final_sol - Solutions to Final Exam May 24, 2011 problem 1...

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Solutions to Final Exam May 24, 2011 problem 1 a At steady state, we have T xx + C = 0 , for 0 < x < L T ( x = 0) = 0 , T x ( x = L ) = 0 The general solution to the ODE is T = - C 2 x 2 + ax + b The boundary conditions give us b = 0, a = CL , so we Fnally have T S ( x ) = Cx ( L - x 2 ) which plots as the left half of a down-facing parabola. Note that the hottest section of the rod, at x = L , is at a temperature of CL 2 / 2. b DeFning T T T - T S , where we have T S from (a), gives for T T (referred to as just T for simplicity) T xx = 1 α 2 T t , for 0 < x < L, t > 0 T ( x = 0 , t ) = 0 , T x ( x = L, t ) = 0 , T ( x, t = 0) = f ( x ) f ( x ) - Cx ( L - x 2 ) , lim t →∞ T ( x, t ) = 0 1
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c Assuming solutions of the form X ( x )Θ( t ), we have from the PDE X ′′ X = 1 α 2 Θ Θ ≡ - k 2 giving the general solutions X ′′ + k 2 X = 0 X = c 1 sin kx + c 2 cos( kx ) Θ + α 2 k 2 Θ = 0 Θ = de k 2 α 2 t The boundary conditions at the ends of the rod give us a Storm-Liouville
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final_sol - Solutions to Final Exam May 24, 2011 problem 1...

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