Differential Equations Lecture Work Solutions 306

Differential Equations Lecture Work Solutions 306 - tu t +...

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Divide by 2(∆ x ) 2 we get u n j +1 2 u n j + u n j 1 2(∆ x ) 2 = 1 2 u xx t 4 u txx + (∆ t ) 2 16 u ttxx + (∆ x ) 2 24 u xxxx + ··· Now the right hand side become α ( u xx + (∆ t ) 2 8 u ttxx + 1 12 (∆ x ) 2 u xxxx + ··· ) Combine LHS and RHS u t + (∆ t ) 2 24 u ttt + ··· = α ( u xx + (∆ t ) 2 8 u ttxx + 1 12 (∆ x ) 2 u xxxx + ··· ) Thus the truncation error is O ((∆ t ) 2 , (∆ x ) 2 ) Compare these results with the T.E. obtained from Taylor series expansion about the point ( n, j ) . To do that we need to exapnd about n, j u n +1 j = u +∆ tu t + (∆ t ) 2 2 u tt + ··· All terms on the right are now at n, j . LHS = u t + t 2 u tt + ··· For the right hand side u n +1 j +1 = u +∆ tu t +∆ xu x + (∆ t ) 2 2 u tt +∆ t xu tx + (∆ x ) 2 2 u xx + (∆ t ) 3 6 u ttt +3 (∆ t ) 2 6 xu ttx +3 (∆ x ) 2 6 tu txx + (∆ x ) 3 6 u xxx + ··· u n +1 j = u +∆
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Unformatted text preview: tu t + (∆ t ) 2 2 u tt + (∆ t ) 3 6 u ttt + ··· u n +1 j − 1 = u + ∆ tu t − ∆ xu x + (∆ t ) 2 2 u tt − ∆ t ∆ xu tx + (∆ x ) 2 2 u xx + (∆ t ) 3 6 u ttt − 3 (∆ t ) 2 6 ∆ xu ttx + 3 (∆ x ) 2 6 ∆ tu txx − (∆ x ) 3 6 u xxx ± ··· u n +1 j +1 − 2 u n +1 j + u n +1 j − 1 = (∆ x ) 2 u xx + ∆ t (∆ x ) 2 u txx + ··· So u n +1 j +1 − 2 u n +1 j + u n +1 j − 1 2(∆ x ) 2 = 1 2 u xx + ∆ t 2 u txx + ··· 306...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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