Differential Equations Lecture Work Solutions 137

# Differential Equations Lecture Work Solutions 137 - n − 1...

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1. u t = ku xx + x u ( x, 0) = x ( L x ) a. u x (0 ,t )=1 w = x + t L u ( L, t )= t Solve v t = kv xx 1+ x (see 1a last section) v x (0 ,t )=0 v ( x, 0) = x ( L x ) ( x L )=( x +1 )( t x ) v ( L, t )=0 eigenvalues: h n 1 2 π L i 2 n =1 , 2 , ··· eigenfunctions : cos ( n 1 2 ) π L xn =1 , 2 , ··· v X n =1 v n ( t )cos n 1 2 ± π L x 1+ x = X n =1 s n cos n 1 2 ± π L x s n = R L 0 ( 1+ x )cos n 1 2 π L xdx R L 0 cos 2 n 1 2 π L xdx X n =1 ˙ v n ( t )cos n 1 2 ± π L x = k X n =1 ² n 1 2 ± π L 2 v n cos n 1 2 ± π L x + X n =1 s n cos n 1 2 ± π L x Compare coeﬃcients ˙ v n ( t )+ k  n 1 2 π L 2 v n = s n v n = v n (0) e [ (
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Unformatted text preview: n − 1 2 ) π L ] 2 kt + s n Z t e − [ ( n − 1 2 ) π L ] 2 k ( t − τ ) dτ | {z } see (8 . 2 . 39) v n (0) = coeﬃcients of expansion of (1 + x ) ( L − x ) v n (0) = R L (1 + x ) ( L − x ) cos n − 1 2 π L x dx R L cos 2 n − 1 2 π L x dx u = v + w 137...
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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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