Differential Equations Lecture Work Solutions 301

Differential Equations Lecture Work Solutions 301 - 1 − h...

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2. Let h = x i x i 1 and h + = x i +1 x i ,then u xij = u i +1 j h + h 2 u ij 1 1 h + h 2 ± u ij h h + h 2 + h + Now take Taylor series expansions u i +1 j = u ij + h + u xij + h 2 + 2 u xx i j + h 3 + 6 u xxx i j + h 4 + 24 u xxxx i j + ··· u ij 1 = u ij h u xij + h 2 2 u xx i j h 3 6 u xxx i j + h 4 24 u xxxx i j + ··· So numerator = 1 h + h ! 2 1+ h + h ! 2 | {z } =0 u ij + h + + h h + h ! 2 u xij + h 2 + 2 h 2 2 h + h ! 2 | {z } =0 u xx i j + h 3 + 6 + h 3 6 h + h ! 2 | {z } = h 2 + ( h + + h ) 6 u xxx i j ±··· Divide by the coeﬃcient of u xij we get u xij = u i +1 j h + h 2 u i 1 j
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Unformatted text preview: 1 − h + h − 2 ± u i j h + + h − h + h − 2 − h 2 + ( h + + h − ) 6( h + + h − h + h − 2 ) u xxx i j The error term can be manipulated − h 2 + ( h + + h − ) 6( h + + h − h + h − 2 ) = − h 2 + ( h + + h − ) 6 h − h + + h 2 + h − = − h − h + 6 301...
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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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