differential equations level 2

Partial Differential Equations part 2

... equation (19.1 .20 ) isrn+1j+1 /2 − rnj+1 /2 ∆t=sn+1 /2 j+1− sn+1 /2 j∆xsn+1 /2 j− sn−1 /2 j∆t= vrnj+1 /2 − rnj −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) 836Chapter 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)Thenξ =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,...

Tài liệu Integration of Ordinary Differential Equations part 2 pptx

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... 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 ... Publications, New York),§ 25 .5. [1]Gear, C.W. 1971,Numerical Initial Value Problems in Ordinary Differential Equations (EnglewoodCliffs, NJ: Prentice-Hall), Chapter 2. [2] Shampine, L.F., and...

Báo cáo " Periodic solutions of some linear evolution systems of natural differential equations on 2-dimensional tore " ppt

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...

introduction to stochastic differential equations v1.2 (berkeley lecture notes) - l. evans

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introduction to stochastic differential equations 1.2 - evans l c

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stochastic differential equations and applications vol 2 - friedman a

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Partial Differential Equations part 1

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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 prototypicalparabolicequation is the diffusion equation∂u∂t=∂∂xD∂u∂x(19.0 .2) where ... representation (seeFigure 19.0 .2) ,uj+1,l− 2uj,l+ uj−1,l∆ 2 +uj,l+1− 2uj,l+ uj,l−1∆ 2 = ρj,l(19.0.5)or equivalentlyuj+1,l+ uj−1,l+ uj,l+1+ uj,l−1− 4uj,l=∆ 2 ρj,l(19.0.6)To...

Free grammar ebook level 2

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... 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...

Pictures of English Tenses Level 2 Lower Intermediate

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Tài liệu Integration of Ordinary Differential Equations part 1 doc

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... 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, ... 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 Partial Differential Equations part 3 pptx

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... (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 ... = D(u)du (19 .2. 23)analytically for z(u), then the right-hand side of (19 .2. 1) becomes ∂ 2 z/∂x 2 ,whichwe difference implicitly aszn+1j+1− 2zn+1j+ zn+1j −1(∆x) 2 (19 .2. 24)Now linearize...

Tài liệu Partial Differential Equations part 4 ppt

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... ∆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 ... second-order accurateand unitary:e−iHt1 −1 2 iH∆t1+1 2 iH∆t(19 .2. 35)In other words,1+1 2 iH∆tψn+1j=1 −1 2 iH∆tψnj(19 .2. 36)On replacing H by its ﬁnite-difference approximation ... L1piece; likewise U 2 , Um. Then a method of getting fromunto un+1isun+1/m= U1(un, ∆t/m)un +2/ m= U 2 (un+1/m, ∆t/m)···un+1= Um(un+(m−1)/m, ∆t/m)(19.3 .22 )The timestep for...

Tài liệu Solution of Linear Algebraic Equations part 2 ppt

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... 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 ... vector):a11a 12 a13a140 a 22 a 23 a 24 00a33a34000a44·x1x 2 x3x4=b1b 2 b3b4 (2. 2.1)Here the primes signify...

Tài liệu Partial Differential Equations part 5 ppt

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... 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 ... 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 ... 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−...

Tài liệu Partial Differential Equations part 6 doc

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... 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 Root Finding and Nonlinear Sets of Equations part 2 docx

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... 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));elsef2=(*func)(*x2 += FACTOR*(*x2-*x1));}return 0;}Alternatively, you might want to “look inward” on an initial...