# 11_1 - tribution by expanding it in a Fourier sine series f...

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11/1 10.5.12(graded) November 2, 2010 10.5.12 Consider the conduction of heat in a rod 40 cm in length whose ends are maintained at 0 C for all t > 0. Find an expression for the temperature u ( x, t ) given the initial temp dist. and α 2 = 1. u ( x, 0) = x, 0 < x < 40 (1) The heat conduction problem with α 2 = 1 is, u xx = u t , 0 < x < 40 , t > 0 (2) u (0 , t ) = 0 , u (40 , t ) = 0 , t > 0 (3) u ( x, 0) = x, 0 < x < 40 (4) Employing separation of variables assume a solution of the form, u ( x, t ) = X ( x ) T ( t ). This leads to the follow- ing pair of O.D.E.s, X 00 + λ X = 0 , X (0) = X (40) = 0 (5) T 0 + λT = 0 (6) Solving the eigenvalue problem on X we ﬁnd eigenval- ues exist for λ > 0 with eigenvalues and eigenfunctions given by, λ n = n 2 π 2 1600 (7) X n = sin nπx 40 (8) Solutions to the O.D.E. on T are T n = e - λ n t . Hence the general solution is, u ( x, t ) = X n =1 C n e - n 2 π 2 t/ 16000 sin nπx 40 (9) To solve for C n we impose the initial temperature dis-
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Unformatted text preview: tribution by expanding it in a Fourier sine series. f ( x ) = x, < x < 40 , ( period = 80) (10) f ( x ) = ∞ X n =1 b n sin nπx 40 (11) Solving for b n b n = 1 20 Z 40 x sin nπx 40 dx (12) = 1 20 ± 1600 sin nπx 40 n 2 π 2-40 x cos nπx 40 nπ ² ³ ³ ³ π (13) = 80 nπ ´ 1 , n = odd-1 , n = even (14) = 80(-1) n +1 nπ (15) the initial temperature dist. is therefore given by, u ( x, 0) = 80 ∞ X n =1 (-1) n +1 nπ sin nπx 40 (16) Applying this initial temp. dist. to (9) gives C n , C n = 80(-1) n +1 nπ (17) The solution to the given heat conduction problem is therefore, u ( x, t ) = 80 π ∞ X n =1 (-1) n +1 n e-n 2 π 2 t/ 16000 sin nπx 40 (18) 1...
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