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# Conversely assuming v x t solves problem b show that

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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) d dt L 0 u ( x, t ) dx = L 0 ∂u ∂t ( x, t ) dx. (1) With these preliminaries out of the way, consider the ICBC problem u t ( x, t ) = ku xx ( x, t ) + q ( x ) , 0 < x < L, t > 0 , u x (0 , t ) = α, u x ( L, t ) = β, t > 0 , u ( x, 0) = f ( x ) , 0 < x < L, where α, β are constants. This will be called Problem C. (a) Show that if a steady state (equilibrium) solution exists it is necessary that k ( β - a ) + L 0 q ( x ) dx = 0 . (2) (b) Show that if and only if condition (2) holds, there is a solution u of u ′′ ( x ) = - 1 k k ( x ) , 0 x L, u (0) = α, u ( L ) = β. (c) Assuming (2) let u be the solution of u ′′ ( x ) = - 1 k k ( x ) , 0 x L, u (0) = α, u ( L ) = β that satisfies L 0 u ( x

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