On the other hand this expression for potential

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Unformatted text preview: Neumann boundary condition ∂u ∂u (0, t) = (L, t) = 0. ∂x ∂x (4.14) (Physically, this corresponds to insulated endpoints, from which no heat can enter or escape.) In this case separation of variables leads to a slightly different eigenvalue problem, which consists of finding the nontrivial solutions to f (x) = d2 (f (x)) = λf (x), dx2 f (0) = 0 = f (L). a. Solve this eigenvalue problem. (Hint: The solution should involve cosines instead of sines.) b. Find the general solution to the heat equation ∂u ∂2u = ∂t ∂x2 subject to the Neumann boundary condition (4.14). 91 c. Find the function u(x, t), defined for 0 ≤ x ≤ π and t ≥ 0, such that: ∂u ∂2u , = ∂t ∂x2 ∂u ∂u (0, t) = (π, t) = 0, ∂x ∂x u(x, 0) = 3 cos x + 7 cos 2x. 4.2.9. We can also treat a mixture of Dirichlet and Neumann conditions, say u(0, t) = ∂u (L, t) = 0. ∂x (4.15) In this case separation of variables leads to the eigenvalue problem which consists of finding the nontrivial solutions to f (x) = d2 (f (x)) = λf (x), dx2 f (0) = 0 = f (L). a. Solve t...
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This document was uploaded on 01/12/2014.

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