Chem Differential Eq HW Solutions Fall 2011 46

# Chem Differential Eq HW Solutions Fall 2011 46 - 46 Chapter...

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46 Chapter 3 Partial Differential Equations in Rectangular Coordinates Solving the equation for T , we find T ( t ) = e t ; thus we have the product solution c 0 e - x e t , where, for convenience, we have used c 0 as an arbitrary constant. (d) If k = - α 2 < 0, then X + μ 2 X = 0 X = c 1 cos μx + c 2 sin μx ; X (0) = - X (0) μc 2 = - c 1 X (1) = - X (1) - μc 1 sin μ + μc 2 cos μ = - c 1 cos μ - c 2 sin μ - μc 1 sin μ - c 1 cos μ = - c 1 cos μ - c 2 sin μ - μc 1 sin μ = - c 2 sin μ - μc 1 sin μ = c 1 μ sin μ. Since μ = 0, take c 1 = 0 (otherwise you will get a trivial solution) and divide by c 1 and get μ 2 sin μ = - sin μ sin μ = 0 μ = nπ, where n is an integer. So X = c 1 cos nπx + c 2 sin nπx . But c 1 = - c 2 μ , so X = - c 1 ( cos nπx - sin nπx ) . Call X n = cos nπx - sin nπx . (e) To establish the orthogonality of the X n ’s, treat the case k = 1 separately. For k = - μ 2 , we refer to the boundary value problem X + μ 2 n X = 0 , X (0) = - X (0) , X (1) = - X (1) , that is satisfied by the X n ’s, where μ n = . We establish orthogonality using a trick from Sturm-Liouville theory (Chapter 6, Section 6.2). Since X m = μ 2 m X m and X n = μ 2 n X n , multiplying the first equation by X n and the second by
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