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# Solution bycranknicolsonequation1canbeapproximatedas

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Unformatted text preview: )u i, j1 ru i1, j1 ru i1, j (2 2r)u i, j ru i1, j (2) As r= 0.75 and x 0.1 Put r =0.75 in equation (2),we get 3u i1, j1 14u i, j1 3u i1, j1 3u i1, j 2u i, j 3u i1, j (3) Now I.C. is u(x,0) x i.e. u 0,0 0,u1,0 .1,u 2,0 .2,u 3,0 .3,u 4,0 .4,u 5,0 .5 u 6,0 .6,u 7,0 .7,u 8,0 .8,u 9,0 .9,u10,0 1.0 And B.C. is u(o, t) 0, u 0,0 u 0,1 u 0,2 u 0,3 ...................... u 0, j 0 u(1, t) 1 i.e. u10,0 u10,1 u10,2 u10,3 ........................ u10, j 1 Now for Ist time level we use C‐N scheme 3u 2 14u1 0.8 3u1 14u 2 3u 3 1.3 3u 2 14u 3 3u 4 3u 3 14u 4 2.4 3u 5 3u 4 14u 5 3.2 3u 6 3u 5 14u 6 4 3u 7 3u 6 14u 7 4.8 3u 8 3u 7 14u 8 5.6 3u 9 6.4 3u 8 14u 9 10.2 u1 u1,1 0.0949, u 2 u 2,1 0.1763, u 3 u 3,1 0.2947 u 4 u 4,1 0.3988, u 5 u 5,1 0.4997, u 6 u 6,1 0.5999 u 7 u 7,1 0.7000 , u 8 u 8,1 0.8000, u 9 u 9,1 0.9000 (i) Dufort Frankel Explicit Scheme is u i, j1 1 2r 2r u i, j1 u i1, j u i1, j) 1 2r 1 2r Put r=.75 u i, j1 1 3 u i, j1 u i1, j u i1, j (3) 5 5 For 2nd time level Put j=1 in (3) u 0,2 0,u10,2 1.000 1 3 u1,0 (u 0,1 u 2,1 ) .08578 5 5 3 1 u 2,2 u 2,0 (u1,1 u 3,1 ) .19376 5...
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