Differential Equations Lecture Work Solutions 157

# Differential - d utt c2 uxx = xt u(x 0 = sin x ut(x 0 = 0 ux(0 t = 0 ux(L t = 1 w(x t = x2 2L wt = 0 wtt = 0 Thus w(x 0 = x2 2L vtt c2 vxx = xt wxx

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d. u tt c 2 u xx = xt u ( x, 0) = sin x u t ( x, 0) = 0 u x (0 ,t )=0 u x ( L, t )=1 w ( x, t )= x 2 2 L ; w t =0 w tt w xx = 1 L Thus w ( x, 0) = x 2 2 L ,w t ( x, 0) = 0 v tt c 2 v xx = xt + c 2 L | {z } s ( x,t ) v x (0 v x ( L, t λ n = L 2 φ n =co s L xn , 1 , 2 , ··· v ( x, 0) = sin x x 2 2 L v t ( x, 0) = 0 v ( x, t 1 2 v 0 ( t )+ X n =1 v n ( t )cos L x s ( x, t 1 2 s 0 ( t X n =1 s n ( t L x s n ( t 1 L Z L 0 xt + c 2 L ! cos L xdx n , 1 , 2 , v ( x, 0) = sin x x 2 2 L = 1 2 v 0 (0) + X n =1 v n (0) cos L x v n (0) = 1 L Z L 0 sin x x 2 2 L ! cos L n , 1 , 2 , v t ( x, 0) = 0 = 1 2 ˙ v 0 (0) + X n =1 ˙ v n (0) cos L x ˙ v n (0) = 0
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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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