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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 ) = n =0 u n cos n π x L e k ( n π L ) 2 t u ( x, 0) = n =0 u n cos n π x L f ( x ) u n are the coefficients of expanding f ( x ) in terms of Fourier cosine series. u 0 = 1 L L 0 f ( x ) dx u n = 2 L 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 fix this
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