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# It also helps to know that under fairly general

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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 ) dx = L 0 f ( x ) dx,

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3 where f ( x ) is as in Problem C. Show that if u solves Problem C , then v ( x, t ) = u ( x, t ) - u ( x ) solves v t ( x, t ) = kv xx ( x, t ) , 0 < x < L, t > 0 , v x (0 , t ) = 0 , v x ( L, t ) = 0 , t > 0 , v ( x, 0) = f ( x ) - u ( x ) , 0 < x < L, and use this to explain (prove?) that lim t →∞ u ( x, t ) = u ( x ). (d) Calculus question: In equation (1), on the left hand side the derivative is denoted by d dt , on the right hand side by ∂t . Why the change in notation?
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It also helps to know that under fairly general conditions...

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