M257-316Notes_Lecture19

# M257-316Notes_Lecture19 - Chapter 15 Lecture 19 Further...

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Chapter 15 Lecture 19 Further Heat Conduction Problems: Inhomogeneous BC Example 15.1 Specifed Temperatures u t = α 2 u xx 0 <x<L, t> 0 (15.1) BC: u (0 ,t )= u 0 u ( L, t u 1 u 0 ,u 1 constants (15.2) u ( x, 0) = g ( x ) . (15.3) Firstly consider the steady-state solution (i.e., when u t = 0) which we denote by u α ( x ). In this case (15.1) becomes α 2 u 0 ( x )=0 u ( x A 0 x + B 0 u (0) = B 0 = u 0 u ( L A 0 L + u 0 = u 1 u ( x ( u 1 u 0 L ) x + u 0 steady state solution . Let u ( x, t u ( x )+ v ( x, t ). Substitute into (15.1) u t = ( u ( x v ( x, t ) ) t = α 2 ( u ( x v ( x, t ) ) xx v t = α 2 v xx (15.4) since ( u ( x ) ) xx = 0. Substitute into (15.2) u (0 u 0 = u (0) + v (0 u 0 + v (0 ) v (0 u ( L, t u 1 = u ( L v ( L, t u 1 + v ( L, t ) v ( L, t . 95

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Lecture 19 Further Heat Conduction Problems: Inhomogeneous BC Substitute into (15.3) u ( x, 0) = g ( x )= u ( x )+ v ( x, 0) v ( x, 0) = g ( x ) u ( x ) . Thus we have to solve a new problem for v which has zero BC: v t = α 2 v xx v (0 ,t )=0 = v ( L, t ) v ( x, 0) = g ( x ) u ( x ) . (15.5) Separate variables: v ( x, t X ( x ) T ( t ). ˙ T ( t ) α 2 T ( t ) = X 0 ( x ) X ( x ) = λ 2 = const T ( t c e λ 2 α 2 t X 0 + λ 2 X =0 X (0) = 0 = X ( L ) X n ( x )=sin ³ nπx L ´ L λ n = nπ, n =1 ,... v ( x, t X n =1 b n e α 2 ( L ) 2 t sin ³ nπx L ´ (15.6) v ( x, 0) = g ( x ) u ( x X n =1 b n sin ³ nπx L ´ b n = 2 L L Z 0 { g ( x ) u ( x ) } sin ³ nπx L ´ dx Thus the solution to the inhomogeneous problem is: u ( x, t u ( x v
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## This note was uploaded on 05/04/2011 for the course MATH 25 taught by Professor Lo during the Spring '11 term at BC.

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M257-316Notes_Lecture19 - Chapter 15 Lecture 19 Further...

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