Bc2 q s h e l integrating once gives 42 x 2 d q s

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BC2: q s H e L = 0 Integrating once gives (42) x 2 d q s ÅÅÅÅÅÅÅÅÅ d x = C 1 Integrating a second time gives (43) q s H x L = - C 1 ÅÅÅÅÅÅ x + C 2 Applying BC1 gives (44) „q s ÅÅÅÅÅÅÅÅÅ „x H x L = C 1 ÅÅÅÅÅÅ x 2 ï „q s ÅÅÅÅÅÅÅÅÅ „x H 1 L = C 1 = 0 Applying BC 2 gives (45) C 2 = 0 Thus the steady state solution is (46) q s H x L = 0 Note: We could have deduced this result immediately as the ODE and BCs for q are homogeneous! (iii) Since the problem is homogeneous, we let q H x , t L = G H t L F H x L and find (47) dG ÅÅÅÅÅÅÅ d t = -l 2 G fl G H t L ~ Exp H -l 2 t L This study resource was shared via CourseHero.com
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and (48) 1 ÅÅÅÅÅÅ x 2 ÅÅÅÅÅÅÅ „x J x 2 „ F ÅÅÅÅÅÅÅÅ „x N + l 2 F = 0 „F ÅÅÅÅÅÅÅ „x H 1 L = 0, F H e L = 0 The equation for F H x L is a Sturm-Liouville eigenvalue problem, with the following orthogonality properties for the eigenfunctions (49) 1 e F n H x L F m H x L x 2 „x = 0, n m The general solution for q H x , t L is (50) q H x , t L = n = 1 A n G n H t , l n L F n H x , l n L where (51) G n H t , l n L = Exp H -l n 2 t L To find the A n we make use of the orthogonality condition. Thus (52) q H x , 0 L = 1 = n = 1 A n F n H x , l n L Hence (53) A n = Ÿ 1 e F n H x L x 2 „x ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅ Ÿ 1 e F 2 n H x L x 2 „x Note in this case f 1 H x L in the problem statement is zero! ECH259MT1_04Soln.nb 7 This study resource was shared via CourseHero.com Powered by TCPDF ()
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  • Summer '14
  • Trigraph, sh, ed d, Partial differential equation, Sturm–Liouville theory, HxL

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