ECH140_FirstExam_2013_Soln

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BC2: U s H L L = 0 Integrating the ODE twice gives U s H x L = Q ` 0 12 x 4 + C 1 x + C 2 Applying the BCs gives BC1: U s H 0 L = a ï C 2 = a BC2: U s H L L = 0 ï 0 = Q ` 0 12 L 4 + C 1 L + a Thus the steady state solution is U s H x L = Q ` 0 12 I x 4 - L 3 x M + a H 1 - x ê L L (ii) We assume the solution can be decomposed as follows u H x, t L = U s H x L + w H x, t L Then the PDE for w(x,t) is ¶∂ w ¶∂ t = k ¶∂ 2 w ¶∂ x 2 IC: w H x, 0 L = 6 + 4Sin H 3 p x ê L L - a H 1 - x ê L L BC1: w H 0, t L = 0 BC2: w H L, t L = 0 Separating variables we seek a solution given by w H x, t L = f H x L G H t L Substituting this expression into the PDE and reorganizing the terms gives G' kG = f '' f = -l 2 Consider first the equation for G (t) : dG dt = -l 2 kG Solving for G H t L gives G H t L = AExp A -l 2 kt E Consider next the ODE for f (x). d 2 f dx 2 + l 2 f = 0 The general solution is f H x L = C 1 Cos H l x L + C 2 Sin H l x L 4 ECH140FirstExam_2013Soln.nb This study resource was shared via CourseHero.com
Next we apply the BCs: Thus BC1: f H 0 L = 0 fl C 1 = 0 and BC2: f H L L = 0 fl C 2 Sin H l L L = 0 For a non trivial solution we require l L = n p , wheren = 1, 2, 3 Note that n = 0 , gives the trivial solution If we solve for l we get l = K n p L O 2 , n = 1, 2, 3, .... with f n = C 2 Sin H n p x ê L L The general solution is then w H x, t L = n = 1 A n Sin H n p x ê L L Exp A - k H n p ê L L 2 t E TodeterminethecoefficientsA n wemakeuseoftheinitialcondition w H x, 0 L = 4Sin H 3 p x ê L L - a H 1 - x ê L L = n = 1 A n Sin H n p x ê L L We use the orthogonality of the eigenfunctions to find the A n 0 L 8 4Sin H 3 p x ê L L - a H 1 - x ê L L< Sin H m p x ê L L x = n = 1 A n 0 L Sin H m p x ê L L Sin H n p x ê L L x we find A n = 2 L 0 L 8 4Sin H 3 p x ê L L - a H 1 - x ê L L< Sin H m p x ê L L x

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• Summer '14
• Trigraph, LL, HxL

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