Differential Equations Lecture Work Solutions 85

# Differential Equations Lecture Work Solutions 85 - tions,...

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17. u t = ku xx 0 <x<L Boundary conditions (inhomogeneous) u (0 ,t )=1 u x ( L, t )=1 Initial condition u ( x, 0) = 0 Let w ( x, t ) satisfes the inhomogeneous boundary conditions w (0 ,t )=1 w x ( L, t )=1 For example, we can take a linear ±unction in x w ( x, t )= αx + β Using the boundary conditions we get α = β =1 and so w ( x, t )= x +1 Now let v ( x, t )= u ( x, t ) w ( x, t ), then clearly v will satis±y homogeneous boundary condi-
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Unformatted text preview: tions, and the PDE becomes: v t + w t = k ( v xx + w xx ) Since w t = w xx = 0, we can write the PDE v t = kv xx The initial condition is v ( x, 0) + w ( x, 0) = 0 or v ( x, 0) = − x − 1 So the problem is now: v t = kv xx v (0 , t ) = 0 v x ( L, t ) = 0 v ( x, 0) = − x − 1 85...
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## This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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