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6 consider the icbc problem u t x t ku xx x t q x x l

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6. Consider the ICBC problem u t ( x, t ) = ku xx ( x, t ) + q ( x ) , 0 < x < L, t > 0 , u (0 , t ) = α, u ( L, t ) = β, t > 0 , u ( x, 0) = f ( x ) , 0 < x < L, where α, β are constants. I will call this Problem A.
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2 (a) Let u ( x ) satisfy u ′′ ( x ) = - 1 k k ( x ) , 0 x L, u (0) = α, u ( L ) = β. Assuming u solves the first ICBC problem, show that the function v defined by v ( x, t ) = u ( x, t ) - u ( x ) solves the following Problem B: v t ( x, t ) = kv xx ( x, t ) , 0 < x < L, t > 0 , v (0 , t ) = 0 , v ( L, t ) = 0 , t > 0 , v ( x, 0) = f ( x ) - u ( x ) , 0 < x < L, Conversely , assuming v ( x, t ) solves problem B, show that u ( x, t ) = v ( x, t ) + u ( x ) solves problem A. (b) Solve Problem B (to the extent that it is possible without having a concrete value for q ( x ) and f ( x )) and use the solution of Problem B, and its relation to the solution u of Problem A to SHOW that lim t →∞ u ( x, t ) = u ( x ). Believe it or not, this is NOT a hard exercise. 7. For part a of this exercise you will need to use the fact that the total heat energy at time t in a bar stretching from 0 to L whose temperature distribution is given by u ( x, t ) is proportional to L 0 u ( x, t ) dx. It also helps to know that under fairly general conditions (which we, in an applied course, may assume hold)
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6 Consider the ICBC problem u t x t ku xx x t q x x L t u 0...

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