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Unformatted text preview: What are p, q, σ ? Solution. d dx ( (1 + x 2 ) y ′ ) ′ + λ (1 + x 2 ) y = 0 , y ′ (0) = y ′ (1) = 0 . Thus: p ( x ) = 1 + x 2 , q ( x ) = 0, σ ( x ) = 1 + x 2 . (b) (5 points) Explain why there canNOT be any negative eigenvalues. Solution. The Rayleigh quotient works out to λ = ∫ 1 (1 + x 2 ) y ′ ( x ) 2 dx ∫ 1 (1 + x 2 ) y ( x ) 2 dx , which can clearly not be negative. (c) (5 points) Can λ = 0 be an eigenvalue? JUSTIFY! Solution. The answer is YES. y = 1 is a nonzero solution of the problem for λ = 0. 3. Solve by separation of variables the following problems. You do not have to go through a detailed analysis of any SturmLiouville problem that you already solved several times before, assuming you remember what you did. Or have it on your cheat sheet. But explain from where it comes! (For example, Write, “As seen in class, the solution of this problem is given by. . . ”) (a) (20 points) u t = u xx , < x < π, t > , u (0 , t ) = , u ( π, t ) = 0 , t > u ( x, 0) = 3 sin 5 x − 4 sin 7 x, < x < π. Solution. The solution of the PDE solving the boundary conditions obtained by separation of variables is u ( x, t ) = ∞ ∑ n =1 c n e − n 2 t sin nx....
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 Spring '13
 Schonbek
 Sin, Boundary value problem, Partial differential equation, Sturm–Liouville theory

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