{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Differential Equations Lecture Work Solutions 306

Differential Equations Lecture Work Solutions 306 - tu...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 + ∆
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: tu t + (βˆ† t ) 2 2 u tt + (βˆ† t ) 3 6 u ttt + Β·Β·Β· u n +1 j βˆ’ 1 = u + βˆ† tu t βˆ’ βˆ† xu x + (βˆ† t ) 2 2 u tt βˆ’ βˆ† t βˆ† xu tx + (βˆ† x ) 2 2 u xx + (βˆ† t ) 3 6 u ttt βˆ’ 3 (βˆ† t ) 2 6 βˆ† xu ttx + 3 (βˆ† x ) 2 6 βˆ† tu txx βˆ’ (βˆ† x ) 3 6 u xxx Β± Β·Β·Β· u n +1 j +1 βˆ’ 2 u n +1 j + u n +1 j βˆ’ 1 = (βˆ† x ) 2 u xx + βˆ† t (βˆ† x ) 2 u txx + Β·Β·Β· So u n +1 j +1 βˆ’ 2 u n +1 j + u n +1 j βˆ’ 1 2(βˆ† x ) 2 = 1 2 u xx + βˆ† t 2 u txx + Β·Β·Β· 306...
View Full Document

{[ snackBarMessage ]}