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Differential Equations Lecture Work Solutions 50

# Differential Equations Lecture Work Solutions 50 - scribed...

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(a) u xx + u yy = 1 (b) u xy + u yy = u (c) x 2 yu xx + y 4 u yy = 4 u (d) u t + uu x = 0 (e) u tt + f ( t ) u t = u xx (f) x 2 y 2 u xxx = u y 16. (a) Solve the one dimensional heat equation in a bar u t = ku xx 0 < x < L which is insulated at either end, given the initial temperature distribution u ( x, 0) = f ( x ) (b) What is the equilibrium temperature of the bar? and explain physically why your answer makes sense. 17. Solve the 1-D heat equation u t = ku xx 0 < x < L subject to the nonhomogeneous boundary conditions u (0) = 1 u x ( L ) = 1 with an initial temperature distribution u ( x, 0) = 0. (Hint: First solve for the equilibrium temperature distribution v ( x ) which satisfies the steady state heat equation with the pre-
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Unformatted text preview: scribed boundary conditions. Once v is found, write u ( x, t ) = v ( x ) + w ( x, t ) where w ( x, t ) is the transient response. Substitue this u back into the PDE to produce a new PDE for w which now has homogeneous boundary conditions. 18. Solve Laplace’s equation, ∇ 2 u = 0 , ≤ x ≤ π, ≤ y ≤ π subject to the boundary conditions u ( x, 0) = sin x + 2 sin 2 x u ( π, y ) = 0 u ( x, π ) = 0 u (0 , y ) = 0 19. Repeat the above problem with u ( x, 0) = − π 2 x 2 + 2 πx 3 − x 4 50...
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