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Differential Equations Lecture Work Solutions 306

# Differential Equations Lecture Work Solutions 306 - tu...

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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 + (β t ) 2 2 u tt + (β t ) 3 6 u ttt + Β·Β·Β· u n +1 j β 1 = u + β tu t β β xu x + (β t ) 2 2 u tt β β t β xu tx + (β x ) 2 2 u xx + (β t ) 3 6 u ttt β 3 (β t ) 2 6 β xu ttx + 3 (β x ) 2 6 β tu txx β (β x ) 3 6 u xxx Β± Β·Β·Β· u n +1 j +1 β 2 u n +1 j + u n +1 j β 1 = (β x ) 2 u xx + β t (β x ) 2 u txx + Β·Β·Β· So u n +1 j +1 β 2 u n +1 j + u n +1 j β 1 2(β x ) 2 = 1 2 u xx + β t 2 u txx + Β·Β·Β· 306...
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