for any ≤ t ≤ T, we have the ... T], L2 [0, 1]) (3: 3) Therefore, the generic constant C2(T) depending only on the norm of initial data (r0, u0) in H2, the norms of f in the class of functions in (3. 2)- (3. 3) and time T Huang and ... 0t (3: 11) Inserting (3. 11) into (3. 9), by virtue of Lemmas 2.1-2 .3 and assumption (3. 1), we obtain (3. 6) We infer from (1.9), ut = −γ ρ γ −1 ρx + ρ θ +1 uxx + (θ + 1)ρ θ ρx ux + f (r, t) (3: 12)...

Ngày tải lên: 20/06/2014, 22:20

... ; Ls R3 with r 3 < 2, < s ≤ ∞, s 1.8 then u is smooth Berselli and Galdi 11 have extended the range of r and s to 2/r 3/ s and 3/ 2 < s ≤ ∞ When s ∞, Chen and Zhang 12 also see Fan et al 13 refined ... in L2 as t −→ 1. 13 The energy inequality is u t t L2 ∇u s ds L2 ≤ u0 for any t ∈ 0, T , L2 1.14 Our main result reads as follows Theorem 1.2 Let u0 ∈ L2 ∩ L4 R3 with div u0 in R3 Suppose that ... R3 , s ≥ 3; then there exists T and a unique classical solution u ∈ L∞ ∩ C 0, T ; Ls Moreover, let 0, T ∗ be the maximal interval such that u solves 1.1 – 1 .3 in C 0, T ∗ ; Ls , s > Then, for...

Ngày tải lên: 22/06/2014, 02:20

... project timeline Porting Complete Typical Project Timeline OS Customization on SDB OS Optimization on H/W Driver N Driver B Driver A Porting OAL Boot loader Time For Your Information For more information ... platform development cycle is to use Platform Builder to create a base configuration You can use the New Platform Wizard in Platform Builder to quickly and easily configure your platform Platform ... C++ for creating native applications Native applications must be rebuilt for each new CPU and each new platform Applications built for the desktop must be rebuilt for the target device Use NET for...

Ngày tải lên: 18/10/2013, 17:15

... instance For these reasons, Gauss-Jordan elimination should usually not be your method of first choice, either for solving linear equations or for matrix inversion The decomposition methods in §2 .3 are ... at the same time, and (ii) that when the inverse matrix is not desired, Gauss-Jordan is three times slower than the best alternative technique for solving a single linear set (§2 .3) The method’s ... need to resort to sophisticated methods even for the case of N = 10 (though rarely for N = 5) Singular value decomposition (§2.6) is a technique that can sometimes turn singular problems into...

Ngày tải lên: 15/12/2013, 04:15

... d[n]; for (i=n-1;i>=1;i ) { for (sum=0.0,j=i+1;j

Ngày tải lên: 15/12/2013, 04:15

... http://www.nr.com or call 1-800-872-74 23 (North America only),or send email to trade@cup.cam.ac.uk (outside North America) a x2 x3 b x1 d c f e 35 1 9.1 Bracketing and Bisection 35 2 Chapter Root Finding and ... 0-521- 431 08-5) Copyright (C) 1988-1992 by Cambridge University Press.Programs Copyright (C) 1988-1992 by Numerical Recipes Software Permission is granted for internet users to make one paper copy for ... 0-521- 431 08-5) Copyright (C) 1988-1992 by Cambridge University Press.Programs Copyright (C) 1988-1992 by Numerical Recipes Software Permission is granted for internet users to make one paper copy for...

Ngày tải lên: 15/12/2013, 04:15

... (2 .3. 1) would look like this:       α11  α21 31 α41 α22 32 α42 0 33 α 43 β11   · 0 α44 β12 β22 0 β 13 β 23 33 β14 β24  34 β44 a11 a =  21 a31 a41 a12 a22 a32 a42 a 13 a 23 a 33 a 43 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 solving for the vector y such that L·y=b (2 .3. 4) U·x=y ... Backsubstitution But how we solve for the x’s? The last x (x4 in this example) is already isolated, namely x4 = b4 /a44 (2.2.2) x3 = [b − x4 a34 ] a 33 and then proceed with the x before that one The typical...

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  α21  31 α41 β12 β22 32 α42 β 13 β 23 33 α 43  β14 β24   34 β44 (2 .3. 14) by ... 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 summation term is taken ... } big=0.0; Initialize for the search for largest pivot element for (i=j;i

Ngày tải lên: 24/12/2013, 12:16

... mm=m1+m2+1; l=m1; for (i=1;i

Ngày tải lên: 24/12/2013, 12:16

... Mathematik, vol 13, pp 35 4 35 6 [1] Kronsjo, L 1987, Algorithms: Their Complexity and Efficiency, 2nd ed (New York: Wiley) ¨ Winograd, S 1971, Linear Algebra and Its Applications, vol 4, pp 38 1 38 8 Pan, ... 2.7.22 and 2.7.25): R1 = Inverse(a11 ) R2 = a21 × R1 R3 = R1 × a12 R4 = a21 × R3 R5 = R4 − a22 R6 = Inverse(R5 ) c12 = R3 × R6 c21 = R6 × R2 R7 = R3 × c21 c11 = R1 − R7 c22 = −R6 (2.11.6) Sample page ... (2.11 .3) –(2.11.4)? Don’t they outweigh the advantage of the fewer multiplications? For large N , it turns out that there are six times as many additions as multiplications implied by (2.11 .3) –(2.11.4)...

Ngày tải lên: 24/12/2013, 12:16

... lubksb(alud,n,indx,r); Solve for the error term, for (i=1;i

Ngày tải lên: 24/12/2013, 12:16

... ++a[i][i]; } for (k=n;k>=1;k ) { Diagonalization of the bidiagonal form: Loop over for (its=1;its< =30 ;its++) { singular values, and over allowed iterations flag=1; for (l=k;l>=1;l ) { Test for splitting ... { g=1.0/g; for (j=l;j

Ngày tải lên: 24/12/2013, 12:16

... Gradient Method for a Sparse System So-called conjugate gradient methods provide a quite general means for solving the N × N linear system A·x =b (2.7.29) The attractiveness of these methods for large ... robust of these methods Note that equation (2.7 .38 ) gives Φ(x) = AT · (A · x − b) (2.7 .39 ) For any nonsingular matrix A, AT · A is symmetric and positive definite You might therefore be tempted ... (see [1] for details) The Sherman-Morrison and Woodbury formulas, discussed in §2.7, can sometimes be used to convert new special forms into old ones Reference [2] gives some other special forms...

Ngày tải lên: 24/12/2013, 12:16

... PrenticeHall), pp 163ff [5] Bunch, J.R 1985, SIAM Journal on Scientific and Statistical Computing, vol 6, pp 34 9 36 4 [6] de Hoog, F 1987, Linear Algebra and Its Applications, vol 88/89, pp 1 23 138 [7] 2.9 ... ···  RN −2 RN −1 R−1 R0 R1 R−2 R−1 R0 RN 3 RN −2 RN −4 RN 3 ··· ··· ··· ··· ··· ··· R−(N −2) R−(N 3) R−(N −4) R0 R1 R−(N −1)  R−(N −2)  R−(N 3)     R−1 R0 (2.8.8) The linear Toeplitz ... related to those of L by LT = Lji ij (2.9 .3) Writing out 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...

Ngày tải lên: 21/01/2014, 18:20

... (see [1] for details) The Sherman-Morrison and Woodbury formulas, discussed in §2.7, can sometimes be used to convert new special forms into old ones Reference [2] gives some other special forms ... PrenticeHall), pp 163ff [5] Bunch, J.R 1985, SIAM Journal on Scientific and Statistical Computing, vol 6, pp 34 9 36 4 [6] de Hoog, F 1987, Linear Algebra and Its Applications, vol 88/89, pp 1 23 138 [7] 2.9 ... ···  RN −2 RN −1 R−1 R0 R1 R−2 R−1 R0 RN 3 RN −2 RN −4 RN 3 ··· ··· ··· ··· ··· ··· R−(N −2) R−(N 3) R−(N −4) R0 R1 R−(N −1)  R−(N −2)  R−(N 3)     R−1 R0 (2.8.8) The linear Toeplitz...

Ngày tải lên: 21/01/2014, 18:20

... Strong Schemes 8.9 Approximation of Pure Jump Processes 8.10 Exercises 34 7 34 7 35 0 35 5 35 6 35 9 36 1 36 2 36 8 37 5 ... Jump-Adapted Weak Schemes 13. 7 Numerical Results on Weak Schemes 13. 8 Exercises 5 23 5 23 529 530 533 534 5 43 548 569 ... 30 9 30 9 31 6 32 1 32 6 33 1 34 6 Contents XI Jump-Adapted Strong Approximations 8.1 Introduction to Jump-Adapted Approximations 8.2 Jump-Adapted Strong Taylor Schemes...

Ngày tải lên: 19/02/2014, 22:20