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### AM255-HW2

Course: AM 255, Fall 2009
School: Sanford-Brown Institute
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Word Count: 506

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2 HW Solutions 2.1.2 Modify the scheme (2.1.11) such that it approximates ut = ux . Prove that the conditions (2.1.14) and (2.1.15) are also necessary for stability in this case. Solution: Similar to (2.1.11), a scheme approximating ut = ux can be written as n+1 n n 0 vj = (1 kD0 )vj + khD+ D vj , vj = fj , (1) in which is a positive constant. Fourier transform of equation (1) gives v n+1 () = 1 i sin(h) 4...

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2 HW Solutions 2.1.2 Modify the scheme (2.1.11) such that it approximates ut = ux . Prove that the conditions (2.1.14) and (2.1.15) are also necessary for stability in this case. Solution: Similar to (2.1.11), a scheme approximating ut = ux can be written as n+1 n n 0 vj = (1 kD0 )vj + khD+ D vj , vj = fj , (1) in which is a positive constant. Fourier transform of equation (1) gives v n+1 () = 1 i sin(h) 4 sin2 h 2 v v n () = Q()n (), (2) where is a wavenumber, Q() is the amplication factor and = k/h. For the stability of the numerical scheme, it requires that, |Q|2 1. That is |Q|2 = 1 4 sin 2 h 2 2 + 2 sin2 (h) 1. (3) (i) When 2 1, the scheme is stable if 0 8 42 . . That is 0 < 2 1. 1 (5) (4) (ii) When 2 1, (16 2 4)2 > 0. Therefore, 2 1. (6) In conclusion, the conditions of stability for ut = ux is the same as the conditions for ut = ux . 2.1.3 Choose in Eq.(2.1.11) such that Q uses only two gridpoints. What is stability condition. Solution: Expanding Eq.(2.1.11) gives n+1 n vj = vj + k n k n n n n (vj+1 vj1 ) + (vj+1 2vj + vj1 ) 2h h 1 1 n n = + vj+1 + (1 2) vj + vn . 2 2 j1 (7) There are only three cases which use only two grid points. (i) = 1/2 and = 1: Equation (7) becomes n+1 n n vj = vj+1 + (1 ) vj . (8) From the discussion in p. 45, the scheme is stable if the following conditions is satised: 0 < 1. Since = 1, the condition is 0 < < 1. 2 (10) (9) (ii) = 1/2 Equation (7) becomes 1 1 n+1 n n vj = (1 + )vj+1 + (1 )vj1 . 2 2 (11) Again the scheme is stable if and only if it satises one of the following conditions: 0 < 2 1. 2 1 and 2 1. this In case, the scheme is stable if < 1. (iii) = 1/2: If is negative, the problem is ill posed and, hence, the numerical scheme is unstable. ComputationalP roblem Implement the Lax-Friedrich (p.46) and the Lax-Wendro (p.46) method for solving ut = ux . Use u(x, 0) = f (x) given on p.46 and recall that u is 2-periodic. Use the codes to reproduce Figures 2.1.4 and 2.1.5 in the text. Now choose k/h = 1. Run the two schemes for one of the previous examples and monitor the error. What do you observe and how do you explain it? Solution: u =u , < x < , 0 t, t x u(x, 0) = f (x), < x < , 3 (12) (a) 2 (b) 0.6 N=10 N=10 0.4 1.5 ||e|| 1 N=100 ||e|| 0.2 0.5 N=100 0 0 2 4 t 6 8 10 0 0 2 4 t 6 8 10 Figure 1: The maximum absolute errors of (a) Lax-Friedrichs and (b) LaxWendro schemes. where x, f or 0 x , f (x) = 2 x, f or x 2. The solution of equation (12) is u(x, t) = f (x + t). (13) Let h be the grid space and k be time step...

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