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 ≤ 1, or v∆t ≤ ∆x as usual. Incidentally, ... |ξ| 2 =1−(1 − α 2 )(k∆x) 2 + ... (19.1.44) for small k∆x. 846 Chapter 19. Partial Differential Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -43108-5)...

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 differential equations (PDEs) are ... Equations (19.0.1) and (19.0 .2) both deﬁne initial value or Cauchy problems: If information on u (perhaps including time derivative information) is 827 828 Chapter 19. Partial Differential Equations...

Tài liệu Integration of Ordinary Differential Equations part 2 pptx

... the whole interval. Figure 16.1 .2 illustrates theidea. In equations, k1= hf(xn,yn)k 2 =hfxn+1 2 h, yn+1 2 k1yn+1= yn+ k 2 + O(h3)(16.1 .2) As indicated in the error term, ... organization about it:k1= hf(xn,yn)k 2 =hf(xn+h 2 ,yn+k1 2 )k3=hf(xn+h 2 ,yn+k 2 2)k4=hf(xn+ h, yn+ k3)yn+1= yn+k16+k 2 3+k33+k46+ O(h5)(16.1.3)The ... about the7 12 Chapter 16. Integration of Ordinary Differential 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...

Tài liệu Partial Differential Equations part 3 pptx

... equation (19 .2. 4) isξ =1−4D∆t(∆x) 2 sin 2 k∆x 2 (19 .2. 5)The requirement |ξ|≤1leads to the stability criterion2D∆t(∆x) 2 ≤ 1(19 .2. 6)848Chapter 19. Partial Differential Equations Sample ... (19 .2. 19) writeDj+1 /2 =1 2 D(unj+1)+D(unj)(19 .2. 22) Implicit schemes are not as easy. The replacement (19 .2. 22) with n → n +1leavesus with a nasty set of coupled nonlinear equations ... of (19 .2. 8) leads toiψn+1j− ψnj∆t= −ψn+1j+1− 2 n+1j+ ψn+1j −1(∆x) 2 + Vjψn+1j(19 .2. 27)for whichξ =11+i4∆t(∆x) 2 sin 2 k∆x 2 + Vj∆t(19 .2. 28)This...

Tài liệu Partial Differential Equations part 4 ppt

... ∆t /2. In each substep, a different dimension is treated implicitly:un+1 /2 j,l= unj,l+1 2 αδ 2 xun+1 /2 j,l+ δ 2 yunj,lun+1j,l= un+1 /2 j,l+1 2 αδ 2 xun+1 /2 j,l+ δ 2 yun+1j,l(19.3.16)The ... usun+1j,l= unj,l+1 2 αδ 2 xun+1j,l+ δ 2 xunj,l+ δ 2 yun+1j,l+ δ 2 yunj,l(19.3.13)Hereα ≡D∆t∆ 2 ∆ ≡ ∆x =∆y (19.3.14)δ 2 xunj,l≡ unj+1,l− 2unj,l+ unj−1,l(19.3.15)and ... 1-800-8 72- 7 423 (North America only),or send email to trade@cup.cam.ac.uk (outside North America).The expression for |ξ| 2 can be manipulated into the form|ξ| 2 =1−(sin 2 kx∆+sin 2 ky∆)1 2 −...

Tài liệu Partial Differential Equations part 5 ppt

... then adding the three equations, we getuj 2 + T(1)· uj+ uj +2 = g(1)j∆ 2 (19.4. 32) This is an equation of the same form as (19.4 .29 ), withT(1) =21 −T 2 g(1)j=∆ 2 (gj−1−T·gj+gj+1)(19.4.33)After ... Similarly,ρjl=1JLJ−1m=0L−1n=0ρmne 2 ijm/Je 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umne 2 im/J+ e 2 im/J+ e 2 in/L+ e 2 in/L− 4=ρmn∆ 2 (19.4.4)orumn=ρmn∆ 2 2cos 2 mJ+cos 2 nL− ... writing down three successive equations like(19.4 .29 ):uj 2 + T · uj−1+ uj= gj−1∆ 2 uj−1+ T · uj+ uj+1= gj∆ 2 uj+ T · uj+1+ uj +2 = gj+1∆ 2 (19.4.31)Matrix-multiplyingthe...

Tài liệu Partial Differential Equations part 6 doc

... turns out to beρs 1 −π 2 2J 2 (19.5.11)The number of iterations r required to reduce the error by a factor of 10−pis thusr 2pJ 2 ln 10π 2 1 2 pJ 2 (19.5. 12) In other words, the number ... (19.5.31)implicitly in two half-steps:un+1 /2 − un∆t /2 = −Lxun+1 /2 + Lyun∆ 2 − ρun+1− un+1 /2 ∆t /2 = −Lxun+1 /2 + Lyun+1∆ 2 − ρ(19.5.35)(cf. equation 19.3.16). Here we ... that λ(r)kwill involve terms likecos (2 k/L) − 2 raised to a power. Solve the tridiagonal systems forukjat the levelsj =2 r ,2 2 r,4 2 r, , J − 2 r. Fourier synthesize to get the y-values...

Tài liệu Partial Differential Equations part 7 doc

... (19.6. 32) givesτhLH(u+h 2 u 2 )−Lh(u+h 2 u 2 )=LH(u)−Lh(u)+h 2 [LH(u 2 )−Lh(u 2 )] + ···=(H 2 −h 2 )τ 2 +O(h4)(19.6.38)For the usual case of H =2hwe therefore haveτ 13τh13τh(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 (ISBN 0- 521 -43108-5)Copyright (C) 1988-19 92 by Cambridge ... i,ipass,isw,j,jsw=1;double foh2,h,h2i,res;h=1.0/(n-1);h2i=1.0/(h*h);foh2 = -4.0*h2i;for (ipass=1;ipass< =2; ipass++,jsw=3-jsw) { Red and black sweeps.isw=jsw;for (j =2; j<n;j++,isw=3-isw) {for...

Tài liệu Integration of Ordinary Differential Equations part 1 doc

... 16.6 of this chapter treats the subject of stiff equations, relevant both toordinary differential equations and also to partial differential equations (Chapter 19). ... involving ordinary differential equations (ODEs) can always bereduced to the study of sets of ﬁrst-order differential equations. For example thesecond-order equationd 2 ydx 2 + q(x)dydx= ... generic problem in ordinary differential equations is thus reduced to thestudy of a set of N coupled ﬁrst-order differential equations for the functionsyi,i=1 ,2, ,N, having the general formdyi(x)dx=...

Tài liệu Solution of Linear Algebraic Equations part 2 ppt

... equationa11a 12 a13a14a 21 a 22 a 23 a 24 a31a 32 a33a34a41a 42 a43a44·x11x 21 x31x41x 12 x 22 x 32 x 42 x13x 23 x33x43y11y 12 y13y14y 21 y 22 y 23 y 24 y31y 32 y33y34y41y 42 y43y44=b11b 21 b31b41b 12 b 22 b 32 b 42 b13b 23 b33b431000010000100001 (2. 1.1)Here ... equationa11a 12 a13a14a 21 a 22 a 23 a 24 a31a 32 a33a34a41a 42 a43a44·x11x 21 x31x41x 12 x 22 x 32 x 42 x13x 23 x33x43y11y 12 y13y14y 21 y 22 y 23 y 24 y31y 32 y33y34y41y 42 y43y44=b11b 21 b31b41b 12 b 22 b 32 b 42 b13b 23 b33b431000010000100001 (2. 1.1)Here ... j and i  =2, 41 if i =2, j=41 if i =4,j =2 0 otherwise (2. 1.5)effects the interchange of rows 2 and 4. Gauss-Jordan elimination by row operations alone(including the possibility of partial pivoting)...

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