Differential Equations Lecture Work Solutions 84

Differential Equations Lecture Work Solutions 84 - 16. ut =...

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16. u t = ku xx u x (0 ,t )=0 u x ( L, t )=0 u ( x, 0) = f ( x ) u ( x, t )= X n =0 u n cos nπx L e k ( L ) 2 t u ( x, 0) = X n =0 u n cos nπx L f ( x ) u n are the coefficients of expanding f ( x ) in terms of Fourier cosine series. u 0 = 1 L Z L 0 f ( x ) dx u n = 2 L Z L 0 f ( x )cos L xdx b. The equilibrium is when u xx = 0 subject to the same boundary conditions. The solution is then obtained by integration with respect to x u x = K and K = 0 because of the boundary conditions. Now integrate again to get u = C . This means that the temperature is constant. There is no other condition to ±x this
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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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