hw 12-b - 642:527 SOLUTIONS: ASSIGNMENT 12 FALL 2007 12.B...

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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 12 FALL 2007 12.B We want to find the eigenvalues and eigenfunctions for y ′′ + λy = 0, y (0) = −y (1), y ′ (0) = −y ′ (1). As usual we have several cases. √ (I) λ > 0, y (x) = A cos κx + B sin κx with κ = λ. We have to solve A = −(A cos κ + B sin κ), or 1 + cos κ sin κ − sin κ 1 + cos κ A = 0. B κB = −(−Aκ sin κ + Bκ cos κ) The determinant of the coefficient matrix must vanish to have an eigenvaluue, so the eigenvalues satisfy 1 + cos κ = 0, i.e., κn = (2n − 1)π , n = 1, 2, . . .; the eigenvalues are λn = ((2n − 1)π )2 . For each such λn the matrix is identically zero so A and B can be chosen freely; thus there are two eigenfunctions, cos κn x and sin κn x. (II) λ = 0. Here y = A + Bx and the boundary conditions imply A = −(A + B ),, B = −B ; these equations hoave no nonzero solution so λ = 0 is not an eigenvalue. √ (III) λ < 0. As in (I) we find the condition 1 + cosh −λ = 0, which has no solutions. For any function f (x), 0 < x < 1, we expect an expansion ∞ f ( x) = 1 an cos κn x + bn sin κn x) , (12.1) with cos κn x, f (x) an = = cos κn x, cos κn x bn = sin κn x, f (x) = sin κn x, sin κn x 1 0 f (x) cos(κn x) dx 1 0 1 cos(κn x)2 dx =2 0 1 f (x) cos(κn x) dx, f (x) sin(κn x) dx. 0 1 0 f (x) sin(κn x) dx 1 0 sin(κn x)2 dx =2 If f (x) = 1 these give an = 0, bn = 4/κn = 4/((2n − 1)π ). As may be checked, (12.1) is in general just the usual Fourier series of a certain period-2 extension g of f : g (x) = f (x) for 0 < x < 1, g (x) = −f (x − 1) for 1 < x < 2, and g (x) then extended to have period 2. 1 ...
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This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.

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