hw 12-b

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.

Ask a homework question - tutors are online