differential equations level 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 ... propagation v ∂ 2 u ∂t 2 = v 2 ∂ 2 u ∂x 2 (19.1 .2) 836 Chapter 19. Partial Differential Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -43108-5) Copyright ... ≡ v∆t ∆x (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 ≤ 1, or v∆t ≤ ∆x as usual. Incidentally,...

Ngày tải lên: 07/11/2013, 19:15

... 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 + k 3 3 + k 4 6 + O(h 5 )(16.1.3) The ... Publications, New York), § 25 .5. [1] Gear, C.W. 1971, Numerical Initial Value Problems in Ordinary Differential Equations (Englewood Cliffs, NJ: Prentice-Hall), Chapter 2. [2] Shampine, L.F., and...

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... velocity of propagation v ∂ 2 u ∂t 2 = v 2 ∂ 2 u ∂x 2 (19.1 .2) Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -43108-5) Copyright (C) 1988-19 92 by Cambridge University ... equation ∂ 2 u ∂t 2 = v 2 ∂ 2 u ∂x 2 (19.0.1) where v = constant is the velocity of wave propagation. The prototypicalparabolic equation is the diffusion equation ∂u ∂t = ∂ ∂x  D ∂u ∂x  (19.0 .2) where ... representation (see Figure 19.0 .2) , u j+1,l − 2u j,l + u j−1,l ∆ 2 + u j,l+1 − 2u j,l + u j,l−1 ∆ 2 = ρ j,l (19.0.5) or equivalently u j+1,l + u j−1,l + u j,l+1 + u j,l−1 − 4u j,l =∆ 2 ρ j,l (19.0.6) To...

Ngày tải lên: 28/10/2013, 22:15

... lived here since 20 04.” • “I’ve lived here for 8 years.” Since is used with a point in time, and means “from that point in time until the present.” Use since with dates (20 11, January, Tuesday, ... sentences and present perfect continuous sentences. “I’ve been married for over 20 years.” “So have I.” ~ 29 ~ www.espressoenglish.net “We’ve cleaned the bathroom and the kitchen.” ... here to take the quiz! http://www.espressoenglish.net/prepositions-of-time-in-english#quiz ~ 22 ~ www.espressoenglish.net Simple Past and Past Continuous The past continuous is often...

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Ngày tải lên: 14/12/2013, 15:53

... 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, ... 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 = ... generic problem in ordinary differential equations is thus reduced to the study of a set of N coupled first-order differential equations for the functions y i ,i=1 ,2, ,N, having the general form dy i (x) dx =...

Ngày tải lên: 15/12/2013, 04:15

... (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 set of coupled nonlinear equations ... of (19 .2. 8) leads to i  ψ n+1 j − ψ n j ∆t  = −  ψ n+1 j+1 − 2 n+1 j + ψ n+1 j −1 (∆x) 2  + V j ψ n+1 j (19 .2. 27) for which ξ = 1 1+i  4∆t (∆x) 2 sin 2  k∆x 2  + V j ∆t  (19 .2. 28) This ... = D(u)du (19 .2. 23) analytically for z(u), then the right-hand side of (19 .2. 1) becomes ∂ 2 z/∂x 2 ,which we difference implicitly as z n+1 j+1 − 2z n+1 j + z n+1 j −1 (∆x) 2 (19 .2. 24) Now linearize...

Ngày tải lên: 15/12/2013, 04:15

... ∆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 ... second-order accurate and unitary: e −iHt  1 − 1 2 iH∆t 1+ 1 2 iH∆t (19 .2. 35) In other words,  1+ 1 2 iH∆t  ψ n+1 j =  1 − 1 2 iH∆t  ψ n j (19 .2. 36) On replacing H by its finite-difference approximation ... L 1 piece; likewise U 2 , U m . Then a method of getting from u n to u n+1 is u n+1/m = U 1 (u n , ∆t/m) u n +2/ m = U 2 (u n+1/m , ∆t/m) ··· u n+1 = U m (u n+(m−1)/m , ∆t/m) (19.3 .22 ) The timestep for...

Ngày tải lên: 15/12/2013, 04:15

... 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      b 13 b 23 b 33 b 43      1000 0100 0010 0001     (2. 1.1) Here ... 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      b 13 b 23 b 33 b 43      1000 0100 0010 0001     (2. 1.1) Here ... vector):    a  11 a  12 a  13 a  14 0 a  22 a  23 a  24 00a  33 a  34 000a  44    ·    x 1 x 2 x 3 x 4    =    b  1 b  2 b  3 b  4    (2. 2.1) Here the primes signify...

Ngày tải lên: 15/12/2013, 04:15

... 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 ... that λ (r) k will involve terms like cos (2 k/L) − 2 raised to a power. Solve the tridiagonal systems foru k j at the levels j =2 r ,2 2 r ,4 2 r , , J − 2 r . Fourier synthesize to get the y-values ... 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), we find u mn  e 2 im/J + e 2 im/J + e 2 in/L + e 2 in/L − 4  =ρ mn ∆ 2 (19.4.4) or u mn = ρ mn ∆ 2 2  cos 2 m J +cos 2 n L −...

Ngày tải lên: 15/12/2013, 04:15

... 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. equation 19.3.16). Here we ... that λ (r) k will involve terms like cos (2 k/L) − 2 raised to a power. Solve the tridiagonal systems foru k j at the levels j =2 r ,2 2 r ,4 2 r , , J − 2 r . Fourier synthesize to get the y-values...

Ngày tải lên: 15/12/2013, 04:15

... f1,f2; if (*x1 == *x2) nrerror("Bad initial range in zbrac"); f1=(*func)(*x1); f2=(*func)(*x2); for (j=1;j<=NTRY;j++) { if (f1*f2 < 0.0) return 1; if (fabs(f1) < fabs(f2)) f1=(*func)(*x1 ... 3 52 Chapter 9. Root Finding and Nonlinear Sets of Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -43108-5) Copyright (C) 1988-19 92 by Cambridge ... < 0.0) return 1; if (fabs(f1) < fabs(f2)) f1=(*func)(*x1 += FACTOR*(*x1-*x2)); else f2=(*func)(*x2 += FACTOR*(*x2-*x1)); } return 0; } Alternatively, you might want to “look inward” on an initial...

Ngày tải lên: 15/12/2013, 04:15

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